23 Chapter Solutions
10.0 grams electrons 24 23 atoms 47.0 = 2.62 10 N= 6.02 10 atom mol 107.87 grams mol # electrons added = Q 1.00 10 -3 C = = 6.25 1015 e 1.60 10 -19 C electron
23.1
(a)
(b)
or
2.38 electrons for every 10 9 already present
23.2
(a)
8.99 10 9 N m 2/ C 2 1.60 10 -19 C k q1q Fe = e 2 2 = r (3.80 10 - 10 m)2
(
)(
)
2
= 1.59 10 - 9 N
(repulsion)
(b)
6.67 10 -11 N m 2 kg 2 (1.67 10 - 27 kg)2 G m1m2 Fg = = = 1.29 10 - 45 N r2 (3.80 10 -10 m)2 The electric force is larger by 1.24 10 36 times
(
)
(c)
If ke q = m
q1q2 mm = G 12 2 2 r r G = ke
with q1 = q2 = q and m1 = m2 = m, then
6.67 10 -11 N m 2 / kg 2 = 8.61 10 -11 C / kg 8.99 10 9 N m 2 / C 2
23.3
If each person has a mass of 70 kg and is (almost) composed of water, then each person contains 70, 000 grams protons 23 molecules 10 N 2.3 10 28 protons 6.02 10 molecule mol 18 grams mol With an excess of 1% electrons over protons, each person has a charge q = (0.01)(1.6 10 -19 C)(2.3 10 28 ) = 3.7 107 C So F = ke q1q2 (3.7 107 )2 = (9 10 9 ) N = 4 10 25 N ~ 1026 N r2 0.6 2
This force is almost enough to lift a "weight" equal to that of the Earth: Mg = (6 10 24 kg)(9.8 m s 2 ) = 6 10 25 N ~ 1026 N
2000 by Harcourt, Inc. All rights reserved.
2
Chapter 23 Solutions
23.4
We find the equal-magnitude charges on both spheres: F = ke q1q2 q2 = ke 2 2 r r so q=r F 1.00 10 4 N = (1.00 m ) = 1.05 10 -3 C ke 8.99 10 9 N m 2/ C 2
The number of electron transferred is then N xfer = 1.05 10 -3 C
(
) (1.60 10 (
-19
C / e - = 6.59 1015 electrons
)
The whole number of electrons in each sphere is 10.0 g 23 - 24 - Ntot = 6.02 10 atoms / mol 47 e / atom = 2.62 10 e 107.87 g / mol
)(
)
The fraction transferred is then f= N xfer Ntot 6.59 1015 = = 2.51 109 2.62 1024 = 2.51 charges in every billion
23.5
8.99 10 9 N m 2 C 2 1.60 10 -19 C qq F = ke 1 2 2 = 2 r 2(6.37 106 m)
(
[
)(
]
) (6.02 10 )
2
23 2
= 514 kN
*23.6
(a)
The force is one of attraction. The distance r in Coulomb's law is the distance between centers. The magnitude of the force is F=
-9 -9 ke q1q2 N m 2 12.0 10 C 18.0 10 C = 8.99 10 9 = 2.16 10 - 5 N r2 C2 (0.300 m)2
(
)(
)
(b)
The net charge of - 6.00 10 -9 C will be equally split between the two spheres, or - 3.00 10 -9 C on each. The force is one of repulsion, and its magnitude is
2 3.00 10 -9 C 3.00 10 -9 C ke q1q2 9 Nm F= = 8.99 10 = r2 C2 (0.300 m)2
(
)(
)
8.99 10 -7 N
Chapter 23 Solutions q1q2 (8.99 10 9 N m 2/ C 2 )(7.00 10 -6 C)(2.00 10 -6 C) = = 0.503 N r2 (0.500 m)2 q1q2 (8.99 10 9 N m 2 / C 2 )(7.00 10 -6 C)(4.00 10 -6 C) = = 1.01 N (0.500 m)2 r2
3
23.7
F1 = ke
F2 = k e
Fx = (0.503 + 1.01) cos 60.0 = 0.755 N Fy = (0.503 - 1.01) sin 60.0 = - 0.436 N F = (0.755 N)i - (0.436 N)j = 0.872 N at an angle of 330
Goal Solution Three point charges are located at the corners of an equilateral triangle as shown in Figure P23.7. Calculate the net electric force on the 7.00- C charge. G: Gather Information: The 7.00- C charge experiences a repulsive force F1 due to the 2.00- C charge, and an attractive force F 2 due to the -4.00- C charge, where F2 = 2F1. If we sketch these force vectors, we find that the resultant appears to be about the same magnitude as F2 and is directed to the right about 30.0 below the horizontal. Organize : We can find the net electric force by adding the two separate forces acting on the 7.00- C charge. These individual forces can be found by applying Coulomb's law to each pair of charges. Analyze: The force on the 7.00- C charge by the 2.00- C charge is
9
O:
A:
F1
(8.99 10 =
N m 2/ C 2 7.00 10 -6 C 2.00 10 -6 C
(0.500 m)
)(
)(
2
) (cos60i + sin 60j) = F
1
= (0.252i + 0.436j) N
Similarly, the force on the 7.00- C by the -4.00- C charge is
-6 -6 N m 2 7.00 10 C - 4.00 10 C F 2 = - 8.99 10 9 (cos60i - sin 60 j) = (0.503i - 0.872j) N C2 (0.500 m)2
(
)(
)
Thus, the total force on the 7.00- C , expressed as a set of components, is F = F1 + F 2 = (0.755 i - 0.436 j) N = 0.872 N at 30.0 below the +x axis L: Learn: Our calculated answer agrees with our initial estimate. An equivalent approach to this problem would be to find the net electric field due to the two lower charges and apply F=qE to find the force on the upper charge in this electric field.
2000 by Harcourt, Inc. All rights reserved.
4 *23.8
Chapter 23 Solutions Let the third bead have charge Q and be located distance x from the left end of the rod. This bead will experience a net force given by F= ke ( 3q)Q x2 i+ ke ( q)Q
( d - x )2
( -i)
1 x 3 = , or d - x = 3 x 2 ( d - x )2 x = 0.634d
The net force will be zero if
This gives an equilibrium position of the third bead of The equilibrium is
stable if the third bead has positive charge .
*23.9
(a)
F=
k ee 2 (1.60 1019 C)2 = (8.99 109 N m2/C 2) = 8.22 108 N 2 r (0.529 1010 m)2 mv 2 r Fr = m
(b)
We have F =
from which v =
(8.22 10
-8
N 0.529 10 -10 m
-31
9.11 10
)(
kg
)=
2.19 106 m/s
23.10
The top charge exerts a force on the negative charge to the left, at an angle of tan -1 ( d / 2x ) to the x-axis. force (- x)i d2 4 + x2 = ma 1/2
( d 2 )2 + x 2
ke qQ
which is directed upward and
The two positive charges together exert
2 k qQ 2 e d 4 + x2
(
) (
)
or for x << d 2,
a
- 2 ke qQ x md 3 / 8
(a)
The acceleration is equal to a negative constant times the excursion from equilibrium, as i n 16 ke qQ a = - 2 x, so we have Simple Harmonic Motion with 2 = . md 3 T= 2 =
2
md 3 ke qQ
, where m is the mass of the object with charge -Q.
(b)
vmax = A =
4a
ke qQ md 3
Chapter 23 Solutions For equilibrium, F e = -F g , or qE = -mg( -j) . Thus, E = (a) E= mg j. q
5
23.11
(9.11 10 -31 kg)(9.80 m s 2 ) mg j= j = - ( 5.58 10 -11 N C) j q -1.60 10 -19 C
(
)
(b)
E=
1.67 10 -27 kg 9.80 m s 2 mg j= j= q 1.60 10 -19 C
(
(
)(
)
)
(1.02 10
-7
NC j
)
23.12
Fy = 0:
QE j + mg(- j) = 0 m= (24.0 10 -6 C)(610 N / C) QE = = 1.49 grams g 9.80 m / s 2
*23.13
The point is designated in the sketch. The magnitudes of the electric fields, E1, (due to the 2.50 106 C charge) and E2 (due to the 6.00 106 C charge) are E1 = E2 = k eq (8.99 109 N m2/C 2)(2.50 106 C) = r2 d2 k eq (8.99 109 N m2/C 2)(6.00 106 C) = r2 (d + 1.00 m)2 (1)
(2)
Equate the right sides of (1) and (2) to get or which yields
(d + 1.00 m)2 = 2.40d 2 d + 1.00 m = 1.55d d = 1.82 m or d = 0.392 m
The negative value for d is unsatisfactory because that locates a point between the charges where both fields are in the same direction. Thus, d = 1.82 m to the left of the -2.50 C charge.
23.14
If we treat the concentrations as point charges, E + = ke q N m 2 ( 40.0 C) = 8.99 10 9 (-j) = 3.60 105 N / C (-j) (downward) 2 r C 2 (1000 m )2 q N m 2 ( 40.0 C) = 8.99 10 9 (-j) = 3.60 105 N / C (-j) (downward) r2 C 2 (1000 m )2
E - = ke
E = E + + E - = 7.20 10 5 N / C downward
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6
Chapter 23 Solutions 8.99 10 9 7.00 10 -6 ke q E1 = 2 = = 2.52 10 5 N C 2 r (0.500) E2 = 8.99 10 9 4.00 10 -6 ke q = = 1.44 10 5 N C r2 (0.500)2
*23.15 (a)
(
)(
)
(
)(
)
Ex = E2 - E1 cos 60 = 1.44 10 5 - 2.52 10 5 cos 60.0 = 18.0 10 3 N C Ey = -E1 sin 60.0 = -2.52 10 5 sin 60.0 = -218 10 3 N C E = [18.0i - 218 j] 10 3 N C = [18.0i - 218 j] kN C (b) F = qE = 2.00 10 -6 C (18.0i - 218 j) 10 3 N C = ( 36.0i - 436 j) 10 -3 N =
(
)
(36.0i - 436 j ) mN
*23.16 (a)
E1 =
ke q1
2 r1
(8.99 10 )(3.00 10 ) (- j) = - (2.70 10 (- j) =
9 -9
(0.100)2
9
3
NC j
)
E2 =
ke q2
2 r2
(- i) =
(8.99 10 )(6.00 10 ) (- i) = - (5.99 10
-9
(0.300)2
2
NC i
)
E = E 2 + E1 = - 5.99 10 2 N C i - 2.70 10 3 N C j (b) F = qE = 5.00 10 -9 C ( -599i - 2700 j) N C F = - 3.00 10 -6 i - 13.5 10 -6 j N =
(
) ( )
)
(
)
(
(- 3.00 i - 13.5 j ) N
23.17
(a)
The electric field has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the three charges individually produce fields that cancel out. You may need to review vector addition in Chapter Three.
(b)
The magnitude of the field at point P due to each of the charges along the base of the triangle is E = ke q a 2 . The direction of the field in each case is along the line joining the charge in question to point P as shown in the diagram at the right. The x components add to zero, leaving E= ke q kq sin 60.0) j + e2 (sin 60.0) j = 2 ( a a 3 ke q j a2
Chapter 23 Solutions
7
Goal Solution Three equal positive charges q are at the corners of an equilateral triangle of side a, as shown in Figure P23.17. (a) Assume that the three charges together create an electric field. Find the location of a point (other than ) where the electric field is zero. (Hint: Sketch the field lines in the plane of the charges.) (b) What are the magnitude and direction of the electric field at P due to the two charges at the base? G: The electric field has the general appearance shown by the black arrows in the figure to the right. This drawing indicates that E = 0 at the center of the triangle, since a small positive charge placed at the center of this triangle will be pushed away from each corner equally strongly. This fact could be verified by vector addition as in part (b) below. The electric field at point P should be directed upwards and about twice the magnitude of the electric field due to just one of the lower charges as shown in Figure P23.17. For part (b), we must ignore the effect of the charge at point P , because a charge cannot exert a force on itself. O: The electric field at point P can be found by adding the electric field vectors due to each of the two lower point charges: E = E1 + E2 (b) The electric field from a point charge is As shown in the solution figure above, E1 = ke E 2 = ke q to the right and upward at 60 a2
A:
q to the left and upward at 60 a2 q q q E = E1 + E2 = ke 2 (cos60i + sin 60j) + ( - cos60i + sin 60 j) = ke 2 2(sin 60j) = 1.73ke 2 j a a a
[
]
[
]
L:
The net electric field at point P is indeed nearly twice the magnitude due to a single charge and is entirely vertical as expected from the symmetry of the configuration. In addition to the center of the triangle, the electric field lines in the figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically.
23.18
(a)
E=
ke q (8.99 10 9 )(2.00 10 -6 ) = = 14, 400 N C (1.12)2 r2 and Ey = 2(14, 400) sin 26.6 = 1.29 10 4 N C E = 1.29 10 4 j N C
Ex = 0 so (b)
F = Eq = (1.29 10 4 j)(-3.00 10 -6 ) = -3.86 10 -2 j N
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8
Chapter 23 Solutions k ( 2q) k ( 3q) k ( 4q) ke q1 k q k q ~ + e 22 ~2 + e 23 ~3 = e 2 i + e 2 (i cos 45.0 + j sin 45.0) + e 2 j 2 1 r1 r2 r3 a 2a a ke q kq kq i + 5.06 e2 j = 5.91 e2 at 58.8 a2 a a ke q 2 at 58.8 a2
23.19
(a)
E=
E = 3.06
(b)
F = qE = 5.91
23.20
The magnitude of the field at (x, y) due to charge q at (x0 , y0 ) is given by E = ke q r 2 where r is the distance from (x0 , y0 ) to (x, y). Observe the geometry in the diagram at the right. From triangle ABC, r 2 = (x - x0 )2 + (y - y0 )2 , or
r = (x - x0 )2 + (y - y0 )2 , Ex = Ecos =
sin =
(y - y0 ) , r
and
cos =
(x - x0 ) r
Thus,
ke q(x - x0 ) ke q (x - x0 ) = 2 r r [(x - x0 )2 + (y - y0 )2 ]3/2 ke q(y - y0 ) ke q (y - y0 ) = r r2 [(x - x0 )2 + (y - y0 )2 ]3/2
and
Ey = Esin =
23.21
The electric field at any point x is
E=
k eq (x a)2
k eq (x (a))2 (4a)(k eq) x3
=
k eq(4ax) (x 2 a 2)2
When x is much, much greater than a, we find
E
23.22
(a)
One of the charges creates at P a field at an angle to the x-axis as shown.
E=
ke Q/n R 2 + x2
When all the charges produce field, for n > 1, the components perpendicular to the x-axis add to zero. The total field is (b) nke (Q/n)i keQxi cos = 2 2 2 R +x (R + x 2)3/2
A circle of charge corresponds to letting n grow beyond all bounds, but the result does not depend on n. Smearing the charge around the circle does not change its amount or its distance from the field point, so it does not change the field. .
Chapter 23 Solutions
9
23.23
E=
- ke qi 2 ke q ke q kq kq kq 1 1 ~ = e2 (- i)+ e 2 (- i) + e 2 (- i) + . . . = 1+ 2 + 3 + ... = - i 2 2 r a (2a) (3a) 6 a2 2 3 a
23.24
E=
(8.99 109)(22.0 106) k (Q / l)l ke l keQ = e = = (0.290)(0.140 + 0.290) d(l+ d) d(l+ d) d(l+ d) directed toward the rod .
E = 1.59 106 N/C ,
23.25
E=
ke dq x2
where dq = 0 dx
E = ke 0
dx 1 2 = k e x x0 x
=
x0
k e 0 x0
The direction is i or left for 0 > 0
23.26
E = dE =
x0
1 ke 0 x0 dx( -i ) = - ke 0 x0 i x -3 dx = - ke 0 x0 i - 2 2x x0 x3
x0
=
ke 0 (- i) 2x0
23.27 (a) (b) (c)
E=
(8.99 109)(75.0 106)x 6.74 105 x k exQ = 2 2 2 3/2 = 2 3/2 (x + a ) (x + 0.100 ) (x + 0.0100)3/2
2
At x = 0.0100 m, At x = 0.0500 m, At x = 0.300 m,
E = 6.64 106 i N/C = 6.64 i MN/C E = 2.41 107 i N/C = 24.1 i MN/C E = 6.40 106 i N/C = 6.40 i MN/C E = 6.64 105 i N/C = 0.664 i MN/C
(d) At x = 1.00 m,
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10
Chapter 23 Solutions ke Q x (x + a2 )3/2
2
23.28
E=
For a maximum,
3x 2 dE 1 = Qke 2 - 2 =0 2 3/2 2 5/2 dx (x + a ) (x + a ) or x= a 2
x 2 + a2 - 3x 2 = 0
Substituting into the expression for E gives E= 2ke Q ke Qa k Q = e = 3 3 3 a2 2( 2 a2 )3/2 3 3 a2 2
=
Q 6 3 e0 a2
23.29
x E = 2 ke 1 - 2 2 x +R E = 2 8.99 10 9 7.90 10 -3 1 -
(
)(
)
= 4.46 108 1 - 2 x 2 + (0.350) x
x 2 + 0.123 x
(a) (b) (c)
At x = 0.0500 m, At x = 0.100 m, At x = 0.500 m,
E = 3.83 108 N C = 383 MN C E = 3.24 108 N C = 324 MN C E = 8.07 107 N C = 80.7 MN C E = 6.68 108 N C = 6.68 MN C
(d) At x = 2.00 m,
23.30
(a)
From Example 23.9:
E = 2 ke 1 -
x2 + R2 x
=
Q = 1.84 10 -3 C m 2 R 2
E = (1.04 108 N C)(0.900) = 9.36 107 N C = 93.6 MN/C appx: E = 2 ke = 104 MN/C (about 11% high) E = (1.04 108 N / C) 1 - appx: E = ke 8 = (1.04 10 N C)(0.00496) = 0.516 MN/C 2 2 30.0 + 3.00 cm 30.0 cm
(b)
5.20 10 -6 Q = (8.99 10 9 ) = 0.519 MN/C (about 0.6% high) r2 (0.30)2
Chapter 23 Solutions Ex = 2 ke 1 - Ex = 2 ke 1 - 1+ x +R x
2 2
11
23.31
The electric field at a distance x is
This is equivalent to
1 1+ R 2
x2
For large x, R 2 x 2 << 1 and
R2 R2 1+ 2 x2 2x
so
1 Ex = 2 ke 1 - 1+ R 2 (2x 2 )
[
]
1+ R 2 (2x 2 ) - 1 = 2 ke 1+ R 2 (2x 2 )
(
[
]
)
Substitute = Q/ R2,
Ex =
[1+ R
keQ 1 x 2
2
(
R2 = keQ x 2 + 2 (2x 2 )
)
]
But for x > > R,
1 1 2 , so 2 x +R 2 x
2
Ex
keQ for a disk at large distances x2
23.32
The sheet must have negative charge to repel the negative charge on the Styrofoam. The magnitude of the upward electric force must equal the magnitude of the downward gravitational force for the Styrofoam to "float" (i.e., Fe = F g ). Thus, 2e0 mg -qE = mg, or -q = mg which gives = - q 2e0
23.33
Due to symmetry Ey = dEy = 0, and E x = dE sin = ke where dq = ds = r d , so that, q where = L Solving, Since the rod has a negative charge, L . Thus, Ex = ke r
dq sin r2
0
sin d =
ke 2k (- cos ) = e r r 0
and r =
Ex =
2ke q 2(8.99 109 N m2/C 2)(7.50 106 C) = L2 (0.140 m)2
E = Ex = 2.16 107 N/C E = (2.16 107 i) N/C = 21.6 i MN/C
2000 by Harcourt, Inc. All rights reserved.
12 23.34
Chapter 23 Solutions (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has charge Qdx/h and produces, at the chosen point, a field dE = Q dx ke x i 2 3/2 h (x + R )
2
The total field is E=
all charge
dE
=
d+h d
keQ x dx k Qi d + h 2 i = e (x + R 2 )- 3/2 2x dx 2 2 3/2 2h x = d h(x + R )
d+h x=d
k Q i (x 2 + R 2 )- 1/2 E= e 2h (- 1/ 2) (b)
keQ i 1 1 = - (d 2 + R 2 )1/2 h (d + h)2 + R 2
(
)
1/2 charge Q dx/h, and charge-
Think of the cylinder as a stack of disks, each with thickness dx, per-area = Q dx / R 2 h. One disk produces a field dE = 2 keQ dx R2h E= x 1 - (x 2 + R 2 )1/2 i dE =
d+h
So,
all charge
x=d
2 keQ dx R2 h
x 1 - (x 2 + R 2 )1/2 i
d + h d
E= E=
d+ h 1 (x 2 + R 2 )1/2 2 keQ i d + h 1 d + h(x 2 + R 2 )- 1/2 2x dx = 2 keQ i x dx - 2 - x=d 2 1/ 2 R 2 h d R2h d 2 keQ i R 2h 2 keQ i R 2h d + h - d - (d + h)2 + R 2
(
)
1/2
+ (d 2 + R 2 )1/2
E=
h + (d 2 + R 2 )1/2 - (d + h)2 + R 2
(
)
1/2
23.35
(a)
The electric field at point P due to each element of length dx, is dE =
ke dq and is directed (x + y 2 ) along the line joining the element of length to point P . By symmetry,
2
Ex = dEx = 0
and since dq = dx, y (x + y 2 )1 2
2
E = Ey = dEy = dE cos where cos = Therefore, dx = (x + y 2 )3 2
2
2ke sin 0 y 2ke y
(b)
For a bar of infinite length, 90 and Ey =
Chapter 23 Solutions The whole surface area of the cylinder is A = 2 r 2 + 2 rL = 2 r(r + L) . Q = A = 15.0 10 -9 C m 2 2 (0.0250 m )[0.0250 m + 0.0600 m ] = 2.00 10 -10 C (b) For the curved lateral surface only, A = 2 rL. Q = A = 15.0 10 -9 C m 2 2 (0.0250 m )(0.0600 m ) = 1.41 10 -10 C (c) Q = V = r 2 L = 500 10 -9 C m 3 (0.0250 m ) (0.0600 m ) = 5.89 10 -11 C
2
13
*23.36 (a)
( (
) )
(
)
*23.37 (a)
Every object has the same volume, V = 8(0.0300 m ) = 2.16 10 -4 m 3 .
3
For each, Q = V = 400 10 -9 C m 3 2.16 10 -4 m 3 = 8.64 10 -11 C (b) We must count the 9.00 cm 2 squares painted with charge: (i) 6 4 = 24 squares Q = A = 15.0 10 -9 C m 2 24.0 9.00 10 -4 m 2 = 3.24 10 -10 C (ii) 34 squares exposed Q = A = 15.0 10 -9 C m 2 34.0 9.00 10 -4 m 2 = 4.59 10 -10 C (iii) 34 squares Q = A = 15.0 10 -9 C m 2 34.0 9.00 10 -4 m 2 = (iv) 32 squares Q = A = 15.0 10 -9 C m 2 32.0 9.00 10 -4 m 2 = (c) (i) total edge length: = 24 (0.0300 m )
(
)(
)
( ( ( (
) (
)
) ( ) ( ) (
) )
4.59 10 -10 C
)
4.32 10 -10 C
Q = = 80.0 10 -12 C m 24 (0.0300 m ) = (ii) (iii)
(
)
5.76 10 -11 C
Q = = 80.0 10 -12 C m 44 (0.0300 m ) = 1.06 10 -10 C Q = = 80.0 10 -12 C m 64 (0.0300 m ) = 1.54 10 -10 C
( (
) )
2000 by Harcourt, Inc. All rights reserved.
14
Chapter 23 Solutions Q = = 80.0 10 -12 C m 40 (0.0300 m ) = 0.960 10 -10 C
(iv)
(
)
Chapter 23 Solutions
15
22.38
22.39
23.40
(a)
q1 - 6 = = q2 18
-
1 3
(b)
q1 is negative, q2 is positive
23.41
F = qE = ma a = v=
qE m qEt m (1.602 10 -19 )(520)(48.0 10 -9 ) = 4.39 106 m/s 9.11 10 -31
v = v i + at
electron:
ve =
in a direction opposite to the field proton: vp = (1.602 10 -19 )(520)(48.0 10 -9 ) = 2.39 103 m/s 1.67 10 -27
in the same direction as the field
23.42
(a) (b)
a =
qE (1.602 10 -19 )(6.00 10 5 ) = = 5.76 1013 m s so -27 m (1.67 10 )
a=
-5.76 1013 i m s 2
v = vi + 2a(x - xi ) 0 = vi 2 + 2(-5.76 1013 )(0.0700) v i = 2.84 106 i m s
(c)
v = vi + at 0 = 2.84 106 + (-5.76 1013 )t t = 4.93 10 -8 s
2000 by Harcourt, Inc. All rights reserved.
16
Chapter 23 Solutions 1.602 10 -19 (640) qE a= = = m 1.67 10 -27
23.43
(a)
(
(
)
)
6.14 1010 m/s2
(b)
v = v i + at 1.20 106 = (6.14 1010)t t = 1.95 10-5 s
(c)
1 x - xi = 2 ( vi + v )t 1 x = 2 1.20 106 1.95 10 -5 = 11.7 m
(
)(
)
(d)
1 K = 2 mv 2 = 1 (1.67 10 - 27 kg)(1.20 106 m / s)2 = 1.20 10-15 J 2
23.44
The required electric field will be in the direction of motion . We know that Work = K So, Fd = 2 m v i
1
2
(since the final velocity = 0)
1 2
1 This becomes Eed = mvi 2 2 E = 1.60 1017 J (1.60 1019 C)(0.100 m)
mv i ed
2
or
E=
= 1.00 103 N/C (in direction of electron's motion)
23.45
The required electric field will be in the direction of motion . Work done = K so, Fd = 2 m v i K E=ed
1
2
(since the final velocity = 0)
which becomes eEd = K
and
Chapter 23 Solutions
17
Goal Solution The electrons in a particle beam each have a kinetic energy K . What are the magnitude and direction of the electric field that stops these electrons in a distance of d? G: We should expect that a larger electric field would be required to stop electrons with greater kinetic energy. Likewise, E must be greater for a shorter stopping distance, d. The electric field should be i n the same direction as the motion of the negatively charged electrons in order to exert an opposing force that will slow them down. The electrons will experience an electrostatic force F = qE. Therefore, the work done by the electric field can be equated with the initial kinetic energy since energy should be conserved. The work done on the charge is and Assuming v is in the + x direction, E is therefore in the direction of the electron beam: W = F d = qE d Ki + W = K f = 0 K + ( -e )E di = 0 eE ( di ) = K K E= i ed
O:
A:
L:
As expected, the electric field is proportional to K , and inversely proportional to d. The direction of the electric field is important; if it were otherwise the electron would speed up instead of slowing down! If the particles were protons instead of electrons, the electric field would need to be directed opposite to v in order for the particles to slow down.
23.46
The acceleration is given by Solving, Now F = ma:
v 2 = v i + 2a(x xi)or v2 a = 2h mgj + qE = mv 2 j 2h
2
v 2 = 0 + 2a(h)
Therefore
mv2 qE = 2h + m g j
(a)
Gravity alone would give the bead downward impact velocity 2 9.80 m / s 2 ( 5.00 m ) = 9.90 m / s To change this to 21.0 m/s down, a downward force. electric field must exert a downward electric
(
)
(b)
m v 2 q = E 2h g
=
1.00 103 kg N s2 (21.0 m/s)2 kg m 2(5.00 m) 9.80 m/s2 = 3.43 C 4 1.00 10 N/C
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18
Chapter 23 Solutions x 0.0500 = = 1.11 10-7 s = 111 ns v 4.50 10 5 qE (1.602 10 -19 )(9.60 10 3 ) = = 9.21 1011 m / s 2 m (1.67 10 - 27 )
23.47
(a)
t=
(b)
ay =
1 y - yi = v y i t + 2 ay t 2 1 y = 2 (9.21 1011 )(1.11 10 -7 )2 = 5.67 10-3 m = 5.67 mm
(c)
v x = 4.50 105 m/s vy = vy i + ay = (9.21 1011)(1.11 10-7) = 1.02 105 m/s
23.48
ay =
qE (1.602 10 -19 )(390) = = 6.86 1013 m / s 2 - 31 m (9.11 10 ) from projectile motion equations
(a)
t=
2vi sin ay
t=
2(8.20 10 5 )sin 30.0 = 1.20 10-8 s) = 12.0 ns 6.86 1013 vi 2 sin 2 (8.20 10 5 )2 sin 2 30.0 = = 1.23 mm 2ay 2(6.86 1013 ) vi 2 sin 2 (8.20 10 5 )2 sin 60.0 = = 4.24 mm 2ay 2(6.86 1013 )
(b)
h=
(c)
R=
23.49
vi = 9.55 103 m/s (a) ay = eE (1.60 10 -19 )(720) = = 6.90 1010 m s 2 m (1.67 10 -27 ) vi 2 sin 2 = 1.27 10-3 m ay so that 90.0 = 53.1 If = 36.9, t = 167 ns If = 53.1, t = 221 ns (9.55 10 3 )2 sin 2 = 1.27 10 -3 6.90 1010
R=
sin 2 = 0.961 (b) t= R R = vix vi cos
= 36.9
Chapter 23 Solutions The field, E1, due to the 4.00 109 C charge is in the x direction. ke q r2 (8.99 109 N m2/C 2)(- 4.00 109 C) i = -5.75i N/C (2.50 m)2
19
*23.50 (a)
E1 =
=
Likewise, E2 and E3, due to the 5.00 109 C charge and the 3.00 109 C charge are E2 = N/C E3 = (8.99 109 N m2/C 2)(3.00 109 C) i = 18.7 N/C (1.20 m)2 in +x direction. ke q r2 = (8.99 109 N m2/C 2)(5.00 109 C) i (2.00 m)2 = 11.2
ER = E1 + E2 + E3 = 24.2 N/C (b) E1 = ke q r2 ke q r2 ke q r2
= ( -8.46 N / C)(0.243i + 0.970j) = (11.2 N / C)( +j) = ( 5.81 N / C)( -0.371i + 0.928j) Ey = E1y + E2y + E3y = 8.43j N/C
E2 =
E3 =
Ex = E1x + E3x = 4.21i N/C ER = 9.42 N/C
= 63.4 above x axis
23.51
1.60 10 -19 C (640 N / C) qE The proton moves with acceleration ap = = = 6.13 1010 m s 2 -27 m 1.67 10 kg
(
)
while the e- has acceleration
ae =
(1.60 10
-19
C (640 N/C)
-31
9.11 10
)
kg
= 1.12 1014 m s 2 = 1836 ap
(a)
1 We want to find the distance traveled by the proton (i.e., d = 2 apt 2 ), knowing: 1 1 1 4.00 cm = 2 apt 2 + 2 ae t 2 = 1837 2 apt 2
(
)
4.00 cm 1 Thus, d = 2 apt 2 = = 1837
21.8 m
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20
Chapter 23 Solutions
(b)
The distance from the positive plate to where the meeting occurs equals the distance the 1 sodium ion travels (i.e., dNa = 2 aNat 2 ). This is found from: 1 eE 2 1 eE 2 1 1 t + t 4.00 cm = 2 aNat 2 + 2 aClt 2 : 4.00 cm = 2 22.99 u 2 35.45 u This may be written as so
1 1 1 4.00 cm = 2 aNat 2 + 2 (0.649aNa )t 2 = 1.65 2 aNat 2 1 dNa = 2 aNat 2 =
(
)
4.00 cm = 2.43 cm 1.65
23.52
From the free-body diagram shown, and So From Fx = 0, we have or q= qE = T sin 15.0
Fy = 0 T cos 15.0 = 1.96 102 N T = 2.03 102 N
T sin 15.0 (2.03 102 N) sin 15.0 = = 5.25 106 C = 5.25 C E 1.00 103 N/C
23.53
(a)
Let us sum force components to find Fx = qEx T sin = 0, and Fy = qEy + T cos mg = 0
Combining these two equations, we get q= (b) mg (Ex cot + Ey) = (1.00 10-3)(9.80) = 1.09 108 C = 10.9 nC (3.00 cot 37.0 + 5.00) 105 Free Body Diagram for Goal Solution
From the two equations for Fx and Fy we also find T= qEx 3 sin 37.0 = 5.44 10 N = 5.44 mN
Chapter 23 Solutions
21
Goal Solution A charged cork ball of mass 1.00 g is suspended on a light string in the presence of a uniform electric field, as shown in Fig. P23.53. When E = ( 3.00i + 5.00j) 10 5 N / C , the ball is in equilibrium at = 37.0. Find (a) the charge on the ball and (b) the tension in the string. G: (a) Since the electric force must be in the same direction as E, the ball must be positively charged. If we examine the free body diagram that shows the three forces acting on the ball, the sum of which must be zero, we can see that the tension is about half the magnitude of the weight. The tension can be found from applying Newton's second law to this statics problem (electrostatics, in this case!). Since the force vectors are in two dimensions, we must apply F = ma to both the x and y directions. Applying Newton's Second Law in the x and y directions, and noting that F = T + qE + F g = 0, Fx = qEx - T sin 37.0 = 0 Fy = qEy + T cos 37.0 - mg = 0 (1) (2)
O:
A:
We are given Ex = 3.00 10 5 N / C and Ey = 5.00 10 5 N / C; substituting T from (1) into (2): q= mg Ex Ey + tan 37.0 = (1.00 10 -3 kg)(9.80 m / s 2 ) = 1.09 10 -8 C 3.00 5 5.00 + tan 37.0 10 N / C T= qEx = 5.44 10 -3 N sin 37.0
(b) Using this result for q in Equation (1), we find that the tension is L:
The tension is slightly more than half the weight of the ball ( F g = 9.80 10 -3 N) so our result seems reasonable based on our initial prediction.
23.54
(a)
Applying the first condition of equilibrium to the ball gives: Fx = qEx - T sin = 0 and Fy = qEy + T cos - mg = 0 or or T= qEx qA = sin sin
qB + T cos = mg
Substituting from the first equation into the second gives: q( A cot + B) = mg , or (b) q=
( A cot + B)
mg
Substituting the charge into the equation obtained from Fx yields T= mgA mg A = sin A cos + Bsin ( A cot + B)
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22
Chapter 23 Solutions
Goal Solution A charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field, as shown in Figure P23.53. When E = ( Ai + Bj) N / C , where A and B are positive numbers, the ball is i n equilibrium at the angle . Find (a) the charge on the ball and (b) the tension in the string. G: This is the general version of the preceding problem. The known quantities are A , B, m, g , and . The unknowns are q and T . O: The approach to this problem should be the same as for the last problem, but without numbers to substitute for the variables. Likewise, we can use the free body diagram given in the solution to problem 53. A: Again, Newton's second law: and (a) Substituting T = qA , into Eq. (2), sin -T sin + qA = 0 + T cos + qB - mg = 0 qA cos + qB = mg sin q= T= mg A cot + B) ( (1) (2)
Isolating q on the left, (b) Substituting this value into Eq. (1),
( A cos + Bsin )
mgA
L : If we had solved this general problem first, we would only need to substitute the appropriate values in the equations for q and T to find the numerical results needed for problem 53. If you find this problem more difficult than problem 53, the little list at the Gather step is useful. It shows what symbols to think of as known data, and what to consider unknown. The list is a guide for deciding what to solve for in the Analysis step, and for recognizing when we have an answer.
23.55
F=
ke q1q2 r2
15.0 tan = 60.0
= 14.0
F1 = F3 = F2 =
(8.99 109)(10.0 106)2 = 40.0 N (0.150)2 (8.99 109)(10.0 106)2 = 2.50 N (0.600)2 (8.99 109)(10.0 106)2 = 2.35 N (0.619)2
Fx = F3 F2 cos 14.0 = 2.50 2.35 cos 14.0 = 4.78 N Fy = F1 F2 sin 14.0 = 40.0 2.35 sin 14.0 = 40.6 N Fnet = Fx + Fy =
2 2
( 4.78)2 + ( 40.6)2 = 40.9 N
Fy 40.6 tan = F = 4.78 x
= 263
Chapter 23 Solutions 15.0 cm cos 30.0
23
23.56
From Fig. A:
d cos 30.0 = 15.0 cm,
or
d=
From Fig. B:
= sin -1
Fq mg
d 15.0 cm = sin -1 = 20.3 50.0 cm 50.0 cm(cos 30.0)
= tan Figure A (1)
or
Fq = mg tan 20.3 ke q 2 Fq = 2F cos 30.0 = 2 cos 30.0 2 (0.300 m )
From Fig. C:
(2)
Equating equations (1) and (2), mg(0.300 m ) tan 20.3 2ke cos 30.0
2
ke q 2 2 cos 30.0 = mg tan 20.3 2 (0.300 m )
Figure B
q2 =
q2 =
(2.00 10 kg)(9.80 m s )(0.300 m) tan 20.3 2(8.99 10 N m C ) cos 30.0
-3 2 2 9 2 2
q = 4.2 0 10 -14 C 2 = 2.05 10 -7 C= 0.205 C
Figure C
23.57
Charge Q/2 resides on each block, which repel as point charges: F= ke(Q/2)(Q/2) = k(L L i) L2 k ( L - Li ) = 2(0.400 m ) ke
Q = 2L
(8.99 10
(100 N / m)(0.100 m)
9
N m 2 / C2
)
= 26.7 C
23.58
Charge Q /2 resides on each block, which repel as point charges: F = k ( L - Li ) ke
ke (Q 2)(Q 2) L2
= k(L - L i )
Solving for Q ,
Q = 2L
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24 *23.59
Chapter 23 Solutions According to the result of Example 23.7, the lefthand rod creates this field at a distance d from its righthand end: k eQ E = d(2a + d) dx keQQ dF = 2a d(d + 2a) F= k eQ 2 2a
x = b 2a x(x + 2a)
dx
b
=
k eQ 2 1 2a + x b ln 2a 2a x b 2a
F =
+k eQ 2 2a + b b k eQ 2 b2 k e Q 2 b2 ln + ln = 2 ln (b 2a)(b + 2a) = 2 ln 2 2 b b 2a 4a 2 4a 4a b 4a
*23.60
The charge moves with acceleration of magnitude a given by qE 1.60 1019 C (1.00 N/C) = 1.76 1011 m/s2 a= m = 9.11 1031 kg Then v = v i + at = 0 + at gives
F = ma = q E
(a)
v 3.00 107 m/s t=a = = 171 s 1.76 1011 m/s2
(b) (c)
vm (3.00 107 m/s)(1.67 1027 kg) v = 0.313 s t = a = qE = (1.60 1019 C)(1.00 N/C) vm From t = qE , as E increases, t gets shorter in inverse proportion.
23.61
Q = dl = 90.0 0 cos Rd = 0 R sin Q = 12.0 C = (2 0 )(0.600) m = 12.0 C
90.0
90.0 90.0
= 0 R [1 (1)] = 2 0 R
so
0 = 10.0 C/m
( 3.00 C) 0 cos 2 Rd 1 ( 3.00 C)( dl) 1 dFy = cos = 4 e0 4 e0 R2 R2
(
)
90.0 N m2 (3.00 106 C)(10.0 106 C/m) Fy = 90.0 8.99 109 cos2 d (0.600 m) C2
Fy =
8.99( 30.0) 10 -3 N 0.600
(
)(
- /2
/2
1 2
+ cos 2 d
1 2
)
Fy = (0.450 N )
(
1 2
+ sin 2 - /2 = 0.707 N
1 4
/2
)
Downward.
Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0.
Chapter 23 Solutions
25
23.62
At equilibrium, the distance between the charges is
r = 2(0.100 m )sin10.0 = 3.47 10 -2 m
Now consider the forces on the sphere with charge +q , and use Fy = 0: Fy = 0: T cos 10.0 = mg, or T = Fnet = F2 - F1 = T sin10.0 mg cos 10.0 (1) (2) Eliminate T
Fx = 0:
Fnet is the net electrical force on the charged sphere. from (2) by use of (1). Fnet =
mg sin 10.0 = mg tan 10.0 = 2.00 10 -3 kg 9.80 m / s 2 tan 10.0 = 3.46 10 -3 N cos 10.0
(
)(
)
Fnet is the resultant of two forces, F1 and F 2 . F 1 is the attractive force on +q exerted by q, and F 2 is the force exerted on +q by the external electric field. Fnet = F2 F1 or F2 = Fnet + F1 F1 = 8.99 10 N m
9
(
2
(5.00 10 C)(5.00 10 /C ) (3.47 10 m)
-8 2 -3 2
-8
C
) = 1.87 10
-2
N
Thus, F2 = Fnet + F1 yields F2 = 3.46 10 -3 N + 1.87 10 -2 N = 2.21 10 - 2 N and F2 = qE, or E= F2 2.21 10 - 2 N = = 4.43 105 N/C = 443 kN/C q 5.00 10 - 8 C
23.63
(a)
From the 2Q charge we have Combining these we find
Fe - T 2 sin 2 = 0
and
mg - T 2 cos 2 = 0
T sin 2 Fe = 2 = tan 2 mg T 2 cos 2 Fe - T1 sin 1 = 0 and mg - T1 cos 1 = 0 T sin 1 Fe = 1 = tan 1 mg T1 cos 1 or
From the Q charge we have Combining these we find ke 2QQ 2keQ 2 = r2 r2
2 = 1
(b)
Fe =
If we assume is small then equation found in part (a) and solve for r. Fe = tan mg
. Substitute expressions for F e and tan into either
then
and solving for r we find
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26 23.64
Chapter 23 Solutions At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be located by determining the angle corresponding to equilibrium. In terms of lengths s, 1 a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an attractive force 2
1 keQ q (s + 2 a 3)2 . The other two charges exert equal repulsive forces of magnitude keQq r 2 . The horizontal components of the two repulsive forces add, balancing the attractive force,
2 cos 1 =0 Fnet = keQq - 2 1 a 3)2 (s + 2 r From Figure P23.64, r= sin
1a 2
1 s = 2 a cot
The equilibrium condition, in terms of , is Thus the equilibrium value of is
Fnet =
1 4 k Qq 2 cos sin 2 - =0 a2 e ( 3 + cot )2
2 cos sin 2 ( 3 + cot )2 = 1.
One method for solving for is to tabulate the left side. To three significant figures the value of corresponding to equilibrium is 81.7. The distance from the origin to the equilibrium 1 position is x = 2 a( 3 + cot 81.7) = 0.939a
60 70 80 90 81 81.5 81.7
2 cos sin 2 ( 3 + cot )2 4 2.654 1.226 0 1.091 1.024 0.997
23.65
(a)
The distance from each corner to the center of the square is
( L 2 )2 + ( L 2 )2 = L
2
The distance from each positive charge to -Q is then z 2 + L2 2 . Each positive charge exerts a force directed along the line joining q and -Q, of magnitude z
keQq z + L2 2
2
The line of force makes an angle with the z-axis whose cosine is The four charges together exert forces whose x and y components add to zero, while the z-components add to
z + L2 2
2
F= -
(z
4keQ q z
2
+ L2 2
)
32
k
Chapter 23 Solutions 4( 2)3 2 keQq = - z = maz L3
27
(b)
For z << L, the magnitude of this force is
Fz -
( L 2)
2
4keQqz
32
Therefore, the object's vertical acceleration is of the form with 2 = 4( 2) keQq keQq 128 = mL3 mL3
32
az = - 2 z
Since the acceleration of the object is always oppositely directed to its excursion from equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion with a period given by T= 2 2 = (128)1 4 mL3 = keQq
(8)1 4
mL3 keQq
23.66
(a)
The total non-contact force on the cork ball is:
qE , F = qE + mg = m g + m
which is constant and directed downward. Therefore, it behaves like a simple pendulum i n the presence of a modified uniform gravitational field with a period given by: T = 2 L qE g + m = 2 9.80 m / s 2
(2.00 10 +
0.500 m
-6
C 1.00 10 5 N / C
1.00 10 - 3 kg T = 2 L qE m
)(
)
= 0.307 s
(b)
Yes . Without gravity in part (a), we get 0.500 m
T = 2
(2.00 10
-6
C 1.00 10 5 N / C 1.00 10 - 3 kg
)(
)
= 0.314 s (a 2.28% difference).
23.67
(a)
Due to symmetry the field contribution from each negative charge is equal and opposite to each other. Therefore, their contribution to the net field is zero. The field contribution of the +q charge is E= ke q ke q 4k q = = e r 2 ( 3 a 2 4) 3a 2 - 4ke q j 3a2
y
x
in the negative y direction, i.e., E =
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28
Chapter 23 Solutions (b) If Fe = 0, then E at P must equal zero. In order for the field to cancel at P , the - 4q must be above + q on the y-axis. Then, E=0=-
(1.00 m)
ke q
2
+
ke (4q) , which reduces to y 2 = 4.00 m 2 . y2
Thus, y = 2.00 m . Only the positive answer is acceptable since the - 4q must be located above + q. Therefore, the - 4q must be placed 2.00 meters above point P along the + y - axis .
23.68
The bowl exerts a normal force on each bead, directed along the radius line or at 60.0 above the horizontal. Consider the free-body diagram of the bead on the left: Fy = nsin 60.0 -mg = 0 , or Also, or n= mg sin 60.0
n Fe 60.0 mg
Fx = -Fe + ncos 60.0 = 0, ke q 2 mg mg = ncos 60.0 = = 2 tan 60.0 3 R q= mg R ke 3
12
Thus,
23.69
(a)
There are 7 terms which contribute: 3 are s away (along sides) 3 are 1 is 2 s away (face diagonals) and sin = 3 s away (body diagonal) and sin = 1 3 1 = cos 2
The component in each direction is the same by symmetry. F= ke q 2 s2 2 1 1+ 2 2 + 3 3 (i + j + k) = ke q 2 (1.90)(i + j + k) s2
(b)
F= F2 +F2 +F2 = x y z
3.29
ke q 2 away from the origin s2
Chapter 23 Solutions 23.70 (a) Zero contribution from the same face due to symmetry, opposite face contributes 4 ke q sin r2 where r= s s + + s2 = 1.5 s = 1.22 s 2 2
2 2
29
E=4
ke q s kq 4 ke q = = 2.18 e2 s (1.22)3 s2 r3
sin = s/r
(b)
The direction is the k direction.
*23.71
dE =
k -x i + 0.150 m j dx -x i + 0.150 m j ) e ( = 2 3 2 2 2 x + (0.150 m ) x 2 + (0.150 m )2 x 2 + (0.150 m ) ke dq
[
]
dE
y x
0.150 m x dq
E=
all charge
dE = ke
0.400 m
x=0
[x
(-x i + 0.150 m j)dx
2
+ (0.150 m )
2 3 2
]
+i E = ke 2 2 x + (0.150 m )
0.400 m
0
(0.150 m) j x + (0.150 m)2 x 2 + (0.150 m)2
0.400 m
0
N m2 -9 C E = 8.99 10 9 35.0 10 m [ i ( 2.34 - 6.67 ) m + j (6.24 - 0) m ] C2 E = ( -1.36i + 1.96 j ) 10 3 N C =
(-1.36i + 1.96 j ) kN C
23.72
By symmetry
Ex = 0.
q
Using the distances as labeled, + q 2q sin - 2 2 (a + y ) y
2
Ey = ke (a2 + y 2 ) sin
But sin = y
y 1 , so E = Ey = 2ke q 2 - 2 2 32 y (a 2 +y 2 ) (a + y )
Expand (a2 +y 2 )- 3 2 as (a2 + y 2 )- 3 2 = y -3 - (3 2)a2 y -5 + . . . Therefore, for a << y, we can ignore terms in powers higher than 2, 1 3 a2 1 or E = 2ke q 2 - - 2 y4 y2 y k 3qa 2 E = - e 4 j y
and we have
2000 by Harcourt, Inc. All rights reserved.
30
Chapter 23 Solutions ke xQ (x 2 + a2 )3/2
23.73
The field on the axis of the ring is calculated in Example 23.8,
E = Ex =
The force experienced by a charge q placed along the axis of the ring is x F = -keQq 2 2 3/2 (x + a ) and when x << a, this becomes F= keQq x a3
This expression for the force is in the form of Hooke's law, with an effective spring constant of Since = 2 f = k m , we have k = keQq a3 f= 1 2 keQq ma3
23.74
The electrostatic forces exerted on the two charges result in a net torque = -2Fa sin = -2Eqa sin . For small , sin and using p = 2qa, we have The torque produces an angular acceleration given by
= Ep.
= I = I
d 2 dt 2
Combining these two expressions for torque, we have
d 2 Ep + =0 dt 2 I Ep d 2 = - 2 where 2 = 2 I dt
This equation can be written in the form This is the same form as Equation 13.17 and the frequency of oscillation is found by comparison with Equation 13.19, or
f=
1 2
pE = I
1 2
2qaE I
Chapter 24 Solutions
24.1 (a) (b) (c) E = EA cos = (3.50 103)(0.350 0.700) cos 0 = 858 N m2/C
= 90.0
E = 0
E = (3.50 103)(0.350 0.700) cos 40.0 = 657 N m2/C
24.2
E = EA cos = (2.00 104 N/C)(18.0 m2)cos 10.0 = 355 kN m2/C
24.3
E = EA cos A = r 2 = (0.200)2 = 0.126 m2 5.20 105 = E (0.126) cos 0 E = 4.14 106 N/C = 4.14 MN/C
24.4
The uniform field enters the shell on one side and exits on the other so the total flux is zero .
24.5
(a)
A = (10.0 cm)( 30.0 cm) A = 300 cm 2 = 0.0300 m 2 E, A = EA cos E, A = 7.80 10 4 (0.0300) cos 180 E, A = - 2.34 kN m 2 C
3 0.0 cm
(
)
0.0 cm
0.0
(b)
E, A = EA cos = 7.80 10 4 ( A) cos 60.0 10.0 cm 2 2 A = ( 30.0 cm )( w ) = ( 30.0 cm ) = 600 cm = 0.0600 m cos 60.0 E, A = 7.80 10 4 (0.0600) cos 60 =
(
)
(
)
+ 2.34 kN m 2 C
(c)
The bottom and the two triangular sides all lie parallel to E, so E = 0 for each of these. Thus,
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Chapter 24 Solutions E, total = - 2.34 kN m 2 C + 2.34 kN m 2 C + 0 + 0 + 0 = 0
33
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34
Chapter 24 Solutions
24.6
(a) (b) (c)
E = E A = (ai + b j) A i = aA E = (ai + bj) Aj = bA E = (ai + bj) Ak = 0
24.7
Only the charge inside radius R contributes to the total flux. E = q / e0
24.8
E = EA cos through the base E = ( 52.0)( 36.0) cos 180 = 1.87 kN m2/C Note the same number of electric field lines go through the base as go through the pyramid's surface (not counting the base). For the slanting surfaces, E = +1.87 kN m 2 / C
24.9
The flux entering the closed surface equals the flux exiting the surface. The flux entering the left side of the cone is E = E dA = E R h . This is the same as the flux that exits the right side of the cone. Note that for a uniform field only the cross sectional area matters, not shape.
*24.10 (a)
E=
k eQ r2 (8.99 109)Q , (0.750)2 But Q is negative since E points inward.
8.90 102 =
Q = 5.56 108 C = 55.6 nC (b) The negative charge has a spherically symmetric charge distribution.
24.11
(a) (b)
E =
qin ( +5.00 C - 9.00 C + 27.0 C - 84.0 C) = = 6.89 106 N m2/C = 6.89 MN m2/C e0 8.85 10 -12 C 2 / N m 2
Since the net electric flux is negative, more lines enter than leave the surface.
Chapter 24 Solutions qin e0 E = -2Q + Q = e0 - Q e0
35
24.12
E =
Through S1
Through S2 E =
+ Q-Q = 0 e0 E = -2Q + Q - Q 2Q = - e0 e0
Through S3
Through S4
E = 0
24.13
(a)
One-half of the total flux created by the charge q goes through the plane. Thus, E, plane = q 1 1 q E, total = = 2e0 2 2 e0
(b)
The square looks like an infinite plane to a charge very close to the surface. Hence, E, square E, plane = q 2e0
(c)
The plane and the square look the same to the charge.
24.14
The flux through the curved surface is equal to the flux through the flat circle, E0 r 2 .
24.15
(a)
+Q 2 e0 Q 2 e0
Simply consider half of a closed sphere.
(b)
(from , total = , dome + , flat = 0)
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36
Chapter 24 Solutions
Goal Solution A point charge Q is located just above the center of the flat face of a hemisphere of radius R, as shown i n Figure P24.15. What is the electric flux (a) through the curved surface and (b) through the flat face? G: From Gauss's law, the flux through a sphere with a point charge in it should be Q e0 , so we should expect the electric flux through a hemisphere to be half this value: curved = Q 2e0 . Since the flat section appears like an infinite plane to a point just above its surface so that half of all the field lines from the point charge are intercepted by the flat surface, the flux through this section should also equal Q 2e0 . O: We can apply the definition of electric flux directly for part (a) and then use Gauss's law to find the flux for part (b). A : (a) With very small, all points on the hemisphere are nearly at distance R from the charge, so the field everywhere on the curved surface is keQ / R 2 radially outward (normal to the surface). Therefore, the flux is this field strength times the area of half a sphere: Q curved = E dA = Elocal Ahemisphere = ke 2 R
1 ( 2 )(4R2 ) = 41e
Q( 2 ) =
0
Q 2e0
(b) The closed surface encloses zero charge so Gauss's law gives curved + flat = 0 or flat = - curved = -Q 2e0
L : The direct calculations of the electric flux agree with our predictions, except for the negative sign i n part (b), which comes from the fact that the area unit vector is defined as pointing outward from an enclosed surface, and in this case, the electric field has a component in the opposite direction (down).
24.16
(a) (b) (c)
E, shell =
qin 12.0 10 -6 = = 1.36 106 N m 2 / C = 1.36 MN m2/C e0 8.85 10 -12
1 E, half shell = 2 (1.36 106 N m 2 / C) = 6.78 10 5 N m 2 / C = 678 kN m2/C
No, the same number of field lines will pass through each surface, no matter how the radius changes.
24.17
From Gauss's Law, E = E dA = E =
qin . e0 5.22 kN m 2 C
Thus,
Q 0.0462 10 -6 C = = e0 8.85 10 -12 C 2 N m 2
Chapter 24 Solutions If R d, the sphere encloses no charge and E = qin / e0 = 0 If R > d, the length of line falling within the sphere is 2 R 2 - d 2 so = 2 R 2 - d 2 e0
37
24.18
24.19
The total charge is Q - 6 q . The total outward flux from the cube is Q - 6 q / e0 , of which one-sixth goes through each face:
(
)
(E )one face = (E )one face =
Q-6 q 6e0 Q - 6 q (5.00 - 6.00) 10 - 6 C N m 2 = = - 18.8 kN m 2/C 6e0 6 8.85 10 -12 C 2
24.20
The total charge is Q - 6 q . The total outward flux from the cube is Q - 6 q / e0 , of which one-sixth goes through each face:
(
)
(E )one face =
Q-6 q 6e0
24.21
When R < d, the cylinder contains no charge and = 0 . E = qin L = e0 e0
When R > d,
24.22
E, hole = E A hole =
8.99 10 9 N m 2 C 2 10.0 10 -6 C keQ r2 = 1.00 10 -3 m R2 (0.100 m)2
( ) (
)(
) (
)
2
E, hole = 28.2 N m 2 C
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38
Chapter 24 Solutions
Chapter 24 Solutions qin 170 10 -6 C = = 1.92 107 N m 2 C e0 8.85 10 -12 C 2 N m 2 1.92 107 N m 2 C 6
39
24.23
E =
(a)
1 (E )one face = 6 E =
(E )one face
(b) (c)
= 3.20 MN m 2 C
E = 19.2 MN m 2 C The answer to (a) would change because the flux through each face of the cube would not be equal with an unsymmetrical charge distribution. The sides of the cube nearer the charge would have more flux and the ones farther away would have less. The answer to (b) would remain the same, since the overall flux would remain the same.
24.24
(a)
E =
qin e0 qin 8.85 10 -12
8.60 10 4 =
qin = 7.61 10 - 7 C = 761 nC (b) (c) Since the net flux is positive, the net charge must be positive . It can have any distribution. The net charge would have the same magnitude but be negative.
24.25
No charge is inside the cube. The net flux through the cube is zero. Positive flux comes out through the three faces meeting at g. These three faces together fill solid angle equal to one1 eighth of a sphere as seen from q, and together pass flux (q e0 ) . Each face containing a 8 intercepts equal flux going into the cube: 0 = E, net = 3 E, abcd + q / 8e0 E, abcd = -q / 24e0
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40 24.26
Chapter 24 Solutions The charge distributed through the nucleus creates a field at the surface equal to that of a point charge at its center: E = ke q r 2 E= (8.99 109 Nm2/C 2)(82 1.60 1019 C) [(208)1/3 1.20 1015 m] 2 away from the nucleus
E = 2.33 1021 N/C
24.27
(a)
E=
ke Qr = 0 a3 ke Qr a3 = (8.99 109)(26.0 106)(0.100) = 365 kN/C (0.400)3
(b)
E=
(c)
E=
(8.99 109)(26.0 106) ke Q = = 1.46 MN/C 2 r (0.400)2 ke Q (8.99 109)(26.0 106) = = 649 kN/C 2 r (0.600)2
(d) E =
The direction for each electric field is radially outward.
*24.28
(a)
E=
2ke r 2(8.99 109)(Q/2.40) (0.190)
3.60 104 =
Q = + 9.13 107 C = +913 nC (b) E=0
24.29
qin E dA = e 0
=
dV = e lr2 e0 0
E2 rl =
lr2 e0
Chapter 24 Solutions
41
E = 2 e r away from the axis 0
Goal Solution Consider a long cylindrical charge distribution of radius R with a uniform charge density . electric field at distance r from the axis where r < R.
Find the
G: According to Gauss's law, only the charge enclosed within the gaussian surface of radius r needs to be considered. The amount of charge within the gaussian surface will certainly increase as and r increase, but the area of this gaussian surface will also increase, so it is difficult to predict which of these two competing factors will more strongly affect the electric field strength. O: We can find the general equation for E from Gauss's law. A : If is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length L and radius r , contained inside the charged rod. Its volume is r 2 L and it encloses charge r 2 L. The circular end caps have no electric flux through them; there E dA = EdA cos 90.0 = 0. The curved surface has E dA = EdA cos 0 , and E must be the same strength everywhere over the curved surface. Gauss's law,
E dA =
q , becomes e0
E
Curved Surface
dA =
r 2 L e0
E( 2 r )L =
Now the lateral surface area of the cylinder is 2 rL : Thus, E=
r radially away from the cylinder axis 2e0
r 2 L e0
L : As we expected, the electric field will increase as increases, and we can now see that E is also proportional to r . For the region outside the cylinder ( r > R), we should expect the electric field to decrease as r increases, just like for a line of charge.
24.30
= 8.60 10 -6 C / cm 2
(
) 100 cm m )
2
= 8.60 10 -2 C / m 2
E=
8.60 10 -2 = = 2e0 2 8.85 10 -12
(
4.86 10 9 N / C
The field is essentially uniform as long as the distance from the center of the wall to the field point is much less than the dimensions of the wall.
24.31
(a)
E=0
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42
Chapter 24 Solutions keQ (8.99 10 9 )(32.0 10 -6 ) = = 7.19 MN/C r2 (0.200)2
(b)
E=
Chapter 24 Solutions
43
24.32
The distance between centers is 2 5.90 1015 m. Each produces a field as if it were a point charge at its center, and each feels a force as if all its charge were a point at its center. F= k eq 1q 2 r2 N m2 (46)2 (1.60 1019 C)2 = 8.99 109 = 3.50 103 N = 3.50 kN C 2 (2 5.90 1015 m)2
*24.33
Consider two balloons of diameter 0.2 m , each with mass 1 g , hanging apart with a 0.05 m separation on the ends of strings making angles of 10 with the vertical. (a) Fy = T cos 10 - mg = 0 T = mg cos 10
Fx = T sin 10 - Fe = 0 Fe = T sin 10 , so mg 2 Fe = sin 10 = mg tan 10 = (0.001 kg ) 9.8 m s tan 10 cos 10
(
)
Fe 2 10 -3 N (b) Fe = ke q 2 r2
-3
~10 -3 N or 1 mN
2 10
(8.99 10 N (
9
N m 2 C2 q2
2
(0.25 m)
)
q 1.2 10 -7 C E=
~10 -7 C or 100 nC
(c)
8.99 10 9 N m 2 C 2 1.2 10 -7 C ke q 1.7 10 4 N C 2 r2 0.25 m ) ( q 1.2 10 -7 C = 1.4 10 4 N m 2 C e0 8.85 10 -12 C 2 N m 2
)(
)
~10 kN C
(d)
E =
~ 10 kN m 2 C
24.34
(a)
=
Q
4 a3 3
=
5.70 10 -6 = 2.13 10 -2 C / m 3 4 (0.0400)3 3
3 qin = 4 r 3 = 2.13 10 -2 4 (0.0200) = 7.13 10 -7 C = 713 nC 3 3
( (
) ( ) (
)( ) )( )
(b)
3 qin = 4 r 3 = 2.13 10 -2 4 (0.0400) = 5.70 C 3 3
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44
Chapter 24 Solutions
9 2 2 2.00 10 -6 C 7.00 m 2ke 2 8.99 10 N m C E= = r 0.100 m
24.35
(a)
(
)[(
)
]
E = 51.4 kN/C, radially outward (b) E = EA cos = E(2 r )cos 0 E = 5.14 10 4 N C 2 (0.100 m )(0.0200 m )(1.00) = 646 N m 2 C
(
)
24.36 (a)
Note that the electric field in each case is directed radially inward, toward the filament. E=
9 2 2 -6 2ke 2 8.99 10 N m C 90.0 10 C = = 16.2 MN C r 0.100 m
( ( (
)( )( )(
) ) )
(b)
9 2 2 -6 2ke 2 8.99 10 N m C 90.0 10 C E= = = 8.09 MN C r 0.200 m
(c)
E=
9 2 2 -6 2ke 2 8.99 10 N m C 90.0 10 C = = 1.62 MN C r 1.00 m
24.37
E=
9.00 10 - 6 C / m 2 = = 508 kN/C, upward 2e0 2(8.85 10 -12 C 2 / N m 2 )
24.38
From Gauss's Law, EA =
Q e0
Q = A = e0 E = (8.85 10-12)(130)= 1.15 10-9 C/m2 = 1.15 nC/m2
24.39 (a) (b)
qin E dA = E(2 rl ) = e 0
E=
qin/l 2 e0 r
=
2 e0 r
r = 3.00 cm E = 0 r = 10.0 cm E=
inside the conductor 30.0 109 = 5400 N/C, outward 2 (8.85 1012)(0.100) 30.0 109 = 540 N/C, outward 2 (8.85 1012)(1.00)
(c)
r = 100 cm
E=
Chapter 24 Solutions *24.40
45
Just above the aluminum plate (a conductor), the electric field is E = e0 where the charge Q is divided equally between the upper and lower surfaces of the plate: Thus
=
(Q 2) =
A
Q 2A
and
E=
Q 2e0 A
For the glass plate (an insulator), E = / 2e0 where = Q / A since the entire charge Q is o n the upper surface. Therefore, E = Q 2e0 A
The electric field at a point just above the center of the upper surface is the same for each of the plates. E= Q , vertically upward in each case (assuming Q > 0) 2e0 A
*24.41 (a)
E = e0
= (8.00 104)(8.85 1012) = 7.08 10-7 C/m2
= 708 nC/m2 , positive on one face and negative on the other.
(b) Q = A Q = A = (7.08 107) (0.500)2 C
Q = 1.77 107 C = 177 nC , positive on one face and negative on the other.
24.42
Use Gauss's Law to evaluate the electric field in each region, recalling that the electric field is zero everywhere within conducting materials. The results are: E = 0 inside the sphere and inside the shell E = ke Q between sphere and shell, directed radially inward r2 2Q outside the shell, directed radially inward r2 Q is on the outer surface of the sphere . +Q is on the inner surface of the shell ,
E = ke Charge Charge
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46
Chapter 24 Solutions
and
+2Q is on the outer surface of the shell.
Chapter 24 Solutions 24.43
47
The charge divides equally between the identical spheres, with charge Q/2 on each. Then they repel like point charges at their centers: F= k e Q2 ke(Q/2)(Q/2) 8.99 109 N m2(60.0 10-6 C)2 = = = 2.00 N 4(L + 2R)2 4 C 2(2.01 m)2 (L + R + R)2
*24.44
The electric field on the surface of a conductor varies inversely with the radius of curvature of the surface. Thus, the field is most intense where the radius of curvature is smallest and viseversa. The local charge density and the electric field intensity are related by E= (a)
e0
or
= e0E
Where the radius of curvature is the greatest,
= e0Emin = 8.85 10 -12 C 2 N m 2 2.80 10 4 N C = 248 nC m 2
(b) Where the radius of curvature is the smallest,
(
)(
)
= e0Emax = 8.85 10 -12 C 2 N m 2 5.60 10 4 N C = 496 nC m 2
(
)(
)
24.45
(a)
Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting shell is zero, the total charge inside the gaussian surface must be zero, so the inside charge/length = . 0 = + qin qin =
Outside surface: The total charge on the metal cylinder is 2 l = qin + qout . qout = 2 l + l so the outside charge/length = 3 2ke (3) r 6ke 3 = r 2 e0 r
(b)
E=
=
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48
Chapter 24 Solutions 8.99 10 9 6.40 10 -6 keQ E= 2 = = 2.56 MN/C, radially inward r (0.150)2 E=0
24.46
(a)
(
)(
)
(b)
Chapter 24 Solutions 24.47 (a) The charge density on each of the surfaces (upper and lower) of the plate is:
-8 q 1 (4.00 10 C) = =1 = 8.00 10 - 8 C / m 2 = 80.0 nC / m 2 2 A 2 (0.500 m)2
49
(b)
8.00 10 - 8 C / m 2 E= k= k= -12 2 e0 C / N m2 8.85 10 E=
(9.04 kN / C) k
(c)
( - 9.04 kN / C) k
24. 48
(a)
The charge +q at the center induces charge -q on the inner surface of the conductor, where its surface density is:
a =
(b)
-q 4 a2
The outer surface carries charge Q + q with density
b =
Q+q 4 b2
24.49
(a)
E=0 8.99 10 9 8.00 10 -6 keQ = = 7.99 107 N / C = 79.9 MN/C r2 (0.0300)2
(b)
E=
(
)(
)
(c)
E=0 8.99 10 9 4.00 10 -6 keQ = = 7.34 106 N / C = 7.34 MN/C r2 (0.0700)2
(d)
E=
(
)(
)
24.50
An approximate sketch is given at the right. Note that the electric field lines should be perpendicular to the conductor both inside and outside.
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50
Chapter 24 Solutions
24.51
(a)
Uniform E, pointing radially outward, so E = EA. ds = Rd , and the circumference is 2 r = 2 R sin
The arc length is
A = 2 rds = (2 R sin )Rd = 2 R 2 sin d = 2 R 2 (- cos ) 0 = 2 R 2 (1 - cos )
0 0
E =
1 Q Q 2 R 2 (1 - cos ) = (1 - cos ) 2 4 e0 R 2e0 E =
[independent of R!]
(b)
For = 90.0 (hemisphere):
Q (1 - cos 90) = 2e0 Q (1 - cos 180) = 2e0
Q 2e0 Q e0 [Gauss's Law]
(c)
For = 180 (entire sphere):
E =
*24.52
In general, In the xy plane, z = 0 and E = E dA = ( ay i + cx k ) k dA E = ch
w
E = ay i + bz j + cx k E = ay i + cx k
x=0 x=w
2
z y=0 y=h y dA = hdx
x=0
x dx = ch
x 2
2 w x=0
=
chw 2
x
*24.53 (a) (b)
qin = +3Q - Q = +2Q The charge distribution is spherically symmetric and qin > 0 . radially outward . Thus, the field is directed
(c)
E=
ke qin 2keQ = r2 r2
for r c for b < r < c .
(d) Since all points within this region are located inside conducting material, E = 0 (e) (f) (g) E = E dA = 0 qin = e0 E = 0 qin = + 3Q E= ke qin 3keQ = r2 r2 (radially outward) for a r < b
Chapter 24 Solutions +3Q 4 r3 qin = V = 4 3 r 3 = +3Q 3 a 3 a 3 E= ke qin ke r3 r = 2 +3Q 3 = 3keQ 3 2 a r r a (radially outward) for 0 r a
51
(h)
(i) (j)
From part (d), E = 0 for b < r < c . Thus, for a spherical gaussian surface with b < r < c , qin = +3Q + qinner = 0 where qinner is the charge on the inner surface of the conducting shell. This yields qinner = - 3Q
(k) Since the total charge on the conducting shell is E qnet = qouter + qinner = - Q , we have qouter = - Q - qinner = - Q - ( - 3Q) = +2Q (l) This is shown in the figure to the right.
a b c r
24.54
The sphere with large charge creates a strong field to polarize the other sphere. That means it pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side. This patch of opposite charge is smaller in amount but located in a stronger external field, so it can feel a force of attraction that is larger than the repelling force felt by the larger charge in the weaker field on the other side.
24.55
(a)
E dA = E( 4 r
For r < a,
2
)=q
in
e0
qin =
(
4 r3 3
) so
E=
pr 3e0 Q 4 r 2e0
For a < r < b and c < r, For b r c, (b)
qin = Q
so that E =
E = 0, since E = 0
inside a conductor.
Let q 1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the conductor, the total charge enclosed by a spherical surface of radius b r c must be zero. Therefore, q1 + Q = 0 and
1 =
q1 Q = 4 b 2 4 b 2
Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollow sphere is uncharged, we require q1 + q2 = 0 and
2 =
q1 = 4 c2
Q 4 c2
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52
Chapter 24 Solutions
24.56 (a)
E dA = E( 4 r
2
) = qe )
in 0 2 -12
(-3.60 10
3
N C 4 (0.100 m ) = - 4.00 nC
Q C2 N m 2
8.85 10
( a < r < b)
Q = - 4.00 10 -9 C = (b)
We take Q to be the net charge on the hollow sphere. Outside c,
(+ 2.00 10
(c)
2
N C 4 (0.500 m ) =
2
)
Q + Q 8.85 10 -12 C 2 N m 2 + 9.56 nC
( r > c)
Q + Q = + 5.56 10 -9 C, so Q = + 9.56 10 -9 C =
For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner surface of the hollow sphere. Thus, Q1 = -Q = + 4.00 nC Then, if Q2 is the total charge on the outer surface of the hollow sphere, Q2 = Q - Q1 = 9.56 nC - 4.00 nC = + 5.56 nC
24.57
The field direction is radially outward perpendicular to the axis. The field strength depends on r but not on the other cylindrical coordinates or z. Choose a Gaussian cylinder of radius r and length L. If r < a , E = qin e0 and E( 2 rL) =
L e0
E=
2 r e0
or
E=
2 r e0
(r < a)
If a < r < b,
E( 2 rL) =
L + r 2 - a2 L
e0
(
)
E=
+ r 2 - a2
2 r e0
(
) )
( a < r < b)
If r > b ,
E( 2 rL) =
L + b 2 - a2 L
e0
(
Chapter 24 Solutions
53
E=
+ b 2 - a2
2 r e0
(
)
(r > b)
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54
Chapter 24 Solutions
24.58
Consider the field due to a single sheet and let E+ and E represent the fields due to the positive and negative sheets. The field at any distance from each sheet has a magnitude given by Equation 24.8: E+ (a) = E =
2e0
To the left of the positive sheet, E+ is directed toward the left and E toward the right and the net field over this region is E = 0 . In the region between the sheets, E+ and E are both directed toward the right and the net field is E=
(b)
toward the right e0
(c)
To the right of the negative sheet, E+ and E are again oppositely directed and E = 0 .
24.59
The magnitude of the field due to each sheet given by Equation 24.8 is E=
directed perpendicular to the sheet. 2e0
(a)
In the region to the left of the pair of sheets, both fields are directed toward the left and the net field is E=
to the left e0
(b)
In the region between the sheets, the fields due to the individual sheets are oppositely directed and the net field is E= 0
(c)
In the region to the right of the pair of sheets, both fields are directed toward the right and the net field is E=
to the right e0
Chapter 24 Solutions
55
Goal Solution Repeat the calculations for Problem 58 when both sheets have positive uniform charge densities of value . Note: The new problem statement would be as follows: Two infinite, nonconducting sheets of charge are parallel to each other, as shown in Figure P24.58. Both sheets have positive uniform charge densities . Calculate the value of the electric field at points (a) to the left of, (b) in between, and (c) to the right of the two sheets. G: When both sheets have the same charge density, a positive test charge at a point midway between them will experience the same force in opposite directions from each sheet. Therefore, the electric field here will be zero. (We should ask: can we also conclude that the electron will experience equal and oppositely directed forces everywhere in the region between the plates?) Outside the sheets the electric field will point away and should be twice the strength due to one sheet of charge, so E = / e0 in these regions. O: The principle of superposition can be applied to add the electric field vectors due to each sheet of charge. A : For each sheet, the electric field at any point is E = (2e0 ) directed away from the sheet. (a) At a point to the left of the two parallel sheets (b) At a point between the two sheets (c) At a point to the right of the two parallel sheets E = E1( -i) + E2 ( -i) = 2E( -i) = - E = E1i + E2 ( -i) = 0 E = E1i + E2i = 2E i =
i e0
i e0
L : We essentially solved this problem in the Gather information step, so it is no surprise that these results are what we expected. A better check is to confirm that the results are complementary to the case where the plates are oppositely charged (Problem 58).
24.60
The resultant field within the cavity is the superposition of two fields, one E + due to a uniform sphere of positive charge of radius 2a , and the other E - due to a sphere of negative charge of radius a centered within the cavity. 4 r 3 = 4 r 2E+ 3 e0 so E+ = E- = E- =
r r = 3e0 3e0 r1 (- 3e0
1)=
3 4 r1 2 = 4 r1 E- so 3 e0
- r1 3e0
Since r = a + r 1 , E = E+ + E- =
-(r - a) 3e0
r r a a a - + = = 0i + j 3e0 3e0 3e0 3e0 3e0
and Ey =
Thus,
Ex = 0
a 3e0
at all points within the cavity.
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56 24.61
Chapter 24 Solutions First, consider the field at distance r < R from the center of a uniform sphere of positive charge (Q = +e ) with radius R.
(4 r )E = qe
2
in 0
=
3 4 V +e 3 r =4 e0 3 R 3 e0
so
e E= r directed outward 3 4 e0 R
(a)
The force exerted on a point charge q = -e located at distance r from the center is then e2 e F = qE = - e r = - r = -Kr 3 3 4 e0 R 4 e0 R
(b)
K=
e2 k e2 = e3 4 e0 R 3 R
(c)
k e2 k e2 Fr = me ar = - e 3 r, so ar = - e 3 r = - 2 r R me R Thus, the motion is simple harmonic with frequency f=
1 = 2 2
ke e 2 me R 3
(d)
f = 2.47 10
15
1 Hz = 2
(8.99 10
9
(9.11 10
N m 2 C 2 1.60 10 -19 C
-31
)(
kg R 3
)
)
2
which yields R 3 = 1.05 10 -30 m 3 , or R = 1.02 10 -10 m = 102 pm
24.62
The electric field throughout the region is directed along x; therefore, E will be perpendicular to dA over the four faces of the surface which are perpendicular to the yz plane, and E will be parallel to dA over the two faces which are parallel to the yz plane. Therefore, E = - Ex x=a A + Ex x=a+c A = - 3 + 2a2 ab + 3 + 2(a + c)2 ab = 2abc(2a + c) Substituting the given values for a, b, and c, we find E = 0.269 N m2/C Q = 0 E = 2.38 10-12 C = 2.38 pC
(
) (
)
(
)
(
)
24.63
E dA = E(4 r
(a) For r > R,
2
)=
qin e0 AR 5 5 and E= AR 5 5e0 r 2 Ar 3 5e0
qin = Ar 2 (4 r 2 )dr = 4
0 r
R
(b)
For r < R,
qin = Ar 2 (4 r 2 )dr =
0
4 Ar 5 5
and
E=
Chapter 24 Solutions 24.64
57
The total flux through a surface enclosing the charge Q is Q/ e0 . The flux through the disk is disk = E dA where the integration covers the area of the disk. We must evaluate this integral and set it 1 equal to 4 Q/ e0 to find how b and R are related. In the figure, take dA to be the area of an annular ring of radius s and width ds. The flux through dA is E dA = E dA cos = E (2 sds) cos The magnitude of the electric field has the same value at all points within the annular ring, E= 1 Q 1 Q = 4 e0 r 2 4 e0 s2 + b 2 and cos = b b = r (s2 + b 2 )1/2
Integrate from s = 0 to s = R to get the flux through the entire disk. E, disk = Qb 2e0
0
R
s ds Qb = -(s2 + b 2 )1/2 (s + b 2 )3/2 2e0
2
[
]
R 0
=
Q 2e0
b 1 - 2 (R + b 2 )1/2 1 b = . 2 1/2 2 (R + b )
2
The flux through the disk equals Q/4 e0 provided that
This is satisfied if R =
3b .
24.65
E dA =
E4 r 2 =
qin 1 a = 4 r 2 dr e0 e0 r 0 4 a 4 a r 2 r dr = e0 2 e0 0
r
r
E=
a 2e0
= constant magnitude
(The direction is radially outward from center for positive a; radially inward for negative a.)
2000 by Harcourt, Inc. All rights reserved.
58
Chapter 24 Solutions 1
24.66
In this case the charge density is not uniform, and Gauss's law is written as We use a gaussian surface which is a cylinder of radius r, length charge distribution. (a) When r < R, this becomes E( 2 rl) =
r
E dA = e0 dV .
, and is coaxial with the
0 r a - b dV . The element of volume is a cylindrical e0 0 shell of radius r, length l, and thickness dr so that dV = 2 rl dr.
2 r 2l0 a r E( 2 rl) = 2 - 3b e0 so inside the cylinder, E=
` (b)
0 r 2r a- 3b 2e0
When r > R, Gauss's law becomes E( 2 rl) =
0 r a - b (2 rldr ) e0 0
R
or outside the cylinder,
E=
0 R 2 2R a- 3b 2e0 r
24.67
(a)
Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with one end in the yz plane and the other end containing the point x : Use Gauss's law:
y
E dA = ein 0
q
gaussian surface
By symmetry, the electric field is zero in the yz plane and is perpendicular to dA over the wall of the gaussian cylinder. Therefore, the only contribution to the integral is over the end cap containing the point x :
x z x
E dA = ein 0
q
or EA =
( Ax ) e0
E=
so that at distance x from the mid-line of the slab, F ( -e )E = - e x = me me mee0
x e0
(b)
a=
The acceleration of the electron is of the form
a = - 2 x
with
=
e mee0
Thus, the motion is simple harmonic with frequency
f=
1 = 2 2
e mee0
Chapter 24 Solutions 24.68 (a) Consider the gaussian surface described in the solution to problem 67. For x > d , 2 1 dq = dV = A dx = C Ax 2 dx
59
E dA = e0 dq
EA = CA e0
d/2
x 2 dx =
0
1 CA d 3 3 e0 8 E= d Cd 3 i for x > ; 2 24e0
x
E=
Cd 3 24e0
or
E=-
d Cd 3 i for x < - 2 24e0
(b)
For -
d d <x< 2 2
E dA =
CA 2 C Ax 3 1 dq = x dx = e0 3e0 e0 0 Cx 3 i for x < 0 3e0
E=
C x3 i for x > 0; 3e0
E=-
24.69
(a)
A point mass m creates a gravitational acceleration
g=-
Gm r2
at a distance r. Gm 4 r 2 = - 4 Gm r2
The flux of this field through a sphere is
g dA = -
(
)
Since the r has divided out, we can visualize the field as unbroken field lines. The same flux would go through any other closed surface around the mass. If there are several or no masses inside a closed surface, each creates field to make its own contribution to the net flux according to
g dA = - 4 Gmin
(b) Take a spherical gaussian surface of radius r. The field is inward so
g dA = g 4 r
and
2
cos 180 = - g 4 r 2
- 4 Gmin = - 4G 4 r 3 3 and g = 4 r G 3 MEGr RE
3
Then, - g 4 r 2 = - 4 G 4 r 3 3 Or, since
3 = ME / 4 RE , 3
g=
or
g=
MEGr inward 3 RE
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Chapter 25 Solutions
25.1
V = 14.0 V and Q = N A e = (6.02 1023)(1.60 1019 C) = 9.63 104 C V = W 4 Q , so W = Q(V) = ( 9.63 10 C)(14.0 J/C) = 1.35 J
25.2
K = q V q = 6.41 10-19 C
7.37 10-17 = q(115)
25.3
W = K = q V
1 2
mv 2 = e(120 V) = 1.92 1017 J 3.84 10 -17 J m v= 3.84 10 -17 J = 1.52 105 m/s = 152 km/s 1.67 10 -27 kg 3.84 10 -17 J = 6.49 106 m/s = 6.49 Mm/s 9.11 10 -31 kg
Thus, v =
(a)
For a proton, this becomes
(b)
If an electron,
v=
Goal Solution (a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V. (b) Calculate the speed of an electron that is accelerated through the same potential difference. G: Since 120 V is only a modest potential difference, we might expect that the final speed of the particles will be substantially less than the speed of light. We should also expect the speed of the electron to be significantly greater than the proton because, with me << mp , an equal force on both particles will result in a much greater acceleration for the electron. O: Conservation of energy can be applied to this problem to find the final speed from the kinetic energy of the particles. (Review this work-energy theory of motion from Chapter 8 if necessary.)
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Chapter 25 Solutions A:
57
(a) Energy is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V: Ki + U i + Enc = K f + U f
1 0 + qV + 0 = 2 mvp 2 + 0
1J 1 = (1.67 10 -27 kg)vp 2 (1.60 10 -19 C)(120 V) 1 V C 2 vp = 1.52 10 5 m / s (b) The electron will gain speed in moving the other way, from Vi = 0 to V f = 120 V:
Ki + U i + Enc = K f + U f
1 0 + 0 + 0 = 2 mve 2 + qV 1 0 = 2 (9.11 10 -31 kg)ve 2 + (-1.60 10 -19 C)(120 J / C)
ve = 6.49 106 m / s
L:
Both of these speeds are significantly less than the speed of light as expected, which also means that we were justified in not using the relativistic kinetic energy formula. (For precision to three significant digits, the relativistic formula is only needed if v is greater than about 0.1 c.)
25.4
For speeds larger than one-tenth the speed of light, 2 mv 2 gives noticeably wrong answers for kinetic energy, so we use 2 1 1 K = mc 2 - 1 = 9.11 10 -31 kg 3.00 108 m / s - 1 = 7.47 1015 J 2 2 2 1 - 0.400 1- v /c
1
(
)(
)
Energy is conserved during acceleration: Ki + U i + E = Kf + U f 0 + qV i + 0 = 7.47 1015 J + qV f The change in potential is V f V i : V f Vi = 7.47 1015 J 7.47 1015 J = = + 46.7 kV q 1.60 1019 C
The positive answer means that the electron speeds up in moving toward higher potential.
25.5
W = K = - q V
1 0 - 2 9.11 10 -31 kg 4.20 10 5 m / s
(
)(
)
2
= - - 1.60 10 -19 C V
(
)
From which, V = 0.502 V
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58 *25.6
Chapter 25 Solutions (a) We follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm). U = (work done) U = -( work from origin to (20.0 cm, 0) ) - ( work from (20.0 cm, 0) to (20.0 cm, 50.0 cm)
)
Note that the last term is equal to 0 because the force is perpendicular to the displacement. U = (qEx)(x) = (12.0 106 C)(250 V/m)(0.200 m) = 6.00 104 J V = 6.00 104 J U = = 50.0 J/C = 50.0 V q 12.0 106 C
(b)
*25.7
E=
3 V 25.0 10 J/C = = 1.67 106 N/C = 1.67 MN/C 1.50 102 m d
*25.8
(a) (b)
V = Ed = (5.90 103 V/m)(0.0100 m) = 59.0 V
1 2
mv 2 = q(V); f
1 2
(9.11 1031) vf2 = (1.60 1019)(59.0)
vf = 4.55 106 m/s
25.9
U = - m v f 2 - vi 2 = - U = q V: V = 38.9 V
1 2
(
)
1 9.11 10 -31 2
kg 1.40 10 5 m / s
(
) - (3.70 10
2
6
2 m / s = 6.23 10 -18 J
)
+ 6.23 1018 = (1.60 1019)V The origin is at higher potential.
Chapter 25 Solutions
B C B
59
*25.10
VB - V A = - E ds = - E ds - E ds
A A C 0.500
VB - V A = (-Ecos180)
-0.300
dy - (Ecos 90.0)
0.400 -0.200
dx
VB VA = (325)(0.800) = + 260 V 25.11 (a) Arbitrarily choose V = 0 at x = 0 . Then at other points, V = - Ex and U e = QV = -QEx . Between the endpoints of the motion, (K + U s + U e )i = (K + U s + U e ) f
1 2 0 + 0 + 0 = 0 + 2 kxmax - QExmax
so the block comes to rest when the spring is stretched by an amount xmax =
-6 5 2QE 2 50.0 10 C 5.00 10 V m = = 0.500 m k 100 N m
(
)(
)
(b)
At equilibrium, x=
Fx = - Fs + Fe = 0 or kx = QE. Thus, the equilibrium position is at
50.0 10 -6 C 5.00 10 5 N C QE = = 0.250 m k 100 N m
(
)(
)
(c)
The equation of motion for the block is Fx = -kx + QE = m so the equation of motion becomes: d 2 ( x + QE k ) QE + QE = m -k x + , or k dt 2 d2x k x =- 2 m dt
d2x QE QE . Let x = x - , or x = x + 2 k k dt
This is the equation for simple harmonic motion ax = - 2 x , with = k m . The period of the motion is then T= 4.00 kg 2 m = 2 = 2 = 1.26 s k 100 N m
(
)
(d)
(K + U s + U e )i + E = (K + U s + U e ) f
1 2 0 + 0 + 0 - k mgxmax = 0 + 2 kxmax - QExmax
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60
Chapter 25 Solutions
-6 5 2 2(QE - k mg) 2 50.0 10 C 5.00 10 N C - 0.200( 4.00 kg ) 9.80 m s = = = 0.343 m 100 N m k
xmax
[(
)(
)
(
)]
25.12
(a)
Arbitrarily choose V = 0 at 0. Then at other points V = - Ex and U e = QV = -QEx . Between the endpoints of the motion, (K + U s + U e )i = (K + U s + U e ) f
1 2 0 + 0 + 0 = 0 + 2 kxmax - QExmax
so
xmax =
2QE k QE k
(b)
At equilibrium,
Fx = - Fs + Fe = 0 or kx = QE. So the equilibrium position is at x =
(c)
The block's equation of motion is Fx = - kx + QE = m the equation of motion becomes: d 2 ( x + QE k ) QE + QE = m -k x + , or k dt 2 d2x k x =- 2 m dt
QE d2x QE . Let x = x - , or x = x + , so 2 k k dt
This is the equation for simple harmonic motion ax = - 2 x , with = k m The period of the motion is then (d) (K + U s + U e )i + E = (K + U s + U e ) f
1 2 0 + 0 + 0 - k mgxmax = 0 + 2 kxmax - QExmax
(
)
T=
2 m = 2 k
xmax =
2(QE - k mg) k
25.13
For the entire motion,
1 y - yi = vyi t + 2 ay t 2 1 0 - 0 = vi t + 2 ay t 2
so
ay = -
2vi t
Fy = may :
-mg - qE = - E=
2mvi t and E =- m 2vi -g j q t
m 2vi -g q t
For the upward flight:
2 2 vyf = vyi + 2ay (y - yi )
2v 0 = vi2 + 2 - i (ymax - 0) t
and
1 ymax = 4 vi t
Chapter 25 Solutions
y max 0
61
V = V =
y max
0
E dy = +
m 2vi -g y q t
=
m 2vi 1 - g 4 vi t q t
(
)
2.00 kg 2(20.1 m s) 1 - 9.80 m s 2 4 (20.1 m s)( 4.10 s) = 40.2 kV 5.00 10 -6 C 4.10 s
[
]
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62 25.14
Chapter 25 Solutions Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length, V = - Ed and U e = - LEd (a) (K + U)i = (K + U) f
1 0 + 0 = 2 Lv 2 - LEd
v=
2 Ed =
2(40.0 10 - 6 C / m)(100 N / C)(2.00 m) = 0.400 m/s (0.100 kg / m)
(b)
The same.
25.15
Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the original position of the charge is E s = EL cos . At the final point a, V = EL. Suppose the table is frictionless: (K + U)i = (K + U) f
1 0 - qEL cos = 2 mv 2 - qEL
v= =
2 qEL(1 - cos ) = m
2(2.00 10 - 6 C)(300 N / C)(1.50 m)(1 - cos 60.0) = 0.300 m/s 0.0100 kg
*25.16 (a)
The potential at 1.00 cm is q (8.99 109 N m2/C 2)(1.60 1019 C) V 1 = ke r = = 1.44 107 V 1.00 102 m
(b)
The potential at 2.00 cm is q (8.99 109 N m2/C 2)(1.60 1019 C) V 2 = ke = = 0.719 107 V r 2.00 102 m Thus, the difference in potential between the two points is V = V 2 V 1 = 7.19 108 V
(c)
The approach is the same as above except the charge is 1.60 1019 C. This changes the sign of all the answers, with the magnitudes remaining the same. That is, the potential at 1.00 cm is 1.44 107 V The potential at 2.00 cm is 0.719 107 V, so V = V 2 V 1 = 7.19 108 V .
Chapter 25 Solutions
63
25.17
(a) (b)
Since the charges are equal and placed symmetrically, F = 0 Since F = qE = 0, E = 0 q N m2 2.00 106 C V = 2k e r = 2 8.99 109 C 2 0.800 m V = 4.50 104 V = 45.0 kV
(c)
25.18
(a)
Ex =
ke q1 ke q2 + 2 x (x 2.00)2
=0
becomes
2q +q E x = ke 2 + =0 (x 2.00)2 x 2qx2 = q(x 2.00)2 x2 + 4.00x 4.00 = 0
Dividing by ke,
Therefore
E = 0 when x =
-4.00 16.0 + 16.0 = 4.83 m 2
(Note that the positive root does not correspond to a physically valid situation.) (b) ke q2 ke q1 + =0 V= x (2.00 x) Again solving for x, For 0 x 2.00 and or +q 2q V = k e x (2.00 x) = 0 2qx = q(2.00 x) V = 0 when x = 0.667 m q x = 2q 2x
For x < 0
x = 2.00 m
25.19
(a)
U=
ke q1q2 -(8.99 10 9 )(1.60 10 -19 )2 = r 0.0529 10 -9 ke q1q2 r =
= 4.35 1018 J = 27.2 eV
(b)
U=
(8.99 109)(1.60 1019)2 = 6.80 eV 22(0.0529 109)
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64
Chapter 25 Solutions ke q1q2 r ke e 2
(c)
U=
=
= 0
Goal Solution The Bohr model of the hydrogen atom states that the single electron can exist only in certain allowed orbits around the proton. The radius of each Bohr orbit is r = n2 (0.0529 nm) where n = 1, 2, 3, . . . . Calculate the electric potential energy of a hydrogen atom when the electron is in the (a) first allowed orbit, n = 1; (b) second allowed orbit, n = 2; and (c) when the electron has escaped from the atom ( r = ). Express your answers in electron volts. G: We may remember from chemistry that the lowest energy level for hydrogen is E1 = -13.6 eV, and higher energy levels can be found from En = E1 / n2 , so that E2 = -3.40 eV and E = 0 eV . (see section 42.2) Since these are the total energies (potential plus kinetic), the electric potential energy alone should be lower (more negative) because the kinetic energy of the electron must be positive. O: The electric potential energy is given by A: (a) U = ke q1q2 r
For the first allowed Bohr orbit, - 4.35 10 -18 J N m 2 (-1.60 10 -19 C)(1.60 10 -19 C) U = 8.99 10 9 = - 4.35 10 -18 J = = -27.2 eV (0.0529 10 -9 m) C2 1.60 10 -19 J / eV For the second allowed orbit, (-1.60 10 -19 C)(1.60 10 -19 C) U = (8.99 10 9 N m 2 / C 2 ) = -1.088 10 -18 J = -6.80 eV 2 2 (0.0529 10 -9 m) When the electron is at r = ,
(b)
(c)
U = 8.99 10 9 N m 2/ C 2
(
-1.60 10 )(
-19
C 1.60 10 -19 C
)(
)=0J
L : The potential energies appear to be twice the magnitude of the total energy values, so apparently the kinetic energy of the electron has the same absolute magnitude as the total energy. *25.20 (a) U= (5.00 109 C)( 3.00 109 C)(8.99 109 V m/C) qQ = = 3.86 107 J (0.350 m) 4 e0 r
The minus sign means it takes 3.86 107 J to pull the two charges apart from 35 cm to a much larger separation. (b) V= Q1 Q2 (5.00 10 -9 C)(8.99 10 9 V m / C) (-3.00 10 -9 C)(8.99 10 9 V m / C) + = + 0.175 m 0.175 m 4 e0 r1 4 e0 r2
V = 103 V
Chapter 25 Solutions qi ri
65
25.21
V= k
i
1 1 1 V = (8.99 109)(7.00 106) + 0.0100 0.0387 0.0100 V = 1.10 107 C = 11.0 MV
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66
Chapter 25 Solutions 1 q 1 q2 q3 + + 4 0 r 1 r 2 r 3 1 1 N m 2/ C 2 + + 0.600 m 0.150 m
*25.22
U e = q4V 1 + q4V 2 + q4V 3 = q4
U e = 10.0 10 -6 C
(
) (8.99 10
2
9
)
(0.600 m)2 + (0.150 m)2 1
U e= 8.95 J
25.23
U = U1 + U2 + U3 + U4 U = 0 + U 12 + (U 13 + U 23) + (U 14 + U 24 + U 34) U =0+
2 keQ 2 keQ 2 1 1 kQ + + 1 + e 1 + + 1 2 s s s 2 2
U=
ke Q keQ 2 2 4 + = 5.41 s s 2
An alternate way to get the term 4 + 2 diagonal pairs.
(
2 is to recognize that there are 4 side pairs and 2 face
)
*25.24 (a)
V=
8.99 10 9 N m 2 C 2 2.00 10 -6 C ke q1 ke q2 k q + = 2 e = 2 r r1 r2 (1.00 m)2 + (0.500 m)2
2.00 C
(
)(
)
y P (0, 0.500 m) 2.00 C (1.00 m, 0) x
V = 3.22 10 4 V = 32.2 kV (b) J = -9.65 10 -2 J U = qV = -3.00 10 -6 C 3.22 10 4 C
(-1.00 m, 0)
(
)
*25.25
Each charge creates equal potential at the center. The total potential is: k ( -q) 5ke q V = 5 e = - R R
Chapter 25 Solutions *25.26 (a)
67
Each charge separately creates positive potential everywhere. The total potential produced by the three charges together is then the sum of three positive terms. There is no point located at a finite distance from the charges, where this total potential is zero.
(b)
V=
2ke q ke q ke q + = a a a
25.27
(a)
Conservation of momentum: By conservation of energy, and v1 = 2m2 ke q1q2 1 1 - m1 ( m1 + m2 ) r1 + r2 d
m 1v 1 0 = m 1 v 1 i + m 2 v 2 (i) or v 2 = m 2 0+ k e (q 1 )q 2 d ke q1q2 r1 + r2 = 2 m 1v 1 + 2 m 2v 2 ke q1q2 d =
1 2
2 m 1v 1
1
2
1
2
+ +
k e(q 1 )q 2 (r1 + r2)
2 2
1 m 1v 1 2 m2
v1 =
2(0.700 kg ) 8.99 10 9 N m 2/ C 2 2 10 -6 C 3 10 -6 C 1 1 - = 10.8 m/s 8 10 -3 m 1.00 m 0.100 kg )(0.800 kg ) (
(
)(
)(
)
m 1v 1 (0.100 kg)(10.8 m/s) = = 1.55 m/s v2 = m 0.700 kg 2 (b) If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .
25.28
(a)
Conservation of momentum: By conservation of energy,
0 = m1v1 i + m2 v2 (- i) 0+
or
v2 = m1v1 / m2
ke (- q1 )q2 1 k (- q1 )q2 2 1 = 2 m1v 1 + 2 m2 v 2 + e 2 d (r1 + r2 )
and 2m2 ke q1q2 1 1 r + r - d m1(m1 + m2 ) 1 2
2 2 ke q1q2 kqq 1 1 m1 v1 2 - e 1 2 = m1v 1 + 2 2 m2 r1 + r2 d
v1 = (b)
m v2 = 1 v1 = m2
2m1ke q1q2 1 1 r + r - d m2 (m1 + m2 ) 1 2
If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .
2000 by Harcourt, Inc. All rights reserved.
68
Chapter 25 Solutions 8.99 10 9 N m 2 C 2 8.00 10 -9 C ke Q 72.0 V m r= = = V V V r = 0.720 m, 1.44 m, and 2.88 m to the potential.
25.29
k Q V= e r
so
(
)(
)
For V = 100 V, 50.0 V, and 25.0 V, The radii are inversely proportional
25.30
(a)
V (x) =
ke ( +Q) ke ( +Q) ke Q1 ke Q2 + = + 2 r1 r2 x 2 + a2 x 2 + ( -a) ke Q = a x 2 + a2 2ke Q 2 2 ( x a) + 1 2
V (x) =
V (x) = ( ke Q a) V( y) = V( y) =
( x a) 2 + 1
(b)
ke Q1 ke Q2 ke ( +Q) ke ( -Q) + = + r1 r2 y-a y+a ke Q 1 1 y a - 1 - y a + 1 a
1 V( y) 1 = - ( ke Q a) y a - 1 y a + 1
25.31
Using conservation of energy, we have K f + U f = Ki + U i . But Ui = ke q qgold ri , and ri . Thus, U i = 0. ke q qgold rmin
Also K f = 0 ( v f = 0 at turning point), so U f = Ki ,
or
2 1 = 2 m v
rmin =
2ke q qgold
2 m v
=
2(8.99 10 9 N m 2 / C 2 )(2)(79) 1.60 10 -19 C (6.64 10 - 27 kg)(2.00 107 m / s)2
(
)
2
= 2.74 10 -14 m = 27.4 fm
Chapter 25 Solutions 25.32 Using conservation of energy we have: ke eQ k eQ 1 = e + 2 mv 2 r1 r2 v= 1 2ke eQ 1 - r2 m r1
69
which gives:
or
v=
1 (2)(8.99 10 9 N m 2 / C 2 )(-1.60 10 -19 C)(10 - 9 C) 1 - -31 0.0300 m 0.0200 m 9.11 10 kg 7.26 106 m / s
Thus,
v=
25.33
U=
ke qi q j ri j
, summed over all pairs of (i, j ) where i j q( 3q) a2 + b 2 2q( - 2q) a2 + b 2
q( - 2q) ( - 2q)( 3q) ( 2q)( 3q) q( 2q) U = ke + + + + a b a b
+
6 6 2 3 4 -2 U = ke q 2 - + + + - 0.400 0.200 0.400 0.200 0.447 0.447
2 4 4 1 = 3.96 J U = 8.99 10 9 6.00 10 - 6 - - 0.400 0.200 0.447
(
)(
)
25.34
Each charge moves off on its diagonal line. All charges have equal speeds.
(K + U)i
0+
= (K + U) f
4 ke q 2 2 ke q 2 4 k q 2 2 ke q 2 1 + = 4 2 mv 2 + e + L 2L 2L 2 2L
(
)
1 ke q 2 = 2 mv 2 2 + 2 L 1 ke q 2 v = 1 + 8 mL
2000 by Harcourt, Inc. All rights reserved.
70
Chapter 25 Solutions A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 6 = 12 face diagonal pairs separated by U= ke q 2 s 2 s, and 4 interior diagonal pairs separated 3 s.
25.35
k q2 12 4 12 + + = 22.8 e s 2 3
25.36 (a)
V = a + bx = 10.0 V + ( -7.00 V m )x At x = 0, V = 10.0 V At x = 3.00 m, V = At x = 6.00 m, V = (b) E=- - 11.0 V - 32.0 V
dV = -b = -( -7.00 V m ) = 7.00 N C in + x direction dx
25.37
V = 5x 3x2y + 2yz2 Ex =
Evaluate E at (1, 0 2) = 5 + 6(1)(0) = 5
V = 5 + 6xy x
Ey =
V = +3x 2 2z2 = 3(1)2 2(2)2 = 5 y V = 4yz z
= 4(0)(2) = 0
Ez =
E = Ex 2 + Ey 2 + Ez 2 =
( -5)2 + ( -5)2 + 02
= 7.07 N/C
25.38
(a)
For r < R Er = - dV = dr
V= 0 V=
keQ R
(b)
For r R Er = -
keQ r keQ r2
dV k Q = - - e2 = r dr
Chapter 25 Solutions
2 2 V k Q l+ l + y = - e ln y l y y
71
25.39
Ey = -
Ey =
y2 keQ keQ 1 - = 2 2 2 2 ly l + y + l l + y y l2 + y 2
25.40
Inside the sphere, Outside, So Ex = -
Ex = Ey = Ez = 0 .
V =- V0 - E0 z + E0 a3 z(x 2 + y 2 + z 2 )- 3/2 x x
(
)
Ex = - 0 + 0 + E0 a3 z(- 3 / 2)(x 2 + y 2 + z 2 )- 5/2 (2x)
[
]
= 3E0 a3 xz(x 2 + y 2 + z 2 )- 5/2
Ey = -
V =- V0 - E0 z + E0 a3 z(x 2 + y 2 + z 2 )- 3/2 y y
(
)
- 5/2
Ey= - E0 a3 z(- 3 / 2)(x 2 + y 2 + z 2 )- 5/2 2y = 3E0 a3 yz(x 2 + y 2 + z 2 ) Ez = -
V = E0 - E0 a3 z(- 3 / 2)(x 2 + y 2 + z 2 )- 5/2 (2z) - E0 a3 (x 2 + y 2 + z 2 )- 3/2 z
Ez = E0 + E0 a3 (2z 2 - x 2 - y 2 )(x 2 + y 2 + z 2 )- 5/2
*25.41
V = V 2R - V0 =
keQ R + ( 2R)
2 2
-
k eQ keQ keQ 1 = - 1 = 0.553 R 5 R R
*25.42
V = dV =
dq 1 r 4 e0
All bits of charge are at the same distance from O, so
2 -7.50 10 -6 C 1 Q 9 Nm V= = 8.99 10 = 1.51 MV 4 e0 R C 2 (0.140 m / )
(
)
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72
Chapter 25 Solutions C [] = x = m
25.43
(a)
1 C = m m2
(b)
L dx x dx L d q V = ke = ke r = ke = ke L - d ln 1 + r d (d + x) 0
25.44
V=
x dx ke dq = ke 2 2 r b + (L 2 - x)
L - x. 2 Then x= L - z, 2 2 and dz b2 + z2
L 0
Let z =
dx = -dz + ke + ke z dz b2 + z2 = - ke L ln(z + z 2 + b 2 ) + ke z 2 + b 2 2
L 0
V = ke V=-
(L 2 - z)(-dz) = - ke L
b2 + z2
ke L ln ( L 2 - x ) + 2
( L 2 - x )2 + b 2
( L 2 - x )2 + b 2
2 2 ke L L 2 - L + ( L 2 ) + b + ke V=- ln 2 2 2 L 2 + ( L 2) + b 2 2 ke L b + (L 4) - L ln 2 b 2 + (L2 4) + L
( L 2 - L)2 + b 2 - ( L 2 )2 + b 2
V= -
2 2
25.45
dV =
ke dq r2 + x2
b
where dq = dA = 2 r dr r dr 2 ke x 2 + b 2 - x 2 + a 2
V = 2 ke
a
r +x
2
2
=
25.46
V = ke
all charge
- R dx 3R dx ds dq = ke + ke + ke - 3R - x semicircle R R r x -R - 3R 3R ke R + ke ln x R R
V = - ke ln(- x) V = ke ln
+
3R + ke + ke ln 3 = ke ( + 2 ln 3) R
Chapter 25 Solutions (8.99 109 Nm2/C2) q (0.300 m)
73
25.47
Substituting given values into
V=
ke q , r
7.50 103 V =
Substituting
q = 2.50 10 -7 C,
N=
2.50 10-7 C = 1.56 1012 electrons 1.60 10-19 C/e-
25.48
q1 + q2 = 20.0 C q1 r1 =r q2 2 Therefore Solving, E1 = q2 = 12.0 C
so so
q1 = 20.0 C q2 20.0 C q2 4.00 cm = q2 6.00 cm 6.00(20.0 C q2) = 4.00q2 ;
and
q1 = 20.0 C 12.0 C = 8.00 C
(a)
8.99 10 9 8.00 10 -6 ke q1 = = 4.50 107 V / m = 45.0 MV/m 2 r12 0.0400) (
(
)(
)
8.99 10 9 12.0 10 -6 ke q2 E2 = 2 = = 3.00 107 V / m = 30.0 MV/m 2 r2 (0.0600) (b) V1 = V 2 = ke q2 = 1.80 MV r2
(
)(
)
25.49
(a)
E= 0 ;
V=
ke q (8.99 10 9 )(26.0 10 -6 ) = = 1.67 MV R 0.140 away
(b)
E=
ke q (8.99 10 9 )(26.0 10 -6 ) = = 5.84 MN/C (0.200)2 r2 ke q (8.99 10 9 )(26.0 10 -6 ) = = 1.17 MV r (0.200)
V=
(c)
E= V=
ke q (8.99 10 9 )(26.0 10 -6 ) = = 11.9 MN/C away R2 (0.140)2 ke q = 1.67 MV R
25.50
No charge stays on the inner sphere in equilibrium. If there were any, it would create an electric field in the wire to push more charge to the outer sphere. Charge Q is on the outer sphere. Therefore, zero charge is on the inner sphere and 10.0 C is on the outer sphere .
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74
Chapter 25 Solutions keQ keQ 1 1 = = Vmax r r r r2
25.51
(a)
Emax = 3.00 106 V / m =
Vmax = Emax r = 3.00 106 (0.150) = 450 kV (b) keQmax = Emax r2 Qmax = keQmax = Vmax or r
Emax r 2 3.00 106 (0.150)2 = = 7.51 C ke 8.99 10 9
Goal Solution Consider a Van de Graaff generator with a 30.0-cm-diameter dome operating in dry air. (a) What is the maximum potential of the dome? (b) What is the maximum charge on the dome? G: Van de Graaff generators produce voltages that can make your hair stand on end, somewhere on the order of about 100 kV (see the Puzzler at beginning of Chapter 25). With these high voltages, the maximum charge on the dome is probably more than typical point charge values of about 1 C . The maximum potential and charge will be limited by the electric field strength at which the air surrounding the dome will ionize. This critical value is determined by the dielectric strength of air which, from page 789 or from Table 26.1, is Ecritical = 3 106 V / m . An electric field stronger than this will cause the air to act like a conductor instead of an insulator. This process is called dielectric breakdown and may be seen as a spark. O: From the maximum allowed electric field, we can find the charge and potential that would create this situation. Since we are only given the diameter of the dome, we will assume that the conductor is spherical, which allows us to use the electric field and potential equations for a spherical conductor. With these equations, it will be easier to do part (b) first and use the result for part (a). A : (b) For a spherical conductor with total charge Q , Q= V= E = keQ r2
3.00 106 V / m (0.150 m ) Er 2 = (1 N m / V C) = 7.51 C ke 8.99 10 9 N m 2 / C 2
2
(
)
(a)
keQ (8.99 10 9 N m 2 / C 2 )(7.51 10 -6 C) = = 450 kV r 0.150 m
L : These calculated results seem reasonable based on our predictions. The voltage is about 4000 times larger than the 120 V found from common electrical outlets, but the charge is similar in magnitude to many of the static charge problems we have solved earlier. This implies that most of these charge configurations would have to be in a vacuum because the electric field near these point charges would be strong enough to cause sparking in air. (Example: A charged ball with Q = 1 C and r = 1 mm would have an electric field near its surface of 9 10 9 N m 2/ C 2 1 10 -6 C keQ E= 2 = = 9 10 9 V / m r (0.001 m)2 which is well beyond the dielectric breakdown of air!)
(
)(
)
Chapter 25 Solutions ke q r ke q r2 V , r and
75
25.52
V=
and E =
Since E =
(b)
r=
6.00 105 V V = = 0.200 m E 3.00 106 V/m Vr k e = 13.3 C
(a) q =
25.53
U = qV = k e
q1q2 r 12
= (8.99 109)
(38)(54)(1.60 1019)2 (5.50 + 6.20) 1015
= 4.04 1011 J = 253 MeV
*25.54 (a)
To make a spark 5 mm long in dry air between flat metal plates requires potential difference V = Ed = 3.0 106 V m 5.0 10 -3 m = 1.5 10 4 V ~10 4 V
(
)(
)
(b)
Suppose your surface area is like that of a 70-kg cylinder with the density of water and radius 12 cm. Its length would be given by 70 10 3 cm 3 = (12 cm ) l
2
l = 1.6 m
The lateral surface area is A = 2 r l = 2 (0.12 m )(1.6 m ) = 1.2 m 2 The electric field close to your skin is described by E= Q = , so e0 Ae0 ~10 -5 C
N C2 1.2 m 2 8.85 10 -12 Q = EA 0 = 3.0 106 C N m2
(
)
25.55
(a)
1 2 1 V = ke Q x + x a x + a x(x a) 2(x + a)(x a) + x(x + a) 2k e Qa 2 = 3 V = ke Q x(x + a)(x a) x xa 2 2k e Qa 2 x3 a for x << 1
(b)
V=
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76
Chapter 25 Solutions
25.56
(a)
Ex = -
dV d 2k Qa2 (2keQa2 )(3x 2 - a2 ) = - 3 e 2 = dx dx x - xa (x 3 - xa 2 )2
and
Ey = Ez = 0
(b)
Ex =
2(8.99 10 9 N m 2/ C 2 )(3 10 -6 C)(2 10 -3 m)2 3(6 10 -3 m)2 - (2 10 -3 m)2
[(6 10
[
-3
m) - (6 10
3
-3
m)(2 10
-3
m)
2 2
]
]
Ex = 609 106 N/C = 609 MN/C
25.57
(a)
E=
Q 4 0 r 2 Q 4 0 r V 3000 V = = 6.00 m 500 V / m E Q 4 0 (6.00 m)
V=
r=
(b)
V = -3000 V =
Q=
-3000 V (6.00 m) = 2.00 C (8.99 10 9 V m / C)
25.58
From Example 25.5, the potential created by the ring at the electron's starting point is Vi = keQ xi2 +a
2
=
ke ( 2 a) xi2 + a2 V f = 2 ke . From conservation of energy,
while at the center, it is
1 0 + ( -eVi ) = 2 me v 2 + -eV f f
(
)
a 1 - xi2 + a2
v2 = f
4 eke 2e V f - Vi = me me
(
)
v2 f
4 1.60 10 -19 8.99 10 9 1.00 10 -7 = 1 - 9.11 10 -31
(
)(
)(
)
(0.100)2 + (0.200)2 0.200
v f = 1.45 107 m/s
Chapter 25 Solutions 25.59 (a)
77
Take the origin at the point where we will find the potential. One ring, of width dx, has charge Q dx/h and, according to Example 25.5, creates potential dV = keQ dx h x 2 + R2
The whole stack of rings creates potential V = dV (b) =
d+h
keQ dx h x2 + R2
all charge
d
=
ke Q ln x + x 2 + R 2 h
d+h d
=
2 2 ke Q d + h + (d + h) + R ln h d + d2 + R2
A disk of thickness dx has charge Q dx/h and charge-per-area Q dx/ R 2 h. Example 25.6, it creates potential dV = 2 ke Q dx x 2 + R 2 - x R2h
According to
Integrating, V=
d+ h
d
2 ke Q 2keQ 2 x + R 2 dx - x dx = 2 2 R h R h
1 x2 R2 2 2 ln x + x 2 + R 2 - x x +R + 2 2 2 d
d+h
V=
d + h + (d + h)2 + R 2 keQ (d + h) (d + h)2 + R 2 - d d 2 + R 2 - 2 dh - h 2 + R 2 ln R 2h d + d2 + R2
25.60
The positive plate by itself creates a field
E=
36.0 109 C/m2 kN = = 2.03 C 12 2 2 2 0 2(8.85 10 C /N m )
away from the + plate. The negative plate by itself creates the same size field and between the plates it is in the same direction. Together the plates create a uniform field 4.07 kN/C in the space between. (a) Take V = 0 at the negative plate. The potential at the positive plate is then V - 0 = -
12.0 cm
0
( -4.07 kN / C) dx
The potential difference between the plates is V = (4.07 103 N/C)(0.120 m) = 488 V (b) 1 mv 2 + q V 2
1 = 2 mv 2 + q V i 1
f 2
qV = (1.60 1019 C)(488 V) = 2 m v f = 7.81 1017 J (c) v f = 306 km/s
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78
Chapter 25 Solutions (d) v f 2 = v i
2
+ 2a(x xi)
(3.06 105 m/s)2 = 0 + 2a(0.120 m) a = 3.90 1011 m/s2 (e) F = ma = (1.67 1027 kg)(3.90 1011 m/s2) = 6.51 1016 N 6.51 1016 N F = 4.07 kN/C E=q = 1.60 1019 C
(f)
25.61
ke q W = V dq where V = R ;
0
Q
Therefore,
k eQ 2 W = 2R
25.62
(a)
charged rod (where r > radius of charged rod) is 2k E= = e 2 e0 r r
VB - V A = - E ds and the field at distance r from a uniformly
A
B
ra rb
-
In this case, the field between the central wire and the coaxial cylinder is directed perpendicular to the line of charge so that VB - V A = - (b)
rb
ra
r 2ke dr = 2ke ln a , r rb
or
r V = 2ke ln a rb
From part (a), when the outer cylinder is considered to be at zero potential, the potential at a distance r from the axis is V = 2ke ln ra r r r a 2k V = - 2ke - 2 = e r r ra r Therefore, E= V 1 ln(r a rb ) r
The field at r is given by E = -
But, from part (a), 2ke =
V . ln(r a rb )
Chapter 25 Solutions
r2 r2
79
25.63
V 2 - V1 = - E dr = -
r1
r1
dr 2 0r
V2 V1 =
r - ln 2 2 0 r1
25.64
For the given charge distribution,
V ( x, y, z) = r1 =
ke ( q) ke ( -2q) + r1 r2 and r2 = x 2 + y 2 + z 2
where The surface on which is given by
( x + R )2 + y 2 + z 2
V ( x, y, z) = 0 1 2 ke q - = 0, or 2r1 = r2 r1 r2 4( x + R) + 4y 2 + 4z 2 = x 2 + y 2 + z 2
2
This gives: which may be written in the form:
x2 + y 2 + z2 +
4 2 8 R x + (0) y + (0) z + R =0 3 3
[1]
The general equation for a sphere of radius a centered at ( x0 , y0 , z0 ) is:
(x - x0 )2 + ( y - y0 )2 + ( z - z0 )2 - a2 = 0
or
2 2 2 x 2 + y 2 + z 2 + ( -2x0 )x + ( -2y0 ) y + ( -2z0 )z + x0 + y0 + z0 - a2 = 0
(
)
[2]
Comparing equations [1] and [2], it is seen that the equipotential surface for which V = 0 is indeed a sphere and that: -2x0 = 8 R; 3 -2y0 = 0; y0 = z0 = 0, -2z0 = 0; and a2 =
2 2 2 x0 + y0 + z0 - a2 =
4 2 R 3
4 Thus, x0 = - R, 3
16 4 2 4 2 - R = R . 9 3 9 4 - R, 0, 0 3 , having a radius 2 R 3
The equipotential surface is therefore a sphere centered at
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80
Chapter 25 Solutions
25.65
(a)
From Gauss's law, EB = ke qA (1.00 10 - 8 ) = (8.99 10 9 ) = r2 r2
EA = 0
(no charge within)
89.9 V/m r2 45.0 - 2 V/m r 45.0 V - r
EC = ke
(qA + qB ) (- 5.00 10 - 9 ) = (8.99 10 9 ) = r2 r2 (qA + qB ) (- 5.00 10 - 9 ) = (8.99 10 9 ) = r r 45.0 = - 150 V 0.300
r
(b)
VC = k e
At r2 , V = -
Inside r2 , VB = - 150 V + At r1 , V = - 450 +
r2
89.9 1 89.9 1 = - 450 + V dr = - 150 + 89.9 - 2 r 0.300 r r so V A = + 150 V
89.9 = + 150 V 0.150
25.66
From Example 25.5, the potential at the center of the ring is Vi = keQ R and the potential at an infinite distance from the ring is V f = 0. Thus, the initial and final potential energies of the point charge are: U i = QVi = keQ 2 R and U f = QV f = 0
From conservation of energy, K f + U f = Ki + U i or
1 Mv 2 f 2
+0=0+
keQ 2 R
giving
vf =
2keQ 2 MR
25.67
The sheet creates a field E1 = field E2 =
i for x > 0 . Along the x - axis, the line of charge creates a 2 0
away = (- i) for x < 3.00 m 2 r 0 2 0 (3.00 m - x)
The total field along the x - axis in the region 0 < x < 3.00 m is then E = E1 + E2 = - i 2 0 2 0 ( 3.00 - x )
Chapter 25 Solutions (a) The potential at point x follows from V - V0 = - V = V0 -
81
0
x
E idx = -
0 2 0 - 2 0 (3.00 - x) dx
x
x x - ln 1 - 3.00 2 0 2 0
(25.0 10 -9 C / m 2 )x 80.0 10 -9 C / m x - ln 1 - -12 2 2 -12 2 2 3.00 C / N m ) 2 (8.85 10 C /N m ) 2(8.85 10
V = 1.00 kV -
x kV x - (1.44 kV) ln 1.00 - V = 1.00 kV - 1.41 m 3.00 m (b) At x = 0.800 m, and V = 316 V
U = QV = 2.00 10 -9 C ( 316 J C) = 6.33 10 -7 J = 633 nJ
(
)
25.68
V = ke
a+L
a
= ke ln x + (x 2 + b 2 ) 2 2 x +b
dx
a+L a
a + L + ( a + L)2 + b 2 = ke ln a + a2 + b 2
25.69
(a)
Er = -
2ke p cos V = r r3 1 . r
In spherical coordinates, the component of the gradient is E = - k p sin 1 V = e 3 r r 2ke p r3 and Er (90) = 0 ,
Therefore,
For r >> a, Er (0) =
E (0) = 0
and
E (90) =
ke p r3
These results are reasonable for r >> a . However, for r 0, E(0) . ke py (x 2 + y 2 )3/2 Ex = - 3ke pxy V = x (x 2 + y 2 )5/2
(b)
V=
and
Ey = -
k p(2y 2 - x 2 ) V = e 2 y (x + y 2 )5/2
2000 by Harcourt, Inc. All rights reserved.
82
Chapter 25 Solutions V s down
25.70
(a)
EA > EB since E = V s
(b) (c)
EB =
=
(6 2) V = 200 N/C 2 cm
The figure is shown to the right, with sample field lines sketched in.
25.71
For an element of area which is a ring of radius r and width dr, dq = dA = Cr (2 r dr) and V = C(2 ke )
R 0
dV =
ke dq r2 + x2
r 2 dr r2 + x2
x = C( ke )R R 2 + x 2 + x 2 ln R + R2 + x2
25.72
dU = V dq where the potential V =
ke q r . or dq = (4 r 2 dr) and the
The element of charge in a shell is dq = (volume element) charge q in a sphere of radius r is
r 4 r 3 q = 4 r 2 dr = 3 0
Substituting this into the expression for dU, we have dU = 4 r 3 1 16 2 2 4 ke q (4 r 2 dr) = ke dq = ke r dr r 3 3 r
16 2 2 R 4 16 2 2 5 U = dU = ke r dr = ke 15 R 3 0 But the total charge, Q = 4 R 3 . 3 Therefore, U= 3 keQ 2 5 R
Chapter 25 Solutions
83
*25.73 (a)
From Problem 62,
E=
V 1 ln(r a rb ) r V 50.0 10 3 V 1 = m 0.850 m rb ln rb
We require just outside the central wire
5.50 106
or
(110 m )r ln 0.850 m = 1 r
-1 b b
We solve by homing in on the required value rb (m) 0.0100 4.89 0.00100 0.740 0.00150 1.05 0.00145 1.017 0.00143 1.005 0.00142 0.999
(110 m )
-1
0.850 m rb ln rb
Thus, to three significant figures, (b) At r a , E =
rb = 1.42 mm
50.0 kV 1 = 9.20 kV m ln(0.850 m 0.00142 m ) 0.850 m
2000 by Harcourt, Inc. All rights reserved.
Chapter 26 Solutions
*26.1
(a) (b)
Q = C (V) = (4.00 106 F)(12.0 V) = 4.80 105 C = 48.0 C Q = C (V) = (4.00 106 F)(1.50 V) = 6.00 106 C = 6.00 C
26.2
(a)
C=
10.0 10 - 6 C Q = = 1.00 10 - 6 F = 1.00 F V 10.0 V Q 100 10 - 6 C = = 100 V C 1.00 10 - 6 F
(b)
V =
26.3
E=
ke q : r2
q=
(4.90 10 4 N / C)(0.210 m)2 = 0.240 C 8.99 10 9 N m 2 / C 2
(a) (b)
=
q 0.240 10 -6 = = 1.33 C/m2 A 4 (0.120)2
C = 4 e0 r = 4 (8.85 10 -12 )(0.120) = 13.3 pF
26.4
(a)
C = 4 e0 R R= C = ke C = (8.99 109 N m2/C 2)(1.00 1012 F) = 8.99 mm 4 e0 4 (8.85 1012 C 2)(2.00 103 m) = 0.222 pF N m2
(b) (c)
C = 4 e0 R =
Q = CV = (2.22 1013 F)(100 V) = 2.22 1011 C
26.5
(a)
Q1 R1 = Q2 R2 R Q1 + Q2 = 1 + 1 Q2 = 3.50Q2 = 7.00 C R2 Q2 = 2.00 C Q1= 5.00 C
(b)
V1 = V 2 =
5.00 C Q1 Q2 = = = 8.99 10 4 V = 89.9 kV -1 9 C1 C2 8.99 10 m F (0.500 m)
(
)
2000 by Harcourt, Inc. All rights reserved.
Chapter 26 Solutions
85
*26.6
C=
e0 A (1.00)(8.85 1012 C 2)(1.00 103 m)2 = = 11.1 nF d N m2(800 m)
The potential between ground and cloud is V = Ed = (3.00 106 N/C)(800 m) = 2.40 109 V Q = C (V) = (11.1 10-9 C/V)(2.40 109 V) = 26.6 C
26.7
(a)
V = Ed E= E= 20.0 V 1.80 103 m = 11.1 kV/m
(b)
e0
= (1.11 104 N/C)(8.85 1012 C 2/N m2) = 98.3 nC/m2
8.85 10 -12 C 2 / N m 2 7.60 cm 2 (1.00 m / 100 cm ) e0 A C= = = 3.74 pF d 1.80 10 -3 m
2
(c)
(
)(
)
Q (d) V = C
Q = (20.0 V)(3.74 1012 F) = 74.7 pC
26.8
C=
e0 A = 60.0 10 -15 F d
-12 21.0 10 -12 e0 A (1) 8.85 10 = C 60.0 10 -15
d=
(
)(
)
d = 3.10 10 -9 m = 3.10 nm
26.9
Q=
e0 A ( V ) d
e ( V ) Q = = 0 A d
8.85 10 -12 C 2 N m 2 (150 V ) e0 ( V ) d= = = 4.42 m 30.0 10 - 9 C cm 2 1.00 10 4 cm 2 m 2
(
(
)(
)
)
2000 by Harcourt, Inc. All rights reserved.
86 26.10
Chapter 26 Solutions With = , the plates are out of mesh and the overlap area is zero. With = 0, the overlap area is that of a semi-circle, R 2 2 . By proportion, the effective area of a single sheet of charge is ( - )R 2 2. When there are two plates in each comb, the number of adjoining sheets of positive and negative charge is 3, as shown in the sketch. When there are N plates on each comb, the number of parallel capacitors is 2N - 1 and the total capacitance is C = ( 2N - 1) e0 Aeffective
distance
d
+ + + +
-
-
+ + + +
+ + + +
-
=
(2N - 1)e0 ( - )R 2
d2
2
=
(2N - 1) 0 ( - )R 2
d
26.11
(a)
C=
l
b 2k e ln a
=
50.0 7.27 2(8.99 109) ln 2.58
= 2.68 nF
(b)
Method 1:
b V = 2k e ln a 8.10 106 C = 1.62 107 C/m 50.0 m
=q/l =
7.27 V = 2(8.99 10 9)(1.62 107) ln 2.58 = 3.02 kV Method 2: Q 8.10 106 V = C = = 3.02 kV 2.68 109
26.12
Let the radii be b and a with b = 2a. Put charge Q on the inner conductor and Q on the outer. Electric field exists only in the volume between them. The potential of the inner sphere is V a = keQ a; that of the outer is Vb = keQ b . Then V a - Vb = Here C = ke Q ke Q Q b - a - = a b 4 e0 ab 4 e0 2a2 = 8 e0 a a a= and C 8 e0
4 3 4 3
C=
4 e0 ab Q = b-a V a - Vb
The intervening volume is
Volume = b 3 - a3 = 7
(
4 a3 3
C ) = 7( ) 8 e
4 3 3 3 3
0
3
=
7C 3 384 2e0 3
Volume = The outer sphere is 360 km in diameter.
7 (20.0 10 - 6 C 2 / Nm)3 = 2.13 1016 m3 384 2 (8.85 10 -12 C 2 / Nm 2 )3
Chapter 26 Solutions 26.13 Fy = 0: T cos - mg = 0 Dividing, tan = V = Ed = Eq , mg so E= mg tan q Fx = 0: T sin - Eq = 0
87
mgd tan (350 10 - 6 kg)(9.80 m / s 2 )(4.00 10 - 2 m) tan 15.0 = = 1.23 kV q 30.0 10 -9 C
26.14
Fy = 0: T cos - mg = 0 Dividing, tan = Eq , mg so E= mg tan q
Fx = 0: T sin - Eq = 0 and V = Ed = mgd tan q
26.15
(a) (b)
C= C=
ab (0.0700)(0.140) = = 15.6 pF ke (b - a) (8.99 10 9 )(0.140 - 0.0700) Q V V = Q 4.00 10 -6 C = = 256 kV C 15.6 10 -12 F
Goal Solution An air-filled spherical capacitor is constructed with inner and outer shell radii of 7.00 and 14.0 cm, respectively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a charge of 4.00 C on the capacitor? G: Since the separation between the inner and outer shells is much larger than a typical electronic capacitor with d ~ 0.1 mm and capacitance in the microfarad range, we might expect the capacitance of this spherical configuration to be on the order of picofarads, (based on a factor of about 700 times larger spacing between the conductors). The potential difference should be sufficiently low to prevent sparking through the air that separates the shells. O: The capacitance can be found from the equation for spherical shells, and the voltage can be found from Q = CV . A : (a) For a spherical capacitor with inner radius a and outer radius b, C= ab (0.0700 m)(0.140 m) = = 1.56 10 -11 F = 15.6 pF k(b - a) 8.99 10 9 N m 2 C 2 (0.140 - 0.0700) m
(
)
(b)
V =
Q (4.00 10 -6 C) = = 2.56 10 5 V = 256 kV C 1.56 10 -11 F
L : The capacitance agrees with our prediction, but the voltage seems rather high. We can check this voltage by approximating the configuration as the electric field between two charged parallel plates separated by d = 7.00 cm, so E~ V 2.56 10 5 V = = 3.66 106 V / m d 0.0700 m
This electric field barely exceeds the dielectric breakdown strength of air 3 106 V / m , so it may not even be possible to place 4.00 C of charge on this capacitor!
(
)
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88
Chapter 26 Solutions C = 4 e0 R = 4 8.85 10 -12 C N m 2 6.37 106 m = 7.08 10 -4 F
26.16
(
)(
)
*26.17 (a)
Capacitors in parallel add. Thus, the equivalent capacitor has a value of Ceq = C1 + C2 = 5.00 F + 12.0 F = 17.0 F
(b)
The potential difference across each branch is the same and equal to the voltage of the battery. V = 9.00 V
(c)
Q 5 = C (V) = (5.00 F)(9.00 V) = 45.0 C
and
Q12 = C (V) = (12.0 F)(9.00 V) = 108 C
*26.18 (a)
In series capacitors add as 1 1 1 1 1 C eq = C 1 + C 2 = 5.00 F + 12.0 F and C eq = 3.53 F
(c)
The charge on the equivalent capacitor is Q eq = C eq (V) = (3.53 F)(9.00 V) = 31.8 C Each of the series capacitors has this same charge on it. So Q 1 = Q 2 = 31.8 C
(b)
The voltage across each is Q1 31.8 C V 1 = C = 5.00 F = 6.35 V 1 and V 2 = Q2 31.8 C = 12.0 F = 2.65 V C2
26.19
1 Cp = C1 + C2 C
s
1 1 = C +C 1 2 1 1 Cp C1 + C1 1 = C + C = C Cs 1 p C1 1(C p C 1)
Substitute C 2 = C p C 1
2
Simplifying, C 1 C 1 C p + C p C s = 0 C1 = Cp Cp 2 - 4CpCs 2 = Cp
1 2 1 C 2 4 p
- CpCs
We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus sign, we would get the same two answers with their names interchanged.) C1 = Cp +
1 2 1 C 2 4 p 1 2
- CpCs =
1 C 2 4 p
1 2
(9.00 pF ) +
1
1 4
(9.00 pF )2 - (9.00 pF )(2.00 pF ) =
6.00 pF
C2 = Cp - C1 = Cp -
- CpCs = 2 (9.00 pF) 1.50 pF = 3.00 pF
Chapter 26 Solutions 1 1 1 = + Cs C1 C2 Cp - C1 + C1 1 1 1 = + = Cs C1 Cp - C1 C1 Cp - C1
2 C1 - C1Cp + CpCs = 0 2 Cp Cp - 4CpCs
89
26.20
Cp = C1 + C2 Substitute Simplifying,
and C2 = Cp - C1 :
and
C1 =
2
=
1 2
Cp +
1 4
2 Cp - CpCs
where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values for the capacitances, with the names reversed). Then, from C2 = Cp - C1 C2 =
1 2
Cp -
1 4
Cp - CpCs
2
26.21
(a)
1 1 1 Cs = 15.0 + 3.00 Cs = 2.50 F Cp = 2.50 + 6.00 = 8.50 F 1 1 Ceq = + 8.50 F 20.0 F
-1
= 5.96 F on 20.0 F
(b)
Q = ( V )C = (15.0 V)(5.96 F) = 89.5 C V = Q 89.5 C = = 4.47 V C 20.0 F
15.0 4.47 = 10.53 V Q = ( V )C= (10.53)(6.00 F) = 63.2 C on 6.00 F 89.5 63.2 = 26.3 C on 15.0 F and 3.00 F
26.22
The circuit reduces first according to the rule for capacitors in series, as shown in the figure, then according to the rule for capacitors in parallel, shown below. 1 1 11 = C = 1.83C Ceq = C 1 + + 2 3 6
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90
Chapter 26 Solutions Q V Q 20.0
26.23
C=
so
6.00 106 = Q C
and
Q = 120 C
Q1 = 120 C Q2 120 Q 2 Q2 =C C1 2
and
V =
or
120 Q2 Q2 = 3.00 6.00
(3.00)(120 Q2) = (6.00)Q2 Q2 = 360 = 40.0 C 9.00
Q1 = 120 C 40.0 C = 80.0 C
*26.24 (a)
In series , to reduce the effective capacitance: 1 1 1 = + 32.0 F 34.8 F Cs Cs = 1 = 398 F 2.51 10 -3 F
(b)
In parallel , to increase the total capacitance: 29.8 F + Cp = 32.0 F Cp = 2.20 F
26.25 (a) (b)
With switch closed, distance d' = 0.500d and capacitance C = Q = C (V) = 2C(V) = 2(2.00 10 - 6 F)(100 V) = 400 C The force stretching out one spring is F= 2C 2 (V) 2 2C(V) 2 Q2 4C 2 (V) 2 = = = 2e0 A 2e0 A (e0 A / d)d d
e0 A 2e0 A = = 2C d d
One spring stretches by distance x = d /4, so k= F 2C(V) 2 4 8C(V) 2 8(2.00 10 - 6 F)(100 V) 2 = = = = 2.50 kN/m d x d d2 (8.00 10 - 3 m) 2
Chapter 26 Solutions 26.26 Positive charge on A will induce equal negative charges on B, D, and F, and equal positive charges on C and E. The nesting spheres form three capacitors in series. From Example 26.3, CAB = CCD = CEF = Ceq = R(2R) 2R ab = = ke (b - a) ke R ke
91
(3R)( 4R) = 12R
ke R ke
( 5R)( 6R)
ke R
=
30R ke
60 R 1 = ke / 2R + ke / 12R + ke / 30R 37 ke
26.27
nC =
=
100 n/C
nC =
100C n
so
n2 = 100
and
n = 10
Goal Solution A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group? G: Since capacitors in parallel add and ones in series add as inverses, 2 capacitors in parallel would have a capacitance 4 times greater than if they were in series, and 3 capacitors would give a ratio Cp / Cs = 9 , so maybe n = Cp / Cs = 100 = 10. O: The ratio reasoning above seems like an efficient way to solve this problem, but we should check the answer with a more careful analysis based on the general relationships for series and parallel combinations of capacitors. A : Call C the capacitance of one capacitor and n the number of capacitors. The equivalent capacitance for n capacitors in parallel is Cp = C1 + C2 + . . . + Cn = nC The relationship for n capacitors in series is Cp Cs = nC = n2 C/n n= Cp Cs 1 1 1 n 1 = + +...+ = Cn C Cs C1 C2 = 100 = 10
Therefore
or
L : Our prediction appears to be correct. A qualitative reason that Cp / Cs = n2 is because the amount of charge that can be stored on the capacitors increases according to the area of the plates for a parallel combination, but the total charge remains the same for a series combination.
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92
Chapter 26 Solutions
-1
26.28
Cs =
1 1 + 5.00 10.0
= 3.33 F
Cp1 = 2(3.33) + 2.00 = 8.66 F Cp2 = 2(10.0) = 20.0 F Ceq = 1 1 + 8.66 20.0
-1
= 6.04 F
26.29
Qeq = Ceq ( V ) = 6.04 10 -6 F (60.0 V ) = 3.62 10 -4 C Qp1 = Qeq , so V p1 = Qeq Cp1 = 3.62 10 -4 C = 41.8 V 8.66 10 -6 F
(
)
Q3 = C3 V p1 = 2.00 10 -6 F ( 41.8 V ) = 83.6 C
(
) (
)
26.30
Cs =
1 1 + 5.00 7.00
-1
= 2.92 F
Cp = 2.92 + 4.00 + 6.00 = 12.9 F
*26.31 (a) (b)
U = 2 C (V)2 = 2 (3.00 F)(12.0 V) 2 = 216 J U = 2 C (V)2 = 2 (3.00 F)(6.00 V) 2 = 54.0 J
1 1
1
1
*26.32
U = 2 C (V)2 The circuit diagram is shown at the right. (a) Cp = C1 + C2 = 25.0 F + 5.00 F = 30.0 F U = 2 (30.0 106)(100) 2 = 0.150 J (b) 1 1 Cs = C + C 1 2 U = 2 C (V)2 V = 2U = C 4.17 10 -6
1
1
1
1
1 1 = 25.0 F + 5.00 F
1
= 4.17 F
(0.150)(2)
= 268 V
Chapter 26 Solutions 1 Q2 2 C e0 A d Therefore, the stored energy doubles .
93
*26.33
Use U =
and C = 1 C . 2 1
If d2 = 2d1, C2 =
26.34
u=
U 1 = e0E 2 V 2
1.00 10 -7 1 = (8.85 10 -12 )(3000)2 2 V V = 2.51 10 -3 m 3 = 2.51 10 -3 m 3
(
) 1000 L = m
3
2.51 L
26.35
W = U = F dx
so
F=
Q2 d Q2 d Q2x dU = = = dx dx 2c dx 2e0 A 2 e0 A
26.36
Plate a experiences force kx i from the spring and force QE i due to the electric field created by plate b according to E = / 2e0 = Q / 2 Ae0 . Then, Q2 kx = 2A e 0 x= Q2 2Ae0 k
where A is the area of one plate.
26.37
1 1 The energy transferred is W = 2 Q( V ) = 2 (50.0 C)(1.00 108 V) = 2.50 10 9 J and 1% of this (or W' = 2.50 107 J) is absorbed by the tree. If m is the amount of water boiled away, then
W' = m(4186 J/kg C)(100 C 30.0 C) + m(2.26 106 J/kg) = 2.50 107 J giving m = 9.79 kg
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94
Chapter 26 Solutions 1 R kQ kQ 2 C( V ) where C = 4 e0 R = and V = e - 0 = e 2 ke R R 1 R keQ k Q2 = e 2 ke R 2R
2
26.38
U=
U=
26.39
keQ 2 = mc 2 2R R= ke e 2 (8.99 10 9 N m 2 / C)(1.60 10 -19 C)2 = = 1.40 fm 2 2 mc 2(9.11 10 - 31 kg)(3.00 108 m / s)2
*26.40
C=
0 A 4.90(8.85 1012 F/m)(5.00 104 m2) = = 1.08 1011 F = 10.8 pF d 2.00 103 m
*26.41 (a) (b)
C=
0 A 2.10(8.85 1012 F/m)(1.75 104 m2) = = 8.13 1011 F = 81.3 pF d 4.00 105 m
Vmax = Emax d = (60.0 106 V/m)(4.00 105 m) = 2.40 kV
*26.42
Qmax = C (Vmax), but Vmax = Emax d
0 A d 0 A Thus, Qmax = d (Emax d) = 0 AEmax
Also, C = (a) With air between the plates, = 1.00 and Emax = 3.00 106 V/m. Therefore, Qmax = 0 AEmax = (8.85 1012 F/m)(5.00 104 m2)(3.00 106 V/m) = 13.3 nC (b) With polystyrene between the plates, = 2.56 and Emax = 24.0 106 V/m. Qmax = 0 AEmax = 2.56(8.85 1012 F/m)(5.00 104 m2)(24.0 106 V/m) = 272 nC
26.43
C=
0 A d
or
Chapter 26 Solutions
95
= 1.04 m
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96 *26.44
Chapter 26 Solutions Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between them. Suppose the plastic has 3, Emax ~ 107 V/m and thickness 1 mil = 2.54 cm/1000. Then, C=
0 A 3(8.85 1012 C2/N m2)(0.4 m2) ~ ~ 106 F d 2.54 105 m
V 2.54 10 -5 m ~ 10 2 V Vmax = Emax d ~ 107 m
(
)
*26.45 (a) (b)
With air between the plates, we find
C0 =
Q V
=
48.0 C = 4.00 F 12.0 V
When Teflon is inserted, the charge remains the same (48.0 C) because the plates are isolated. However, the capacitance, and hence the voltage, changes. The new capacitance is C' = C 0 = 2.10(4.00 F) = 8.40 F
(c)
The voltage on the capacitor now is and the charge is 48.0 C
Q 48.0 C V' = C' = 8.40 F = 5.71 V
26.46 (a)
Originally, C =0 A / d = Q /(V)i The charge is the same before and after immersion, with value Q = 0 A(V)i / d . Q= (b) (8.85 10 -12 C 2 )(25.0 10 - 4 m 2 )(250 V) = 369 pC N m 2 (1.50 10 - 2 m) C f = 0 A / d = Q /(V)f
Finally, Cf =
80.0(8.85 10 -12 C 2 )(25.0 10 - 4 m 2 ) = 118 pF N m 2 (1.50 10 - 2 m)
(V) f = (c)
A(V)i d (V)i 250 V Qd = 0 = = = 3.12 V 80.0 0 A 0 A d
1 U = 2 C(V)i2 =
2
Originally, Finally,
0 A(V)i2 2d 0 A(V) i2 0 A(V) i2 = 2d 2d 2
1 U f = 2 C f (V) f =
So,
U = U f - U = -
0 A(V)i2 ( - 1) 2d
U =
(- 8.85 10 -12 C 2)( 25.0 10 - 4 m 2 )(250 V)2 (79.0) = 45.5 nJ N m 2 2(1.50 10 - 2 m)80
Chapter 26 Solutions k (b - a) ke (c - b) 1 1 1 = + = e + C 1ab 2bc 1ab 2bc k (b - a) k (c - b) e e C=
97
26.47
1 2 abc 1 = = ke (b - a) ke (c - b) ke 2 (bc - ac) + ke 1 (ac - ab) + 1 ab 2bc
4 1 2 abc0 2bc - 1ab + ( 1 - 2 )ac
26.48
(a) (b)
C = C0 =
0 A (173)(8.85 10 -12 )(1.00 10 -4 m 2 ) = = 1.53 nF d 0.100 10 -3 m
The battery delivers the free charge Q = C (V) = (1.53 10-9 F)(12.0 V) = 18.4 nC
(c)
The surface density of free charge is
=
Q 18.4 10 -9 C = = 1.84 10-4 C/m2 A 1.00 10 -4 m 2
The surface density of polarization charge is 1 1 p = 1 - = 1 - = 1.83 10-4 C/m2 173 (d) We have E = E0/ and E0 = V/d ; hence, E= V 12.0 V = = 694 V/m d (173)(1.00 10 -4 m)
26.49
The given combination of capacitors is equivalent to the circuit diagram shown to the right. Put charge Q on point A . Then, Q = (40.0 F)V AB = (10.0 F)VBC = (40.0 F)VCD So, VBC = 4 V AB = 4 VCD , and the center capacitor will break down first, at VBC = 15.0 V. When this occurs,
1 VAB = VCD = 4 ( VBC ) = 3.75 V
and V AD = V AB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V
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98
Chapter 26 Solutions The displacement from negative to positive charge is 2a = ( -1.20 i + 1.10 j) mm - (1.40 i - 1.30 j) mm = ( -2.60i + 2.40 j) 10 -3 m The electric dipole moment is p = 2aq = 3.50 10 -9 C ( -2.60i + 2.40 j) 10 -3 m = (b)
*26.50 (a)
(
)
(-9.10i + 8.40 j) 10-12 C m
= p E = ( -9.10i + 8.40 j) 10 -12 C m (7.80i - 4.90 j) 10 3 N C = ( + 44.6 k - 65.5k ) 10 -9 N m = -2.09 10 -8 N m k
[
] [
] ]
(c)
U = -p E = - ( -9.10i + 8.40 j) 10 -12 C m (7.80i - 4.90 j) 10 3 N C U = (71.0 + 41.2) 10 -9 J = 112 nJ
[
][
(d)
p = E =
(9.10)2 + (8.40)2 10 -12 (7.80)2 + ( 4.90)2 103
E = 114 nJ,
C m = 12.4 10 -12 C m
N C = 9.21 10 3 N C
U max = p
U min = - 114 nJ
U max - U min = 228 nJ
*26.51 (a)
Let x represent the coordinate of the negative charge. Then x + 2acos is the coordinate of the positive charge. The force on the negative charge is F - = -qE( x ) i. The force on the positive charge is dE F + = +qE( x + 2acos ) i qE( x ) i + q ( 2acos ) i dx The force on the dipole is altogether F = F- + F+ = q ke q i. x2
E
p F-
F+
dE dE (2acos ) i = p cos i dx dx
(b)
The balloon creates field along the x - axis of dE ( -2)ke q = dx x3
Thus,
At x = 16.0 cm ,
9 -6 MN dE ( -2) 8.99 10 2.00 10 = = - 8.78 3 Cm dx (0.160)
(
)(
)
N cos 0 i = 55.3 i mN F = 6.30 10 -9 C m - 8.78 106 C m
(
)
Chapter 26 Solutions qin 0
r2 r1
99
26.52
2 r E =
so r dr = ln 1 2 r 0 2 0 r2
E=
2 r 0
V = - E dr =
r1
r2
max = Emax rinner 2 0
V 25.0 (0.100 10 -3 m) ln V = 1.20 106 0.200 m Vmax = 579 V
*26.53 (a)
Consider a gaussian surface in the form of a cylindrical pillbox with ends of area A << A parallel to the sheet. The side wall of the cylinder passes no flux of electric field since this surface is everywhere parallel to the field. Gauss's law becomes EA + EA = Q A , A so E= Q 2A directed away from the positive sheet.
(b)
In the space between the sheets, each creates field Q 2 A away from the positive and toward the negative sheet. Together, they create a field of E= Q A
(c)
Assume that the field is in the positive x - direction. Then, the potential of the positive plate relative to the negative plate is V = -
+ plate - plate
E ds = -
+ plate
- plate
Q i ( - i dx ) = A
+
Qd A
(d) Capacitance is defined by: C =
Q Q A 0 A = = = V Qd A d d
26.54
(a) (c)
1 1 C = 3.00 + 6.00
1
1 1 + 2.00 + 4.00
1
= 3.33 F
Qac = Cac (Vac) = (2.00 F)(90.0 V) = 180 C Therefore, Q3 = Q6 = 180 C
Qdf = Cdf V df = (1.33 F )(90.0 V ) = 120 C
(
)
2000 by Harcourt, Inc. All rights reserved.
100 Chapter 26 Solutions
(b)
V 3 = V 6 = V 2 = V 4 =
Q3 180 C = = 60.0 V C3 3.00 F Q6 180 C = = 30.0 V C6 6.00 F Q2 120 C = = 60.0 V C2 2.00 F Q4 120 C = = 30.0 V C4 4.00 F
1
(d) UT = 2 C eq (V)2 = 2 (3.33 106)(90.0 V) 2 = 13.4 mJ
1
*26.55
The electric field due to the charge on the positive wire is perpendicular to the wire, radial, and of magnitude E+ =
2 0 r
The potential difference between wires due to the presence of this charge is V1 = -
+wire - wire
E dr = -
2 0
D-d
d
dr D - d = ln r 2 0 d
The presence of the linear charge density - on the negative wire makes an identical contribution to the potential difference between the wires. Therefore, the total potential difference is V = 2( V1 ) =
D - d ln 0 d
and the capacitance of this system of two wires, each of length , is
The capacitance per unit length is:
Chapter 26 Solutions 101 26.56 (a) We use Equation 26.11 to find the potential energy. As we will see, the potential difference V changes as the dielectric is withdrawn. The initial and final energies are 1 Q 2 Ui = 2 C i and Uf = 1 Q 2 2 Cf 1 Q 2 Uf = 2 Ci
But the initial capacitance (with the dielectric) is Ci = C f. Therefore,
Since the work done by the external force in removing the dielectric equals the change i n potential energy, we have Q 2 1 W = Uf Ui = 2 C i 1 Q 2 2 Ci = 1 Q 2 ( 1) 2 Ci
To express this relation in terms of potential difference V i , we substitute Q = Ci (V i), and evaluate: W = 2 C i (V i)2( 1) = 2 (2.00 109 F)(100 V) 2(5.00 1.00) = 4.00 105 J The positive result confirms that the final energy of the capacitor is greater than the initial energy. The extra energy comes from the work done o n the system by the external force that pulled out the dielectric. (b) Q The final potential difference across the capacitor is V f = C f Substituting C f = Ci and Q = Ci (Vi) gives Vf = Vi = (5.00)(100 V) = 500 V
1 1
Even though the capacitor is isolated and its charge remains constant, the potential difference across the plates does increase in this case.
26.57
= 3.00, Emax = 2.00 108 V/m = Vmax /d
For C= A=
e0 A = 0.250 106 F, d
(0.250 106)(4000) Cd C( Vmax ) = = = 0.188 m2 (3.00)(8.85 1012)(2.00 108) e0 e0Emax
26.58
(a)
C1 =
1 e0 A/2 ; d
1
C2 =
2 e0 A/2 ; d/2
C3 =
3 e0 A/2 d/2
1 + 1 C 2 C 3
e0 A 2 3 C 2C 3 = C = d 2 + C3 2 + 3
1
1 1 C = C1 + + C C 2 3
=
e0 A 1 2 3 d 2 + 2 + 3
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102 Chapter 26 Solutions Ctotal = 1.76 1012 F = 1.76 pF
(b)
Using the given values we find:
Chapter 26 Solutions 103
26.59
The system may be considered to be two capacitors in series: C1 = e0 A t1 and C2 = e0 A t2
1 1 1 t1 + t2 C = C1 + C2 = e0 A C= e0 A e0 A t1 + t2 = s d
Goal Solution A conducting slab of a thickness d and area A is inserted into the space between the plates of a parallelplate capacitor with spacing s and surface area A, as shown in Figure P26.59. The slab is not necessarily halfway between the capacitor plates. What is the capacitance of the system? G: It is difficult to predict an exact relationship for the capacitance of this system, but we can reason that C should increase if the distance between the slab and plates were decreased (until they touched and formed a short circuit). So maybe C 1/ ( s - d) . Moving the metal slab does not change the amount of charge the system can store, so the capacitance should therefore be independent of the slab position. The slab must have zero net charge, with each face of the plate holding the same magnitude of charge as the outside plates, regardless of where the slab is between the plates. O: If the capacitor is charged with + Q on the top plate and -Q on the bottom plate, then free charges will move across the conducting slab to neutralize the electric field inside it, with the top face of the slab carrying charge -Q and the bottom face carrying charge + Q. Then the capacitor and slab combination is electrically equivalent to two capacitors in series. (We are neglecting the slight fringing effect of the electric field near the edges of the capacitor.) Call x the upper gap, so that s - d - x is the distance between the lower two surfaces. A : For the upper capacitor, and the lower has So the combination has C1 = 0 A x C2 = C=
0 A s-d-x
A 1 1 = = 0 1 1 x s-d-x s-d + + C1 C2 0 A 0 A
L : The equivalent capacitance is inversely proportional to ( s - d) as expected, and is also proportional to A . This result is the same as for the special case in Example 26.9 when the slab is just halfway between the plates; the only critical factor is the thickness of the slab relative to the plate spacing.
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104 Chapter 26 Solutions 26.60 (a) Put charge Q on the sphere of radius a and Q on the other sphere. Relative to V = 0 at infinity, k eQ k eQ the potential at the surface of a is Va = a d and the potential of b is The difference in potential is Vb = k eQ k eQ + b d k eQ k eQ k eQ k eQ + a b d d
Va Vb =
and
C=
Q 4 0 V a V b = (1 a) + (1 b) - ( 2 d)
(b)
As d , 1/d becomes negligible compared to 1/a. Then, C= 4 0 1 a+1 b and 1 1 1 = + C 4 0 a 4 0 b
as for two spheres in series.
26.61
Note that the potential difference between the plates is held constant at Ci = q0 Vi and C f = so qf Vi = q0 + q Vi
Vi by the battery.
But C f = Ci , Thus, =
q q0 + q = 0 Vi Vi
q0 + q q or = 1 + q0 q0
26.62
(a)
(b)
1 U = 2 C( V )2 =
(c)
F = -
dU = dx
to the left
Chapter 26 Solutions 105 (2000)2 (8.85 10 -12 )(0.0500)(4.50 - 1) = 1.55 10 -3 N 2(2.00 10 -3 )
(d)
F=
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106 Chapter 26 Solutions *26.63 The portion of the capacitor nearly filled by metal has capacitance 0 ( x) d and stored energy Q 2 2C 0 . The unfilled portion has capacitance 0 ( -x ) d. The charge on this portion is Q = ( -x)Q0 / . (a) The stored energy is
(b)
F= -
dU = dx
F=
to the right
(c)
(d)
26.64
Gasoline:
1.00 m 3 1.00 gal Btu J 7 J 126 000 gal 1054 Btu = 5.25 10 kg -3 3 3.786 10 m 670 kg
Battery:
(12.0
J C)(100 C s)( 3600 s) = 2.70 10 5 J kg 16.0 kg F )(12.0 V )
2
Capacitor:
1 0.100 2(
0.100 kg
= 72.0 J kg
Gasoline has 194 times the specific energy content of the battery and 727 000 times that of the capacitor
Chapter 26 Solutions 107 26.65 Call the unknown capacitance Cu Q = Cu (Vi ) = (Cu + C)(V f ) C(V f ) (Vi ) - (V f ) (10.0 F)( 30.0 V) = 4.29 F (100 V - 30.0 V)
Cu =
=
Goal Solution An isolated capacitor of unknown capacitance has been charged to a potential difference of 100 V. W h e n the charged capacitor is then connected in parallel to an uncharged 10.0- F capacitor, the voltage across the combination is 30.0 V. Calculate the unknown capacitance. G: The voltage of the combination will be reduced according to the size of the added capacitance. (Example: If the unknown capacitance were C = 10.0 F, then V1 = 50.0 V because the charge is now distributed evenly between the two capacitors.) Since the final voltage is less than half the original, we might guess that the unknown capacitor is about 5.00 F. O: We can use the relationships for capacitors in parallel to find the unknown capacitance, along with the requirement that the charge on the unknown capacitor must be the same as the total charge o n the two capacitors in parallel. A : We name our ignorance and call the unknown capacitance Cu . The charge originally deposited o n each plate, + on one, - on the other, is Q = Cu V = Cu (100 V ) Now in the new connection this same conserved charge redistributes itself between the two capacitors according to Q = Q1 + Q2 . Q1 = Cu ( 30.0 V ) and Q2 = (10.0 F )( 30.0 V ) = 300 C
We can eliminate Q and Q1 by substitution: Cu (100 V ) = Cu ( 30.0 V ) + 300 C so Cu = 300 C = 4.29 F 70.0 V
L : The calculated capacitance is close to what we expected, so our result seems reasonable. In this and other capacitance combination problems, it is important not to confuse the charge and voltage of the system with those of the individual components, especially if they have different values. Careful attention must be given to the subscripts to avoid this confusion. It is also important to not confuse the variable " C " for capacitance with the unit of charge, " C" for coulombs.
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108 Chapter 26 Solutions
26.66
Put five 6.00 pF capacitors in series. The potential difference across any one of the capacitors will be: V = Vmax 1000 V = = 200 V 5 5
and the equivalent capacitance is: 1 1 = 5 Ceq 6.00 pF 26.67 or Ceq = 6.00 pF = 1.20 pF 5 Q2 Q3 = C2 C3
When V db = 0, V bc = V dc , and Also, V ba = V da Q1 Q4 = C1 C4
or
C Q Q From these equations we have C2 = 3 2 4 C1 C4 Q1 Q3 However, from the properties of capacitors in series, we have Therefore, C 9.00 C2 = 3 C1 = (4.00 F) = 3.00 F 12.0 C4 Q 1 = Q 2 and Q 3 = Q 4
26.68
Let C = the capacitance of an individual capacitor, and Cs represent the equivalent capacitance of the group in series. While being charged in parallel, each capacitor receives charge Q = C Vchg = (5.00 10-4 F)(800 V) = 0.400 C While being discharged in series, Vdisch = Q Q 0.400 C = = = 8.00 kV Cs C / 10 5.00 10 -5 F
or 10 times the original voltage.
26.69
(a)
C0 =
0 A Q0 = d V0
When the dielectric is inserted at constant voltage,
Chapter 26 Solutions 109 C0 (V0 )2 2 and U = U0
C = C0 =
Q ; V0
U0 =
U=
C(V0 )2 C0 (V0 )2 = 2 2
The extra energy comes from (part of the) electrical work done by the battery in separating the extra charge. (b) Q0 = C0 V0 and Q = C V0 = C0 V0 so Q /Q 0 =
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110 Chapter 26 Solutions A slice of width (dx) at coordinate x in 0 x L has thickness x d /L filled with dielectric 2, and d x d /L is filled with the material having constant 1. This slice has a capacitance given by 1 1 xd dL - xd 1 xd + 2 dL - 2 xd = + = + = 1 2 e0 (dx)W 1 e0 (dx)W 2 e0 W L(dx) 1 e0 W L(dx) dC 1 2 e0WL( dx ) xd / L d - xd / L dC =
26.70
(a)
1 2 e0 W L(dx) 2 dL + ( 1 - 2 )xd
The whole capacitor is all the slices in parallel: C = dC = C= (b)
L
x=0
1 2 e0W L(dx) e WL -1 = 1 2 0 ( 2Ld + ( 1 - 2 )xd) ( 1 - 2 )d(dx) 2 Ld + ( 1 - 2 )xd ( 1 - 2 )d x=0 1 2 e0 W L ln 1 2 ( 1 - 2 )d
L
L 1 2 e0 W L e WL ln [ 2 Ld + ( 1 - 2 )xd] = 1 2 0 [ln 1Ld - ln 2Ld] = 0 ( 1 - 2 )d ( 1 - 2 )d
To take the limit 1 2 , write 1 = 2 (1 + x) and let x 0. Then C=
2 2 (1 + x)e0 W L ln (1 + x) ( 2 + 2 x - 2 )d
Use the expansion of ln(1 + x) from Appendix B.5. C=
2 2 (1 + x) 0 W L (1 + x) 0 W L 1 1 1 (x - 2 x 2 + 3 x 3 . . . ) = 2 (1 - 2 x + . . . ) 2 xd d
x0
lim C =
2 0 W L 0 A = d d
26.71
The vertical orientation sets up two capacitors in parallel, with equivalent capacitance Cp =
0 ( A 2) 0 ( A 2) + 1 0 A + = 2 d d d
where A is the area of either plate and d is the separation of the plates. The horizontal orientation produces two capacitors in series. If f is the fraction of the horizontal capacitor filled with dielectric, the equivalent capacitance is fd (1 - f )d = f + (1 - f ) d , or 1 = + Cs 0 A 0 A 0 A Requiring that Cp = Cs gives 0 A Cs = f + (1 - f ) d
+1 = , or ( + 1)[ f + (1 - f )] = 2 2 f + (1 - f )
For = 2.00, this yields 3.00[2.00 - (1.00) f ] = 4.00 , with the solution f = 2 / 3 .
Chapter 26 Solutions 111 26.72 Initially (capacitors charged in parallel), q1 = C1(V) = (6.00 F)(250 V) = 1500 C q2 = C2(V) = (2.00 F)(250 V) = 500 C After reconnection (positive plate to negative plate), qtotal = q1 q2 = 1000 C Therefore, q1 = C1(V ) = (6.00 F)(125 V) = 750 C q2 = C2 (V ) = (2.00 F)(125 V) = 250 C and V = qtotal 1000 C = = 125 V Ctotal 8.00 F
26.73
Emax occurs at the inner conductor's surface. Emax = 2ke from Equation 24.7. a b from Example 26.2 a
V = 2ke ln Emax =
V a ln(b / a) b 3.00 = (18.0 106 V / m)(0.800 10 -3 m) ln = 19.0 kV 0.800 a
Vmax = Emax a ln
26.74
E=
2 ; a
V = 2 ln b a
b a
Vmax = Emax a ln
b dVmax 1 b + a = Emax ln =0 - a b / a a2 da ln b =1 a or b = e1 a b so a = e
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112 Chapter 26 Solutions
26.75
Assume a potential difference across a and b, and notice that the potential difference across the 8.00 F capacitor must be zero by symmetry. Then the equivalent capacitance can be determined from the following circuit:
Cab = 3.00 F
26.76
By symmetry, the potential difference across 3C is zero, so the circuit reduces to Ceq = 4 (2C)(4C) 8 = C= 3C 2C + 4C 6
Chapter 27 Solutions
Q t
27.1
I=
Q = I t = (30.0 106 A)(40.0 s) = 1.20 103 C
N=
Q 1.20 103 C = = 7.50 1015 electrons e 1.60 1019 C/electron
*27.2
The atomic weight of silver = 107.9, and the volume V is V = (area)(thickness) = (700 10-4 m2)(0.133 10-3 m) = 9.31 10-6 m3 The mass of silver deposited is mAg = V = 10.5 10 3 kg / m 3 9.31 10 -6 m 3 = 9.78 10 -2 kg. and the number of silver atoms deposited is N = (9.78 10-2 kg) 6.02 1026 atoms = 5.45 1023 107.9 kg
(
)(
)
V 12.0 V I= R = = 6.67 A = 6.67 C/s 1.80 t = Ne (5.45 1023)(1.60 10-19 C) Q = I = = 1.31 104 s = 3.64 h I 6.67 C/s
27.3 (a) (b) (c)
Q(t) = 0 Idt = I0 (1 e t/ ) Q() = I0 (1 e1) = (0.632)I0 Q(10 ) = I0 (1 e10) = (0.99995)I0 Q() = I0 (1 e ) = I0
t
27.4
(a) (b)
Using
ke e 2 mv 2 = , we get: r r2
v=
ke e 2 = 2.19 106 m / s . mr
The time for the electron to revolve around the proton once is: t= 2 r 2 (5.29 10 -11 m) = = 1.52 10 -16 s v (2.19 106 m / s)
The total charge flow in this time is 1.60 10 -19 C, so the current is
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112 Chapter 27 Solutions 1.60 10 -19 C = 1.05 10 -3 A = 1.05 mA 1.52 10 -16 s
I=
Chapter 27 Solutions 113 2 T
27.5
=
I=
where T is the period.
q (8.00 10 - 9 C)(100 rad / s) q = = = 4.00 10 - 7 A = 400 nA T 2 2
27.6
The period of revolution for the sphere is T = revolving charge is I = q q = . T 2
2 , and the average current represented by this
27.7
q = 4t + 5t + 6 I (1.00 s) = J= dq dt
3
1.00 m -4 2 A = (2.00 cm ) = 2.00 10 m 100 cm
2
2
(a)
= 12t 2 + 5
t=1.00 s
(
)
t=1.00 s
= 17.0 A
(b)
I 17.0 A = = 85.0 kA / m 2 A 2.00 10 - 4 m 2
27.8
I=
dq dt
1/240 s 0
q = dq = I dt = q=
(100 A) sin (120 t / s)dt
-100 C +100 C [cos( / 2) - cos 0] = 120 = 0.265 C 120
27.9
(a)
J=
I 5.00 A = = 99.5 kA/m2 A (4.00 10 -3 m)2 1 J1 ; 4 1 A2 4 1 I I = A2 4 A1 so
(b)
J2 = A1 =
(4.00 10 -3 )2 =
1 2 r2 4
r2 = 2(4.00 10 -3 ) = 8.00 10 -3 m = 8.00 mm
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114 Chapter 27 Solutions
1
27.10
(a)
The speed of each deuteron is given by (2.00 106)(1.60 1019 J) = 2 (2 1.67 1027 kg) v2
1
K = 2 mv2 and v = 1.38 107 m/s I = q /t t = 1.60 1014 s
The time between deuterons passing a stationary point is t in 10.0 106 C/s = 1.60 1019 C/t or
So the distance between them is vt = 1.38 107 m / s 1.60 10 -14 s = 2.21 107 m (b) One nucleus will put its nearest neighbor at potential V= k eq (8.99 109 N m2/C 2)(1.60 10-19 C) = = 6.49 103 V r 2 .21 107 m
(
)(
)
This is very small compared to the 2 MV accelerating potential, so repulsion within the beam is a small effect.
27.11
(a)
J=
8.00 10 -6 A I = A 1.00 10 -3 m
(
)
2
= 2.55 A / m 2 2.55 A / m 2 J = = evd 1.60 10 -19 C 3.00 108 m / s
(b)
From J = nevd , we have
n=
(
)(
)
5.31 1010 m - 3
(c)
From I = Q / t , we have (This is about 381 years!)
6.02 10 23 1.60 10 -19 C Q N A e t = = = = 1.20 1010 s I I 8.00 10 -6 A
(
)(
)
*27.12
where n is the number of charge carriers per unit volume, and is We use I = nqAv d identical to the number of atoms per unit volume). We assume a contribution of 1 free electron per atom in the relationship above. For aluminum, which has a molecular weight of 27, we know that Avogadro's number of atoms, N A , has a mass of 27.0 g. Thus, the mass per atom is 27.0 g 27.0 g = = 4.49 1023 g/atom NA 6.02 1023 Thus, n = density of aluminum 2.70 g/cm3 = mass per atom 4.49 1023 g/atom n = 6.02 1022 Therefore, or, atoms atoms = 6.02 1028 cm 3 m3 = 1.30 104 m/s
I 5.00 A vd = n q A = (6.02 1028 m3)(1.60 1019 C)(4.00 106 m2) vd = 0.130 mm/s
Chapter 27 Solutions 115 V 120 V = = 0.500 A = 500 mA R 240
*27.13
I=
27.14
(a)
Applying its definition, we find the resistance of the rod, R= V 15.0 V = = 3750 = 3.75 k I 4.00 10 -3 A
(b)
The length of the rod is determined from Equation 27.11: R = / A. Solving for and substituting numerical values for R, A, and the values of given for carbon in Table 27.1, we obtain = RA (3.75 10 3 )(5.00 10 -6 m 2 ) = = 536 m (3.50 10 -5 m)
27.15
V = IR V = I l : A
and
R=
l : A
1.00 m A = 0.600 mm2 1000 mm I=
2
= 6.00 107 m2
(0.900 V)(6.00 107 m2) VA = (5.60 108 m)(1.50 m) l
I = 6.43 A
27.16
J=
3.00 A I = E = = (120 N/C) 2 (0.0120 m)2 r 1 = 0.0181 m
= 55.3( m)-1 =
27.17
(a)
Given M = dV = d Al
where d mass density ,
we obtain: A =
M dl
Taking r resistivity,
R=
r l
A
=
r l
M dl
=
r dl2 M
Thus,
l=
1.00 10 -3 (0.500) MR = = 1.82 m r d (1.70 10 -8 )(8.92 10 3 )
(
)
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116 Chapter 27 Solutions M , or d M d 1.00 10 -3 = 1.40 10 -4 m (8.92 10 3 )(1.82) diameter = 280 m
(b)
V=
r 2l =
Thus, r =
M = dl
The diameter is twice this distance:
*27.18 (a)
Suppose the rubber is 10 cm long and 1 mm in diameter.
13 -1 l 4 l 4 10 m 10 m R= = ~ = ~1018 2 -3 A d2 10 m
(
(
)(
)
)
(b)
R=
4(1.7 108 m)(103 m) 4 l ~ ~ 107 (2 102 m)2 d2
(c)
10 2 V V I= R ~ ~ 1016 A 1018 I~ 10 2 V ~ 10 9 A 107
27.19
The distance between opposite faces of the cube is
90.0 g l= 3 10.5 g cm
13
= 2.05 cm
(a)
R=
l l 1.59 10 -8 m = = = = 7.77 10 -7 = 777 n A l2 l 2.05 10 -2 m
V 1.00 10 -5 V = = 12.9 A R 7.77 10 -7 10.5 g cm 3 electrons 6.02 10 23 mol 107.87 g mol
(b)
I=
n=
electrons 1.00 106 cm 3 28 3 n = 5.86 10 22 = 5.86 10 m cm 3 1.00 m 3 I = nqvA and v= 12.9 C s I = = 3.28 m/s 2 28 3 nqA 5.86 10 m 1.60 10 -19 C (0.0205 m )
(
)(
)
Chapter 27 Solutions 117
27.20
Originally,
R=
l A (l/ 3) l R = = 3A 9A 9
Finally,
Rf =
27.21
The total volume of material present does not change, only its shape. Thus, A f lf = A f (1.25li ) = Aili giving A f = Ai 1.25 The final resistance is then: R f =
lf
Af
=
(1.25li ) l = 1.56 i = 1.56R Ai 1.25 Ai
27.22
(r Al )
All
2
=
(rCu )
Cul
2
rAl Al = = rCu Cu
2.82 10 -8 = 1.29 1.70 10 -8
27.23
J = E
so
=
J 6.00 10 -13 A / m 2 = = 6.00 10-15 ( m)-1 100 V / m E
27.24
R=
1l1
A1
+
2l 2
A2
= (1l1 + 2l 2 ) / d 2
R=
(4.00 10 - 3 m)( 0.250 m) + ( 6.00 10 - 3 m)( 0.400 m) = 378 (3.00 10 - 3 m)2
27.25
=
m nq 2 qE m
so
=
m 9.11 1031 = = 2.47 1014 s nq 2 (1.70 108)(8.49 1028)(1.60 1019)2 (1.60 1019)E(2.47 1014) 9.11 1031
vd =
so
7.84 104 =
Therefore
E = 0.181 V/m
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118 Chapter 27 Solutions Goal Solution If the drift velocity of free electrons in a copper wire is 7.84 10 -4 m / s , what is the electric field in the conductor? G: For electrostatic cases, we learned that the electric field inside a conductor is always zero. On the other hand, if there is a current, a non-zero electric field must be maintained by a battery or other source to make the charges flow. Therefore, we might expect the electric field to be small, but definitely not zero. O: The drift velocity of the electrons can be used to find the current density, which can be used with Ohm's law to find the electric field inside the conductor. A : We first need the electron density in copper, which from Example 27.1 is n = 8.49 10 28 e - / m 3 . current density in this wire is then J = nqvd = (8.49 10 28 e - / m 3 )(1.60 10 -19 C / e - )(7.84 10 -4 m / s) = 1.06 107 A / m 2 Ohm's law can be stated as J = E = E / where = 1.7 10 -8 m for copper, so then The
E = J = (1.70 10 -8 m)(1.06 107 A / m 2 ) = 0.181 V / m L : This electric field is certainly smaller than typical static values outside charged objects. The direction of the electric field should be along the length of the conductor, otherwise the electrons would be forced to leave the wire! The reality is that excess charges arrange themselves on the surface of the wire to create an electric field that "steers" the free electrons to flow along the length of the wire from low to high potential (opposite the direction of a positive test charge). It is also interesting to note that when the electric field is being established it travels at the speed of light; but the drift velocity of the electrons is literally at a "snail's pace"!
27.26
(a) (b) (c)
n is unaffected J = I I so it doubles A
J = nevd so v d doubles m is unchanged nq 2 as long as does not change due to heating.
(d) =
27.27
From Equation 27.17,
=
me 9.11 10 -31 = = 2.47 10 -14 s 28 -19 2 -8 nq 2 8.49 10 1.60 10 1.70 10
(
)(
)(
)
l = v = 8.60 10 5 m / s 2.47 10 -14 s = 2.12 10 - 8 m = 21.2 nm
(
)(
)
Chapter 27 Solutions 119 V = R0 [1 + (TC - T0 )] where T0 = 20.0C IC V V = = R0 [1 + (T h - T0 )] Ih 1A = 1 + (3.90 103)(38.0) 1 + (3.90 103)(108)
27.28
At the low temperature TC we write At the high temperature T h , Then and
RC = Rh =
(V)/(1.00 A) (V)/IC
IC = (1.00 A)(1.15/0.579) = 1.98 A
*27.29
R = R0 [1 + (T)] Solving,
gives
140 = (19.0 ) 1 + (4.50 10 -3 /C)T T = 1.42 103 C = T 20.0 C T = 1.44 103 C
[
]
And, the final temperature is
27.30
R = R c + R n = R c [1 + c (T - T0 )] + R n [1 + n (T - T0 )] 0 = Rc c (T T 0) + R n n (T T 0) R = R n so R c = R n
n c
n + Rn c
R c = R(1 c / n)1
1
R n = R(1 n / c)1
(0.400 103/C) Rn = 10.0 k 1 ( 0.500 103/C) Rn = 5.56 k and
Rc = 4.44 k
27.31
(a) (b)
= 0 [1+ (T - T0 )] = (2.82 10 -8 m) 1+ 3.90 10 -3 (30.0) = 3.15 10-8 m
J= 0.200 V / m E = = 6.35 106 A / m 2 3.15 10 - 8 m
[
]
(c) (d)
I = JA = n=
d2 (1.00 10 -4 m)2 J= (6.35 106 A / m 2 ) = 49.9 mA 4 4
6.02 10 23 electrons = 6.02 10 28 electrons / m 3 26.98 g 2.70 106 g / m 3 (6.35 106 A / m 2 ) J = = 659 m/s 28 ne (6.02 10 electrons / m 3 )(1.60 10 -19 C) = (0.200 V / m)(2.00 m) = 0.400 V
vd = (e)
V = E
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120 Chapter 27 Solutions
*27.32
For aluminum, R=
E = 3.90 103/C
(Table 27.1) = 24.0 106/C
(Table 19.2)
(1 + E T) (1.39) l 0 (1 + E T )(1 + T ) = = R0 = (1.234 ) = 1.71 2 (1.0024) (1 + T) A A(1 + T )
27.33
R = R0[1 + T] R R 0 = R 0 T R R0 = T = (5.00 10-3)25.0 = 0.125 R0
27.34
Assuming linear change of resistance with temperature, R = R 0(1 + T) R
77 K =
(1.00 )[1 + (3.92 10-3 )(-216C)] =
0.153
27.35
= 0 (1 + T ) or
TW =
1 W
W - 1 0W
Require that W = 40 Cu so that Therefore,
-8 4 1.70 10 1 TW = - 1 = 47.6 C -3 -8 4.50 10 /C 5.60 10
(
)
TW = 47.6 C + T0 = 67.6C
27.36
=
1 1 R 1 2R0 - R0 = = T - T0 R0 T R0 T - T0 T= 1 + T0 and 1 T= +20.0 C -3 -1 0.400 10 C so T = 2.52 103 C
so,
*27.37
I=
600 W P = 120 V = 5.00 A V 120 V V = 5.00 A = 24.0 I
and R =
Chapter 27 Solutions 121 P = 0.800(1500 hp)(746 W/hp) = 8.95 105 W P = I (V) 8.95 105 = I(2000) I = 448 A
27.38
27.39
The heat that must be added to the water is Q = mc T = (1.50 kg)(4186 J/kgC)(40.0C) = 2.51 105 J Thus, the power supplied by the heater is W Q 2.51 105 J P = t = t = = 419 W 600 s and the resistance is R=
( V )2
P
(110 V)2 = 419 W = 28.9
27.40
The heat that must be added to the water is Thus, the power supplied by the heat is
Q = mc(T 2 T 1) P= W Q mc(T 2 - T1 ) = = t t t
and the resistance is
R=
( V )2
P
=
mc(T 2 - T1 )
( V )2 t
27.41
2 (V)2 / R V P 140 = = = 1.361 = 2 120 P 0 (V0 ) / R V0
2
P - P0 P % = - 1 (100%) = (1.361 - 1)100 = 36.1% (100%) = P0 P0
2000 by Harcourt, Inc. All rights reserved.
122 Chapter 27 Solutions Goal Solution Suppose that a voltage surge produces 140 V for a moment. By what percentage does the power output of a 120-V, 100-W light bulb increase? (Assume that its resistance does not change.) G: The voltage increases by about 20%, but since the voltage: , the power will increase as the square of
or a 36.1% increase.
O: We have already found an answer to this problem by reasoning in terms of ratios, but we can also calculate the power explicitly for the bulb and compare with the original power by using Ohm's law and the equation for electrical power. To find the power, we must first find the resistance of the bulb, which should remain relatively constant during the power surge (we can check the validity of this assumption later). A : From , we find that If = V f R = 140 V = 0.972 A 144
The final current is,
The power during the surge is So the percentage increase is L: 136 W - 100 W = 0.361 = 36.1% 100 W
Our result tells us that this 100 - W light bulb momentarily acts like a 136 - W light bulb, which explains why it would suddenly get brighter. Some electronic devices (like computers) are sensitive to voltage surges like this, which is the reason that surge protectors are recommended to protect these devices from being damaged. In solving this problem, we assumed that the resistance of the bulb did not change during the voltage surge, but we should check this assumption. Let us assume that the filament is made of tungsten and that its resistance will change linearly with temperature according to equation 27.21. Let us further assume that the increased voltage lasts for a time long enough so that the filament comes to a new equilibrium temperature. The temperature change can be estimated from the power surge according to Stefan's law (equation 20.18), assuming that all the power loss is due to radiation. so that a 36% change in power should correspond to only about a 8% increase i n By this law, temperature. A typical operating temperature of a white light bulb is about 3000 C, so T 0.08( 3273 C) = 260 C. Then the increased resistance would be roughly R = R0 (1 + (T - T0 )) = (144 ) 1 + 4.5 10 -3 ( 260) 310 It appears that the resistance could change double from 144 . On the other hand, if the voltage surge lasts only a very short time, the 136 W we calculated originally accurately describes the conversion of electrical into internal energy in the filament.
(
)
Chapter 27 Solutions 123 (V)2 = 500 W R
27.42
P = I (V) = R= (a) (b) R=
(110 V)2 = 24.2 (500 W)
l A
so
l=
RA (24.2 ) (2.50 10 -4 m)2 = = 3.17 m 1.50 10 -6 m
R = R 0 [1 + T] = 24.2 1 + (0.400 10 -3 )(1180) = 35.6 P= (V)2 (110)2 = = 340 W R 35.6
[
]
27.43
R=
6 l (1.50 10 m)25.0 m = = 298 (0.200 103 m)2 A
V = IR = (0.500 A)(298 ) = 149 V (a) (b) (c) E= 149 V V = 25.0 m = 5.97 V/m l
P = (V)I = (149 V)(0.500 A) = 74.6 W R = R0 [1 + (T - T0 )] = 298 1 + (0.400 10 -3 / C)320 C = 337 (149 V) V = 0.443 A I= R = (337 ) P = (V)I = (149 V)(0.443 A) = 66.1 W
[
]
27.44
(a) (b)
1J 1Ws 1C U = q (V) = It (V) = (55.0 A h)(12.0 V) 1 A s 1 V C 1 J = 660 W h = 0.660 kWh $0.0600 Cost = 0.660 kWh 1 kWh = 3.96
27.45
P = I (V) P =
V = IR
(V)2 (10.0)2 = = 0.833 W R 120
2000 by Harcourt, Inc. All rights reserved.
124 Chapter 27 Solutions
27.46
The total clock power is From e =
(270 10
6
J s 3600 s = 2.43 1012 J h clocks 2.50 clock 1 h
)
W out , the power input to the generating plants must be: Qin
Qin W out t 2.43 1012 J h = = = 9.72 1012 J h t e 0.250 and the rate of coal consumption is 1.00 kg coal metric ton 5 kg coal Rate = 9.72 1012 J h = 295 = 2.95 10 h h 33.0 106 J
(
)
27.47
P = I ( V ) = (1.70 A )(110 V ) = 187 W Energy used in a 24-hour day = (0.187 kW)(24.0 h) = 4.49 kWh $0.0600 cost = 4.49 kWh k W h = $0.269 = 26.9
27.48
P = I (V) = (2.00 A)(120 V) = 240 W U = (0.500 kg)(4186 J/kgC)(77.0C) = 161 kJ = 1.61 105 J = 672 s 240 W
27.49 (a) (b)
At operating temperature, P = I (V) = (1.53 A)(120 V) = 184 W Use the change in resistance to find the final operating temperature of the toaster. R = R 0(1 + T) 120 120 -3 1.53 = 1.80 1 + (0.400 10 )T T = 441C T = 20.0C + 441C = 461C
[
]
Chapter 27 Solutions 125 Goal Solution A certain toaster has a heating element made of Nichrome resistance wire. When the toaster is first connected to a 120-V source of potential difference (and the wire is at a temperature of 20.0 C), the initial current is 1.80 A. However, the current begins to decrease as the resistive element warms up. When the toaster has reached its final operating temperature, the current has dropped to 1.53 A. (a) Find the power the toaster consumes when it is at its operating temperature. (b) What is the final temperature of the heating element? G: Most toasters are rated at about 1000 W (usually stamped on the bottom of the unit), so we might expect this one to have a similar power rating. The temperature of the heating element should be hot enough to toast bread but low enough that the nickel-chromium alloy element does not melt. (The melting point of nickel is 1455 C , and chromium melts at 1907 C.) O: The power can be calculated directly by multiplying the current and the voltage. The temperature can be found from the linear conductivity equation for Nichrome, with = 0.4 10 -3 C -1 from Table 27.1. A : (a) (b) P = ( V ) I = (120 V )(1.53 A ) = 184 W The resistance at 20.0 C is At operating temperature, Neglecting thermal expansion, R0 = R= R= V 120 V = = 66.7 I 1.80 A
120 V = 78.4 1.53 A
l 0 (1 + (T - T0 ))l = = R0 (1 + (T - T0 )) A A
R R0 - 1 78.4 66.7 - 1 = 20.0 C + = 461 C 0.4 10 -3 C -1
T = T0 +
L : Although this toaster appears to use significantly less power than most, the temperature seems high enough to toast a piece of bread in a reasonable amount of time. In fact, the temperature of a typical 1000-W toaster would only be slightly higher because Stefan's radiation law (Eq. 20.18) tells us that , so that the temperature might be about 700 C. (assuming all power is lost through radiation) In either case, the operating temperature is well below the melting point of the heating element.
27.50
P = (10.0 W / ft 2 )(10.0 ft)(15.0 ft) = 1.50 kW Energy = P t = (1.50 kW)(24.0 h) = 36.0 kWh Cost = (36.0 kWh)($ 0.0800 / kWh) = $2.88
*27.51
Consider a 400-W blow dryer used for ten minutes daily for a year. The energy converted is 1 kWh P t = ( 400 J s)(600 s d)( 365 d) 9 107 J 20 kWh 3.6 106 J We suppose that electrical energy costs on the order of ten cents per kilowatt-hour. Then the cost of using the dryer for a year is on the order of Cost ( 20 kWh )($ 0.100 kWh ) = $ 2 ~$1
2000 by Harcourt, Inc. All rights reserved.
126 Chapter 27 Solutions V R (V)2 R and = (120 V)2 = 144 100 W
*27.52 (a)
I=
so
P = (V)I =
=
(120 V)2 = 576 25.0 W
(b)
I=
=
25.0 W Q 1.00 C = 0.208 A = = 120 V t t The charge has lower potential energy .
t=
1.00 C 0.208 A = 4.80 s U 1.00 J = t t
(c)
P = 25.0 W =
1.00 J t = 25.0 W = 0.0400 s (d)
The energy changes from electrical to heat and light .
U = P t = (25.0 J/s)(86400 s/d)(30.0 d) = 64.8 106 J The energy company sells energy . $0.0700 k Ws h Cost = 64.8 106 J kWh 1000 J 3600 s = $1.26 $0.0700 k W h 8 Cost per joule = k W h 6 = $1.94 10 /J 3.60 10 J
*27.53
We find the drift velocity from vd = I n q r
2
I = nqv d A = nqv d r 2 1000 A = 2.49 104 m/s
=
8.00 10
28
m
3
(1.60 1019 C) (102 m) 2
x v= t
x 200 103 m t= v = = 8.04 108 s = 25.5 yr 2.49 104 m/s
*27.54
0.500 The resistance of one wire is m i (100 mi) = 50.0 The whole wire is at nominal 700 kV away from ground potential, but the potential difference between its two ends is IR = (1000 A)(50.0 ) = 50.0 kV Then it radiates as heat power P = (V)I = (50.0 103 V)(1000 A) = 50.0 MW
Chapter 27 Solutions 127 1 d dT
27.55
We begin with the differential equation
=
(a)
Separating variables,
d T 0 = T0 d T
ln = (T T 0) 0 (b)
and
= 0e(T T0)
From the series expansion e x 1 + x, (x << 1),
0 [1 + (T - T0 )]
*27.56
Consider a 1.00-m length of cable. The potential difference between its ends is = 6.67 mV V 6.67 103 V = = 22.2 I 300 A
The resistance is
R=
Then
gives
1.56 cm
27.57 (m) 0.540 1.028 1.543 R() 10.4 21.1 31.8
( m) 1.41 106 1.50 106 1.50 106
(in agreement with tabulated value) (Table 27.1)
= 1.47 106 m
1.50 106 m
2000 by Harcourt, Inc. All rights reserved.
128 Chapter 27 Solutions 2 wires R= (a) (b) (c)
27.58
= 100 m
0.108 300 m (100 m) = 0.0360
(V)home = (V)line IR = 120 (110)(0.0360) = 116 V P = I (V) = (110 A)(116 V) = 12.8 kW P wires = I 2R = (110 A)2(0.0360 ) = 436 W
*27.59 (a)
E=-
dV (0 - 4.00) V = 8.00i V m i=- dx (0.500 - 0) m
(b)
R=
-8 l 4.00 10 m (0.500 m ) = = 0.637 2 A 1.00 10 -4 m
(
(
)
)
(c)
I= J=
V 4.00 V = = 6.28 A R 0.637 I 6.28 A i= A 1.00 10 -4 m
(d)
(
)
2
= 2.00 108 i A m 2 = 200i MA m 2
(e)
J = 4.00 10 -8 m 2.00 108 i A m 2 = 8.00i V m = E
(
)(
)
*27.60 (a)
V dV(x) E = dx i = L i 4L l = d2 A
(b)
R=
(c)
V d 2 V I= R = 4L I V J = A i= i L V J = L i= E
(d)
(e)
Chapter 27 Solutions 129 1R 1 I - 1 = T0 + 0 - 1 R0 I
27.61
R = R0 [1 + (T - T0 )] I0 , 10
so
T = T0 +
In this case,
I=
so
T = T0 +
1 9 (9) = 20 + = 2020 C 0.00450/C
27.62
R=
12.0 6.00 V = = I (I 3.00) I
thus
12.0I 36.0 = 6.00I
and
I = 6.00 A
Therefore, R =
12.0 V 6.00 A = 2.00
27.63
(a)
= I ( V )
so
667 A d = vt = ( 20.0 m / s) 2.50 10 3 s = 50.0 km
(b)
and
(
)
27.64
(a)
We begin with
R=
l 0 [1 + (T - T0 )] l0 [1 + (T - T0 )] = , A A0 [1 + 2 (T - T0 )]
R0 [1 + (T - T0 )] [1 + (T - T0 )] [1 + 2 (T - T0 )]
-3
which reduces to
R=
(b)
For copper:
0 = 1.70 10 - 8 m , = 3.90 10
C -1 , and = 17.0 10
-6
C -1
R0 =
0l0 (1.70 10 - 8 )(2.00) = = 1.08 A0 (0.100 10 - 3 )2
The simple formula for R gives: R = (1.08 ) 1 + (3.90 10 -3 C -1 )(100C - 20.0C) = 1.420 while the more complicated formula gives: R=
[
]
(1.08 )[1+ (3.90 10 -3 C -1 )(80.0C)][1+ (17.0 10 -6 C -1 )(80.0C)]
[1 + 2(17.0 10
-6
C -1 )(80.0C)
]
= 1.418
2000 by Harcourt, Inc. All rights reserved.
130 Chapter 27 Solutions Let be the temperature coefficient at 20.0C, and be the temperature coefficient at 0 C. Then = 0 [1 + (T - 20.0C)], and = [1 + (T - 0C)] must both give the correct resistivity at any temperature T. That is, we must have: 0 [1 + (T - 20.0C)] = [1 + (T - 0C)] (1) Setting T = 0 in equation (1) yields: and setting T = 20.0C in equation (1) gives:
27.65
= 0 [1 - (20.0C)] , 0 = [1 + (20.0C)] 0 = 0 [1 - (20.0C)][1 + (20.0C)]
Put from the first of these results into the second to obtain:
Therefore which simplifies to
1 + ( 20.0C) =
=
[1 - (20.0 C)]
1 1 - ( 20.0C)
From this, the temperature coefficient, based on a reference temperature of 0C, may be computed for any material. For example, using this, Table 27.1 becomes at 0C : Material Silver Copper Gold Aluminum Tungsten Iron Platinum Lead Nichrome Carbon Germanium Silicon Temp Coefficients at 0C 4.1 10- 3/C 4.2 10- 3/C 3.6 10- 3/C 4.2 10- 3/C 4.9 10- 3/C 5.6 10- 3/C 4.25 10- 3/C 4.2 10- 3/C 0.4 10- 3/C -0.5 10- 3/C -24 10- 3/C -30 10- 3/C
27.66
(a)
R=
l L = 2 2 A rb - r a
(
) )
2
(b)
R=
(0.0120 m ) - (0.00500 m )
2
(3.50 10
5
[
m (0.0400 m )
]
= 3.74 107 = 37.4 M
(c)
, so R =
rb dr = 2 L ra r
6
r ln b 2 L ra
(d)
R=
(3.50 10
5
m
2 (0.0400 m )
) ln 1.20 = 1.22 10
0.500
= 1.22 M
Chapter 27 Solutions 131 27.67 Each speaker receives 60.0 W of power. Using I= = 60.0 W 4.00 = 3.87 A = I 2 R, we then have
The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less .
27.68
V = E l
or
dV = E dx
V = IR = E l I= dq E l A A dV = = E l = E = - A = A R dx dt l dV dx
Current flows in the direction of decreasing voltage. Energy flows as heat in the direction of decreasing temperature.
27.69
R= R=
dx dx = A wy
L
where
y = y1 +
y2 - y1 x L
L 0
y - y1 dx L = ln y1 + 2 x y2 - y1 w0y + L x w(y2 - y1 ) 1 L y L ln 2 w(y2 - y1 ) y1
R =
27.70
From the geometry of the longitudinal section of the resistor shown in the figure, we see that (b - r) (b - a) = h y From this, the radius at a distance y from the base is r = (a - b) y +b h dy [(a - b)(y / h) + b]2 . 0
h
For a disk-shaped element of volume dR =
dy : r 2
1
R=
Using the integral formula
(au + b)2 = - a(au + b) ,
du
R=
h ab
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132 Chapter 27 Solutions I = I 0 exp( e V / kBT ) - 1 with
27.71
[
]
and
R=
V I
I 0 = 1.00 10 -9 A, e = 1.60 10 -19 C, and kB = 1.38 10 -23 J K .
The following includes a partial table of calculated values and a graph for each of the specified temperatures. (i) For T = 280 K : V ( V ) 0.400 0.440 0.480 0.520 0.560 0.600 I (A) 0.0156 0.0818 0.429 2.25 11.8 61.6 R ( ) 25.6 5.38 1.12 0.232 0.0476 0.0097
(ii) For T = 300 K : V ( V ) 0.400 0.440 0.480 0.520 0.560 0.600 I (A) 0.005 0.024 0.114 0.534 2.51 11.8 R ( ) 77.3 18.1 4.22 0.973 0.223 0.051
(iii) For T = 320 K : V ( V ) 0.400 0.440 0.480 0.520 0.560 0.600 I (A) 0.0020 0.0084 0.0357 0.152 0.648 2.76 R ( ) 203 52.5 13.4 3.42 0.864 0.217
Chapter 28 Solutions
28.1
(a) P =
( V )2
R
becomes so so
20.0 W =
(11.6 V)2 R and
so
R = 6.73
(b) V = IR
11.6 V = I (6.73 )
I = 1.72 A
= IR + Ir
r = 1.97
15.0 V = 11.6 V + (1.72 A)r Figure for Goal Solution
Goal Solution A battery has an emf of 15.0 V. The terminal voltage of the battery is 11.6 V when it is delivering 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance of the battery? G: The internal resistance of a battery usually is less than 1 , with physically larger batteries having less resistance due to the larger anode and cathode areas. The voltage of this battery drops significantly (23%), when the load resistance is added, so a sizable amount of current must be drawn from the battery. If we assume that the internal resistance is about 1 , then the current must be about 3 A to give the 3.4 V drop across the battery's internal resistance. If this is true, then the load resistance must be about R 12 V / 3 A = 4 . O: We can find R exactly by using Joule's law for the power delivered to the load resistor when the voltage is 11.6 V. Then we can find the internal resistance of the battery by summing the electric potential differences around the circuit. A : (a) R= Combining Joule's law, P = VI , and the definition of resistance, V = IR , gives V 2 (11.6 V ) = = 6.73 P 20. W 0
2
(b) The electromotive force of the battery must equal the voltage drops across the resistances: = IR + Ir , where I = V R. r=
- IR = ( - V )R = (15.0 V - 11.6 V)(6.73 ) = 1.97
I V 11.6 V
L : The resistance of the battery is larger than 1 , but it is reasonable for an old battery or for a battery consisting of several small electric cells in series. The load resistance agrees reasonably well with our prediction, despite the fact that the battery's internal resistance was about twice as large as we assumed. Note that in our initial guess we did not consider the power of the load resistance; however, there is not sufficient information to accurately solve this problem without this data.
2000 by Harcourt, Inc. All rights reserved.
Chapter 28 Solutions 135 (a) Vterm = IR becomes so (b) Vterm = Ir becomes so 10.0 V = (1.79 A)(0.200 ) 10.4 V 10.0 V = I (5.60 ) I = 1.79 A
28.2
=
28.3
3.00 V The total resistance is R = 0.600 A = 5.00 (a) Rlamp = R rbatteries = 5.00 0.408 = 4.59
2 P batteries (0.408 )I = 2 = 0.0816 = 8.16% (5.00 )I P total
(b)
28.4
(a) Here Then,
= I(R + r),
so I =
R + r
=
12.6 V = 2.48 A (5.00 + 0.0800 )
V = IR = ( 2.48 A )( 5.00 ) = 12.4 V
(b) Let I1 and I 2 be the currents flowing through the battery and the headlights, respectively. Then, so giving Thus, I1 = I 2 + 35.0 A , and
- I1r - I 2 R = 0
= (I 2 + 35.0 A)(0.0800 ) + I 2 (5.00 ) = 12.6 V
I 2 = 1.93 A V 2 = (1.93 A)(5.00 ) = 9.65 V
28.5
V = I 1R1 = (2.00 A)R 1 and V = I 2(R 1 + R 2) = (1.60 A)(R 1 + 3.00 ) Therefore, (2.00 A)R 1 = (1.60 A)(R 1 + 3.00 ) or R 1 = 12.0
2000 by Harcourt, Inc. All rights reserved.
136 Chapter 28 Solutions 1
28.6
(a) Rp =
(1 7.00 ) + (1 10.0 )
= 4.12
Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 (b) V = IR 34.0 V = I (17.1 ) I = 1.99 A for 4.00 , 9.00 resistors Applying V = IR, (1.99 A)(4.12 ) = 8.18 V 8.18 V = I (7.00 ) 8.18 V = I (10.0 ) so so I = 1.17 A for 7.00 resistor I = 0.818 A for 10.0 resistor
*28.7
If all 3 resistors are placed in parallel , 1 2 5 1 = 500 + 250 = 500 R and R = 100
*28.8
For the bulb in use as intended, I= P 75.0 W = = 0.625 A V 120 V and
V 120 V R = I = 0.625 A = 192
Now, presuming the bulb resistance is unchanged, I= 120 V = 0.620 A 193.6
Across the bulb is V = IR = 192 (0.620 A) = 119 V so its power is P = (V)I = 119 V(0.620 A) = 73.8 W
Chapter 28 Solutions 137 28.9 If we turn the given diagram on its side, we find that it is the same as Figure (a). The 20.0- and 5.00- resistors are in series, so the first reduction is as shown in (b). In addition, since the 10.0-, 5.00-, and 25.0- resistors are then in parallel, we can solve for their equivalent resistance as: Req =
(
1
1 10.0
+
1 5.00
+
1 25.0
)
= 2.94
This is shown in Figure (c), which in turn reduces to the circuit shown in (d). Next, we work backwards through the diagrams applying I = V/R and V = IR. The 12.94- resistor is connected across 25.0-V, so the current through the battery in every diagram is 25.0 V V = 1.93 A I= R = 12.94 In Figure (c), this 1.93 A goes through the 2.94- equivalent resistor to give a potential difference of: V = IR = (1.93 A)(2.94 ) = 5.68 V From Figure (b), we see that this potential difference is the same across Vab, the 10- resistor, and the 5.00- resistor. (b) Therefore, V ab = 5.68 V (a) Since the current through the 20.0- resistor is also the current through the 25.0- line ab, 5.68 V Vab = = 0.227 A = 227 mA I =R ab 25.0 (c) (d) (a)
(b)
28.10
l l l l (120 V ) 120 V = IReq = I + + + , or I l = 1 A1 A2 A3 A4 1 1 1 + + + A1 A2 A3 A4 V 2 = I l = A2
(120 V ) = 29.5 V 1 1 1 1 A2 + + + A1 A2 A3 A4
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138 Chapter 28 Solutions 28.11 (a) Since all the current flowing in the circuit must pass through the series 100- resistor, P = RI 2 P max = RI max 2 so I max =
1
P = R
25.0 W = 0.500 A 100
1 1 R eq = 100 + + 100 100 Vmax = R eq I max = 75.0 V
= 150
(b) P = (V)I = (75.0 V)(0.500 A) = 37.5 W P 1 = 25.0 W
total power
P 2 = P 3 = RI 2 = (100 )(0.250 A)2 = 6.25 W
28.12
Using 2.00-, 3.00-, 4.00- resistors, there are 7 series, 4 parallel, and 6 mixed combinations: Series 2.00 3.00 4.00 5.00 Parallel Mixed 0.923 1.56 1.20 2.00 1.33 2.22 1.71 3.71 4.33 5.20 The resistors may be arranged in patterns:
6.00 7.00 9.00
28.13
The potential difference is the same across either combination. V = IR = 3I R 1 + 500 = 3
(
1 R
1 1 + 500
)
so
R
1 1 =3 + R 500
and
R = 1000 = 1.00 k
28.14
If the switch is open, If the switch is closed, Then,
I = / ( R + R) I = / (R + R / 2)
and and
P = 2 R / ( R + R)2 P = 2 ( R / 2) / (R + R / 2)2
2 R
( R + R)
2
=
2 R
2(R + R / 2)2
2R 2 + 2RR + R 2 / 2 = R 2 + 2RR + R 2 The condition becomes R 2 = R 2 / 2 so R = 2 R = 2 (1.00 ) = 1.41
Chapter 28 Solutions 139
-1
28.15
Rp =
1 1 + 3.00 1.00
= 0.750
Rs = ( 2.00 + 0.750 + 4.00) = 6.75 I battery = P = I 2 R: V 18.0 V = = 2.67 A Rs 6.75 P 2 = ( 2.67 A ) ( 2.00 )
2
P 2 = 14.2 W in 2.00 P 4 = ( 2.67 A ) ( 4.00 ) = 28.4 W
2
in 4.00
V 2 = ( 2.67 A )( 2.00 ) = 5.33 V, V p = 18.0 V - V 2 - V 4 = 2.00 V P3 =
V 4 = ( 2.67 A )( 4.00 ) = 10.67 V
(= V3 = V1 )
in 3.00
( V3 )2 = (2.00 V)2
R3 3.00
2 ( V1 )2 = (2.00 V)
= 1.33 W
P1 =
R1
1.00
= 4.00 W
in 1.00
28.16
Denoting the two resistors as x and y, 1 1 1 x + y = 690, and 150 = x + y 1 1 1 (690 x) + x 150 = x + 690 x = x(690 x) x 2 690x + 103,500 = 0 x= 690 (690)2 414,000 2 y = 220
x = 470
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140 Chapter 28 Solutions 28.17 (a) V = IR : 33.0 V = I1 (11.0 ) 33.0 V = I 2 ( 22.0 ) I1 = 3.00 A P = I 2 R: I 2 = 1.50 A
2
P 1 = ( 3.00 A ) (11.0 ) P 1 = 99.0 W
P 2 = (1.50 A ) ( 22.0 )
2
P 2 = 49.5 W
The 11.0- resistor uses more power. (b) P 1 + P 2 = 148 W P = I ( V ) = ( 4.50)( 33.0) = 148 W
(c) Rs = R1 + R2 = 11.0 + 22.0 = 33.0 V = IR : P = I 2 R: 33.0 V = I ( 33.0 ), so I = 1.00 A P 1 = (1.00 A ) (11.0 )
2
P 2 = (1.00 A ) ( 22.0 )
2
P 1 = 11.0 W The 22.0- resistor uses more power.
P 2 = 22.0 W
(d) P 1 + P 2 = I 2 ( R1 + R2 ) = (1.00 A ) ( 33.0 ) = 33.0 W
2
P = I ( V ) = (1.00 A )( 33.0 V ) = 33.0 W (e) The parallel configuration uses more power.
28.18
+15.0 (7.00)I1 (2.00)(5.00) = 0 5.00 = 7.00I1 I3 = I1 + I2 = 2.00 A 0.714 + I2 = 2.00 so I2 = 1.29 A so I1 = 0.714 A
+ 2.00(1.29) (5.00)(2.00) = 0
= 12.6 V
Chapter 28 Solutions 141 28.19 We name the currents I1 , I 2 , and I 3 as shown. From Kirchhoff's current rule, I3 = I1 + I2
Applying Kirchhoff's voltage rule to the loop containing I2 and I3 , 12.0 V (4.00)I3 (6.00)I2 4.00 V = 0 8.00 = (4.00)I3 + (6.00)I2 Applying Kirchhoff's voltage rule to the loop containing I1 and I2 , (6.00)I2 4.00 V + (8.00)I1 = 0 (8.00)I1 = 4.00 + (6.00)I2 Solving the above linear systems, I1 = 846 mA, I2 = 462 mA, I3 = 1.31 A
All currents flow in the directions indicated by the arrows in the circuit diagram.
*28.20 The solution figure is shown to the right.
*28.21
We use the results of Problem 19. (a) By the 4.00-V battery: By the 12.0-V battery: (b) By the 8.00 resistor: By the 5.00 resistor: By the 1.00 resistor: By the 3.00 resistor: By the 1.00 r e s i s t o r : (c) 222 J + 1.88 kJ = 1.66 kJ U = (V)It = 4.00 V( 0.462 A)120 s = 222 J 12.0 V (1.31 A) 120 s = 1.88 kJ I 2 Rt = (0.846 A)2(8.00 ) 120 s = 687 J (0.462 A)2(5.00 ) 120 s = 128 J (0.462 A)2(1.00 ) 120 s = 25.6 J (1.31 A)2(3.00 ) 120 s = 616 J (1.31 A)2(1.00 ) 120 s = 205 J from chemical to electrical.
687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from electrical to heat.
2000 by Harcourt, Inc. All rights reserved.
142 Chapter 28 Solutions 28.22 We name the currents I1 , I 2 , and I 3 as shown. [1] [2] [3] 70.0 60.0 I2 (3.00 k) I1 (2.00 k) = 0 80.0 I3 (4.00 k) 60.0 I2 (3.00 k) = 0 I2 = I1 + I3
(a) Substituting for I2 and solving the resulting simultaneous equations yields I1 = 0.385 mA (through R 1) I3 = 2.69 mA (through R 3) I2 = 3.08 mA (through R 2) (b) V cf = 60.0 V (3.08 mA)(3.00 k) = 69.2 V Point c is at higher potential.
28.23
Label the currents in the branches as shown in the first figure. Reduce the circuit by combining the two parallel resistors as shown in the second figure. Apply Kirchhoff's loop rule to both loops in Figure (b) to obtain:
(2.71R)I1 + (1.71R)I 2 = 250
and
(1.71R)I1 + (3.71R)I 2 = 500
(a)
With R = 1000 , simultaneous solution of these equations yields: I1 = 10.0 mA and I 2 = 130.0 mA From Figure (b), Vc - V a = ( I1 + I 2 )(1.71R) = 240 V Thus, from Figure (a), I 4 = Vc - V a 240 V = = 60.0 mA 4000 4R (b)
Finally, applying Kirchhoff's point rule at point a in Figure (a) gives: I = I 4 - I1 = 60.0 mA - 10.0 mA = + 50.0 mA , or I = 50.0 mA flowing from point a to point e .
Chapter 28 Solutions 143 28.24 Name the currents as shown in the figure to the right. w + x + z = y. Loop equations are 200w 40.0 + 80.0x = 0 80.0x + 40.0 + 360 20.0y = 0 + 360 20.0y 70.0z + 80.0 = 0 Eliminate y by substitution. x = 2.50 w + 0.500 400 - 100 x - 20.0 w - 20.0 z = 0 440 - 20.0 w - 20.0 x - 90.0 z = 0 350 - 270 w - 20.0 z = 0 430 - 70.0 w - 90.0 z = 0 430 - 70.0 w - 1575 + 1215 w = 0 w = 70.0/70.0 = 1.00 A upward in 200 Now z = 4.00 A upward in 70.0 x = 3.00 A upward in 80.0 y = 8.00 A downward in 20.0 and for the 200 , V = IR = (1.00 A)(200 ) = 200 V Then
Eliminate x : Eliminate z = 17.5 13.5w to obtain
28.25
Using Kirchhoff's rules, 12.0 - (0.0100) I1 - (0.0600) I 3 = 0 10.0 + (1.00) I 2 - (0.0600) I 3 = 0 and I1 = I 2 + I 3 12.0 - (0.0100) I 2 - (0.0700) I 3 = 0 10.0 + (1.00) I 2 - (0.0600) I 3 = 0 Solving simultaneously, dead battery, I 2 = 0.283 A downward in the
and I 3 = 171 A downward in the starter.
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144 Chapter 28 Solutions 28.26 V ab = (1.00) I1 + (1.00)( I1 - I 2 ) V ab = (1.00) I1 + (1.00) I 2 + ( 5.00)( I - I1 + I 2 ) V ab = ( 3.00)( I - I1 ) + ( 5.00)( I - I1 + I 2 ) Let I = 1.00 A , I1 = x, and I 2 = y Then, the three equations become: V ab = 2.00 x - y , or y = 2.00 x - V ab V ab = - 4.00 x + 6.00 y + 5.00 and V ab = 8.00 - 8.00 x + 5.00 y
Substituting the first into the last two gives: 7.00V ab = 8.00 x + 5.00 and 6.00V ab = 2.00 x + 8.00 Solving these simultaneously yields V ab = Then, Rab = V ab 27 17 V = I 1.00 A 27 V 17 Rab = 27 17
or
28.27
We name the currents I1 , I 2 , and I 3 as shown. (a) I1 = I2 + I3 Counterclockwise around the top loop, 12.0 V (2.00 )I3 (4.00 )I1 = 0 Traversing the bottom loop, 8.00 V (6.00 )I2 + (2.00 )I3 = 0 I1 = 3.00 2 I3
1
I2 = 3 + 3 I 3
4
1
and
I3 = 909 mA
(b) Va (0.909 A)(2.00 ) = Vb V b V a = 1.82 V
Chapter 28 Solutions 145 28.28 We apply Kirchhoff's rules to the second diagram. 50.0 2.00I1 2.00I2 = 0 20.0 2.00I3 + 2.00I2 = 0 I1 = I2 + I3 (1) (2) (3)
Substitute (3) into (1), and solve for I1, I2, and I3 I1 = 20.0 A; I2 = 5.00 A; I3 = 15.0 A
Then apply P = I 2R to each resistor: (2.00 )1 : (4.00 ) : P = I12 (2.00 ) = (20.0 A)2 (2.00 ) = 800 W
2 5.00 P = 2 A (4.00 ) = 25.0 W (Half of I2 goes through each)
(2.00 )3 :
P = I 3 2 (2.00 ) = (15.0 A)2(2.00 ) = 450 W
28.29
(a) RC = (1.00 106 )(5.00 106 F) = 5.00 s (b) Q = C = (5.00 106 C)(30.0 V) = 150 C (c) I(t) =
e -t/RC =
R
30.0 -10.0 exp = 4.06 A 6 6 -6 1.00 10 (1.00 10 )(5.00 10 )
28.30
(a) I(t) = I0et/RC Q 5.10 106 C I0 = RC = = 1.96 A (1300 )(2.00 109 F) 9.00 106 s I(t) = (1.96 A) exp = 61.6 mA 9 (1300 )(2.00 10 F) 8.00 106 s (b) q(t) = Qet/RC = (5.10 C) exp = 0.235 C 9 (1300 )(2.00 10 F) (c) The magnitude of the current is I0 = 1.96 A
2000 by Harcourt, Inc. All rights reserved.
146 Chapter 28 Solutions
2 U = 1 C ( V )
28.31
2
and
V = Q C
Therefore, U = Q 2 2C and when the charge decreases to half its original value, the stored energy is one-quarter its original value: U f = 1 U0
4
28.32
(a) = RC = (1.50 105 )(10.0 106 F) = 1.50 s (b) = (1.00 105 )(10.0 106 F) = 1.00 s (c) The battery carries current 10.0 V = 200 A 50.0 103 10.0 V t/ 1.00 s I = I0et/RC = 3 e 100 10 200 A + (100 A)et / 1.00 s
The 100 k carries current of magnitude
So the switch carries downward current
28.33
(a) Call the potential at the left junction V L and at the right V R . "long" time, the capacitor is fully charged. VL = 8.00 V because of voltage divider: IL = 10.0 V 5.00 = 2.00 A
After a
VL = 10.0 V (2.00 A)(1.00 ) = 8.00 V Likewise, 2.00 VR = 10.0 V = 2.00 V 2.00 + 8.00 IR = 10.0 V = 1.00 A 10.0
or
VR = (10.0 V) (8.00 )(1.00 A) = 2.00 V Therefore, (b) Redraw the circuit V = VL VR = 8.00 2.00 = 6.00 V R= 1 = 3.60 (1/ 9.00 ) + (1/ 6.00 )
RC = 3.60 106 s and et/RC = 1 10 so t = RC ln 10 = 8.29 s
Chapter 28 Solutions 147 (a) = RC = 4.00 106 3.00 10 -6 F = 12.0 s (b) I =
28.34
(
)(
)
e - t/RC
R
=
12.0 e - t/12.0 s 6 4.00 10
q = C 1 - e - t/RC = 3.00 10 - 6 (12.0) 1 - e - t/12.0 q = 36.0 C 1 - e - t/12.0
[
]
[
)
[
]
I = 3.00 Ae - t/12.0
28.35
V0 =
Q C q(t) = Qe - t/RC V(t) = V0 e - t/RC V (t ) = e - t/RC V0 4.00 1 = exp - R 3.60 10 - 6 2
Then, if
Therefore
(
)
ln
4.00 1 =- 2 R 3.60 10 - 6
(
)
R = 1.60 M
28.36
V0 =
Q C V(t) = ( V0 ) e -t RC
Then, if q(t) = Q e -t RC and 1 ( V0 ) , then 2
( V0 )
V(t)
= e -t RC 1 2
When V(t) =
e -t RC = -
t 1 = - ln 2 = ln 2 RC t C(ln 2)
Thus,
R=
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148 Chapter 28 Solutions
28.37
q(t) = Q 1 - e -t/RC
[
]
so or thus
q(t) = 1 - e -t/RC Q e - 0.900/RC = 1 - 0.600 = 0.400 RC = - 0.900 = 0.982 s ln(0.400)
0.600 = 1 - e - 0.900/RC - 0.900 = ln(0.400) RC
28.38
Applying Kirchhoff's loop rule,
-I g (75.0 ) + I - I g Rp = 0
(
)
Therefore, if I = 1.00 A when I g = 1.50 mA , Rp = I g (75.0 ) =
(I - Ig )
(1.50 10
-3
A (75.0 )
1.00 A - 1.50 10 -3 A
)
= 0.113
28.39
Series Resistor Voltmeter V = IR: Solving, 25.0 = 1.50 10-3(Rs + 75.0) Rs = 16.6 k Figure for Goal Solution
Goal Solution The galvanometer described in the preceding problem can be used to measure voltages. In this case a large resistor is wired in series with the galvanometer in a way similar to that shown in Figure P28.24b This arrangement, in effect, limits the current that flows through the galvanometer when large voltages are applied. Most of the potential drop occurs across the resistor placed in series. Calculate the value of the resistor that enables the galvanometer to measure an applied voltage of 25.0 V at full-scale deflection. G: The problem states that the value of the resistor must be "large" in order to limit the current through the galvanometer, so we should expect a resistance of k to M. O: The unknown resistance can be found by applying the definition of resistance to the portion of the circuit shown in Figure 28.24b. Rg = 75.0 . For the two resistors in series, A : V ab = 25.0 V; From Problem 38, I = 1.50 mA and Req = Rs + Rg so the definition of resistance gives us: V ab = I(Rs + Rg ) Therefore, Rs = V ab 25.0 V - Rg = - 75.0 = 16.6 k I 1.50 10 -3 A
L : The resistance is relatively large, as expected. It is important to note that some caution would be necessary if this arrangement were used to measure the voltage across a circuit with a comparable resistance. For example, if the circuit resistance was 17 k, the voltmeter in this problem would cause a measurement inaccuracy of about 50%, because the meter would divert about half the current that normally would go through the resistor being measured. Problems 46 and 59 address a similar concern about measurement error when using electrical meters.
Chapter 28 Solutions 149 28.40 We will use the values required for the 1.00-V voltmeter to obtain the internal resistance of the galvanometer. V = Ig (R + rg) Solve for rg : rg = V 1.00 V R= 900 = 100 Ig 1.00 10-3 A
We then obtain the series resistance required for the 50.0-V voltmeter: R= V 50.0 V Ig rg = 1.00 10-3 A 100 = 49.9 k
28.41
V = I g r g = I - I g Rp , or Rp =
(
)
(I - Ig )
I grg
=
I g (60.0 )
(I - Ig )
Therefore, to have I = 0.100 A = 100 mA when I g = 0.500 mA : Rp =
(0.500 mA)(60.0 ) =
99.5 mA
0.302
Figure for Goal Solution
Goal Solution Assume that a galvanometer has an internal resistance of 60.0 and requires a current of 0.500 mA to produce full-scale deflection. What resistance must be connected in parallel with the galvanometer if the combination is to serve as an ammeter that has a full-scale deflection for a current of 0.100 A? G: An ammeter reads the flow of current in a portion of a circuit; therefore it must have a low resistance so that it does not significantly alter the current that would exist without the meter. Therefore, the resistance required is probably less than 1 . O: From the values given for a full-scale reading, we can find the voltage across and the current through the shunt (parallel) resistor, and the resistance value can then be found from the definition of resistance. A : The voltage across the galvanometer must be the same as the voltage across the shunt resistor i n parallel, so when the ammeter reads full scale, V = (0.500 mA )(60.0 ) = 30.0 mV Through the shunt resistor, Therefore, I = 100 mA - 0.500 mA = 99.5 mA R= V 30.0 mV = = 0.302 I 99.5 mA
L : The shunt resistance is less than 1 as expected. It is important to note that some caution would be necessary if this meter were used in a circuit that had a low resistance. For example, if the circuit resistance was 3 , adding the ammeter to the circuit would reduce the current by about 10%, so the current displayed by the meter would be lower than without the meter. Problems 46 and 59 address a similar concern about measurement error when using electrical meters.
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150 Chapter 28 Solutions R 2R 3 R1 R 2R 3 2.50 R 2 1000 = 400 2.50
28.42
Rx =
=
=
28.43
Using Kirchhoff's rules with Rg << 1, - ( 21.0 ) I1 + (14.0 ) I 2 = 0, so I1 = 70.0 - 21.0I1 - 7.00 I1 + I g = 0 , and 70.0 - 14.0I 2 - 7.00 I 2 - I g = 0 The last two equations simplify to 10.0 - 4.00 2 I 2 = I g ,
3
2 I2 3
21.0 I1 G I2 14.0 I2 - Ig Ig
7.00 I1 + Ig
(
)
7.00
(
)
+
70.0 V
( )
and 10.0 - 3.00I 2 = -I g
Solving simultaneously yields: Ig = 0.588 A
28.44
R=
L L and Ri = i A Ai L2 V
and Ri =
But, V = AL = Ai Li , so R =
L2 i V
= Ri [1 + ]
2
2 Li 1 + ( L Li ) ( Li + L) = Therefore, R = V V
[
]
2
where
L L
This may be written as:
R = R i (1 + 2 + 2)
28.45
x = s ; = s Rx x
Rs Rs Rs
48.0 = (1.0186 V) = 1.36 V 36.0
Chapter 28 Solutions 151 *28.46 (a) In Figure (a), the emf sees an equivalent resistance of 200.00 . 6.000 0 V I= = 0.030 000 A 200.00
20.000
20.000 6.0000 V
V
20.000
A
V
A
180.00 (a)
180.00 (b)
180.00 (c)
The terminal potential difference is (b) In Figure (b), The equivalent resistance across the emf is The ammeter reads and the voltmeter reads (c) In Figure (c), Therefore, the emf sends current through The current through the battery is but not all of this goes through the ammeter. The voltmeter reads The ammeter measures current
V = IR = (0.030 000 A )(180.00 ) = 5.400 0 V 1 1 Req = + 180.00 20 000
-1
= 178.39
178.39 + 0.500 00 + 20.000 = 198.89 I=
R
=
6.000 0 V = 0.030 167 A 198.89
V = IR = (0.030 167 A )(178.39 ) = 5.381 6 V 1 1 180.50 + 20 000
-1
= 178.89
Rtot = 178.89 + 20.000 = 198.89 I= 6.000 0 V = 0.030 168 A 198.89
V = IR = (0.030 168 A )(178.89 ) = 5.396 6 V I= V 5.396 6 V = = 0.029 898 A R 180.50
The connection shown in Figure (c) is better than that shown in Figure (b) for accurate readings.
28.47
(a) P = I(V)
So for the Heater,
I=
P 1500 W = = 12.5 A V 120 V
For the Toaster, And for the Grill, (b) 12.5 + 6.25 + 8.33 = 27.1 A sufficient.
750 W I = 120 W = 6.25 A 1000 W I = 120 V = 8.33 A (Grill) The current draw is greater than 25.0 amps, so this would not be
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152 Chapter 28 Solutions
2 -8 l (1.00 A) (1.70 10 m)(16.0 ft)(0.3048 m / ft) = = 0.101 W A (0.512 10 -3 m)2
28.48
(a) P = I 2 R = I 2
(b) P = I 2 R = 100(0.101 ) = 10.1 W
28.49
2 2 I Al RAl = I Cu RCu
so
I Al =
RCu Cu I Cu = I Cu = RAl Al
1.70 (20.0) = 0.776(20.0) = 15.5 A 2.82
*28.50 (a) Suppose that the insulation between either of your fingers and the conductor adjacent is a chunk of rubber with contact area 4 mm 2 and thickness 1 mm. Its resistance is R=
13 -3 l 10 m 10 m 2 1015 A 4 10 -6 m 2
(
)(
)
The current will be driven by 120 V through total resistance (series) 2 1015 + 10 4 + 2 1015 5 1015 It is: I = V 120 V ~ R 5 1015 ~ 10 -14 A
(b) The resistors form a voltage divider, with the center of your hand at potential V h 2 , where V h is the potential of the "hot" wire. The potential difference between your finger and thumb is V = IR ~ 10 -14 A 10 4 ~ 10 -10 V . So the points where the rubber meets your fingers are at potentials of
(
)(
)
~
Vh + 10 -10 V 2
and
~
Vh - 10 -10 V 2
*28.51
The set of four batteries boosts the electric potential of each bit of charge that goes through them by 4 1.50 V = 6.00 V. The chemical energy they store is U = qV = (240 C)(6.00 J/C) = 1440 J The radio draws current So, its power is 6.00 V V = 0.0300 A I= R = 200 P = (V)I = (6.00 V)(0.0300 A) = 0.180 W = 0.180 J/s t= E 1440 J = = 8.00 10 3 s P 0.180 J / s
Then for the time the energy lasts, we have P = E t: We could also compute this from I = Q/t:
240 C Q t = I = 0.0300 A = 8.00 103 s = 2.22 h
Chapter 28 Solutions 153
*28.52
I=
R+r
, so P = I 2 R =
( R + r )2
2R
or
( R + r )2 =
2 R P
Let x
2 ,
P
then
( R + r )2 = xR
or
R 2 + ( 2r - x )R - r 2 = 0 R 2 + ( 2.40 - x )R - 1.44 = 0, R= R= -( 2.40 - x ) + 4.21
With r = 1.20 , this becomes which has solutions of
(2.40 - x)2 - 5.76
2 3.84
(a) With
= 9.20 V and
P = 12.8 W , x = 6.61:
( 4.21)2 - 5.76
2
=
or
0.375 (b) For
= 9.20 V and
P = 21.2 W, x
2 = 3.99
P
R=
+1.59
(1.59)2 - 5.76
2
=
1.59 -3.22 2
The equation for the load resistance yields a complex number, so there is no resistance that will extract 21.2 W from this battery. The maximum power output occurs when R = r = 1.20 , and that maximum is: P max = 2 4r = 17.6 W
28.53
Using Kirchhoff's loop rule for the closed loop,
+12.0 - 2.00 I - 4.00 I = 0,
so
I = 2.00 A
Vb - V a = + 4.00 V - ( 2.00 A )( 4.00 ) - (0)(10.0 ) = - 4.00 V Thus, V ab = 4.00 V and point a is at the higher potential .
28.54
The potential difference across the capacitor Using 1 Farad = 1 s ,
- 3.00 10 5 R
V (t ) = Vmax 1 - e -t RC
[
]
R 10.0 10 -6 s
- ( 3.00 s ) 4.00 V = (10.0 V )1 - e
(
)
Therefore, 0.400 = 1.00 - e
(
)
or
e
- 3.00 10 5 R
(
)
= 0.600
Taking the natural logarithm of both sides,
-
3.00 10 5 = ln(0.600) R
and
R=-
3.00 10 5 = + 5.87 10 5 = 587 k ln(0.600)
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154 Chapter 28 Solutions 28.55 Let the two resistances be x and y. Then, Rs = x + y = Ps 225 W = = 9.00 2 I ( 5.00 A)2 y = 9.00 x
y x y
x
and
Pp xy 50.0 W Rp = = 2 = = 2.00 x+y I ( 5.00 A)2 x(9.00 - x ) = 2.00 x + (9.00 - x ) x 2 - 9.00 x + 18.0 = 0
so
Factoring the second equation, so Then, y = 9.00 - x gives
( x - 6.00)( x - 3.00) = 0
x = 6.00 or x = 3.00 y = 3.00 or y = 6.00
The two resistances are found to be 6.00 and 3.00 .
28.56
Let the two resistances be x and y. Then, Rs = x + y = Pp xy Ps and Rp = = 2. 2 x+y I I Ps - x , and the second I2 Ps Pp Ps or x 2 - 2 x + 4 = 0. I I
2 P s P s - 4P s P p
From the first equation, y = x Ps I 2 - x
becomes
(
x + Ps I - x
2
(
)
)
=
Pp I
2
Using the quadratic formula, x =
2I 2
.
Then, y =
2 P s > P s - 4P s P p Ps . - x gives y = 2I 2 I2 2 P s + P s - 4P s P p 2 P s - P s - 4P s P p
The two resistances are
2I 2
and
2I 2
Chapter 28 Solutions 155
28.57
The current in the simple loop circuit will be I = (a) (b) (c) Vter = Ir = I= R+r R (R + r)2
R+r
R+r
R
and and
Vter as R I
r
as R 0
P = I 2R = 2
-2 2 R dP 2 = 0 = 2 R(-2)(R + r)-3 + 2 (R + r)-2 = + dR (R + r)3 (R + r)2 Then 2R = R + r and R=r
Figure for Goal Solution
Goal Solution A battery has an emf and internal resistance r. A variable resistor R is connected across the terminals of the battery. Determine the value of R such that (a) the potential difference across the terminals is a maximum, (b) the current in the circuit is a maximum, (c) the power delivered to the resistor is a maximum. G: If we consider the limiting cases, we can imagine that the potential across the battery will be a maximum when R = (open circuit), the current will be a maximum when R = 0 (short circuit), and the power will be a maximum when R is somewhere between these two extremes, perhaps when R = r. O: We can use the definition of resistance to find the voltage and current as functions of R, and the power equation can be differentiated with respect to R. A : (a) The battery has a voltage Vterminal = - Ir = I=
R R+r
or as R , Vterminal or as R 0, I
(b) The circuit's current is
R+r
r
(c) The power delivered is
P = I 2R =
2R
(R + r)2
To maximize the power P as a function of R, we differentiate with respect to R, and require that dP /dR = 0 -2 2 R dP 2 = 0 = 2 R(-2)(R + r)-3 + 2 (R + r)-2 = + 3 dR (R + r) (R + r)2 Then 2R = R + r and R=r
L : The results agree with our predictions. Making load resistance equal to the source resistance to maximize power transfer is called impedance matching.
2000 by Harcourt, Inc. All rights reserved.
156 Chapter 28 Solutions 28.58 (a)
- I(R) - (1 + 2 ) = 0
40.0 V - (4.00 A)[(2.00 + 0.300 + 0.300 + R)] - (6.00 + 6.00) V = 0; so 32.0 W R = 4.40
(b) Inside the supply, Inside both batteries together, For the limiting resistor,
P = I 2 R = ( 4.00 A ) ( 2.00 ) =
2 2
P = I 2 R = ( 4.00 A ) (0.600 ) = 9.60 W P = ( 4.00 A ) ( 4.40 ) = 70.4 W
2
(c) P = I( 1 + 2 ) = (4.00 A)[(6.00 + 6.00)V ] = 48.0 W
28.59
Let R m = measured value, R = actual value, IR = current through the resistor R I = current measured by the ammeter. (a) When using circuit (a), IRR = V = 20 000(I IR) V V and IR = R , we have But since I = R m and or I R = 20 000 - 1 IR I IR R = R m (R R m) Rm (b) (a)
Figure for Goal solution
R = 20 000
(1)
When R > R m , we require Therefore, R m R (1 0.0500) and from (1) we find (b) When using circuit (b), But since IR = V Rm ,
(R R m ) 0.0500 R R 1050 IRR = V IR (0.5 ). Rm = (0.500 + R) (R m R) 0.0500 R R 10.0 (2)
When R m > R, we require From (2) we find
Chapter 28 Solutions 157 Goal Solution The value of a resistor R is to be determined using the ammeter-voltmeter setup shown in Figure P28.59. The ammeter has a resistance of 0.500 , and the voltmeter has a resistance of 20 000 . Within what range of actual values of R will the measured values be correct to within 5.00% if the measurement is made using (a) the circuit shown in Figure P28.59a (b) the circuit shown in Figure P28.59b? G: An ideal ammeter has zero resistance, and an ideal voltmeter has infinite resistance, so that adding the meter does not alter the current or voltage of the existing circuit. For the non-ideal meters in this problem, a low values of R will give a large voltage measurement error in circuit (b), while a large value of R will give significant current measurement error in circuit (a). We could hope that these meters yield accurate measurements in either circuit for typical resistance values of 1 to 1 M . O: The definition of resistance can be applied to each circuit to find the minimum and maximum current and voltage allowed within the 5.00% tolerance range. A : (a) In Figure P28.59a, at least a little current goes through the voltmeter, so less current flows through the resistor than the ammeter reports, and the resistance computed by dividing the voltage by the inflated ammeter reading will be too small. Thus, we require that V/ I = 0.950R where I is the current through the ammeter. Call I R the current through the resistor; then I - I R is the current i n the voltmeter. Since the resistor and the voltmeter are in parallel, the voltage across the meter equals the voltage across the resistor. Applying the definition of resistance: V = I R R = ( I - I R )( 20 000 ) Our requirement is IRR 0.95R I R (R + 20 000 ) 20 000 20 000 0.95(R + 20 000 ) = 0.95R + 19000 R 1000 0.95 or R 1.05 k so I= I R (R + 20 000 ) 20 000
Solving, and
(b) If R is too small, the resistance of an ammeter in series will significantly reduce the current that would otherwise flow through R. In Figure 28.59b, the voltmeter reading is I (0.500 ) + IR , at least a little larger than the voltage across the resistor. So the resistance computed by dividing the inflated voltmeter reading by the ammeter reading will be too large. We require Thus, V 1.05R I so that and I (0.500 ) + IR 1.05R I R 10.0
0.500 0.0500R
L : The range of R values seems correct since the ammeter's resistance should be less than 5% of the smallest R value ( 0.500 0.05R means that R should be greater than 10 ), and R should be less than 5% of the voltmeter's internal resistance ( R 0.05 20 k = 1 k ). Only for the restricted range between 10 ohms and 1000 ohms can we indifferently use either of the connections (a) and (b) for a reasonably accurate resistance measurement. For low values of the resistance R, circuit (a) must be used. Only circuit (b) can accurately measure a large value of R.
2000 by Harcourt, Inc. All rights reserved.
158 Chapter 28 Solutions dE E = P = E I = E e -1/RC R dt
28.60
The battery supplies energy at a changing rate
Then the total energy put out by the battery is
dE =
0
t=0
2 exp -
R
t dt RC
dE =
2 (-RC)
R
0
t dt t - exp - = - 2C exp - RC RC RC
= - 2C[0 - 1] = 2 C
The heating power of the resistor is
dE 2 2t = P = V R I = I 2 R = R 2 exp - RC dt R
So the total heat is
dE = 0
2C exp - 2t 2t 2dt exp - - =- RC RC RC 2
1
0
2 exp -
R
2t dt RC
2 - RC dE =
R 2
0
=-
1
2C [0 - 1] = 2C
2 2
The energy finally stored in the capacitor is U = 2 C (V)2 = 2 C 2. Thus, energy is conserved:
2C = 1 2C + 1 2C 2 2
and resistor and capacitor share equally in the energy from the battery.
28.61
(a) q = C( V ) 1 - e -t RC
[
]
= 9.93 C
q = 1.00 10 -6 F (10.0 V ) (b) I = I= dq V -t RC e = dt R 10.0 V -5.00 = 3.37 10 -8 A = 33.7 nA e 2.00 106
(
)
(c)
dU d 1 q 2 q dq q I = = = dt dt 2 C C dt C dU 9.93 10 -6 C = 3.37 10 -8 A = 3.34 10 -7 W = 334 nW dt 1.00 10 -6 C V
(
)
(d) P battery = IE = 3.37 10 -8 A (10.0 V ) = 3.37 10 -7 W = 337 nW
(
)
Chapter 28 Solutions 159 28.62 Start at the point when the voltage has just reached 2 2 V and the switch has just closed. The voltage is V 3 3 and is decaying towards 0 V with a time constant RBC. 2 VC (t) = V e -t/RBC 3 We want to know when V C(t) will reach Therefore, or
1 V. 3
1 1 2 V = V e -t/RBC or e -t/RBC = 3 2 3
t1 = R BC ln 2
1 V, 3
After the switch opens, the voltage is
increasing toward V with time constant ( RA + RB )C :
2 V C (t) = V V e -t/(RA +RB )C 3 When V C (t) = so
2 V, 3
2 2 V = V - Ve -t/(RA +RB )C 3 3 t2 = (R A + R B)C ln 2
or and
e -t/(RA +RB )C =
1 2
T = t1 + t2 = (R A + 2R B)C ln 2
28.63
(a) First determine the resistance of each light bulb: R= (V)2 (120 V)2 = = 240 P 60.0 W
P = (V)2 R
We obtain the equivalent resistance Req of the network of light bulbs by applying Equations 28.6 and 28.7: Req = R1 + 1 = 240 + 120 = 360 (1/ R2 ) + (1/ R3 ) P= (V)2 (120 V)2 = = 40.0 W Req 360 P = Req 40.0 W 1 = A 360 3 1 A (240 ) = 80.0 V 3
The total power dissipated in the 360 is
(b) The current through the network is given by P = I 2 Req : The potential difference across R 1 is
I=
V1 = IR1 =
The potential difference V 23 across the parallel combination of R 2 and R 3 is V 23 = IR23 = 1 1 A = 40.0 V 3 (1/ 240 ) + (1/ 240 )
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160 Chapter 28 Solutions V = IR (a) 20.0 V = (1.00 10-3 A)(R1 + 60.0 ) R1 = 1.994 104 = 19.94 k (b) 50.0 V = (1.00 10-3 A)(R2 + R1 + 60.0 ) (c) 100 V = (1.00 10-3 A)(R3 + R1 + 60.0 ) R 2 = 30.0 k R 3 = 50.0 k
28.64
28.65
Consider the circuit diagram shown, realizing that I g = 1.00 mA. For the 25.0 mA scale:
(24.0 mA)( R1 + R2 + R3 ) = (1.00 mA)(25.0 )
or For the 50.0 mA scale: or For the 100 mA scale: or R1 + R2 + R3 = 25.0 24.0 (1)
( 49.0 mA)( R1 + R2 ) = (1.00 mA)(25.0 + R3 )
49.0( R1 + R2 ) = 25.0 + R3 (2)
(99.0 mA)R1 = (1.00 mA)(25.0 + R2 + R3 )
99.0R1 = 25.0 + R2 + R3 (3)
Solving (1), (2), and (3) simultaneously yields R1 = 0.260 , R2 = 0.261 , R3 = 0.521
28.66
Ammeter: or Voltmeter:
I g r = 0.500 A - I g (0.220 ) I g (r + 0.220 ) = 0.110 V 2.00 V = I g (r + 2500 ) (1) (2)
(
)
Solve (1) and (2) simultaneously to find: Ig = 0.756 mA and r = 145
Chapter 28 Solutions 161
28.67
(a) After steady-state conditions have been reached, there is no DC current through the capacitor. Thus, for R 3: I R3 = 0 (steady-state)
For the other two resistors, the steady-state current is simply determined by the 9.00-V emf across the 12-k and 15-k resistors in series: For R 1 and R 2: I(R1 +R2 ) = R1 + R2
=
9.00 V = 333 A (steady-state) (12.0 k + 15.0 k)
(b) After the transient currents have ceased, the potential difference across C is the same as the potential difference across R 2(= IR 2) because there is no voltage drop across R 3 . Therefore, the charge Q on C is Q = C (V)R2 = C (IR2) = (10.0 F)(333 A)(15.0 k ) = 50.0 C (c) When the switch is opened, the branch containing R 1 is no longer part of the circuit. The capacitor discharges through (R 2 + R 3) with a time constant of (R 2 + R 3)C = (15.0 k + 3.00 k )(10.0 F) = 0.180 s. The initial current Ii in this discharge circuit is determined by the initial potential difference across the capacitor applied to (R2 + R3) in series: Ii = (333 A)(15.0 k) (V)C IR2 = = = 278 A (R2 + R3 ) (R2 + R3 ) (15.0 k + 3.00 k)
Thus, when the switch is opened, the current through R 2 changes instantaneously from 333 A (downward) to 278 A (downward) as shown in the graph. Thereafter, it decays according to I R2 = Ii e -t/(R2 +R3 )C = (278 A)e -t/(0.180 s) (for t > 0) (a)
(d) The charge q on the capacitor decays from Q i to Q i/5 according to q = Qi e -t/(R2 +R3 )C Qi = Qi e(-t/0.180 s) 5 5 = e t/0.180 s t ln 5 = 180 ms t = (0.180 s)(ln 5) = 290 ms
2000 by Harcourt, Inc. All rights reserved.
162 Chapter 28 Solutions V = e -t RC so ln 1 = t V RC
28.68
A plot of ln
1 versus t should be a straight line with slope = . V RC t (s) 0 4.87 11.1 19.4 30.8 46.6 67.3 102.2 V (V) 6.19 5.55 4.93 4.34 3.72 3.09 2.47 1.83 ln( V ) 0 0.109 0.228 0.355 0.509 0.695 0.919 1.219
Using the given data values:
(a) A least-square fit to this data yields the graph to the right.
xi = 282 , Slope =
xi2 = 1.86 10 4 ,
xi yi = 244, = 0.0118
yi = 4.03 ,
2 i
N=8
i i i i
N ( xi yi ) - ( xi )( yi ) N
( ) - (x )
xi2
i
2
(x )(y ) - (x )(x y ) = 0.0882 Intercept = N ( x ) - ( x )
2 i 2 i
The equation of the best fit line is:
ln
= (0.0118) t + 0.0882 V
(b) Thus, the time constant is = RC =
1 1 = = 84.7 s slope 0.0118
and the capacitance is C =
84.7 s = = 8.47 F R 10.0 106
Chapter 28 Solutions 163
28.69
r r a
r r r r r r r r i/3 r r i/3 b a i/3
i/6 i/6 i/6 i/6 i/6 i/6 i/3 i/3 i/3 b
3 junctions at the same potential
another set of 3 junctions at the same potential
28.70
(a) For the first measurement, the equivalent circuit is as shown in Figure 1. Rab = R1 = Ry + Ry = 2Ry so Ry = 1 R1 2 (1)
a Ry
R1 c Rx Figure 1 R2 b Ry
For the second measurement, the equivalent circuit is shown in Figure 2. Thus, Rac = R2 = 1 Ry + Rx 2 (2)
Ry
a
c Rx
Ry
Figure 2
Substitute (1) into (2) to obtain:
R2 =
1 1 R1 + Rx , 22
or
Rx = R2 -
1 R1 4
(b) If R1 = 13.0 and R2 = 6.00 , then
Rx = 2.75
The antenna is inadequately grounded since this exceeds the limit of 2.00 .
28.71
Since the total current passes through R3 , that resistor will dissipate the most power. When that resistor is operating at its power limit of 32.0 W, the current through it is
2 I total =
R1 R3
P 32.0 W = = 16.0 A 2 , or I total = 4.00 A R 2.00
R2 R 1 = R2 = R3 = 2.00
Half of this total current (2.00 A) flows through each of the other two resistors, so the power dissipated in each of them is: P=
(
2 1 I R 2 total
)
= (2.00 A)2 (2.00 ) = 8.00 W
Thus, the total power dissipated in the entire circuit is:
2000 by Harcourt, Inc. All rights reserved.
164 Chapter 28 Solutions
P total = 32.0 W + 8.00 W + 8.00 W = 48.0 W 28.72 The total resistance between points b and c is: R=
.00 k
1
= 2.00 F
(2 .00 k)(3.00 k) = 1.20 k
2 .00 k + 3.00 k
.00 k
2
= 3.00 F 20 V
The total capacitance between points d and e is: C = 2.00 F + 3.00 F = 5.00 F The potential difference between point d and e in this series RC circuit at any time is: V = 1 - e -t RC = (120.0 V ) 1 - e -1000t
[
]
[
6
]
6
Therefore, the charge on each capacitor between points d and e is: q1 = C1 ( V ) = ( 2.00 F )(120.0 V ) 1 - e -1000t
[
]=
(240 C)[1 - e -1000t
6
6
]
6
and q2 = C2 ( V ) = ( 3.00 F )(120.0 V ) 1 - e -1000t
[
]=
(360 C)[1 - e -1000t
]
*28.73 (a) Req = 3R (b) Req = 1 R = (1/ R) + (1/ R) + (1/ R) 3
I= I=
3R 3 R connection.
P series = I = P parallel = I =
2
3R
3 2 R
(c) Nine times more power is converted in the parallel
Chapter 29 Solutions
29.1 (a) (b) up out of the page, since the charge is negative.
(c) no deflection (d) into the page
29.2 (a) (b) (c)
At the equator, the Earth's magnetic field is horizontally north. Because an electron has negative charge, F = q v B is opposite in direction to v B. Figures are drawn looking down. Down North = East, so the force is directed West North North = sin 0 = 0: Zero deflection West North = Down, so the force is directed Up (a) (c) (d)
(d) Southeast North = Up, so the force is Down
29.3
FB = q v B;
F B (j) = e v i B
Therefore, B = B (k) which indicates the negative z direction
*29.4
(a)
FB = qvB sin = (1.60 1019 C)(3.00 106 m/s)(3.00 101 T) sin 37.0 FB = 8.67 1014 N 8.67 1014 N F = 5.19 1013 m/s2 a= m = 1.67 1027 kg
(b)
29.5
F = ma = (1.67 1027 kg)(2.00 1013 m/s2) = 3.34 1014 N = qvB sin 90 F 3.34 1014 N B = qv = = 2.09 102 T (1.60 1019 C)(1.00 107 m/s) The right-hand rule shows that B must be in the y direction to yield a force in the +x direction when v is in the z direction.
2000 by Harcourt, Inc. All rights reserved.
2
Chapter 29 Solutions
*29.6
First find the speed of the electron: 2e( V ) = m
K = 2 mv 2 = e(V) = U
1
v=
2 1.60 10 -19 C ( 2400 J / C)
(
(
)
9.11 10
-31
kg
)
= 2.90 107 m / s
(a) (b)
FB, max = qvB = (1.60 1019 C)(2.90 107 m/s)(1.70 T) = 7.90 1012 N FB, min = 0 occurs when v is either parallel to or anti-parallel to B
29.7
Gravitational force: Electric force: Magnetic force:
Fg = mg = (9.11 1031 kg)(9.80 m/s2) = 8.93 1030 N down Fe = qE = (1.60 1019 C)100 N/C down = 1.60 1017 N up m Ns F B = qv B = -1.60 10 -19 C 6.00 106 E 50.0 10 -6 N s Cm FB = 4.80 1017 N up = 4.80 1017 N down
(
)
29.8
We suppose the magnetic force is small compared to gravity. Then its horizontal velocity component stays nearly constant. We call it v i. From vy 2 = vyi 2 + 2ay ( y - yi ) , the vertical component at impact is - 2gh j. Then, F B = qv B = Q vi - 2gh j Bk = QvB( -j) - Q 2gh Bi FB = QvB vertical + Q 2gh B horizontal FB = 5.00 106 C(20.0 m/s)(0.0100 T) j + 5.00 106 C 2(9.80 m/s2)(20.0 m) (0.0100 T) i FB = (1.00 106 N) vertical + (0.990 106 N) horizontal
(
)
29.9
F B = qvB sin sin = 0.754
so and
8.20 10-13 N = (1.60 10-19 C)(4.00 106 m/s)(1.70 T) sin
= sin-1(0.754) = 48.9 or 131
Chapter 29 Solutions 29.10 qE = -1.60 10 -19 C ( 20.0 N / C)k = -3.20 10 -18 N k
3
(
)
(
)
F = qE + qv B = ma
(-3.20 10
-18
N k 1.60 10-19 C (1.20 104 m/s i) B = (9.11 10-31)(2.00 1012 m/s2)k
)
(3.20 10-18 N)k (1.92 10-15 C m/s)i B = (1.82 10-18 N)k (1.92 10-15 C m/s)i B = (5.02 10-18 N)k The magnetic field may have any x-component . Bz = 0 and By = 2.62 mT
29.11
F B = qv B i j k v B = +2 -4 +1 = (12 - 2) i + (1 + 6) j + ( 4 + 4) k = 10i + 7 j + 8 k +1 +2 -3 v B = 10 2 + 7 2 + 8 2 = 14.6 T m / s F B = q v B = 1.60 10 -19 C (14.6 T m s) = 2.34 10 -18 N
(
)
29.12
F B = qv B = -1.60 10
(
-19
)
i j 0 3.70 10 5 1.40 2.10
k 0 0
F B = -1.60 10 -19 C (0 - 0) i + (0 - 0) j + 0 - (1.40 T ) 3.70 10 5 m s k =
(
)[
(
(
))
]
(8.29 10
-14
k N
)
29.13
FB = ILB sin mg = ILB sin
with so
F B = F g = mg m L g = IB sin 100 cm / m m -2 = (0.500 g / cm ) = 5.00 10 kg / m L 1000 g / kg (5.00 102)(9.80) = (2.00)B sin 90.0
I = 2.00 A Thus
and
B = 0.245 Tesla with the direction given by right-hand rule: eastward
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4
Chapter 29 Solutions
Goal Solution A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward? G: Since I = 2.00 A south, B must be to the right-hand rule for currents in a magnetic field. east to make F upward according to the
The magnitude of B should be significantly greater than the earth's magnetic field (~ 50 T), since we do not typically see wires levitating when current flows through them.
O: The force on a current-carrying wire in a magnetic field is F B = I1 B, from which we can find B. A : With I to the south and B to the east, the force on the wire is simply FB = I lBsin 90, which must oppose the weight of the wire, mg. So, B= g 10 2 cm / m FB mg g m 9.80 m / s 2 = = = 0.500 cm 3 = 0.245 T Il I l 2.00 A Il 10 g / kg
L : The required magnetic field is about 5000 times stronger than the earth's magnetic field. Thus it was reasonable to ignore the earth's magnetic field in this problem. In other situations the earth's field can have a significant effect.
29.14
FB = I L B = (2.40 A)(0.750 m)i (1.60 T)k = (2.88 j) N
29.15
(a) (b) (c)
FB = ILB sin = (5.00 A)(2.80 m)(0.390 T) sin 60.0 = 4.73 N FB = (5.00 A)(2.80 m)(0.390 T) sin 90.0 = 5.46 N FB = (5.00 A)(2.80 m)(0.390 T) sin 120 = 4.73 N
29.16
F B mg I L B = = L L L mg (0.0400 kg/m)(9.80 m/s2) I = BL = = 0.109 A 3.60 T The direction of I in the bar is to the right .
Chapter 29 Solutions 29.17
5
The magnetic and gravitational forces must balance. Therefore, it is necessary to have F B = BIL = mg, or I = (mg/BL) = ( g/B) [ is the mass per unit length of the wire]. Thus, I = (1.00 10-3 kg/m)(9.80 m/s2) = 196 A (5.00 10-5 T) (if B = 50.0 T)
The required direction of the current is eastward , since East North = Up.
29.18
For each segment, I = 5.00 A Segment ab bc cd da L
and
B = 0.0200 N / A m j F B = I (L B) 0 (40.0 mN)( i) (40.0 mN)( k) (40.0 mN)(k + i)
-0.400 m j 0.400 m k 0.400 m i + 0.400 m j 0.400 m i 0.400 m k
29.19
The rod feels force
F B = I (d B) = Id(k ) B( -j) = IdB(i )
The work-energy theorem is
1 1 0 + 0 + Fscos = 2 mv 2 + 2 I 2
(Ktrans + Krot )i + E = (Ktrans + Krot ) f
v 1 1 1 IdBLcos 0 = 2 mv 2 + 2 2 mR 2 R v= 4IdBL = 3m
(
)
2
and
3 IdBL = 4 mv 2
4(48.0 A)(0.120 m)(0.240 T)(0.450 m) = 1.07 m/s 3(0.720 kg)
29.20
The rod feels force The work-energy theorem is
1 1 0 + 0 + Fscos = 2 mv 2 + 2 I 2
F B = I (d B) = Id(k ) B( -j) = IdB(i )
(Ktrans + Krot )i + E = (Ktrans + Krot ) f
v 1 1 1 IdBLcos 0 = 2 mv 2 + 2 2 mR 2 R
(
)
2
and
v=
4IdBL 3m
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6 29.21
Chapter 29 Solutions The magnetic force on each bit of ring is I ds B = I ds B radially inward and upward, at angle above the radial line. The radially inward components tend to squeeze the ring but all cancel out as forces. The upward components I ds B sin all add to I 2 rB sin up .
*29.22
Take the x-axis east, the y-axis up, and the z-axis south. The field is B = ( 52.0 T ) cos 60.0 ( -k ) + ( 52.0 T ) sin 60.0 ( -j) The current then has equivalent length: L = 1.40 m ( -k ) + 0.850 m ( j)
F B = I L B = (0.0350 A )(0.850 j - 1.40 k ) m ( - 45.0 j - 26.0 k ) 10 -6 T F B = 3.50 10 -8 N ( -22.1i - 63.0i ) = 2.98 10 -6 N ( -i ) = 2.98 N west 2 r = 2.00 m so
29.23
(a)
r = 0.318 m
= IA = (17.0 10-3 A) (0.318)2 m 2 = 5.41 mA m2 (b) =B so
[
]
= (5.41 10-3 A m2)(0.800 T) = 4.33 mN m
*29.24
= B sin
so
4.60 10-3 N m = (0.250) sin 90.0
= 1.84 102 A m2 = 18.4 mA m2
29.25
= NBAI sin
= 100(0.800 T ) 0.400 0.300 m 2 (1.20 A ) sin 60
= 9.98 N m
Note that is the angle between the magnetic moment and the B field. The loop will rotate so as to align the magnetic moment with the B field. Looking down along the y-axis, the loop will rotate in a clockwise direction.
(
)
(a)
(b)
Chapter 29 Solutions
7
29.26
(a)
Let represent the unknown angle; L , the total length of the wire; and d, the length of one side of the square coil. Then, use the right-hand rule to find L = NAI = d 2 I 4d At equilibrium, at angle with the horizontal.
= ( B) (r mg) = 0
and mgd ILBd cos sin = 4 2
mgd ILBd sin(90.0 - ) - sin = 0 4 2
ILB -1 (3.40 A)(4.00 m)(0.0100 T) = tan -1 = tan = 3.97 2 2mg 2(0.100 kg)(9.80 m / s ) (b)
m =
1 ILBd cos = (3.40 A)(4.00 m)(0.0100 T)(0.100 m) cos 3.97 = 3.39 mN m 4 4
29.27 (a)
From = B = IA B, the magnitude of the torque is IAB sin 90.0 Each side of the triangle is 40.0 cm/3. Its altitude is
1
13.3 2 6.67 2 cm = 11.5 cm and its area is
A = 2 (11.5 cm)(13.3 cm) = 7.70 10-3 m2 Then = (20.0 A)(7.70 10-3 m2)(0.520 N s/C m) = 80.1 mN m (b) Each side of the square is 10.0 cm and its area is 100 cm2 = 10-2 m2.
= (20.0 A)(10-2 m2)(0.520 T) = 0.104 N m
(c) r = 0.400 m/2 = 0.0637 m A = r 2 = 1.27 10-2 m2
= (20.0 A)(1.27 10-2 m2)(0.520) = 0.132 N m
(d) The circular loop experiences the largest torque. Choose U = 0 when the dipole moment is at = 90.0 to the field. The field exerts torque of magnitude Bsin on the dipole, tending to turn the dipole moment in the direction of decreasing . Its energy is given by U -0=
*29.28
90.0
Bsin d = B( - cos ) 90.0 = - Bcos + 0
or
U=B
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8
Chapter 29 Solutions
*29.29 (a)
The field exerts torque on the needle tending to align it with the field, so the minimum energy orientation of the needle is: pointing north at 48.0 below the horizontal where its energy is U min = - Bcos 0 = - 9.70 10 -3 A m 2 55.0 10 -6 T = - 5.34 10 -7 J
(
)(
)
It has maximum energy when pointing in the opposite direction, south at 48.0 above the horizontal where its energy is (b) U min + W = U max : U max = - B cos 180 = + 9.70 10 -3 A m 2 55.0 10 -6 T = + 5.34 10 -7 J W = U max - U min = + 5.34 10 -7 J - -5.34 10 -7 J = 1.07 J
(
)(
)
(
)
29.30
(a)
= B,
so
= B= B sin = NIAB sin
max = NIABsin 90.0 = 1( 5.00 A ) (0.0500 m )2 3.00 10 -3 T = 118 N m
(b) U = - B, so - B U + B Since B = ( NIA)B = 1( 5.00 A ) (0.0500 m ) the range of the potential energy is:
[
](
)
[
2
](3.00 10 T) = 118 J,
-3
-118 J U +118 J
29.31
(a)
B = 50.0 10-6 T;
v = 6.20 106 m/s
Direction is given by the right-hand-rule: southward F B = qvB sin FB = (1.60 10-19 C)(6.20 106 m/s)(50.0 10-6 T) sin 90.0 = 4.96 10-17 N mv 2 r mv 2 (1.67 10-27 kg)(6.20 106 m/s)2 = = 1.29 km F 4.96 10-17 N
(b)
F=
so
r=
29.32
(a)
1 2
m v 2 = q(V)
1 2
(3.20 10-26 kg) v2 = (1.60 10-19 C)(833 V) mv 2 r
v = 91.3 km/s
The magnetic force provides the centripetal force: qvB sin =
Chapter 29 Solutions mv (3.20 10-26 kg)(9.13 104 m/s) r = qB sin 90.0 = = 1.98 cm (1.60 10-19 C)(0.920 N s/C m)
9
2000 by Harcourt, Inc. All rights reserved.
10
Chapter 29 Solutions mv 2 r eBr m
29.33
For each electron,
q vB sin 90.0 =
and
v=
The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic:
2 2 2 1 1 1 K = 2 mv1i + 0 = 2 mv1f + 2 mv2 f 2 2 2 2 2 2 2 2 1 e B R1 + 1 m e B R2 = e B R 2 + R 2 K = 2 m 1 2 2m m2 2 m2
(
)
K=
e(1.60 10 -19 C)(0.0440 N s / C m)2 (0.0100 m)2 + (0.0240 m)2 = 115 keV -31 2(9.11 10 kg)
[
]
29.34
We begin with qvB =
mv 2 qRB R , so v = m T= 2 R 2 R 2 m = = v qB qRB m
The time to complete one revolution is
2 m Solving for B, B = qT = 6.56 10-2 T
29.35
1 q( V ) = 2 mv 2
or
v=
2q( V ) m mv m 2q( V ) = = qB qB m 2mp ( V ) eB2 2mp ( V ) 2md ( V ) 2 2mp ( V ) 2 = = 2 = 2 rp 2 2 2 qd B eB eB 2mp ( V ) 2m ( V ) 2 4mp ( V ) 2 = = 2 = 2 rp 2 2 2 q B (2e)B eB 2m( V ) qB2
Also, qvB =
mv 2 r
so
r=
Therefore,
2 rp =
2 rd =
( ) (
and
r 2 =
)
The conclusion is:
r = r d = 2 r p
Chapter 29 Solutions
11
Goal Solution 29.35 A proton (charge +e, mass mp ), a deuteron (charge +e, mass 2mp ), and an alpha particle, (charge + 2e , mass 4mp ) are accelerated through a common potential difference V . The particles enter a uniform magnetic field B with a velocity in a direction perpendicular to B. The proton moves in a circular path of radius r p . Determine the values of the radii of the circular orbits for the deuteron rd and the alpha particle r in terms of r p . G: In general, particles with greater speed, more mass, and less charge will have larger radii as they move in a circular path due to a constant magnetic force. Since the effects of mass and charge have opposite influences on the path radius, it is somewhat difficult to predict which particle will have the larger radius. However, since the mass and charge ratios of the three particles are all similar i n magnitude within a factor of four, we should expect that the radii also fall within a similar range. O: The radius of each particle's path can be found by applying Newton's second law, where the force causing the centripetal acceleration is the magnetic force: F = qv B. The speed of the particles can be found from the kinetic energy resulting from the change in electric potential given. A : An electric field changes the speed of each particle according to assuming that the particles start from rest, we can write qV =
1 mv 2 . 2
( K + U )i = ( K + U ) f .
qvBsin 90 =
Therefore,
The magnetic field changes their direction as described by F = ma: thus For the protons, For the deuterons, For the alpha particles,
mv 2 r mv m 2qV 1 2mV r= = = m qB qB B q rp = rd = r = 1 2mp V e B 1 2(2mp )V = 2r p e B 1 2(4mp )V = 2r p 2e B
L:
Somewhat surprisingly, the radii of the deuterons and alpha particles are the same and are only 41% greater than for the protons.
29.36
(a)
We begin with qvB = L = qB
mv 2 , or qRB = mv . But, L = mvR = qR 2B. R
Therefore, R =
(1.60 10 )
4.00 10 -25 J s
-19
C 1.00 10 -3 T
)(
)
= 0.0500 m = 5.00 cm
(b)
Thus, v =
L 4.00 10 -25 J s = = 8.78 106 m s -31 mR kg (0.0500 m ) 9.11 10
(
2000 by Harcourt, Inc. All rights reserved.
12
Chapter 29 Solutions qB (1.60 10 -19 C)(5.20 T) = = 4.98 108 rad/s m 1.67 10 -27 kg
29.37
=
29.38
1 2
mv 2 = q(V) mv qB m 2(V) q B2 qB2 r 2 2(V)
so
v=
2q( V ) m m 2q(V) / m qB m 2(V) q B2
r=
so
r=
r2 =
and
( r )2 = ( m ) =
m=
and
(q)B2 (r )2
2(V)
so
m q ( r ) 2e 2R = 2 = = 8 e R q r m
2 2
29.39
1 E = 2 mv 2 = e(V)
and
evBsin 90 = mv 2 R
B=
mv m 2e(V) 1 2m(V) = = eR eR R m e 2(1.67 10 -27 kg)(10.0 106 V) 1 = 7.88 10-12 T 5.80 1010 m 1.60 10 -19 C
B =
29.40
mv r= qB
so
7.94 10 -3 m 1.60 10 -19 C (1.80 T ) rqB m= = v 4.60 10 5 m s 1u m = 4.97 10 -27 kg = 2.99 u -27 kg 1.66 10
(
)(
)
The particle is singly ionized: either a tritium ion,
3 + 1H
, or a helium ion,
3 + 2 He
.
29.41
FB = Fe
so
qvB = qE where v = 2K / m .
K is kinetic energy of the electrons.
E = vB =
2(750) 1.60 10 -19 2K B= m 9.11 10 -31
(
)
1/2
(0.0150) =
244 kV/m
Chapter 29 Solutions
1
13
29.42
K = 2 mv 2 = q(V) F B = qv B =
so
v=
2q( V ) m
mv 2 mv m 2q(V) / m 1 2m(V) r= = = qB q B q r B
(a) (b)
r238 =
2(238 1.66 10 -27 )2000 1 = 8.28 10 -2 m = 8.28 cm 1.20 1.60 10 -19
r235 = 8.23 cm
r238 = r235
m238 = m235
238.05 = 1.0064 235.04
The ratios of the orbit radius for different ions are independent of V and B.
29.43
In the velocity selector:
v=
E 2500 V m = = 7.14 10 4 m s B 0.0350 T 2.18 10 -26 kg 7.14 10 4 m s mv = = 0.278 m qB 1.60 10 -19 C (0.0350 T )
In the deflection chamber:
r=
(
(
)(
)
)
29.44
K = 2 m v 2: v = 8.07 107 m/s
1
(34.0 10
6
eV 1.60 10 -19 J / eV =
)(
) (1.67 10
1 2
-27
kg v 2
)
mv (1.67 10-27 kg)(8.07 107 m/s) r = qB = = 0.162 m (1.60 10-19 C)(5.20 T)
29.45
(a)
F B = qvB =
mv 2 R
v qBR qB (1.60 10-19 C)(0.450 T) = R = mR = m = = 4.31 107 rad/s 1.67 10-27 kg (b) (1.60 10-19 C)(0.450 T)(1.20 m) qBR = 5.17 107 m/s v= m = 1.67 10-27 kg mv 2 r
29.46
F B = qvB =
mv 4.80 1016 kg m/s B = qr = = 3.00 T (1.60 1019 C)(1000 m)
2000 by Harcourt, Inc. All rights reserved.
14
Chapter 29 Solutions 25.0 = tan-1 10.0 = 68.2 1.00 cm = 1.08 cm sin 68.2
29.47
and
R=
Ignoring relativistic correction, the kinetic energy of the electrons is 1 2q( V ) 2 so v= = 1.33 108 m / s 2 mv = q(V) m From the centripetal force field B= mv (9.11 10 -31 kg)(1.33 108 m / s) = = 70.1 mT qR (1.60 10 -19 C)(1.08 10 -2 m) mv2 R = qvB, we find the magnetic
29.48
(a)
RH
1 nq IB nqt
so
n=
1 1 = = 7.44 10 28 m -3 -19 qRH C 0.840 10 -10 m 3 C 1.60 10
(
)(
)
(b)
VH =
7.44 10 28 m -3 1.60 10 -19 C 0.200 10 -3 m 15.0 10 -6 V nqt( VH ) B= = = 1.79 T I 20.0 A
(
)(
)(
)(
)
29.49
1 t( VH ) (35.0 10 -6 V)(0.400 10 -2 m) = = = 3.70 10-9 m3/C nq (21.0 A)(1.80 T) IB
29.50
IB Since V H = nqt , and given that I = 50.0 A, B = 1.30 T, and t = 0.330 mm, the number of charge carriers per unit volume is n= IB = 1.28 1029 m-3 e(V H )t
The number density of atoms we compute from the density: n0 = 8.92 g 1 mole 6.02 10 23 atoms 106 cm 3 28 3 = 8.46 10 atom / m mole 1 m3 cm 3 63.5 g
So the number of conduction electrons per atom is n 1.28 10 29 = = 1.52 n0 8.46 10 28
Chapter 29 Solutions 8.48 10 28 m -3 1.60 10 -19 C 5.00 10 -3 m 5.10 10 -12 V nqt( VH ) B= = I 8.00 A B = 4.32 10 - 5 T = 43.2 T
15
29.51
(
)(
)(
)(
)
Goal Solution In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.500 cm thick is positioned along an east-west direction. If a current of 8.00 A in the conductor results in a Hall voltage of 5.10 pV, what is the magnitude of the Earth's magnetic field? (Assume that 28 electrons/m3 and that the plane of the bar is rotated to be perpendicular to the direction of n = 8.48 10 B.) G: The Earth's magnetic field is about 50 T (see Table 29.1), so we should expect a result of that order of magnitude. O: The magnetic field can be found from the Hall effect voltage: VH = IB nqt or B= nqtVH I
A : From the Hall voltage,
(8.48 10 B=
28
e - m 3 1.60 10 -19 C e - (0.00500 m ) 5.10 10 -12 V 8.00 A
)(
)
(
) = 4.32 10
-5
T = 43.2 T
L : The calculated magnetic field is slightly less than we expected but is reasonable considering that the Earth's local magnetic field varies in both magnitude and direction.
29.52
(a)
VH =
IB nqt
so
nqt B 0.0800 T T = = = 1.14 10 5 I VH 0.700 10 -6 V V B= nqt ( VH ) I
Then, the unknown field is
B = 1.14 10 5 T V 0.330 10 -6 V = 0.0377 T = 37.7 mT (b) nqt T = 1.14 10 5 I V so T I n = 1.14 10 5 V qt
(
)(
)
0.120 A T n = 1.14 10 5 = 4.29 10 25 m -3 1.60 10 -19 C 2.00 10 -3 m V
(
)(
)
2000 by Harcourt, Inc. All rights reserved.
16
Chapter 29 Solutions mv 2 r v eB = = t r m
29.53 (a)
q vB sin 90 =
=
The time it takes the electron to complete radians is t=
m ( rad)(9.11 1031 kg) = = = 1.79 1010 s eB (1.60 1019 C)(0.100 N s/C m)
q Br ,
(b)
Since v =
m
1 Ke = 2 mv 2 =
q 2B2 r 2 e(1.60 10 -19 C)(0.100 N s / Cm)2 (2.00 10 -2 m)2 = = 351 keV 2m 2(9.11 10 -31 kg)
29.54
Fy = 0: Fx = 0:
+n mg = 0 kn + IBd sin 90.0 = 0 B= k m g 0.100(0.200 kg)(9.80 m/s2) = = 39.2 mT Id (10.0 A)(0.500 m)
29.55
(a)
The electric current experiences a magnetic force . I(h B) in the direction of L.
(b)
The sodium, consisting of ions and electrons, flows along the pipe transporting no net charge. But inside the section of length L, electrons drift upward to constitute downward electric current J (area) = JLw. The current then feels a magnetic force I h B = JLwhB sin 90
This force along the pipe axis will make the fluid move, exerting pressure JLwhB F = hw = JLB area
29.56
The magnetic force on each proton,
F B = qv B = qvB sin 90
downward perpendicular to velocity, supplies centripetal force, guiding it into a circular path of radius r, with qvB = mv 2 r and mv r = qB
1
We compute this radius by first finding the proton's speed: K = 2 m v 2 v= 2K = m 2 5.00 106 eV 1.60 10 -19 J / eV 1.67 10
-27
(
)(
kg
) = 3.10 10
7
m/s
Chapter 29 Solutions mv (1.67 1027 kg)(3.10 107 m/s)(C m) = = 6.46 m qB (1.60 1019 C)(0.0500 N s) sin = 1m 1.00 m = 6.46 m r
17
Now, (b) (a)
r=
From the figure, observe that
= 8.90
The magnitude of the proton momentum stays constant, and its final y component is (1.67 1027 kg)(3.10 107 m/s) sin(8.90) = 8.00 1021 kg m/s
*29.57 (a)
If B = Bx i + By j + Bz k ,
F B = qv B = e( vi i ) Bx i + By j + Bz k = 0 + evi By k - evi Bz j
(
)
Since the force actually experienced is F B = Fi j, observe that Bx could have any value , (b) (c) If v = - vi i , then By = 0 , and Bz = - Fi evi
F B = qv B = e( - vi i ) (Bx i + 0 j - Fi evi k ) = F i j
If q = -e and v = vi i , then F B = qv B = - e( vi i ) (Bx i + 0 j - Fi evi k ) = F i j Reversing either the velocity or the sign of the charge reverses the force.
29.58
A key to solving this problem is that reducing the normal force will reduce the friction force: FB = BIL or B = FB IL When the wire is just able to move, Fy = n + FB cos - mg = 0 so and Also, so FB sin = f : We minimize B by minimizing FB : n = mg - FB cos f = ( mg - FB cos ) Fx = FB sin - f = 0 FB sin = ( mg - FB cos ) and FB =
mg sin + cos
dFB cos - sin = ( mg ) =0 d (sin + cos )2
sin = cos
1 Thus, = tan -1 = tan -1 ( 5.00) = 78.7 for the smallest field, and B=
( m L) FB g = IL I sin + cos
(0.200) 9.80 m s 2 0.100 kg m Bmin = = 0.128 T 1.50 A sin 78.7 + (0.200) cos 78.7 Bmin = 0.128 T pointing north at an angle of 78.7 below the horizontal
(
)
2000 by Harcourt, Inc. All rights reserved.
18 29.59
Chapter 29 Solutions (a) The net force is the Lorentz force given by F = 3.20 10 -19 F = qE + qv B = q(E + v B)
(
)[(4 i - 1j - 2 k) + (2i + 3 j - 1k) (2i + 4 j + 1k)] N
F=
Carrying out the indicated operations, we find: F = cos -1 x = cos -1 F = 24.4 (3.52)2 + (1.60)2 3.52
(3.52i - 1.60 j) 10-18 N
(b)
29.60
r=
mv (1.67 10 -27 )(1.50 108 ) = m = 3.13 104 m = 31.3 km qB (1.60 10 -19 )(5.00 10 -5 )
No, the proton will not hit the Earth .
29.61
Let x1 be the elongation due to the weight of the wire and let x2 be the additional elongation of the springs when the magnetic field is turned on. Then Fmagnetic = 2k x2 where k is the force constant of the spring and can be determined from k = mg/2x1. (The factor 2 is included in the two previous equations since there are 2 springs in parallel.) Combining these two equations, we find mg mg x2 Fmagnetic = 2 ; x2 = x1 2x1 Therefore, where I= but F B = I L B = ILB mg x2 (0.0100)(9.80)(3.00 10 -3 ) = = 0.588 T IL x1 (2.00)(0.0500)(5.00 10 -3 )
24.0 V = 2.00 A, 12.0
B=
*29.62
Suppose the input power is 120 W = (120 V ) I : Suppose rad and the output power is 20 W = = 200 s Suppose the area is about ( 3 cm) ( 4 cm), or From Table 29.1, suppose that the field is
I ~ 1 A = 100 A
= 2000
rad rev 1 min 2 rad ~200 s min 60 s 1 rev
~10 -1 N m
A~10 -3 m 2 B~10 -1 T
Then, the number of turns in the coil may be found from NIAB: C -3 2 -1 N s 10 m 10 giving 0.1 N m ~ N 1 s Cm
(
)
N ~10 3
Chapter 29 Solutions 29.63 Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibrium: Fx = T sin - ILB sin 90.00 = 0 Fy = T cos - mg = 0 , Therefore, tan = ILB IB = mg ( m L) g kg m ) 9.80 m s 2 5.00 A or or or T sin = ILB T cos = mg B=
19
(m L) g tan
I
B=
(0.0100
(
) tan(45.0 ) =
19.6 mT
29.64
Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibrium: Fx = T sin - ILBsin 90.00 = 0 Fy = T cos - mg = 0 , tan = ILB IB = mg ( m L) g mv 2 r or or or T sin = ILB T cos = mg B=
(m L) g tan =
I
g tan I
29.65
F = ma or qvB sin 90.0 =
qB v the angular frequency for each ion is r = = m = 2 f and f = f 12 - f 14 = qB 1 1 (1.60 10 -19 C)(2.40 T) 1 1 - - = 2 m12 m14 2 (1.66 10 -27 kg / u) 12.0 u 14.0 u
f = f12 f14 = 4.38 105 s-1 = 438 kHz
29.66
Let v x and v be the components of the velocity of the positron parallel to and perpendicular to the direction of the magnetic field. (a) The pitch of trajectory is the distance moved along x by the positron during each period, T (see Equation 29.15). 2 m p = vxT = (v cos 85.0) Bq p= (5.00 106 )(cos 85.0)(2 )(9.11 10 -31 ) = 1.04 10-4 m (0.150)(1.60 10 -19 ) mv mv sin 85.0 r = Bq = Bq
(b)
From Equation 29.13, r=
(9.11 10 -31 )(5.00 106 )(sin 85.0) = 1.89 10-4 m -19 (0.150)(1.60 10 )
2000 by Harcourt, Inc. All rights reserved.
20
Chapter 29 Solutions q q T
29.67
= IAB where the effective current due to the orbiting electrons is
and the period of the motion is The electron's speed in its orbit is found by requiring ke q 2 mv 2 = or R R2
I=
t 2 R T= v ke v=q mR T = 2
=
Substituting this expression for v into the equation for T, we find (9.11 10 -31 )(5.29 10 -11 )3 = 1.52 10 -16 s (1.60 10 -19 )2 (8.99 10 9 )
mR 3 q 2 ke
T = 2
Therefore,
=
1.60 10 -19 q AB = (5.29 10 -11 )2 (0.400) = 3.70 10-24 N m -16 T 1.52 10
Goal Solution Consider an electron orbiting a proton and maintained in a fixed circular path of radius R = 5.29 10-11 m by the Coulomb force. Treating the orbiting charge as a current loop, calculate the resulting torque when the system is in a magnetic field of 0.400 T directed perpendicular to the magnetic moment of the electron.
-30 G: Since the mass of the electron is very small (~10 kg), we should expect that the torque on the -30 orbiting charge will be very small as well, perhaps ~10 Nm.
O: The torque on a current loop that is perpendicular to a magnetic field can be found from = IAB sin . The magnetic field is given, = 90, the area of the loop can be found from the radius of the circular path, and the current can be found from the centripetal acceleration that results from the Coulomb force that attracts the electron to proton. A : The area of the loop is A = r 2 = 5.29 10 -11 m
(
)
2
= 8.79 10 -21 m 2 .
If v is the speed of the electron, then the period of its circular motion will be T = 2 R v, and the effective current due to the orbiting electron is I = Q / t = e T . Applying Newton's second law with the Coulomb force acting as the central force gives F = ke q 2 mv 2 = R R2 so that v=q ke mR and T = 2 mR 3 q 2 ke
T = 2
(1.60 10
(9.10 10 -31 kg)(5.29 10 -11 m)3
-19
C
) (8.99 10
2
9
Nm C
2
2
)
= 1.52 10 -16 s
The torque is
=
1.60 1019 C q AB: = ( )(5.29 10 -11 m)2 (0.400 T) = 3.70 10 -24 N m T 1.52 10 -16 s
L : The torque is certainly small, but a million times larger than we guessed. This torque will cause the atom to precess with a frequency proportional to the applied magnetic field. A similar process on the nuclear, rather than the atomic, level leads to nuclear magnetic resonance (NMR), which is used for magnetic resonance imaging (MRI) scans employed for medical diagnostic testing (see Section 44.2).
Chapter 29 Solutions
21
29.68
Use the equation for cyclotron frequency
=
qB m
or
m=
qB qB = 2 f
m=
(1.60 10 -19 C)(5.00 10 -2 T) = 3.82 10-25 kg (2 )(5.00 rev / 1.50 10 -3 s)
29.69
(a)
J 1 K = 2 mv 2 = 6.00 MeV = 6.00 106 eV 1.60 10 -19 eV K = 9.60 10-13 J 2 9.60 10
x
(
)
' x x
x 45 x 45 R x x
x
x
x
x
x x Bin = x x x x x x x x x x x x
x x 1.00 T x x x x x x x x x x x x
v=
1.67 10 -27 kg mv 2 R
(
-13
J
) = 3.39 10
R=
7
ms
x v 45.0 x
FB = qvB =
so
1.67 10 -27 kg 3.39 107 m s mv = = 0.354 m qB 1.60 10 -19 C (1.00 T )
(
(
)(
)
)
Then, from the diagram, x = 2R sin 45.0 = 2(0.354 m ) sin 45.0 = 0.501 m (b) From the diagram, observe that ' = 45.0 .
29.70
(a)
See graph to the right. The Hall voltage is directly proportional to the magnetic field. A least-square fit to the data gives the equation of the best fitting line as: VH = 1.00 10
120.00 100.00 80.00
(
-4
V T B
)
V
H
(V) 60.00
40.00 20.00
(b)
Comparing the equation of the line which fits the data best to I VH = B nqt observe that:
0.00 0.00
0.20
0.40
0.60
0.80
1.00
1.20
B (T)
I I = 1.00 10 -4 V T , or t = nqt nq 1.00 10 -4 V T
(
)
Then, if I = 0.200 A, q = 1.60 10 -19 C, and n = 1.00 10 26 m -3 , the thickness of the sample is t=
(
1.00 10 26 m -3 1.60 10 -19 C 1.00 10 -4 V T
)(
0.200 A
)(
)
= 1.25 10 -4 m = 0.125 mm
2000 by Harcourt, Inc. All rights reserved.
22
Chapter 29 Solutions
*29.71 (a)
The magnetic force acting on ions in the blood stream will deflect positive charges toward point A and negative charges toward point B. This separation of charges produces an electric field directed from A toward B. At equilibrium, the electric force caused by this field must balance the magnetic force, so qvB = qE = q V d
or
v=
V 160 10 -6 V = = 1.33 m/s Bd (0.040 0 T ) 3.00 10 -3 m
(
)
(b)
N o . Negative ions moving in the direction of v would be deflected toward point B, giving A a higher potential than B. Positive ions moving in the direction of v would be deflected toward A, again giving A a higher potential than B. Therefore, the sign of the potential difference does not depend on whether the ions in the blood are positively or negatively charged.
*29.72
When in the field, the particles follow a circular path according to qvB = mv 2 r, so the radius of the path is: r = mv / qB (a) qBh mv , that is, when v = , the m qB particle will cross the band of field. It will move in a full semicircle of radius h, leaving the field at v f = -v j . (2h, 0, 0) with velocity When r = h = When v < qBh mv , the particle will move in a smaller semicircle of radius r = < h. It will m qB v f = -v j .
(b)
leave the field at ( 2r, 0, 0) with velocity (c) When v >
coordinates [r(1 - cos ), h, 0] with velocity v f = v sin i + v cos j .
qBh mv , the particle moves in a circular arc of radius r = > h, centered at (r, 0, 0) . m qB The arc subtends an angle given by = sin -1 ( h r ) . It will leave the field at the point with
Chapter 30 Solutions
30.1
B=
0I 2R
=
0q(v/2 R) = 12.5 T 2R
*30.2
We use the Biot-Savart law. For bits of wire along the straight-line sections, ds is at 0 or 180 to ~, so ds ~= 0. Thus, only the curved section of wire contributes to B at P. Hence, ds is tangent to the arc and ~ is radially inward; so ds ~= ds l sin 90 = d s . All points along the curve are the same distance r = 0.600 m from the field point, so
B=
dB
all current
=
0 I ds ~ I = 0 2 ds 2 r 4 4 r
=
0 I 2 s 4 r
where s is the arclength of the curved wire, 2 = 0.314 m s = r = (0.600 m)30.0 360 Then, B = 107 T m (3.00 A) A (0.600 m)2 (0.314 m)
B = 261 nT into the page
30.3
(a)
B=
40I 3 l cos cos where a = 4 4 2 4 a
is the distance from any side to the center. 4.00 106 2 2 + 2 = 2 2 105 T = 28.3 T into the paper 0.200 2
B=
Figure for Goal Solution
(b)
For a single circular turn with 4 l = 2 R, B= 0I 0 I (4 2 107)(10.0) = 4l = = 24.7 T into the paper 2R 4(0.400)
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Chapter 30 Solutions 191 Goal Solution (a) A conductor in the shape of a square of edge length l = 0.400 m carries a current I = 10.0 A (Fig. P30.3). Calculate the magnitude and direction of the magnetic field at the center of the square. (b) If this conductor is formed into a single circular turn and carries the same current, what is the value of the magnetic field at the center? G: As shown in the diagram above, the magnetic field at the center is directed into the page from the clockwise current. If we consider the sides of the square to be sections of four infinite wires, then we could expect the magnetic field at the center of the square to be a little less than four times the strength of the field at a point l/2 away from an infinite wire with current I. 4 10 -7 T m / A (10.0 A ) 0I = 4 = 40.0 T 2 a 2 (0.200 m )
B<4
(
)
Forming the wire into a circle should not significantly change the magnetic field at the center since the average distance of the wire from the center will not be much different. O: Each side of the square is simply a section of a thin, straight conductor, so the solution derived from the Biot-Savart law in Example 30.1 can be applied to part (a) of this problem. For part (b), the BiotSavart law can also be used to derive the equation for the magnetic field at the center of a circular current loop as shown in Example 30.3. A : (a) We use Equation 30.4 for the field created by each side of the square. Each side contributes a field away from you at the center, so together they produce a magnetic field: B= so at the center of the square,
-6 4 0 I 3 4 4 10 T m / A (10.0 A ) 2 2 = cos - cos + 4 a 4 4 4 (0.200 m ) 2 2
(
)
B = 2.00 2 10 -5 T = 28.3 T perpendicularly into the page
(b) As in the first part of the problem, the direction of the magnetic field will be into the page. The new radius is found from the length of wire: 4 = 2 R, so R = 2 / = 0.255 m. Equation 30.8 gives the magnetic field at the center of a circular current loop: B=
0 I (4 10 -7 T m / A)(10.0 A ) = = 2.47 10 -5 T = 24.7 T 2R 2(0.255 m)
To
Caution! If you use your calculator, it may not understand the keystrokes: . get the right answer, you may need to use
L : The magnetic field in part (a) is less than 40 T as we predicted. Also, the magnetic fields from the square and circular loops are similar in magnitude, with the field from the circular loop being about 15% less than from the square loop. Quick tip: A simple way to use your right hand to find the magnetic field due to a current loop is to curl the fingers of your right hand in the direction of the current. Your extended thumb will then point in the direction of the magnetic field within the loop or solenoid.
2000 by Harcourt, Inc. All rights reserved.
192 Chapter 30 Solutions
30.4
B=
0 I 4 10 - 7 (1.00 A) = = 2.00 10-7 T 2 r 2 (1.00 m)
30.5
For leg 1, ds ~= 0, so there is no contribution to the field from this segment. For leg 2, the wire is only semi-infinite; thus, B= 0I 1 0I 2 x = 4 x into the paper 2
30.6
B=
0I 2R
R=
0 I 20.0 10 - 7 = = 31.4 cm 2B 2.00 10 - 5
30.7
We can think of the total magnetic field as the superposition of the field due to the long straight wire (having magnitude 0 I 2 R and directed into the page) and the field due to the circular loop (having magnitude 0 I 2R and directed into the page). The resultant magnetic field is:
-7 1 0I 1 4 10 T m / A (7.00 A ) B= 1 + = 1 + = 5.80 10 - 5 T 2R 2(0.100 m )
(
)
or
B = 58.0 T
(directed into the page)
30.8
We can think of the total magnetic field as the superposition of the field due to the long straight wire (having magnitude 0 I 2 R and directed into the page) and the field due to the circular loop (having magnitude 0 I 2R and directed into the page). The resultant magnetic field is: 1 0I B= 1 + 2R
(directed into the page)
30.9
For the straight sections ds ~= 0. loop: B= 1 0I 0I = into the paper 4 2R 8R
The quarter circle makes one-fourth the field of a full (4 10 - 7 T m / A)(5.00 A) = 26.2 T into the paper 8(0.0300 m)
B=
Chapter 30 Solutions 193 30.10 Along the axis of a circular loop of radius R, B=
B Along Axis of Circular Loop 1.00
2 x +R
2
(
0 IR 2
2 32
B/B
0
)
0.80 0.60 0.40 0.20 0.00 0.00 1.00 2.00 x/R 3.00 4.00 5.00
or
1 B = B0 ( x R)2 + 1
32
where B0 0 I 2R. xR 0.00 1.00 2.00 3.00 4.00 5.00 B B0 1.00 0.354 0.0894 0.0316 0.0143 0.00754
30.11
dB =
0 I d1 ~ 4 r 2
B=
1 1 0 I 6 2 a 6 2 b - 4 a2 b2
B=
0 I 1 1 directed out of the paper - 12 a b
30.12
Apply Equation 30.4 three times:
B=
0I cos 0 - 4 a 0I 4 d 0I 4 a
toward you d 2 + a2 d away from you d 2 + a2 a
+
a + 2 d + a2
+
-d - cos 180 toward you 2 d + a2
B=
0 I a2 + d 2 - d a2 + d 2
2 a d a2 + d 2
away from you
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194 Chapter 30 Solutions 30.13 The picture requires L = 2R B= 1 2 0I 0I 0I 2 R + 4 R (cos 90.0 - cos 135) + 4 R (cos 45.0 - cos 135) +
0I (cos 45.0 - cos 90.0) into the page 4 R
B=
0I 1 1 0I 4 + 2 = 0.475 R (into the page) R
30.14
Label the wires 1, 2, and 3 as shown in Figure (a) and let the magnetic field created by the currents in these wires be B1 , B2 , and B3 respectively. (a) At Point A : B1 = B2 =
0I 0I and B3 = . 2 ( 3a) 2 a 2
(
)
The directions of these fields are shown in Figure (b). Observe that the horizontal components of B1 and B2 cancel while their vertical components both add to B3 . Therefore, the net field at point A is: BA = B1 cos 45.0 + B2 cos 45.0 + B3 =
-7
Figure (a)
0I 2 1 cos 45.0 + 2 a 2 3
53.3 T Figure (b)
BA
(4 10 T m A)(2.00 A) 2 cos 45 + 1 = = 2 3 2 (1.00 10 m )
-2
(b)
At point B : B1 and B2 cancel, leaving BB = B3 =
0I . 2 ( 2a)
BB =
(4 10
-7
2 ( 2) 1.00 10
(
T m A ( 2.00 A )
-2
)
m
)
= 20.0 T Figure (c)
(c)
0I I and B3 = 0 with the directions shown in Figure (c). Again, 2 a 2 a 2 the horizontal components of B1 and B2 cancel. The vertical components both oppose B3 giving
At point C : B1 = B2 =
(
)
I I 2 cos 45.0 I 0 BC = 2 cos 45.0 - 0 = 0 - 1 = 0 2 a 2 a 2 2 a 2
(
)
Chapter 30 Solutions 195 30.15 Take the x-direction to the right and the y-direction up in the plane of the paper. Current 1 creates at P a field B1 = 2.00 10 -7 T m ( 3.00 A ) 0I = 2 a A(0.0500 m )
(
)
B1 = 12.0 T downward and leftward, at angle 67.4 below the x axis. Current 2 contributes B2 =
(2.00 10
-7
A(0.120 m )
T m ( 3.00 A )
)
clockwise perpendicular to 12.0 cm
B2 = 5.00 T to the right and down, at angle 22.6 Then, B = B1 + B2 = (12.0 T ) ( -i cos 67.4 -j sin 67.4) + ( 5.00 T ) (i cos 22.6 -j sin 22.6) B = ( -11.1 T )j - (1.92 T )j = (13.0 T)j
*30.16
Let both wires carry current in the x direction, the first at y = 0 and the second at y = 10.0 cm . (a) B= 4 10 -7 T m A ( 5.00 A ) 0I k= k 2 r 2 (0.100 m )
(
)
B = 1.00 10 -5 T out of the page (b) F B = I 2L B = (8.00 A ) (1.00 m ) i 1.00 10 -5 T k = 8.00 10 -5 N ( - j ) F B = 8.00 10 -5 N toward the first wire (c) B= 4 10 -7 T m A (8.00 A ) 0I (- k) = (- k) = (1.60 10-5 T) (- k) 2 r 2 (0.100 m )
[
(
)] (
)
(
)
B = 1.60 10 -5 T into the page (d) F B = I1L B = ( 5.00 A ) (1.00 m ) i 1.60 10 -5 T ( - k ) = 8.00 10 -5 N ( + j ) F B = 8.00 10 -5 N toward the second wire
[
(
)
] (
)
30.17
By symmetry, we note that the magnetic forces on the top and bottom segments of the rectangle cancel. The net force on the vertical segments of the rectangle is (using Equation 30.12) FB = 0 I1I2 l 1 1 i c + a c 2
Substituting given values FB = 2.70 105 i N = 27.0 N i
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196 Chapter 30 Solutions Goal Solution In Figure P30.17, the current in the long, straight wire is I 1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and l = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. G: Even though there are forces in opposite directions on the loop, we must remember that the magnetic field is stronger near the wire than it is farther away. By symmetry the forces exerted o n sides 2 and 4 (the horizontal segments of length a) are equal and opposite, and therefore cancel. The magnetic field in the plane of the loop is directed into the page to the right of I1. By the right-hand rule, F = I1 B is directed toward the left for side 1 of the loop and a smaller force is directed toward the right for side 3. Therefore, we should expect the net force to be to the left, possibly in the N range for the currents and distances given. O: The magnetic force between two parallel wires can be found from Equation 30.11, which can be applied to sides 1 and 3 of the loop to find the net force resulting from these opposing force vectors. A : F = F1 + F 2 =
0 I1I 2 l 1 I I l -a 1 i - i= 0 1 2 c + a c 2 2 c( c + a)
N / A 2 ( 5.00 A )(10.0 A)(0.450 m) - 0.150 m i (0.100 m)(0.250 m) 2 or F = 2.70 10 -5 N toward the left
F=
(4 10
-7
)
F = ( -2.70 10 -5 i) N
L : The net force is to the left and in the N range as we expected. The symbolic representation of the net force on the loop shows that the net force would be zero if either current disappeared, if either dimension of the loop became very small ( a 0 or l 0), or if the magnetic field were uniform ( c ) .
30.18
The separation between the wires is a = 2(6.00 cm) sin 8.00 = 1.67 cm. (a) Because the wires repel, the currents are in opposite directions . (b) Because the magnetic force acts horizontally, 0I 2 l FB = = tan 8.00 Fg 2 a mg I2= mg 2 a tan 8.00 l 0 so I = 67.8 A
Chapter 30 Solutions 197 30.19 Each wire is distant from P by (0.200 m) cos 45.0 = 0.141 m Each wire produces a field at P of equal magnitude: BA = 0I 2 a = (2.00 107 T m)(5.00 A) A(0.141 m) = 7.07 T
Carrying currents into the page, A produces at P a field of 7.07 T to the left and down at 135, while B creates a field to the right and down at 45. Carrying currents toward you, C produces a field downward and to the right at 45, while D 's contribution is downward and to the left. The total field is then 4 (7.07 T) sin 45.0 = 20.0 T toward the page's bottom
30.20
Let the current I flow to the right. It creates a field B = 0 I 2 d at the proton's location. And we have a balance between the weight of the proton and the magnetic force mg(- j) + qv(- i)
0I (k) = 0 at a distance d from the wire 2 d
d=
qv 0 I (1.60 10 -19 C)(2.30 10 4 m / s)(4 10 - 7 T m / A)(1.20 10 - 6 A) = = 5.40 cm 2 mg 2 (1.67 10 - 27 kg) (9.80 m / s 2 )
30.21
From Ampre's law, the magnetic field at point a is given by Ba = 0 I a 2 r a , where I a is the net current flowing through the area of the circle of radius r a . In this case, I a = 1.00 A out of the page (the current in the inner conductor), so Ba =
(4 10
-7
T m / A (1.00 A)
2 (1.00 10 -3 m)
)
=
200 T toward top of page
Similarly at point b : Bb = circle having radius rb .
0 Ib , where Ib is the net current flowing through the area of the 2 rb
Taking out of the page as positive, Ib = 1.00 A - 3.00 A = - 2.00 A , or Ib = 2.00 A into the page. Therefore, Bb = (4 10 - 7 T m / A)(2.00 A) = 133 T toward bottom of page 2 (3.00 10 - 3 m)
2000 by Harcourt, Inc. All rights reserved.
198 Chapter 30 Solutions 0I 2 r
*30.22 (a)
In B =
, the field will be one-tenth as large at a ten-times larger distance: 400 cm 4 107 T m (2.00 A) 1 1 = 7.50 nT 0.3985 m 0.4015 m 2 A
(b) (c)
B=
0I 0I k+ (k) 2 r 1 2 r 2
so
B=
Call r the distance from cord center to field point and 2d = 3.00 mm the distance between conductors. B= 1 0I 2d 0I 1 = 2 2 r d r + d 2 2 r d so r = 1.26 m
Tm (3.00 103 m) 7.50 1010 T = 2.00 107 A (2.00 A) 2 r 2.25 106 m 2
The field of the two-conductor cord is weak to start with and falls off rapidly with distance. (d) The cable creates zero field at exterior points, since a loop in Ampre's law encloses zero total current. Shall we sell coaxial-cable power cords to people who worry about biological damage from weak magnetic fields? 0NI 2 r
30.23
(a)
B inner =
= 3.60 T
(b)
Bouter =
0NI = 1.94 T 2 r 0(2.50 A) (0.0125 m) = 10.0 T 2 (0.0250 m)2
*30.24 (a)
B=
0I r for r a 2 a 2 0I 2 B = 0(2.50 A) 2 (10.0 106 T)
so
B=
(b)
r=
= 0.0500 m =
2.50 cm beyond the conductor's surface
30.25
(a)
One wire feels force due to the field of the other ninety-nine. I Within the bundle, B = 0 2 r = 3.17 10 -3 T . 2 R The force, acting inward, is F B = I lB, and the force per unit length is FB 3 l = 6.34 10 N/m inward
(b)
B r, so B is greatest at the outside of the bundle. Since each wire carries the same current, F is greatest at the outer surface . Figures for Goal Solution
Chapter 30 Solutions 199
Goal Solution A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R = 0.500 cm. (a) If each wire carries 2.00 A, what are the magnitude and direction of the magnetic force per unit length acting on a wire located 0.200 cm from the center of the bundle? (b) Would a wire on the outer edge of the bundle experience a force greater or less than the value calculated in part (a)? G: The force on one wire comes from its interaction with the magnetic field created by the other ninetynine wires. According to Ampere's law, at a distance r from the center, only the wires enclosed within a radius r contribute to this net magnetic field; the other wires outside the radius produce magnetic field vectors in opposite directions that cancel out at r . Therefore, the magnetic field (and also the force on a given wire at radius r ) will be greater for larger radii within the bundle, and will decrease for distances beyond the radius of the bundle, as shown in the graph to the right. Applying F = I1 B, the magnetic force on a single wire will be directed toward the center of the bundle, so that all the wires tend to attract each other. O: Using Ampere's law, we can find the magnetic field at any radius, so that the magnetic force F = I1 B on a single wire can then be calculated. A : (a) Ampere's law is used to derive Equation 30.15, which we can use to find the magnetic field at r = 0.200 cm from the center of the cable: 4 10 -7 T m / A (99)( 2.00 A ) 0.200 10 -2 m o I or B= = = 3.17 10 -3 T 2 R2 2 (0.500 10 -2 m)2 This field points tangent to a circle of radius 0.200 cm and exerts a force F = I1 B toward the center of the bundle, on the single hundredth wire: F l = IBsin = ( 2.00 A ) 3.17 10 -3 T (sin 90) = 6.34 mN / m (b) As is shown above in Figure 30.12 from the text, the magnetic field increases linearly as a function of r until it reaches a maximum at the outer surface of the cable. Therefore, the force on a single wire at the outer radius r = 5.00 cm would be greater than at r = 2.00 cm by a factor of 5/2. L : We did not estimate the expected magnitude of the force, but 200 amperes is a lot of current. It would be interesting to see if the magnetic force that pulls together the individual wires in the bundle is enough to hold them against their own weight: If we assume that the insulation accounts for about half the volume of the bundle, then a single copper wire in this bundle would have a cross sectional area of about (1 2)(0.01) (0.500 cm)2 = 4 10-7 m 2 with a weight per unit length of
(
)
(
)
(
)
gA = 8 920 kg / m 3 (9.8 N / kg ) 4 10 -7 m 2 = 0.03 N / m
(
)
(
)
Therefore, the outer wires experience an inward magnetic force that is about half the magnitude of their own weight. If placed on a table, this bundle of wires would form a loosely held mound without the outer sheathing to hold them together.
30.26
From
B d1 = 0 I, I =
2 rB (2)(1.00 10-3)(0.100) = = 500 A 0 4 107
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200 Chapter 30 Solutions
30.27
Use Ampre's law, becomes
B ds = 0 I .
For current density J, this
B ds = 0 J dA
(a) For r1 < R , this gives 2 r1 = 0 B= (b)
r1
0
(br )(2 r dr )
and
2 0br1 3
(for r1 < R or inside the cylinder)
When r2 > R , Ampre's law yields
(2 r2 )B = 0 0 (br )(2 r dr ) = 2 0bR3
R
3,
or
B=
0bR 3 3r2
(for r2 > R or outside the cylinder)
30.28
(a) (b)
See Figure (a) to the right. At a point on the z axis, the contribution from each wire has 0I magnitude B = and is perpendicular to the line from 2 a2 + z 2 this point to the wire as shown in Figure (b). Combining fields, the vertical components cancel while the horizontal components add, yielding 0I z 0I 0I z By = 2 sin = = 2 2 2 2 2 2 a2 + z 2 a +z 2 a + z a +z
(Currents are into the paper) Figure (a)
(
)
The condition for a maximum is: dBy dz = - 0 I z( 2z) +
a2 + z
(
2 2
)
a +z
2
(
0I
2
)
= 0,
or
2 2 0I a - z =0 a2 + z 2 2
( (
) )
Thus, along the z axis, the field is a maximum at d = a .
Figure (b)
Chapter 30 Solutions 201 N B = 0 l I B = 31.8 mA 0n
30.29
so
I=
30.30
(a)
I=
10.0 = 3.98 kA (4 107)(2000)
(b)
FB l = IB = 39.8 kN/m radially outward This is the force the windings will have to resist when the magnetic field in the solenoid is 10.0 T.
30.31
The resistance of the wire is Re =
l = r2 . , so it carries current I = Re l r2
If there is a single layer of windings, the number of turns per length is the reciprocal of the wire diameter: n = 1/ 2r . So, B = n 0I = 0 r 2 = 0 r 2 l = (4 107 T m/A)(20.0 V) (2.00 103 m) 2(1.70 108 m)(10.0 m) = 464 mT
l 2r
*30.32
The field produced by the solenoid in its interior is given by T m 30.0 B = 0nI ( - i ) = 4 10 -7 (15.0 A)( - i) A 10 -2 m B = - 5.65 10 - 2 T i The force exerted on side AB of the square current loop is
(
)
(F B )AB = IL B = (0.200 A)[(2.00 10-2 m) j (5.65 10-2 T)(- i)] (F B )AB = (2.26 10- 4 N) k
Similarly, each side of the square loop experiences a force, lying in the plane of the loop, of 226 N directed away from the center . From the above result, it is seen that the net torque exerted on the square loop by the field of the solenoid should be zero. More formally, the magnetic dipole moment of the square loop is given by
= IA = (0.200 A ) 2.00 10 -2 m
(
) (- i) = - 80.0 A m
2
2
i
The torque exerted on the loop is then
= B = - 80.0 A m 2 i - 5.65 10 - 2 T i = 0
(
) (
)
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202 Chapter 30 Solutions
30.33
(a)
B = B dA = B A = ( 5i + 4 j + 3 k ) T 2.50 10 -2 m i
2
(
)
B = 3.13 10 - 3 T m 2 = 3.13 10 - 3 Wb = 3.13 mWb (b)
(B )total = B dA =
0 for any closed surface (Gauss's law for magnetism)
30.34
(a)
B = B A = BA where A is the cross-sectional area of the solenoid. B = 0 NI r 2 = 7.40 Wb l 0 NI 2 2 r2 - r1 l
( )
(b)
B = B A = BA =
[(
)]
4 10 -7 T m A ( 300)(12.0 A ) 2 (8.00)2 - ( 4.00)2 10 -3 m = 2.27 Wb B = (0.300 m)
(
)
[
](
)
30.35
(a) (b)
(B )flat = B A = BR 2 cos(180 - ) =
B R 2 cos
The net flux out of the closed surface is zero: ( B )flat + ( B )curved = 0
(B )curved =
30.36
B R 2 cos
d E d dQ / dt I = (EA) = = dt dt e0 e0 (a) I dE = = 7.19 1011 V/m s dt e0 A
(b)
B ds = e0 0
B=
E dt
so
2 rB = e0 0
d Q r2 dt e0 A
0 Ir 0 (0.200)(5.00 10 -2 ) = = 2.00 10-7 T 2A 2 (0.100)2
30.37
(a)
d E dQ / dt I (0.100 A) = = = = 11.3 10 9 V m / s e0 e0 8.85 10 -12 C 2 / N m 2 dt I d = e0 d E = I = 0.100 A dt
(b)
Chapter 30 Solutions 203 ev 2 r
30.38
(a)
I=
ev 2 24 2 = IA = r = 9.27 10 A m 2 r The Bohr model predicts the correct magnetic moment. However, the "planetary model" is seriously deficient in other regards. (b) Because the electron is (), its [conventional] current is clockwise, as seen from above, and points downward .
30.39
Assuming a uniform B inside the toroid is equivalent NI to assuming r << R, then B 0 0 2 R and a tightly wound solenoid. B0 = 0 (630)(3.00) = 0.00189 T 2 (0.200) B = 0.191 T
With the steel, B = mB0 = (1 + )B0 = (101)(0.00189 T)
30.40
N B = nI = I 2 r
so I =
(2 r )B =
N
5000 4 10 -7 Wb A m ( 470)
(
2 (0.100 m )(1.30 T )
)
= 277 mA
30.41
B = nIA 500 B = nI = (750 4 10 -7 ) (0.500) = 0.188 T 2 (0.200) A = 8.00 10-4 m2 and B = (0.188 T)(8.00 10-4 m2) = 1.50 10-4 T m2) = 150 T m2
30.42
The period is T = 2 /. The spinning constitutes a current I =
Q Q = . T 2
= IA = =
Q Q R2 R2 = 2 2
in the direction of
(6.00 10 - 6 C)(4.00 / s)(0.0200 m)2 = 4.80 10-9 A m2 2
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204 Chapter 30 Solutions B - M = 2.62 106 A/m 0
30.43
B = 0 (H + M)
so
H=
30.44
B = 0 ( H + M) 2.00 T . 0 But M = xn B where B is the Bohr magneton, n is the number of atoms per unit volume, and x is the number of electrons that contribute per atom. Thus, If 0 M = 2.00 T , then the magnetization of the iron is M = x= 2.00 T 2.00 T M = = = 2.02 n B n B 0 8.50 10 28 m -3 9.27 10 -24 N m T 4 10 -7 T m A
(
)(
)(
)
*30.45 (a)
Comparing Equations 30.29 and 30.30, we see that the applied field is described by B0 = 0H. B C Then Eq. 30.35 becomes M = C 0 = 0 H , and the definition of susceptibility (Eq. 30.32) is T T
=
M C = 0 H T
(b)
2.70 10 - 4 ( 300 K ) KA T C= = = 6.45 10 4 -7 Tm 0 4 10 T m A
(
)
30.46
(a)
Bh = Bcoil =
0 NI (4 10 -7 )(5.00)(0.600) = = 12.6 T 2R 0.300
12.6 T Bh = = 56.0 T sin sin 13.0
(b)
Bh = Bsin B =
30.47
(a)
Number of unpaired electrons =
8.00 1022 A m2 = 8.63 1045 9.27 1024 A m2
Each iron atom has two unpaired electrons, so the number of iron atoms required is
1 2
(8.63 10 ) .
45
(b)
Mass =
(4.31 1045 atoms)(7900 kg/m3) = 4.01 1020 kg 8.50 1028 atoms/m3
Chapter 30 Solutions 205 Goal Solution The magnetic moment of the Earth is approximately 8.00 1022 Am 2. (a) If this were caused by the complete magnetization of a huge iron deposit, how many unpaired electrons would this correspond to? (b) At two unpaired electrons per iron atom, how many kilograms of iron would this correspond to? (Iron has a density of 7 900 kg/m3, and approximately 8.50 1028 atoms/m3.) G: We know that most of the Earth is not iron, so if the situation described provides an accurate model, then the iron deposit must certainly be less than the mass of the Earth ( MEarth = 5.98 10 2 4 kg ). One mole of iron has a mass of 55.8 g and contributes 2(6.02 10 2 3 ) unpaired electrons, so we should expect the total unpaired electrons to be less than 10 50 . O: The Bohr magneton B is the measured value for the magnetic moment of a single unpaired electron. Therefore, we can find the number of unpaired electrons by dividing the magnetic moment of the Earth by B . We can then use the density of iron to find the mass of the iron atoms that each contribute two electrons. 1 A 1T J N m -24 A : (a) B = 9.27 10 -24 1 A m2 = 9.27 10 J N s C m C / s T The number of unpaired electrons is N= 8.00 10 22 A m 2 = 8.63 10 45 e 9.27 10 -24 A m 2
(b) Each iron atom has two unpaired electrons, so the number of iron atoms required is 1 N = 1 (8.63 10 45 ) = 4.31 10 45 iron atoms . 2 2 Thus, MFe
(4.31 10 =
45
atoms 7900 kg / m 3
28
8.50 10
)(
atoms / m
3
) = 4.01 10
20
kg
L : The calculated answers seem reasonable based on the limits we expected. From the data in this problem, the iron deposit required to produce the magnetic moment would only be about 1/15 000 the mass of the Earth and would form a sphere 500 km in diameter. Although this is certainly a large amount of iron, it is much smaller than the inner core of the Earth, which is estimated to have a diameter of about 3000 km.
30.48
B=
0I = 2.00 105 T = 20.0 T 2 R
30.49
B=
0 IR 2 2(R 2 + R 2 )3/2
so
I = 2.00 109 A
flowing west
30.50
(a)
BC =
0I 0(10.0) =0 2 (0.270) 2 (0.0900) 40(10.0) = 88.9 T 2 (0.0900)
so
I = 30.0 A
(b)
BA =
out of paper
2000 by Harcourt, Inc. All rights reserved.
206 Chapter 30 Solutions *30.51 Suppose you have two 100-W headlights running from a 12-V battery, with the whole 200 W = 17 A current going through the switch 60 cm from the compass. Suppose the 12 V dashboard contains little iron, so 0. Model the current as straight. Then, B= 0I 2 r = (4 107)17 2 (0.6) ~ 10 5 T
If the local geomagnetic field is 5 105 T, this is ~101 times as large, compass noticeably.
enough to affect the
30.52
A ring of radius r and width dr has area dA = 2 r dr. The current inside radius r is I = 2 J r dr = 2 J0 r dr - 2 J0 R 2
0 0 r r
(
) r
0
r 3
dr = 2 J0 r 2 2 - 2 J0 R 2 r 4 4
(
)(
)
(a)
Ampre's law says B( 2 r ) = 0 I = 0 J 0 r 2 - r 4 2R 2 , 1 r 1 r 3 - B = 0 J0 R for r R 4 R 2 R B( 2 r ) = 0 I total = 0 J0 R 2 - J0 R 2 2 = 0 J0 R 2 2 B=
(
)
or
and
[
]
or
0 J0 R 2 0 J0 R = for r R 4( r R ) 4r
(b)
0.300 0.250 0.200
B / 0 J 0 R 0.150
0.100 0.050 0.000 0 2
r/R
4
6
(c)
To locate the maximum in the region r R, require that This gives the position of the maximum as r = 2 / 3 R . 1 21 2 1 2 3 2 Here B = 0 J0 R - = 0.272 0 J0 R 4 3 2 3
J r2 dB 0 J0 = - 3 0 02 = 0 dr 2 4R
Chapter 30 Solutions 207 30.53 Consider a longitudinal filament of the strip of width dr as shown in the sketch. The contribution to the field at point P due to the current dI in the element d r is 0dI where dI = I ( dr w ) dB = 2 r
b+w
B = dB =
b
0I dr 0I w k = ln 1 + k b 2 w r 2 w
30.54
We find the total number of turns:
B=
0 NI l
(0.0300 T)(0.100 m)A Bl = 2.39 103 N= I = 0 (4 107 T m)(1.00 A) Each layer contains (10.0 cm/0.0500 cm) = 200 closely wound turns so she needs (2.39 103/200) = 12 layers .
The inner diameter of the innermost layer is 10.0 mm. The outer diameter of the outermost layer is 10.0 mm + 2 12 0.500 mm = 22.0 mm. The average diameter is 16.0 mm, so the total length of wire is (2.39 103) (16.0 103 m) = 120 m
30.55
On the axis of a current loop, the magnetic field is given by
B=
0 IR 2 2(x 2 + R 2 )3/2
where in this case I = strength of B=
q . The magnetic field is directed away from the center, with a (2 /)
0 R 2 q 0 (20.0)(0.100)2 (10.0 10 -6 ) = 1.43 1010 T 2 2 3/2 = 2 2 3/2 4 (x + R ) 4 (0.0500) + (0.100)
[
]
30.56
On the axis of a current loop, the magnetic field is given by q . Therefore, (2 /)
B=
0 IR 2 2(x 2 + R 2 )3/2 0 R 2 q 4 (x 2 + R 2 )3/2
where in this case I =
B=
R when x = 2 , then
B=
0 qR 2
5 4 4 R
(
2 3/2
)
=
0q 2.5 5 R
2000 by Harcourt, Inc. All rights reserved.
208 Chapter 30 Solutions
30.57
(a)
Use Equation 30.7 twice: Bx =
0 IR 2 2(x + R 2 )3/2
2
B = B x1 + B x2 =
0IR 2 1 1 (x 2 + R 2 )3/2 + ((R x)2 + R 2)3/2 2
B=
1 1 0IR 2 2 (x 2 + R 2 )3/2 + (2R 2 + x 2 2xR)3/2 3 2 2 - 2 ( 2x ) x + R
(b)
dB 0 IR 2 = dx 2
(
)
-5 2
-
3 2R 2 + x 2 - 2xR 2 dB =0 dx
(
)
-5 2
(2x - 2R)
Substituting x =
R and cancelling terms, 2
3 IR 2 2 d 2B =- 0 (x + R 2 )-5 2 - 5x 2 (x 2 + R 2 )-7 2 + (2R 2 + x 2 - 2xR)-5 2 - 5(x - R)2 (2R 2 + x 2 - 2xR)-7 2 2 dx 2 Again substituting x = R and cancelling terms, 2 d 2B =0 dx 2
[
]
30.58
"Helmholtz pair" separation distance = radius B=
2 ( R / 2) + R 2
2
[
2 0 IR 2
]
3/2
=
0 IR 2
1 + 1 4
3/2
= R3
0I for 1 turn 1.40R
4 10 -7 100(10.0) 0 NI For N turns in each coil, B = = = 1.80 10- 3 T 1.40R 1.40(0.500)
(
)
Chapter 30 Solutions 209
30.59 (a)
Model the two wires as straight parallel wires (!) FB = 0 I 2L 2 a (Equation 30.12)
FB =
(4 107)(140)22(0.100) = 2.46 N 2 (1.00 103) 2.46 N m loop g = 107 m/s2 m loop
upward
(b)
aloop =
upward
*30.60 (a)
0 Ids ~, the moving charge constitutes a bit of current as in I = nqvA. For a 4 r 2 positive charge the direction of ds is the direction of v , so dB = 0 2 nqA( ds) v ~. Next, A( ds) 4 r is the volume occupied by the moving charge, and nA( ds) = 1 for just one charge. Then,
In dB = B=
0 qv ~ 4 r 2
-7
(b)
B=
(4 10
T m A 1.60 10 -19 C 2.00 107 m s 4 1.00 10
)(
(
-3 2
)
)(
) sin 90.0 = )
3.20 10 -13 T
(c)
FB = q v B = 1.60 10 -19 C 2.00 107 m s 3.20 10 -13 T sin 90.0 FB = 1.02 10 -24 N directed away from the first proton 8.99 10 9 N m 2 C 2 1.60 10 -19 C kqq Fe = qE = e 1 2 = 2 r2 1.00 10 -3
(
)(
)(
(d)
(
(
)(
)
)
2
Fe = 2.30 10 -22 N directed away from the first proton Both forces act together. The electrical force is stronger by two orders of magnitude. It is productive to think about how it would look to an observer in a reference frame moving along with one proton or the other. 4 10 -7 T m A ( 24.0 A ) 0I B= = = 2.74 10 - 4 T 2 r 2 (0.0175 m )
1 At point C, conductor AB produces a field 2 2.74 10 - 4 T ( - j), conductor DE produces a 1 field of 2 2.74 10 -4 T ( - j) , BD produces no field, and AE produces negligible field. The
*30.61 (a) (b)
(
)
(
total field at C is
2.74 10 - 4 T ( - j) .
)
(
)
2000 by Harcourt, Inc. All rights reserved.
210 Chapter 30 Solutions
(c)
F B = IL B = ( 24.0 A )(0.0350 m k ) 5 2.74 10 - 4 T ( - j) = 1.15 10 -3 N i F m a= = = 0.384 2 i -3 m s 3.00 10 kg
[(
) ] (1.15 10
-3
N i
)
(d) (e)
(
)
The bar is already so far from AE that it moves through nearly constant magnetic field. The force acting on the bar is constant, and therefore the bar's acceleration is constant . v 2 = vi2 + 2ax = 0 + 2 0.384 m s 2 (1.30 m ) , so v f = f
(f)
(
)
(0.999
m s) i
Chapter 30 Solutions 211 2 a( m l) g FB 0 I A I B mg = = or I B = l l 2 a 0IA
30.62
At equilibrium,
IB =
2 (0.0250 m )(0.0100 kg m ) 9.80 m s 2
(4 10
-7
T m A (150 A )
)
(
)=
81.7 A
30.63
(a)
The magnetic field due to an infinite sheet of charge (or the magnetic field at points near a large sheet of charge) is given by B = 0 Js 2 . The current density Js = I l and in this case the equivalent current of the moving charged belt is I= dq d = ( lx) = lv; dt dt and dx v = dt 0 v 2
Therefore, Js = v (b)
B=
If the sheet is positively charged and moving in the direction shown, the magnetic field is out of the page, parallel to the roller axes.
30.64
C=
TM (4.00 K)(10.0%)(8.00 10 27 atoms / m 3 )(5.00)(9.27 10 - 24 J / T 2 ) KJ = = 2.97 10 4 2 B 5.00 T T m3
30.65
At equilibrium, or
= + B - mg
L cos 5.00 = 0, 2
B sin 5.00 =
B=
mgL cos 5.00 2
Therefore,
(0.0394 kg) 9.80 m s2 (0.100 m) mgL = 2 tan 5.00 2 (7.65 J T ) tan 5.00
(
)
B = 28.8 mT
30.66
The central wire creates field B = 0 I1 2 R counterclockwise. The curved portions of the loop feels no force since 1 B = 0 there. The straight portions both feel I 1 B forces to the right, amounting to F B = I 2 2L
0 I1 0 I1 I 2 L = to the right 2 R R
2000 by Harcourt, Inc. All rights reserved.
212 Chapter 30 Solutions
30.67
When the conductor is in the rectangular shape shown in figure (a), the segments carrying current straight toward or away from point P1 do not contribute to the magnetic field at P1. Each of the other four setions of length l makes an equal contribution to the total field into the page at P1. To find the contribution of the horizontal section of current in the upper right, we use B=
0I (cos 1 cos 2 ) with a = l, 1 = 90, and 2 = 135 4 a 4 0I 0I 1 0 = 4l 2 2 l
So
B1 =
When the conductor is in the shape of a circular arc, the magnitude or the field at the center is I 4l given by Equation 30.6, B = 0 . From the geometry in this case, we find R = and = . 4 R Therefore, B2 =
0 I I = 0 ; 4 (4l/ ) 16l
so that
B1 8 2 = 2 B2
30.68
I=
3 -8 2 rB 2 9.00 10 1.50 10 = = 675 A 0 4 10 -7
(
)(
)
Flow of positive current is downward or negative charge flows upward .
30.69
By symmetry of the arrangement, the magnitude of the net magnetic field at point P is B = 8B 0x where B 0 is the contribution to the field due to current in an edge length equal to L/2. In order to calculate B0, we use the Biot-Savart law and consider the plane of the square to be the yz-plane with point P on the x-axis. The contribution to the magnetic field at point P due to a current element of length dz and located a distance z along the axis is given by Equation 30.3. B0 =
0 I d1 ~ 4 r 2
From the figure we see that r = x 2 + (L2 / 4) + z 2 and d1 ~ = dz sin = dz L2 / 4 + x 2 L / 4 + x2 + z2
2
By symmetry all components of the field B at P cancel except the components along x (perpendicular to the plane of the square); and
Chapter 30 Solutions 213 L/2 L / 4 + x2
2
B0x = B0 cos
where
L/2
cos =
.
Therefore, B0x
I = 0 4
0
sin cos dz and B = 8B0x . r2
Using the expressions given above for sin cos , and r, we find B=
0 IL2 L2 L2 2 x 2 + x 2 + 4 2
30.70
(a)
From Equation 30.10, the magnetic field produced by one loop at the center of the second loop is given by B =
2 0 IR 2 0 I R = = 0 3 where the magnetic moment of either loop is 3 3 2x 2 x 2 x
(
)
= I R 2 . Therefore,
2 dB 3 3 0 R I = Fx = = 0 2 x 4 dx 2 x 4
(
)
(
)
2
=
3 0 I 2 R 4 2 x4
2
(b)
-7 -3 3 0 I 2 R 4 3 4 10 T m A (10.0 A ) 5.00 10 m Fx = = 4 2 2 x4 5.00 10 -2 m
(
(
)
( )
)
4
= 5.92 10 - 8 N
30.71
There is no contribution from the straight portion of the wire since ds ~= 0. For the field of the spiral, dB =
0 I (ds ~) (4 ) r 2
2
B=
0I 4
=0
2
ds sin ~ r2
=
0I 4
2
=0
(
2 =0
3 1 2 dr sin 2 4 r
)
I B= 0 4
r -2 dr = -
=0
0 I -1 r 4
( )
Substitute r = e:
B=-
0 I - e 4
[ ]
2
0
=-
0 I -2 0I e - e0 = 1 - e -2 4 4
[
]
(
)
(out of the page)
2000 by Harcourt, Inc. All rights reserved.
214 Chapter 30 Solutions 30.72 (a) B = B0 + 0 M M= B - B0 0 and M= B - B0 0 are parallel, this
Assuming that B and B0 becomes M = (B - B0 ) 0
The magnetization curve gives a plot of M versus B 0. (b) The second graph is a plot of the relative permeability (B B0 ) as a function of the applied field B0 .
30.73
Consider the sphere as being built up of little rings of radius r , centered on the rotation axis. The contribution to the field from each ring is dB =
dr
r
2 x +r
2
(
0 r 2 dI
2 32
)
where
dI =
dQ dQ = 2 t
dx x R
dQ = dV = ( 2 r dr )( dx ) dB =
0 r 3 dr dx
2 x2 + r
+R
(
2 32
)
where =
(
Q
4 R3 3
)
B=
x=-R r=0
R 2 -x 2
0 r 3 dr dx 32 2 x2 + r2
(
)
Let v = r 2 + x 2 , B= B= B=
+R R2
dv = 2r dr , and
r2 = v - x2
R2 R 2 -1 2 dv - x 2 2 v -3 2 dv dx v=x 2 v x=-R v=x R
x=-R
2 0 v - x dv v=x 2 2 2 v 3 2 dx = 04 R
(
)
0 4 0 4
2 1 2 R2 2 -1 2 R dx = 0 2v x 2 + 2x v x=-R x2 4
( )
x=-R 2(R - x ) + 2x
R
2
1 1 R - x dx
-R 2 R
R
x
2
2 0 - 4 x + 2R dx = 4
0 2 R
R
x
2
- 4x + 2R dx
B=
2 0 2R 3 4R 2 0R 2 - + 2R 2 = 4 3R 2 3
Chapter 30 Solutions 215 30.74 Consider the sphere as being built up of little rings of radius r , centered on the rotation axis. The current associated with each rotating ring of charge is dI = dQ = ( 2 r dr )( dx ) t 2
dr
r
[
] ]
dx x R
The magnetic moment contributed by this ring is d = A( dI ) = r 2
( 2 r dr )( dx ) = r 3 dr dx 2
[
= =
+R
x=-R
R2 - x2 R2 - x2 R 2 -x 2 3 +R +R r dr dx = dx = r=0 x=-R x=-R 4 4
4
(
)
2
dx
2 2R 5 +R 4 4 2 2 4 2 2R x=-R R - 2R x + x dx = 4 R (2R) - 2R 3 + 5 4
(
)
=
4 2 R 5 16 5 4 R 5 = R 2- + = 3 5 4 4 15 15
up
30.75
Note that the current I exists in the conductor with a current density J = I A , where A = a2 - a2 4 - a2 4 = a2 2 Therefore, J = 2I a2 . To find the field at either point P1 or P2 , find Bs which would exist if the conductor were solid, using Ampre's law. Next, find B1 and B2 that would be due to the conductors of radius a 2 that could occupy the void where the holes exist. Then use the superposition principle and subtract the field that would be due to the part of the conductor where the holes exist from the field of the solid conductor. At point P1 , Bs =
P1 Bs
-B1 -B2
[
]
r Bs a/2 a/2
r 2 + ( a 2)
2
r
P2 - B1
- B2
(a)
0 J a2
2 r
( ),
J ( a 2) J ( a 2) B1 = 0 , and B2 = 0 . 2 (r - a 2) 2 (r + a 2)
2 2
B = Bs - B1 - B2 =
0 J a2 2
1 1 1 - - r 4( r - a 2 ) 4( r + a 2 )
B=
2 2 0 ( 2I ) 4r 2 - a2 - 2r 2 = 0 I 2r - a directed to the left 2 4r r 2 - a2 4 r 4r 2 - a2
(
)
2000 by Harcourt, Inc. All rights reserved.
216 Chapter 30 Solutions
(b)
At point P2 , Bs =
0 J a2
2 r
( )
and B1 = B2 =
0 J ( a 2)
2 2
2 r 2 + ( a 2)
.
The horizontal components of B1 and B2 cancel while their vertical components add. B = Bs - B1 cos - B2 cos =
0 J a2
2 r
( ) - 2
0 J a2 4 r 2 r 2 + a2 4 r 2 + a2 4 0 I 2r 2 + a2 r 4r 2 + a2
directed toward the top of the page
B=
2I 2 0 J a2 r2 1 - = 0 ( ) 1 - 2r = 2 2 r 2 r 2 + a2 4 2 r 4r + a2
( )
(
)
Chapter 31 Solutions
= =
B ( NBA) = = 500 mV t t B (B A ) = = 1.60 mV t t BA cos t I loop =
31.1
31.2
and
R
=
1.60 mV = 0.800 mA 2.00 cos f cos i t
31.3
=
N
=
2 cos
NB
r2
= -25.0 50.0 10 -6 T (0.500 m ) E = + 9.82 mV
(
)
180 - cos 0 0.200 s
31.4
(a)
= =
dB dB ABmax t/ dt = A dt = e (0.160 m2)(0.350 T) 4.00/2.00 e = 3.79 mV 2.00 s
(b) (c)
At t = 0,
=
28.0 mV
31.5
d B (NBA) = N dt = = 3.20 kV t
so
I=
R
= 160 A
2000 by Harcourt, Inc. All rights reserved.
Chapter 31 Solutions 219 Goal Solution A strong electromagnet produces a uniform field of 1.60 T over a cross-sectional area of 0.200 m 2 . A coil having 200 turns and a total resistance of 20.0 is placed around the electromagnet. The current in the electromagnet is then smoothly decreased until it reaches zero in 20.0 ms. What is the current induced in the coil? G: A strong magnetic field turned off in a short time ( 20.0 ms ) will produce a large emf, maybe on the order of 1 kV . With only 20.0 of resistance in the coil, the induced current produced by this emf will probably be larger than 10 A but less than 1000 A. O: According to Faraday's law, if the magnetic field is reduced uniformly, then a constant emf will be produced. The definition of resistance can be applied to find the induced current from the emf. A : Noting unit conversions from F = qv B and U = qV , the induced voltage is
= -N d(B A) = -N 0 - Bi A cos =
dt t I=
+ 200(1.60 T ) 0.200 m 2 (cos 0) 1 N s / C m 1 V C 20.0 10
-3
(
)
s
T
N m
= 3200 V
R
=
3200 V = 160 A 20.0
L : This is a large current, as we expected. The positive sign is indicative that the induced electric field is in the positive direction around the loop (as defined by the area vector for the loop). 31.6
= N
t =
d B N(BA 0) dt = t = NB( r 2) = 500(0.200) (5.00 10-2)2 = 7.85 105 s 10.0 103
NBA
31.7
(a)
=
d(BA) dI = 0.500 0nA dt = 0.480 103 V dt
Iring =
R
=
4.80 10-4 = 1.60 A 3.00 10-4
(b) (c)
Bring =
0I 2rring = 20.1 T
Coil's field points downward, and is increasing, so Bring points upward
31.8
=
d(BA) dI 2 I = 0.500 0nA dt = 0.500 0n r 2 dt t
2000 by Harcourt, Inc. All rights reserved.
220 Chapter 31 Solutions
(a)
Iring =
R
=
0 n r 2 I 2R t
2 2
2
(b)
0 n r2 I 0I B = 2r = 4r R 1 1 t The coil's field points downward, and is increasing, so Bring points upward .
(c)
31.9
(a)
d B = B dA =
0I Ldx : 2 x
B =
h+w 0 IL
x=h
2
IL h + w dx = 0 ln h 2 x
(b)
=
-
d B d IL h + w L h + w dI = - 0 ln = - 0 ln h dt 2 dt 2 h dt
-7
4 10 = -(
T m A (1.00 m ) 2
)
ln
A 1.00 + 10.0 10.0 = s 1.00
- 4.80 V
The long wire produces magnetic flux into the page through the rectangle (first figure, above). As it increases, the rectangle wants to produce its own magnetic field out of the page, which it does by carrying counterclockwise current (second figure, above).
Chapter 31 Solutions 221 31. 10 B = ( 0nI)Asolenoid
2 2 = -N dB = -N 0n( rsolenoid ) dI = -N 0n( rsolenoid )(600
dt
dt
A s) cos(120t)
= -15.0( 4 10-7
T m A 1.00 10 3 m (0.0200 m ) (600 A s) cos(120t)
2
)(
)
E = -14.2 cos(120t) mV
31.11
For a counterclockwise trip around the left-hand loop, with B = At d At(2a2 )cos 0 - I1(5R) - I PQ R = 0 dt and for the right-hand loop, d At a2 + I PQ R - I 2 (3R) = 0 dt where I PQ = I1 - I 2 is the upward current in QP Thus, 2Aa2 - 5R(I PQ + I 2 ) - I PQ R = 0 and Aa2 + I PQ R = I 2 (3R) 2Aa2 - 6RI PQ - 5 (Aa2 + I PQ R) = 0 3 I PQ = Aa2 upward, and since R = (0.100 /m)(0.650 m) = 0.0650 23R (1.00 10 - 3 T / s)(0.650 m)2 = 283 A upward 23(0.0650 )
[ [
]
]
I PQ =
31.12
=
B dB A = N (0.0100 + 0.0800t )A =N dt t
At t = 5.00 s,
= 30.0(0.410 T)[ (0.0400 m)2 ] =
61.8 mV
2000 by Harcourt, Inc. All rights reserved.
222 Chapter 31 Solutions
31.13
B = 0nI = 0n( 30.0 A ) 1 - e -1.60t
(
) ) dA
B = B dA = 0n( 30.0 A ) 1 - e -1.60t B = 0n( 30.0 A ) 1 - e -1.60t R 2
(
(
)
= - N d B = - N 0n(30.0 A) R 2 (1.60)e - 1.60t
dt
= -(250)(4 10-7 N A 2 )(400 m -1 )(30.0 A)[ (0.0600 m)2 ]1.60 s-1 e -1.60t
=
(68.2 mV )e -1.60t
counterclockwise
31.14
B = 0nI = 0nI max (1 - e - t ) B = B dA = 0nI max (1 - e - t ) dA B = 0nI max (1 - e - t ) R 2
= - N d B = - N 0nImax R 2 e - t
dt
= N 0nI max R 2 e - t counterclockwise
31.15
=
l=
Nl2 B cos d (NBl2 cos ) = dt t
N B cos
t
=
(80.0 10 -3 V)(0.400 s) = 1.36 m (50)(600 10 -6 T - 200 10 -6 T)cos(30.0)
Length = 4 lN = 4(1.36 m)(50) = 272 m
Chapter 31 Solutions 223 Goal Solution A coil formed by wrapping 50.0 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0 with the direction of the field. When the magnetic field is increased uniformly from 200 T to 600 T in 0.400 s, an emf of 80.0 mV is induced in the coil. What is the total length of the wire? G: If we assume that this square coil is some reasonable size between 1 cm and 1 m across, then the total length of wire would be between 2 m and 200 m. O: The changing magnetic field will produce an emf in the coil according to Faraday's law of induction. The constant area of the coil can be found from the change in flux required to produce the emf. A : By Faraday's law, For magnitudes,
= -N
d B d dB = -N (BA cos ) = -NA cos dt dt dt B t = 80.0 10 -3 V 600 10 -6 T - 200 10 -6 T 50(cos 30.0) 0.400 s = 1.85 m 2
= NA cos
and the area is
A=
B N cos t
Each side of the coil has length d = A , so the total length of the wire is L = N ( 4d) = 4N A = (4)(50) 1.85 m 2 = 272 m L : The total length of wire is slightly longer than we predicted. With d = 1.36 m , a normal person could easily step through this large coil! As a bit of foreshadowing to a future chapter on AC circuits, an even bigger coil with more turns could be hidden in the ground below high-power transmission lines so that a significant amount of power could be "stolen" from the electric utility. There is a story of one man who did this and was arrested when investigators finally found the reason for a large power loss in the transmission lines!
31.16
The average induced emf is given by Here N = 1, and with
= N
B t
-
B = B(Asquare
Acircle )
Acircle = r 2 = (0.500 m)2 = 0.785 m 2
Also, the circumference of the circle is 2 r = 2 (0.500 m) = 3.14 m Thus, each side of the square has a length and So L= 3.14 m = 0.785 m, 4 Asquare = L2 = 0.617 m 2
B = (0.400 T)(0.617 m 2 - 0.785 m 2 ) = - 0.0672 T m 2
The average induced emf is therefore:
=
- 0.0672 T m 2 = 0.672 V 0.100 s
2000 by Harcourt, Inc. All rights reserved.
224 Chapter 31 Solutions 31.17 In a toroid, all the flux is confined to the inside of the toroid. B=
0 NI 500 0 I = 2 r 2 r
500 0 I max dz dr sin t 2 r
B = B dA = B =
500 0 I max b + R asin t ln R 2 dt 2 R
= N dB = 20 500 0 Imax a ln b + R cos t = 10
rad N ( 3.00 + 4.00) cm 4 10 -7 2 ( 50.0 A ) 377 (0.0200 m)ln cos t = (0.422 V) cos t s 2 4.00 cm A
4
31.18
The field inside the solenoid is: Thus, through the single-turn loop
B = 0nI = 0
N I l N r2 I l -
B = BAsolenoid = 0
( )
and the induced emf in the loop is
= - B = - 0 N ( r 2 ) I =
t l t
0 N r 2 I 2 I1 t l
31.19
= -N dB
dt Idt = - Q=- N d B R
IR = -N
d B dt N
Idt = - R dB
N N B = - A Bf - Bi R R
(
)
200 -4 2 Q = - (100 10 m )(-1.10 - 1.10) T = 0.880 C 5.00
31.20
I=
R
=
Blv R
v = 1.00 m/s
Chapter 31 Solutions 225 31.21 (a) F B = I 1 B = I lB. FB = When I = E / R and
= Blv , we get
Blv B2l2 v (2.50)2 (1.20)2 (2.00) (lB) = = = 3.00 N R 6.00 R
The applied force is 3.00 N to the right (b) P = I 2R = B2l2 v 2 = 6.00 W R or P = Fv = 6.00 W
*31.22
FB = IlB I= E Blv = R R I 2lR lv
and so
E = Blv B= I= IR lv FB v = 0.500 A R
(a) (b) (c)
FB =
and
I 2R = 2.00 W For constant force, P = F v = (1.00 N )( 2.00 m / s) = 2.00 W
31.23
The downward component of B, perpendicular to v, is (50.0 106 T) sin 58.0 = 4.24 105 T E = Blv = 4.24 10 -5 T (60.0 m )( 300 m / s) = 0.763 V The left wing tip is positive relative to the right.
(
)
31.24
= N dt
d
A BA cos = NB cos t (3.00 m 3.00 m sin 60.0) (3.00 m)2 = 1.21 V 0.100 s
= 1(0.100 T) cos 0
I= 1.21 V = 0.121 A 10.0
The flux is into the page and decreasing. The loop makes its own magnetic field into the page by carrying clockwise current.
31.25
= (2.00 rev/s)(2 rad/rev) = (4.00) rad/s
E = B l2 = 2.83 mV
1 2
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226 Chapter 31 Solutions 31.26 (a) Bext = Bext i and Bext decreases; therefore, the induced field is B0 = B 0 i (to the right). Therefore, the current is to the right in the resistor. Bext = Bext (i) increases; therefore, the induced field B0 = B 0 (+ i) is to the right, and the current is to the right in the resistor. Bext = Bext (k) into the paper and Bext decreases; therefore, the induced field is B0 = B 0 (k) into the paper. Therefore, the current is to the right in the resistor.
(b)
(a)
(b)
(c)
(d) By the Lorentz force law, F B = q (v B). Therefore, a positive charge will move to the top of the bar if B is into the paper .
(c)
(d)
31.27
(a)
The force on the side of the coil entering the field (consisting of N wires) is F = N ( ILB) = N ( IwB) The induced emf in the coil is
=N
d(Bwx ) d B =N = NBwv, dt dt
so the current is I =
R
=
NBwv R
counterclockwise.
The force on the leading side of the coil is then: F=N N 2B2 w 2 v NBwv wB = to the left R R
(b)
Once the coil is entirely inside the field, .
B = NBA = constant , so
= 0,
I = 0,
and F = 0
(c)
As the coil starts to leave the field, the flux decreases at the rate Bwv , so the magnitude of the current is the same as in part (a), but now the current flows clockwise. Thus, the force exerted on the trailing side of the coil is:
Chapter 31 Solutions 227
F=
N 2B2 w 2 v to the left again R
31.28
(a)
Motional emf = Bwv appears in the conducting water. Its resistance, if the plates are submerged, is
L w = A ab
Kirchhoff's loop theorem says Bwv IR Bwv R+ I w =0 ab = abvB abR + w
I=
w ab
(b)
Isc =
(100 m)(5.00 m)(3.00 m/s)(50.0 10 6 T) = 0.750 mA 100 m
31.29
Look in the direction of ba. The bar magnet creates a field into the page, and the field increases. The loop will create a field out of the page by carrying a counterclockwise current. Therefore, current must flow from b to a through the resistor. Hence, V a V b will be negative .
31.30
E = B l2 = 0.259 mV
1 2
31.31
Name the currents as shown in the diagram: Left loop: Right loop: At the junction: + Bdv2 - I 2 R 2 - I1R 1 = 0 + Bdv3 - I 3 R 3 + I1R 1 = 0 I 2 = I1 + I 3
Then, Bdv2 - I1R 2 - I 3 R 2 - I1R 1 = 0 I3 = Bdv3 I1R 1 + R3 R3 B dv2 - I1 (R 1 + R 2 ) - B dv3 R 2 R3 - I1R 1R 2 R3 =0
So,
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228 Chapter 31 Solutions v2 R 3 - v3 R 2 I1 = Bd upward R 1R 2 + R 1R 3 + R 2 R 3 (4.00 m / s)(15.0 ) - (2.00 m / s)(10.0 ) I1 = (0.0100 T)(0.100 m) (5.00 )(10.0 ) + (5.00 )(15.0 ) + (10.0 )(15.0 ) upward 31.32 (a) dB 2 dt = 6.00t 8.00t At t = 2.00 s, E = = 145 A
=
dB dt
R 2(dB/dt) 8.00 (0.0250)2 = 2 r 2 2 (0.0500)
clockwise for electron t = 1.33 s
F = qE = 8.00 1021 N (b) When 6.00t2 8.00t = 0,
31.33
dB dt = 0.0600t At t = 3.00 s,
dB = dt
2 dB 3 E = r1 = 1.80 10 N/C perpendicular to r1 and counterclockwise 2 r 1 dt
*31.34
= dB = r 2 dB = E d1
dt dt E(2 R) = r 2 dB , dt or r 2 dB E= 2 R dt dI dB = 0n dt dt dI = 0.600 e 0.200t dt E=
B = 0nI I = 3.00 e 0.200t
At t = 10.0 s,
r2 ( 0n)(0.600 e 0.200t ) 2 R
becomes
E=
(0.0200 m)2 (4 10 -7 N / A 2 )(1000 turns / m)(0.600)e 2.00 = 2.23 10 -5 N / C 2(0.0500 m)
Chapter 31 Solutions 229 d B dt dB dt so E = (9.87 mV/m) cos (100 t) clockwise
31.35
(a)
E d1 =
2 rE = ( r 2 ) (b)
The E field is always opposite to increasing B.
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230 Chapter 31 Solutions rev 2 rad 1 min = 314 rad/s For the alternator, = 3000 m i n 1 rev 60 s
31.36
= -N dB = -250 d [(2.50 10-4 T m 2 )cos(314 t / s)] = +250(2.50 104 T m2)(314/s) sin(314t)
dt dt (a) (b)
= (19.6 V) sin(314t) max = 19.6 V max = NAB = (1000)(0.100)(0.200)(120) = (t) = NBA sin t = NBA sin
31.37
(a) (b)
7.54 kV
is maximal when sin
= 1,
or = 2 ,
so the plane of coil is parallel to B
31.38
Let represent the angle through which the coil turns, starting from = 0 at an instant when the horizontal component of the Earth's field is perpendicular to the area. Then,
=- N
d d BA cos = - NBA cos t = + NBA sin t dt dt
Here sin t oscillates between +1 and 1, so the spinning coil generates an alternating voltage with amplitude
max = NBA = NBA2 f
= 100(2.00 10 -5 T)(0.200 m)2 (1500)
2 rad = 12.6 mV 60.0 s
31.39
B = 0nI = 4 10 -7 T m A 200 m -1 (15.0 A ) = 3.77 10 -3 T For the small coil, Thus, B = NB A = NBA cos t = NB r 2 cos t
(
)(
)
( )
= - dB = NB r 2 sin t
dt
= (30.0)(3.77 10-3 T) (0.0800 m)2 ( 4.00 s-1 ) sin( 4.00 t) = (28.6 mV) sin(4.00 t)
Chapter 31 Solutions 231 31.40 As the magnet rotates, the flux through the coil varies sinusoidally in time with B = 0 at t = 0. Choosing the flux as positive when the field passes from left to right through the area of the coil, the flux at any time may be written as B = - max sin t so the induced emf is given by
1
0.5
I/I
max
0 0 -0.5 0.5 1 1.5 2
= - dB = max cos t .
dt The current in the coil is then I=
-1
t/T = ( t / 2 )
R
=
max cos t = I max cos t R
31.41
(a)
F = NI lB
max = 2Fr = NI lwB = 0.640 N m
(b) P = = (0.640 N m)(120 rad/s) P max = 241 W (about 3 hp)
1
31.42
(a)
1 max = BA = B( 2 R 2 )
rad s
Figure 1
max = (1.30 T) (0.250 m)2 4.00
2
max =
(b) (c)
1.60 V d = BA 2
2 = 0
2
0
2
sin d = 0
The maximum and average
would remain unchanged.
Figure 2
(d) See Figure 1 at the right. (e) See Figure 2 at the right.
31.43
(a)
B = BA cos = BA cos t = (0.800 T)(0.0100 m2) cos 2 (60.0)t = (8.00 mT m2) cos (377t)
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232 Chapter 31 Solutions
(b) (c) (d) (e)
=
dB = (3.02 V) sin (377t) dt
I = R = (3.02 A) sin (377t) P = I 2R = (9.10 W) sin2 (377t) P = Fv = so
=
P = (24.1 mN m) sin2 (377t)
31.44
At terminal speed, the upward magnetic force exerted o n the lower edge of the loop must equal the weight of the loop. That is, Mg = FB = IwB = Thus, B= MgR = w 2 vt B2 w 2 vt Bwvt wB = wB = R R R
(0.150 kg)(9.80 m s2 )(0.750 ) = (1.00 m)2 (2.00 m s)
0.742 T
31.45 (a)
See the figure above with Problem 31.44. At terminal speed, or (b) (c) vt = MgR B2 w 2 Mg = FB = IwB = B2 w 2 vt Bwvt wB = wB = R R R
The emf is directly proportional to vt , but the current is inversely proportional to R. A large R means a small current at a given speed, so the loop must travel faster to get Fm = mg. At given speed, the current is directly proportional to the magnetic field. But the force is proportional to the product of the current and the field. For a small B, the speed must increase to compensate for both the small B and also the current, so vt B2 .
*31.46
The current in the magnet creates an upward magnetic field, so the N and S poles on the solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B increases as the coil approaches, so a current is induced here to create a downward magnetic field. This is clockwise current, so the S pole on the rail is shown correctly. On the rail behind the brake, the upward magnetic flux is decreasing. The induced current in the rail will produce upward magnetic field by being counterclockwise as the picture correctly shows.
Chapter 31 Solutions 233 F = ma = qE + qv B e a = [E + v B] where m i v B = 200 j 0 k 0 = - 200(0.400)j + 200(0.300)k
31.47
0.200 0.300 0.400
a=
1.60 10 -19 [50.0j - 80.0j + 60.0k] = 9.58 107 [-30.0j + 60.0k] 1.67 10 -27 (-2.87 10 9 j + 5.75 10 9 k) m s 2
a = 2.87 10 9 [-j + 2k] m s 2 =
31.48
F = ma = qE + qv B so
-19
-e a = [E + v B] m
i where
j
k
a=
(-1.60 10 ) [2.50i + 5.00 j - 4.00 j] = (-1.76 10 )[2.50i + 1.00 j] 9.11 10
-31 11
v B = 10.0 0 0 = - 4.00j 0 0 0.400
a=
(- 4.39 10
11
i - 1.76 1011 j m s 2
)
*31.49
= -N d (BA cos ) = -N ( r 2 ) cos 0 dB dt dt = - (30.0) (2.70 10-3 m) (1) d [50.0 mT + (3.20 mT) sin (2 dt
2
523t s) s)
]
= - (30.0) (2.70 10-3 m) =
2
(3.20 10 T)(2 )(523 s) cos (2 523t
-3
- 7.22 10 -3 V cos ( 2 523t s)
(
)
*31.50 (a)
Doubling the number of turns. Amplitude doubles: period unchanged Doubling the angular velocity. doubles the amplitude: cuts the period in half Doubling the angular velocity while reducing the number of turns to one half the original value. Amplitude unchanged: cuts the period in half
(b)
(c)
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234 Chapter 31 Solutions
*31.51
= -N (BA cos ) = -N ( r 2 ) cos 0 B = -1(0.00500 m 2 )(1) 1.50 T - 5.00 T = 0.875 V -3 t t
20.0 10 s (a) (b) I=
R
=
0.875 V = 43.8 A 0.020 0
P = EI = (0.875 V )( 43.8 A ) = 38.3 W
31.52
In the loop on the left, the induced emf is
=
d B dB 2 =A = (0.100 m ) (100 T s) = V dt dt
and it attempts to produce a counterclockwise current in this loop. In the loop on the right, the induced emf is
=
d B 2 = (0.150 m ) (100 T s) = 2.25 V dt
and it attempts to produce a clockwise current. Assume that I1 flows down through the 6.00- resistor, I 2 flows down through the 5.00- resistor, and that I 3 flows up through the 3.00- resistor. From Kirchhoff's point rule: Using the loop rule on the left loop: Using the loop rule on the right loop: I 3 = I1 + I 2 6.00 I1 + 3.00I 3 = 5.00 I 2 + 3.00I 3 = 2.25 (1) (2) (3)
Solving these three equations simultaneously, I1 = 0.0623 A , I 2 = 0.860 A , and I 3 = 0.923 A
*31.53
The emf induced between the ends of the moving bar is
= Blv = (2.50 T)(0.350 m)(8.00
m s) = 7.00 V
The left-hand loop contains decreasing flux away from you, so the induced current in it will be clockwise, to produce its own field directed away from you. Let I1 represent the current flowing upward through the 2.00- resistor. The right-hand loop will carry counterclockwise current. Let I 3 be the upward current in the 5.00- resistor.
Chapter 31 Solutions 235 +7.00 V - I1 ( 2.00 ) = 0 +7.00 V - I 3 ( 5.00 ) = 0 I1 = 3.50 A I 3 = 1.40 A
(a)
Kirchhoff's loop rule then gives: and
(b)
The total power dissipated in the resistors of the circuit is P = EI1 + EI 3 = E( I1 + I 3 ) = (7.00 V )( 3.50 A + 1.40 A ) = 34.3 W
(c)
Method 1: The current in the sliding conductor is downward with value I 2 = 3.50 A + 1.40 A = 4.90 A . The magnetic field exerts a force of Fm = IlB = ( 4.90 A )(0.350 m )( 2.50 T ) = 4.29 N directed toward the right on this conductor. A n outside agent must then exert a force of 4.29 N to the left to keep the bar moving. Method 2: The agent moving the bar P = F v = Fv cos 0 . The force required is then: F= P 34.3 W = = 4.29 N v 8.00 m s must supply the power according to
*31.54
Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm. Suppose we use a bar magnet to produce field 10 -3 T through the coil in one direction along its axis. Suppose we then flip the magnet to reverse the flux in 10 -1 s. The average induced emf is then
= - N B = - N [BA cos ] = -NB( r 2 ) cos180 - cos 0
t t t
= - (20)(10-3 T) (0.0150 m)2
+ Induced
R
-2 10 -1 s
~10 - 4 V
31.55
I=
and
Induced = dt (BA)
d
dv F = m dt = IBd dv IBd Bd Bd dt = m = m R ( + Induced) = m R ( Bvd) To solve the differential equation, let 1 du Bd Bd dt = m R u so u = ( Bvd), du dv dt = Bd dt .
u
u
0
t du (Bd)2 = - dt t=0 u mR
Integrating from t = 0 to t = t, e -B
2 2
u ln u 0 u0 =
(Bd)2 = mR
t
or
u u0
=
d t/mR
Since v = 0 when t = 0,
and u = Bvd
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236 Chapter 31 Solutions
Bvd = e -B d t/mR
2 2
and
v=
Bd
(1 - e -B
2 2
d t/mR
)
Chapter 31 Solutions 237
31.56
(a)
For maximum induced emf, with positive charge at the top of the antenna, F+ = q+ (v B), so the auto must move east
(b)
= Blv
65.0 103 m cos 65.0 = 4.58 10 4 V = (5.00 105 T)(1.20 m) 3600 s
31.57
I=
R
=
B A R t (15.0 T)(0.200 m)2 = 1.20 C 0.500
so
q = I t =
Goal Solution The plane of a square loop of wire with edge length a = 0.200 m is perpendicular to the Earth's magnetic field at a point where B = 15.0 T , as shown in Figure P31.57. The total resistance of the loop and the wires connecting it to the galvanometer is 0.500 . If the loop is suddenly collapsed by horizontal forces as shown, what total charge passes through the galvanometer? G: For the situation described, the maximum current is probably less than 1 mA. So if the loop is closed in 0.1 s, then the total charge would be Q = It = (1 mA )(0.1 s) = 100 C O: We do not know how quickly the loop is collapsed, but we can find the total charge by integrating the change in magnetic flux due to the change in area of the loop ( a2 0). A: Q =
Idt =
dt
R
=
1 1 1 B d B - d B = - d(BA) = - dt = - dt R R R R
A2 =0
A1 =a 2
dA
B Q = - A R A
A2 =0
=
2
1 =a
Ba2 (15.0 10 -6 T)(0.200 m)2 = = 1.20 10 -6 C R 0.500
L : The total charge is less than the maximum charge we predicted, so the answer seems reasonable. It is interesting that this charge can be calculated without knowing either the current or the time to collapse the loop. Note: We ignored the internal resistance of the galvanometer. D'Arsonval galvanometers typically have an internal resistance of 50 to 100 , significantly more than the resistance of the wires given in the problem. A proper solution that includes RG would reduce the total charge by about 2 orders of magnitude ( Q ~ 0.01 C).
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238 Chapter 31 Solutions dq I = dt = R
2
*31.58 (a)
where
E = -N
d B dt
so
dq =
Q =
N R
1
dB
and the charge through the circuit will be (b) Q= N BAN = BA cos 0 - BA cos 2 R R B= RQ NA = (200 )(5.00 10 4 C) (100)(40.0 10 4 m 2 )
N (2 1) R
so
= 0.250 T
31.59
(a) (b) (c)
= B lv = 0.360 V
FB = IlB = 0.108 N
I=
R
= 0.900 A
Since the magnetic flux B A is in effect decreasing, the induced current flow through R is from b to a. Point b is at higher potential. N o . Magnetic flux will increase through a loop to the left of ab. Here counterclockwise current will flow to produce upward magnetic field. The in R is still from b to a.
(d)
31.60
= Blv at a distance r from wire
I = 0 lv 2 r
31.61
(a)
At time t , the flux through the loop is At t = 0, B = ar 2
B = BA cos = ( a + bt) r 2 cos 0 = ( a + bt)r 2
( )
(b)
= - dB = - r 2 d( a + bt) =
dt dt I=
- br 2
(c)
R
= -
br 2 R
(d)
br 2 2b 2 r 4 P = I = - - br 2 = R R
(
)
Chapter 31 Solutions 239
31.62 (a) (b) (c)
= - d (NBA) = -1 dB a2 = a2 K
dt dt Q = C = C a2K B into the paper is decreasing; therefore, current will attempt to counteract this. charge will go to upper plate . The changing magnetic field through the enclosed area surrounding the B-field, and this pushes on charges in the wire. Positive ,
induces an electric field
31.63
The flux through the coil is B = B A = BA cos = BA cos t . The induced emf is d cos t ) = - N dB = - NBA ( = NBA sin t. dt dt (a) (b) (c)
max = NBA = 60.0(1.00 T)(0.100 0.200 m 2 )(30.0
d B = , dt N thus d B dt =
max
rad s) = 36.0 V 0.600 Wb/s 35.9 V
max = 36.0 V = 0.600 V =
N 60.0
At t = 0.0500 s, t = 1.50 rad and
= max sin(1.50 rad) = (36.0 V) sin(1.50 rad) =
(d) The torque on the coil at any time is
= B = NIA B = ( NAB) I sin t = max sin t R
=
2 max =
When
= max , sin t = 1.00 and
R
(36.0 V )2 = (30.0 rad s)(10.0 )
4.32 N m
31.64
(a)
We use
= N
B , with N = 1. t
Taking a = 5.00 10- 3 m to be the radius of the washer, and h = 0.500 m, 0I 1 I a2 0 I 1 - 0 ahI - = B = B2 A - B1A = A(B2 - B1 ) = a2 - 0 = h + a a 2 2(h + a) 2 (h + a) 2 a The time for the washer to drop a distance h (from rest) is: t = 2h g
Therefore,
= =
0 ahI 0 ahI = 2(h + a)t 2(h + a)
g 0 aI = 2h 2(h + a)
gh 2 (9.80 m/s 2 )(0.500 m) = 97.4 nV 2
and (b)
( 4 10 -7 T m/A)( 5.00 10 -3 m)(10.0 A) 2(0.500 m + 0.00500 m)
Since the magnetic flux going through the washer (into the plane of the paper) is decreasing in time, a current will form in the washer so as to oppose that decrease. Therefore, the current will flow in a clockwise direction .
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240 Chapter 31 Solutions
31.65
= -N dB = -N d (BA cos ) dt dt = -NBcos A = - 200(50.0 10- 6 T)(cos 62.0 ) 39.0 10
t m2 = 10.2 V 1.80 s
-4
31.66
Find an expression for the flux through a rectangular area "swept out" by the bar in time t. The magnetic field at a distance x from wire is B=
0I and B = BdA. 2 x 0 Ivt 2
r+l
Therefore,
B =
r
dx x
where vt is the distance the bar has moved in time t.
Then,
=
0 Iv l d B = ln 1 + 2 dt r 0I . 2 x
31.67
The magnetic field at a distance x from a long wire is B = flux through the loop. d B =
Find an expression for the
0I (ldx) 2 x
so
B =
0 Il 2
r+w
r
w dx 0 Il = ln 1 + 2 r x w 2 r (r + w) and I=
Therefore,
= - dB = 0 Ilv
dt
R
=
0 Ilv w 2 Rr (r + w)
31.68
As the wire falls through the magnetic field, a motional emf = Blv is induced in it. Thus, a The falling wire counterclockwise induced current of I = R = Blv R flows in the circuit. is carrying a current toward the left through the magnetic field. Therefore, it experiences an upward magnetic force given by FB = IlB = B2l2 v R. The wire will have attained terminal speed when the magnitude of this magnetic force equals the weight of the wire. Thus, mgR B2l2 vt = mg , or the terminal speed is vt = R B2l2
31.69
B = (6.00t 3 18.0t 2) T m2 Maximum
and
=
dB 2 dt = 18.0t + 36.0t
occurs when d
dt
= 36.0t + 36.0 = 0, which gives t = 1.00 s. I=
Therefore, the maximum current (at t = 1.00 s) is
R
=
(18.0 + 36.0)V = 6.00 A 3.00
Chapter 31 Solutions 241 F = Mg T = Ma
31.70
For the suspended mass, M: For the sliding bar, m: B2l2 v = (m + M)a R
F = T I lB = ma, where I =
R
=
Blv R
Mg -
or
a=
Mg dv B2l2 v = - dt m + M R(M + m) Mg M+m and
0
v
t dv = dt 0 ( - v)
where
=
=
B2l2 R(M + m)
Therefore, the velocity varies with time as
v=
2 2 MgR (1 - e - t ) = 1 - e -B l t/R(M+m) 2 2 Bl
*31.71 (a)
= N
d B dB d dt = NA dt = NA dt ( 0nI)
where A = area of coil, N = number of turns in coil, and n = number of turns per unit length in solenoid. Therefore,
(b)
d = N 0 An dt [ 4 sin(120 t)] = N 0An(480 ) cos (120 t) = 40(4 107 ) (0.0500 m)2 (2.00 103)(480) cos(120 t) = (1.19 V) cos(120 t) and P = VI (1.19 V)2 cos2(120 t) (8.00 ) P= 1 (1.19 V ) = 88.5 mW 2 (8.00 )
2
[
]
V I= R
=
From cos2 =
1 1 1 2 2 + 2 cos 2, the average value of cos is 2 , so
31.72
The induced emf is
= Blv where
B=
0I , v = vi + gt = 9.80 m s 2 t , and 2 y
(
)
1 y = yi - 2 gt 2 = 0.800 m - 4.90 m s 2 t 2 .
(
)
=
(4 10
-7
2 0.800 m - 4.90 m s 2 t 2
[
T m A ( 200 A )
(
)
) ]
(0.300 m) 9.80 m s 2 t =
-4
(
)
[0.800 - 4.90t ]
2
(1.18 10 )t
-4
V
At t = 0.300 s ,
=
[
0.800 - 4.90(0.300)
(1.18 10 )(0.300)
2
]
V = 98.3 V
2000 by Harcourt, Inc. All rights reserved.
242 Chapter 31 Solutions 31.73 The magnetic field produced by the current in the straight wire is perpendicular to the plane of the coil at all points within the coil. The magnitude of the field is B = 0 I 2 r . Thus, the flux linkage is N B =
0 NIL h+w dr 0 NI max L h + w sin( t + ) = ln h 2 h 2 r
Finally, the induced emf is
= - 0 NImax L ln 1 + w cos( t + )
2 h 5.00 cm cos( t + ) ln 1 + 5.00 cm
= -( =
4 10 -7 (100)(50.0)(0.200 m)(200 s -1 ) 2 - (87.1 mV ) cos(200 t + )
)
The term sin( t + )in the expression for the current in the straight wire does not change appreciably when t changes by 0.100 rad or less. Thus, the current does not change appreciably during a time interval t< 0.100 = 1.60 10 - 4 s . (200 s -1 )
ct = (3.00 108 m / s)(1.60 10 -4 s) = 4.80 10 4 m equal to the We define a critical length, distance to which field changes could be propagated during an interval of 1.60 10 -4 s. This length is so much larger than any dimension of the coilor its distance from the wire that, although we consider the straight wire to be infinitely long, we can also safely ignore the field propagation effects in the vicinity of the coil. Moreover, the phase angle can be considered to be constant along the wire in the vicinity of the coil. If the frequency were much larger, say, 200 10 5 s -1 , the corresponding critical length would be only 48.0 cm. In this situation propagation effects would be important and the above expression for would require modification. As a "rule of thumb" we can consider field propagation effects for circuits of laboratory size to be negligible for frequencies, f = 2 , that are less than about 106 Hz.
31.74
B = BA cos I sin
dB dt = BA sin ;
IB sin sin2
31.75
The area of the tent that is effective in intercepting magnetic field lines is the area perpendicular to the direction of the magnetic field. This is the same as the base of the tent. In the initial configuration, this is A1 = L(2L cos ) = 2(1.50 m)2 cos 60.0 = 2.25 m 2 After the tent is flattened, The average induced emf is: A2 = L(2L) = 2L2 = 2(1.50 m)2 = 4.50 m 2
= - B = - B( A) = - (
t t
0.300 T )( 4.50 - 2.25) m 2 = 6.75 V 0.100 s
Chapter 32 Solutions
*32.1
= L
I t
= (3.00 10 3 H)
1.50 A - 0.200 A = 1.95 102 V = 19.5 mV 0.200 s
32.2
Treating the telephone cord as a solenoid, we have: L=
0N 2A (4 10 - 7 T m / A)(70.0)2 ( )(6.50 10 - 3 m)2 = = 1.36 H l 0.600 m
32.3
=+L
0.500 A I = (2.00 H) = 100 V t 0.0100 s
32.4
L = 0n2 Al so
n=
L = 7.80 103 turns/m 0 Al
32.5
L=
N B LI B = = 240 nT m2 N I
(through each turn)
32.6
=L
dI dt
where L =
0N 2A l
Thus,
4 10 -7 T m A ( 300) 10 - 4 m 2 N 2 A dI = 0 = (10.0 A s) = 2.37 mV 0.150 m l dt
2
(
)
(
)
32.7
back = = L dt
dI
d = L dt (Imax sin t) = L Imax cos t = (10.0 10-3)(120 )(5.00) cos t (18.8 V) cos (377t)
back = (6.00 ) cos (120 t) =
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Chapter 32 Solutions 243 I , we have t
*32.8
From
=L
L=
t
=
24.0 10 3 V = 2.40 10 3 H 10.0 A/s
From L =
N B , we have I
B =
LI (2.40 10 3 H)(4.00 A) = = 19.2 T m2 N 500
32.9
L=
0(420)2(3.00 10 4) 0 N 2 A = = 4.16 10 4 H 0.160 l dI
6 dI - 175 10 V = = = 0.421 A/s dt 4.16 10 4 H L
= L dt
32.10
The induced emf is dI = - = dt L
= - L dI ,
dt -
where the self-inductance of a solenoid is given by L =
0N 2A . l
Thus,
l
0N 2A
32.11 (a) (b) (c)
dI d = L dt = (90.0 10-3) dt (t 2 6t) V
At t = 1.00 s, At t = 4.00 s,
= =
360 mV 180 mV t = 3.00 s
= (90.0 10-3)(2t 6) = 0 when
32.12
(a)
450 B = 0nI = 0 0.120 (0.0400 mA) = 188 T B = BA = 3.33 10-8 T m2 N B = 0.375 mH I
(b)
(c)
L=
2000 by Harcourt, Inc. All rights reserved.
244 Chapter 32 Solutions
(d)
B and B are proportional to current; L is independent of current
Chapter 32 Solutions 245 0(120)2 ( 5.00 103)2 0 N 2 A = = 15.8 H l 0.0900
32.13
(a)
L=
(b)
m m N 2 A = 800(1.58 10 5 H) = 12.6 mH = B L = B l 0
32.14
L=
NBA NA 0NI 0N 2A N B = = I I I 2 R 2 R
32.15
= 0e - k t = - L dI
dI = -
dt
0 e - k t dt
L I=
If we require I 0 as t , the solution is Q = I dt =
0 e - k t = dq
kL dt k 2L
0
0 e - k t dt = - 0
kL
k 2L
Q =
0
32.16
I=
(1 - e -Rt/L )
R
0.900
R
=
R
[1 - e
-R(3.00 s)/2.50 H
]
R(3.00 s) = 0.100 exp - 2.50 H R= 2.50 H ln 10.0 = 1.92 3.00 s
2000 by Harcourt, Inc. All rights reserved.
246 Chapter 32 Solutions
32.17 (a) (b)
L I = R = 0.200 s: I = 1 et/ max 0.500 = 1 et/0.200 t = ln 2.00 = 0.139 s 0.900 = 1 et/0.200 t = ln 10.0 = 0.461 s Figure for Goal Solution
Goal Solution A 12.0-V battery is about to be connected to a series circuit containing a 10.0- resistor and a 2.00-H inductor. How long will it take the current to reach (a) 50.0% and (b) 90.0% of its final value? G: The time constant for this circuit is = L R = 0.2 s, which means that in 0.2 s, the current will reach 1/e = 63% of its final value, as shown in the graph to the right. We can see from this graph that the time to reach 50% of I max should be slightly less than the time constant, perhaps about 0.15 s, and the time to reach 0.9I max should be about 2.5 = 0.5 s. The exact times can be found from the equation that describes the rising current in the above graph and gives the current as a function of time for a known emf, resistance, and time constant. We set time t = 0 to be the moment the circuit is first connected.
O:
A:
At time t , where, after a long time, At I (t ) = 0.500I max , Isolating the constants on the right, and solving for t , (b) Similarly, to reach 90% of I max ,
I (t ) =
(1 - e -t/ )
R
I max =
(1 - e - ) =
R
R ) so 0.500 = 1 - e -t/0.200 s
(0.500)
= (1 - e
R
-t/0.200 s
R
ln e -t/2.00 s = ln(0.500) - t = -0.693 0.200 s or t = 0.139 s
(
)
0.900 = 1 - e -t/
and t = - ln(1 - 0.900)
Thus,
t = -(0.200 s) ln(0.100) = 0.461 s
L:
The calculated times agree reasonably well with our predictions. We must be careful to avoid confusing the equation for the rising current with the similar equation for the falling current. Checking our answers against predictions is a safe way to prevent such mistakes.
Chapter 32 Solutions 247 dI 1 = I 0 e -t/ - dt
32.18
Taking = L R, IR + L
I = I 0 e -t/ :
dI 1 =0 = 0 will be true if I 0 R e -t/ + L I 0 e -t/ - dt
(
)
Because = L R, we have agreement with 0 = 0
*32.19 (a) (b)
= L R = 2.00 10 3 s = 2.00 ms
I = I max 1 - e -t/ =
(
)
6.00 V 1 - e -0.250/2.00 = 0.176 A 4.00
(
)
(c)
Imax =
R
=
6.00 V = 1.50 A 4.00
(d) 0.800 = 1 et/2.00 ms t = (2.00 ms) ln(0.200) = 3.22 ms
*32.20
I=
120 (1 et/ ) = 9.00 (1 e1.80/7.00) = 3.02 A R
VR = IR = (3.02)(9.00) = 27.2 V VL = VR = 120 27.2 = 92.8 V
32.21
(a)
V R = IR = (8.00 )(2.00 A) = 16.0 V and VL = - V R = 36.0 V - 16.0 V = 20.0 V Therefore, V R 16.0 V = = 0.800 VL 20.0 V Figure for Goal Solution
(b)
V R = IR = (4.50 A)(8.00 ) = 36.0 V VL = - V R = 0
2000 by Harcourt, Inc. All rights reserved.
248 Chapter 32 Solutions Goal Solution For the RL circuit shown in Figure P32.19, let L = 3.00 H, R = 8.00 , and = 36.0 V. (a) Calculate the ratio of the potential difference across the resistor to that across the inductor when I = 2.00 A. (b) Calculate the voltage across the inductor when I = 4.50 A. G: The voltage across the resistor is proportional to the current, VR = IR , while the voltage across the inductor is proportional to the rate of change in the current, L = -L dI dt . When the switch is first closed, the voltage across the inductor will be large as it opposes the sudden change in current. As the current approaches its steady state value, the voltage across the resistor increases and the inductor's emf decreases. The maximum current will be /R = 4.50 A, so when I = 2.00 A, the resistor and inductor will share similar voltages at this mid-range current, but when I = 4.50 A, the entire circuit voltage will be across the resistor, and the voltage across the inductor will be zero. O: We can use the definition of resistance to calculate the voltage across the resistor for each current. We will find the voltage across the inductor by using Kirchhoff's loop rule. A: (a) When I = 2.00 A, the voltage across the resistor is V R = IR = ( 2.00 A )(8.00 ) = 16.0 V
Kirchhoff's loop rule tells us that the sum of the changes in potential around the loop must be zero:
- VR - L = 36.0 V - 16.0 V - L = 0
(b)
so
L = 20.0 V
and
V R
L
=
16.0 V = 0.800 20.0 V
Similarly, for I = 4.50 A , V R = IR = ( 4.50 A )(8.00 ) = 36.0 V so
- VR - L = 36.0 V - 36.0 V - L = 0
L = 0
L : We see that when I = 2.00 A, V R < L , but they are similar in magnitude as expected. Also as predicted, the voltage across the inductor goes to zero when the current reaches its maximum value. A worthwhile exercise would be to consider the ratio of these voltages for several different times after the switch is reopened.
*32.22
After a long time, 12.0 V = (0.200 A)R
L Thus, R = 60.0 . Now, = R gives
L = R = (5.00 10 4 s)(60.0 V/A) = 30.0 mH
32.23
I = I max 1 - e -t/ : L 15.0 H = R = = 0.500 s : 30.0 (a)
(
)
dI 1 = -I max e -t/ - dt dI R t/ dt = L Imax e and Imax =
(
)
R
dI R = 100 V = 6.67 A/s t = 0: dt = L Imax e0 = 15.0 H L dI et/ = (6.67 A/s)e 1.50/(0.500) = (6.67 A/s)e3.00 = 0.332 A/s t = 1.50 s: dt = L
(b)
Chapter 32 Solutions 249 I = I max 1 - e -t/
32.24
(
)
-3
0.980 = 1 - e - 3.00 10 0.0200 = e - 3.00 10
-3
/
/
=-
3.00 10 - 3 = 7.67 10 - 4 s ln(0.0200)
= L R, so L = R = (7.67 10 - 4 )(10.0) = 7.67 mH
32.25
Name the currents as shown. I1 = I 2 + I 3 +10.0 V - 4.00 I1 - 4.00 I 2 = 0 +10.0 V - 4.00 I1 - 8.00 I 3 - (1.00) From (1) and (2), Then (3) becomes
By Kirchhoff's laws: (1) (2) dI 3 = 0 (3) dt and I1 = 0.500 I 3 + 1.25 A dI 3 =0 dt
+10.0 - 4.00 I1 - 4.00 I1 + 4.00 I 3 = 0
10.0 V - 4.00(0.500 I 3 + 1.25 A ) - 8.00 I 3 - (1.00)
(1.00 H)(dI 3
dt ) + (10.0 ) I 3 = 5.00 V
We solve the differential equation using Equations 32.6 and 32.7: I 3 (t ) = 5.00 V - 10.0 ) t 1- e ( 10.0
[
1.00 H
] = (0.500 A)[1 - e ]
-10t/s
I1 = 1.25 + 0.500 I 3 = 1.50 A - (0.250 A )e -10t/s
32.26
(a) (b)
Using = RC =
L , we get R = R
L = C
3.00 H = 1.00 10 3 = 1.00 k 3.00 10 - 6 F
= RC = 1.00 10 3 3.00 10 - 6 F = 3.00 10 -3 s = 3.00 ms
(
)(
)
2000 by Harcourt, Inc. All rights reserved.
250 Chapter 32 Solutions 32.27 For t 0, the current in the inductor is zero. At t = 0, it starts to grow from zero toward 10.0 A with time constant = L R = (10.0 mH) (100 ) = 1.00 10 - 4 s . For 0 t 200 s, At t = 200 s, I = I max 1 - e
-t/
= 10.00 A 1 - e -10 000t/s ( )
(
)
I = (10.00 A ) 1 - e -2.00 = 8.65 A so for t 200 s,
(
)
Thereafter, it decays exponentially as I = I 0 e - t , I = (8.65 A )e
-10 000( t-200 s ) s
= (8.65 A )e -10 000t s + 2.00 = 8.65 e 2.00 A e -10 000t s =
(
)
(63.9 A)e -10 000t s
32.28
(a) (b)
I=
R
=
12.0 V = 1.00 A 12.0 V12 = (1.00 A)(12.00 ) = 12.0 V V 1200 = (1.00 A)(1200 ) = 1.20 kV V L = 1.21 kV
Initial current is 1.00 A, :
(c)
I = Imax eRt/L: Solving Thus,
dI R Rt/L dt = Imax L e 12.0 V = (1212 V)e1212t/2.00 t = 7.62 ms
and so
dI L dt = VL = Imax ReRt/L 9.90 10 3 = e 606t
32.29 (a)
=
L 0.140 = = 28.6 ms; R 4.90
I max =
R
=
6.00 V = 1.22 A 4.90
I = I max 1 - e -t/ e -t/ = 0.820
(
)
so
0.220 = 1.22 1 - e -t/
(
)
t = - ln(0.820) = 5.66 ms
(b)
10.0 - I = I max 1 - e 0.0286 = (1.22 A) 1 - e -350 = 1.22 A
(
)
(c)
I = I max e -t/
and
0.160 = 1.22e -t/
so
t = ln(0.131) = 58.1 ms
Chapter 32 Solutions 251 32.30 (a) For a series connection, both inductors carry equal currents at every instant, so dI/dt is the same for both. The voltage across the pair is L eq (b) L eq dI dI dI = L1 + L 2 dt dt dt dI dI dI = L 1 1 = L 2 2 = VL dt dt dt VL VL VL = + L eq L1 L2 so L eq = L 1 + L 2 and dI dI1 dI 2 = + dt dt dt
where I = I1 + I 2
Thus,
and
1 1 1 = + L eq L 1 L 2
(c)
L eq
dI dI dI + R eq I = L 1 + IR 1 + L 2 + IR 2 dt dt dt
Now I and dI/dt are separate quantities under our control, so functional equality requires both L eq = L 1 + L 2 (d) V = L eq and R eq = R 1 + R 2 where I = I1 + I 2 and dI dI1 dI 2 = + dt dt dt
dI dI dI + R eq I = L 1 1 + R 1I1 = L 2 2 + R 2 I 2 dt dt dt
We may choose to keep the currents constant in time. Then,
1 1 1 = + R eq R 1 R 2 1 1 1 = + L eq L 1 L 2
We may choose to make the current swing through 0.
Then,
This equivalent coil with resistance will be equivalent to the pair of real inductors for all other currents as well.
32.31
L=
200(3.70 10 4) N B = I 1.75
= 42.3 mH
so
U = 2 LI 2 = 2 (0.423 H)(1.75 A) 2 = 0.0648 J
1
1
32.32
(a)
The magnetic energy density is given by u= B2 (4.50 T)2 = = 8.06 106 J/m3 2 0 2(1.26 10 6 T m/A)
(b)
The magnetic energy stored in the field equals u times the volume of the solenoid (the volume in which B is non-zero). U = uV = (8.06 106 J/m3) (0.260 m) (0.0310 m)2 = 6.32 kJ
[
]
2000 by Harcourt, Inc. All rights reserved.
252 Chapter 32 Solutions N 2A (68.0)2 (0.600 10 -2 )2 = 0 = 8.21 H 0.0800 l
32.33
L = 0
1 1 U = 2 LI 2 = 2 (8.21 10 -6 H)(0.770 A)2 = 2.44 J
32.34
(a)
2 2 (0.800)(500)2 1 1 L U = 2 LI 2 = 2 L = = = 27.8 J 2R 8(30.0)2 8R 2
(b)
I=
1 - e -(R/L)t R
[
]
so
2R so
=
1 1 - e -(R/L)t e -(R/L)t = R 2
[
]
R t = ln 2 L
t=
L 0.800 ln 2 = ln 2 = 18.5 ms R 30.0
32.35
u = 0
E2 3 2 = 44.2 nJ/m
B2 u = 2 = 995 J/m3 0
*32.36 (a) (b) (c)
U = 2 LI 2 = 2 (4.00 H)(0.500 A) 2 = 0.500 J dU = LI = (4.00 H)(1.00 A) = 4.00 J/s = 4.00 W dt P = (V)I = (22.0 V)(0.500 A) = 11.0 W
1
1
32.37
From Equation 32.7,
I= I=
(1 - e - R t
R
L
)
(a)
The maximum current, after a long time t , is At that time, the inductor is fully energized and
R
= 2.00 A.
P = I(V) = (2.00 A)(10.0 V) = 20.0 W
(b) (c)
P lost = I 2 R = (2.00 A)2 (5.00 ) = 20.0 W P inductor = I(Vdrop ) = 0 U= LI 2 (10.0 H)(2.00 A)2 = = 20.0 J 2 2
(d)
Chapter 32 Solutions 253 E2 2 B2 2 0
32.38
We have
u = e0
and
u=
Therefore
e0
B2 E2 = 2 2 0
so
B2 = e0 0E 2
B = E e0 0 =
6.80 10 5 V / m = 2.27 10 3 T 3.00 108 m / s
32.39
The total magnetic energy is the volume integral of the energy density, u =
B2 2 0 B 2 R 4 u= 0 2 0 r
Because B changes with position, u is not constant. For B = B0 ( R / r ) ,
2
Next, we set up an expression for the magnetic energy in a spherical shell of radius r and thickness dr. Such a shell has a volume 4 r 2 dr, so the energy stored in it is 2 B0 2 R 4 dr dU = u 4 r 2 dr = 0 r2
(
)
We integrate this expression for r = R to r = to obtain the total magnetic energy outside the sphere. This gives U= 2 B 0 R 3 0
2
=
2 (5.00 105 T)2(6.00 106 m)3 = 2.70 1018 J (1.26 10 6 T m/A)
32.40
I1(t) = I max e - t sin t with I max = 5.00 A, = 0.0250 s -1 , and = 377 rad s . dI1 = I max e - t ( - sin t + cos t ) dt At t = 0.800 s , dI1 = ( 5.00 A s)e - 0.0200 - (0.0250) sin(0.800( 377 )) + 377 cos(0.800( 377 )) dt dI1 = 1.85 10 3 A s dt Thus,
[
]
2 = -M dI1 :
dt
M=
- 2
dI1 dt
=
+ 3.20 V = 1.73 mH 1.85 10 3 A s
2000 by Harcourt, Inc. All rights reserved.
254 Chapter 32 Solutions
32.41
2 = -M dI1 = -(1.00 10- 4 H)(1.00 10 4 A / s) cos(1000t)
dt
( 2 )max =
1.00 V
32.42
M=
2
dI1 dt
=
96.0 mV = 80.0 mH 1.20 A / s
32.43
(a)
M=
N B BA 700(90.0 10 - 6 ) = = 18.0 mH 3.50 IA A 400(300 10 - 6 ) = = 34.3 mH 3.50 IA 9.00 mV
(b)
LA =
(c)
B = -M dI A = -(18.0 mH)(0.500 A / s) =
dt
32.44
M=
N 2 12 N 2 (B1A1 ) N 2 ( 0n1I1 )A 1 = = = N 2 0n1A 1 I1 I1 I1
[
]
M = (1.00) 4 10 -7 T m A
(
70.0 ) 0.0500 m (5.00 10
-3
2 m = 138 nH
)
32.45
B at center of (larger) loop: B1 =
0 I1 2R
(a)
M=
0 r 2 2 B1A 2 ( 0 I1 / 2R)( r 2 ) = = = I1 I1 2R I1 0 (0.0200)2
2(0.200) = 3.95 nH
(b)
M=
Chapter 32 Solutions 255 *32.46 Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at distance x from the wire is B = 0 I 2 x, and this field passes perpendicularly through the plane of the loop. The flux through the loop is B = B dA = B dA = B (ldx ) =
0 I l 1.70 mm dx Il 1.70 = 0 ln 0.400 2 0.400 mm x 2
The mutual inductance between the wire and the loop is then M= 1(4 10 -7 T m A)(2.70 10 -3 m) N 2 12 N 2 0 I l 1.70 N 2 0 l = = ln (1.45) = (1.45) 0.400 I1 2 I 2 2
M = 7.81 10 -10 H = 781 pH
32.47
With I = I1 + I 2 , the voltage across the pair is: V = - L 1 So, dI dI dI1 dI dI - M 2 = - L 2 2 - M 1 = - L eq dt dt dt dt dt - dI1 V M dI 2 = + L 1 L 1 dt dt dI 2 M( V ) M 2 dI 2 + + = V dt L1 L 1 dt dI 2 = V(L1 - M) dt (a) (b)
and
- L2
(-L1L2 + M 2 ) By substitution, -
[1]
dI 2 V M dI1 = + L2 L 2 dt dt dI1 = V (L 2 - M) dt dI = V (L 1 + L 2 - 2M) dt [2]
leads to Adding [1] to [2],
(- L1L 2 + M 2 ) (- L 1L 2 + M 2 )
So,
L eq = -
L 1L 2 - M 2 V = dI / dt L 1 + L 2 - 2M
32.48
At different times,
(UC )max = (U L )max
so
[
1 C V )2 2 ( max
]
1 = 2 LI 2
(
)
max
I max =
1.00 10 -6 F C ( V )max = ( 40.0 V) = 0.400 A L 10.0 10 -3 H
2000 by Harcourt, Inc. All rights reserved.
256 Chapter 32 Solutions
32.49
[
1 C V )2 2 ( max
]
1 = 2 LI 2
(
)
max
so
( VC )max =
L I max = C
20.0 10 -3 H (0.100 A) = 20.0 V 0.500 10 - 6 F
32.50
When the switch has been closed for a long time, battery, resistor, and coil carry constant current I max = / R . When the switch is opened, current in battery and resistor drops to zero, but the coil carries this same current for a moment as oscillations begin in the LC loop. We interpret the problem to mean that the voltage amplitude of 2 1 1 2 these oscillations is V, in 2 C ( V ) = 2 LI max . Then, L =
2 2 0.500 10 - 6 F (150 V ) ( 250 ) C ( V ) C ( V ) R 2 = = = 0.281 H 2 I max 2 ( 50.0 V )2 2 2
(
)
32.51
C =
1 1 = = 608 pF 2 (2 f ) L (2 6.30 106 )2 (1.05 106 )
Goal Solution A fixed inductance L = 1.05 H is used in series with a variable capacitor in the tuning section of a radio. What capacitance tunes the circuit to the signal from a station broadcasting at 6.30 MHz? G: It is difficult to predict a value for the capacitance without doing the calculations, but we might expect a typical value in the F or pF range. O: We want the resonance frequency of the circuit to match the broadcasting frequency, and for a simple RLC circuit, the resonance frequency only depends on the magnitudes of the inductance and capacitance. A : The resonance frequency is f 0 = Thus, C= 1 2 LC 1 1 = = 608 pF 2 (2 f 0 ) L (2 )(6.30 106 Hz) 2 (1.05 10 -6 H)
[
]
L : This is indeed a typical capacitance, so our calculation appears reasonable. However, you probably would not hear any familiar music on this broadcast frequency. The frequency range for FM radio broadcasting is 88.0 108.0 MHz, and AM radio is 535 1605 kHz. The 6.30 MHz frequency falls in the Maritime Mobile SSB Radiotelephone range, so you might hear a ship captain instead of Top 40 tunes! This and other information about the radio frequency spectrum can be found on the National Telecommunications and Information Administration (NTIA) website, which at the time of this printing was at http://www.ntia.doc.gov/osmhome/allochrt.html
Chapter 32 Solutions 257 1 : 2 LC 1 1 = = 0.220 H (2 f )2 C (2 120)2 (8.00 10 - 6 )
32.52
f=
L=
32.53
(a) (b) (c)
f=
1 1 = = 135 Hz 2 LC 2 (0.0820 H)(17.0 10 - 6 F)
Q = Qmax cos t = (180 C) cos(847 0.00100) = 119 C I= dQ = - Qmax sin t = -(847)(180) sin(0.847) = 114 mA dt
32.54
(a) (b) (c)
f=
1 2 LC
=
1 2 (0.100 H)(1.00 10 - 6 F)
= 503 Hz
Q = C = (1.00 10 - 6 F)(12.0 V) = 12.0 C
1C 2
2 =
1 LI 2 2 max
I max =
C = 12 V L
1.00 10 - 6 F = 37.9 mA 0.100 H
1 1 (d) At all times U = 2 C 2 = 2 (1.00 10 - 6 F)(12.0 V)2 = 72.0 J
32.55
=
1 = LC
(3.30 H)(840 10
1
-12
F
)
= 1.899 10 4 rad s
Q = Qmax cos t, I = Q2 UC = = 2C
dQ = - Qmax sin t dt
-6 4
(a)
( [105 10 ]cos[(1.899 10
(
rad s 2.00 10 -3 s
2 840 10 -12
)
)(
)] )
2
= 6.03 J
(b)
2 1 1 U L = 2 LI 2 = 2 L 2Qmax sin 2 ( t ) =
2 Qmax sin 2 ( t ) 2C
UL (c)
(105 10 =
-6
C sin 2 1.899 10 4 rad s 2.00 10 -3 s 2 840 10 -12 F
)
2
[(
(
)
)(
)] =
0.529 J
U total = UC + U L = 6.56 J
2000 by Harcourt, Inc. All rights reserved.
258 Chapter 32 Solutions
32.56
(a)
d =
1 R - = LC 2L fd =
2
(2.20 10 )(1.80 10 )
-3 -6
1
7.60 - 2 2.20 10 -3
(
)
= 1.58 10 4 rad / s
2
Therefore, Rc =
d = 2.51 kHz 2
(b)
4L = 69.9 C
32.57
(a)
0 =
1 1 = = 4.47 krad/s LC (0.500)(0.100 10 -6 ) 1 R - = 4.36 krad/s LC 2L
2
(b)
d =
(c)
= 2.53% lower 0
32.58
Choose to call positive current clockwise in Figure 32.19. It drains charge from the capacitor according to I = dQ/dt. A clockwise trip around the circuit then gives + Q dI - IR - L = 0 C dt Q dQ d dQ + R+L = 0, identical with Equation 32.29. C dt dt dt
+
32.59
(a)
Q = Qmax e 0.500 = e
-
-
Rt 2L
cos dt
so and
I max e
-
Rt 2L
Rt 2L
Rt = - ln(0.500) 2L
t=-
2L 2L ln(0.500) = 0.693 R R and U = 0.500U 0 so (half as long) Q = 0.500 Qmax = 0.707Qmax
(b)
2 U 0 Qmax
t=-
2L 2L ln(0.707) = 0.347 R R
Chapter 32 Solutions 259 32.60 With Q = Qmax at t = 0, the charge on the capacitor at any time is Q = Qmax cos t where = 1 LC . The energy stored in the capacitor at time t is then U=
2 Q 2 Qmax = cos 2 t = U 0 cos 2 t. 2C 2C 1 4
When U = U 0 , Therefore,
cos t = t = LC 3
1 2
and
t = rad 2 t2 = LC 9
L= 9t 2 2C
1 3
or
The inductance is then:
32.61
(a) (b)
d 20.0t ) L = - L dI = -(1.00 mH) ( = dt dt Q = I dt = ( 20.0t )dt = 10.0t 2
t t 0 0
20.0 mV
VC =
-Q -10.0t 2 = = - 10.0 MV s 2 t 2 C 1.00 10 -6 F
(
)
(c)
Q2 1 When LI 2 , 2 2C
or
(-10.0t ) 1 (1.00 10 )(20.0t) , 2 2(1.00 10 )
2 2 -6 -3 2
then 100t 4 400 10 - 9 t 2 . The earliest time this is true is at
(
)
t = 4.00 10 - 9 s = 63.2 s
32.62
(a)
L = - L dI = - L d
dt I= dQ , dt
dt
(Kt) = LK
t t 1 Q = I dt = Kt dt = 2 Kt 2 0 0
(b)
so
and
VC =
-Q Kt2 = - 2C C
(c)
When
1C 2
( VC )2 =
1 LI 2 , 2
1C 2
K2 t 4 1 2 2 4C 2 = 2 L K t
(
)
Thus
t = 2 LC
2000 by Harcourt, Inc. All rights reserved.
260 Chapter 32 Solutions
32.63
1 Q 1 1 Q2 = + LI 2 2C 2 2 2 C
2
so
I=
3Q 2 4CL LI = N Q 2N 3L C
The flux through each turn of the coil is where N is the number of turns.
B =
32.64
Equation 30.16: B =
b
0 NI 2 r
b
(a)
B = B dA =
0 NI NIh dr 0 NIh b h dr = 0 = ln a 2 r 2 r 2 a a
L=
0N 2h b N B = ln a 2 I 0 (500)2 (0.0100) 12.0 = 91.2 H ln 10.0 2 0 N 2 A 0 (500)2 2.00 10 - 4 m 2 = = 90.9 H 2 R 2 0.110
(b)
L=
(c)
Lappx =
*32.65 (a)
At the center,
B=
N 0 IR 2 N 0 I = 2 2 3/2 2R 2(R + 0 ) N 0 I R 2 cos 0 = N 0 IR 2R 2 -L dI dt
So the coil creates flux through itself
B BA cos =
When the current it carries changes,
L = -N d B - N N 0 R dI =
dt 2 dt L
so
2 N 0R 2
(b)
2 r 3(0.3 m), so r 0.14 m;
Tm L 2 12 4 107 A 0.14 m = 2.8 107 H ~ 100 nH
(c)
2.8 107 V s/A L = 1.0 10 9 s ~ 1 ns 270 V/A R
Chapter 32 Solutions 261 32.66 (a) If unrolled, the wire forms the diagonal of a 0.100 m (10.0 cm) rectangle as shown. The length of this rectangle is L =
9.80 m L
0.100 m
(9.80 m)2 - (0.100 m)2
The mean circumference of each turn is C = 2 r , where r = radius of each turn. N= L = C The number of turns is then: = 127
24.0 + 0.644 mm is the mean 2
(9.80 m)2 - (0.100 m)2
2
24.0 + 0.644 10 - 3 m 2
(b)
R=
-8 l 1.70 10 m (10.0 m ) = = 0.522 2 A 0.322 10 -3 m
(
(
)
)
(c)
L=
2 N 2 A 800 0 L = ( r )2 l l C
2 800 4 10 -7 (9.80 m )2 - (0.100 m )2 24.0 + 0.644 10 -3 m L= ( 24.0 + 0.644) 10 -3 m 2 0.100 m
(
)
2
L = 7.68 10 -2 H = 76.8 mH
32.67
From Ampere's law, the magnetic field at distance r R is found as: I B( 2 r ) = 0 J r 2 = 0 r 2 , or 2 R
( )
( )
B=
0 Ir 2 R2
The magnetic energy per unit length within the wire is then
I2 R B2 U = (2 r dr ) = 0 4 0 2 l 4 R 0
0
R
r 3 dr =
0 I 2 R4 = 4 R4 4
0 I 2
16
This is independent of the radius of the wire.
2000 by Harcourt, Inc. All rights reserved.
262 Chapter 32 Solutions 32.68 The primary circuit (containing the battery and solenoid) is an RL circuit with R = 14.0 , and 4 10 -7 (12 500) 1.00 10 - 4 N 2A L= 0 = = 0.280 H l 0.0700
2
(
)
(
)
(a)
The time for the current to reach 63.2% of the maximum value is the time constant of the circuit:
=
L 0.280 H = = 0.0200 s = 20.0 ms R 14.0
(b)
The solenoid's average back emf is
L
=L
I f - 0 I = L t t V 60.0 V = 0.632 = 2.71 A R 14.0 37.9 V
where Thus, (c)
I f = 0.632 I max = 0.632
L = (0.280 H) 2.71 A = 0.0200 s
The average rate of change of flux through each turn of the overwrapped concentric coil is the same as that through a turn on the solenoid: 4 10 -7 T m A (12 500 0.0700 m )( 2.71 A ) 1.00 10 - 4 m 2 B 0n( I )A = = = 3.04 mV t 0.0200 s t
(
)
(
)
(d) The magnitude of the average induced emf in the coil is the average induced current is I=
L
= N ( B t ) and magnitude of
L
R
=
820 N B = 3.04 10 -3 V = 0.104 A = 104 mA R t 24.0
(
)
32.69
Left-hand loop: Outside loop: Eliminating I 2 gives
E - (I + I 2 )R 1 - I 2 R2 = 0 E - (I + I 2 )R 1 - L E - IR - L dI =0 dt I (t ) = E (1 - e - R t L ) R dI =0 dt
This is of the same form as Equation 32.6, so its solution is of the same form as Equation 32.7: But R = R1R2 / ( R1 + R2 ) and E = R2E / ( R1 + R2 ), so
E R 2 / (R 1 + R2 ) E E = = R R 1R2 / (R 1 + R2 ) R 1 I(t) = E (1 - e - R t L ) R1
Thus
Chapter 32 Solutions 263 32.70 When switch is closed, steady current I 0 = 1.20 A. W h e n the switch is opened after being closed a long time, the current in the right loop is I = I0e
- R2 t L
so
e Rt L=
L
=
I0 I
and
Rt I = ln 0 I L
Therefore,
R2 t (1.00 )(0.150 s) = 0.0956 H = 95.6 mH = ln( I 0 I ) ln(1.20 A 0.250 A )
32.71
(a)
While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of the circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer loop of the circuit. Applying Kirchhoff's loop rule to this loop gives: + 0 - ( 2.00 + 6.00) 10 3 9.00 10 -3 A = 0 + 0 = 72 .0 V with end b at the higher potential
[
](
)
(b)
(c)
After the switch is opened, the current around the outer loop decays as I = I max e - R t
L
with
I max = 9.00 mA ,
R = 8.00 k ,
and
L = 0.400 H
Thus, when the current has reached a value I = 2.00 mA , the elapsed time is: t= L I max 0.400 H 9.00 = ln = 7.52 10 -5 s = 75.2 s ln R I 8.00 10 3 2.00
2000 by Harcourt, Inc. All rights reserved.
264 Chapter 32 Solutions 32.72 (a) The instant after the switch is closed, the situation is as shown i n the circuit diagram of Figure (a). The requested quantities are: I L = 0, IC = 0 R , I R = 0 R VL = 0 , VC = 0, V R = 0
Q=0 VC = 0 IC = 0/R
IL = 0
+
VL = 0
-
IR = 0/R
VR = 0
+ 0
Figure (a) VL = 0
-
(b)
After the switch has been closed a long time, the steady-state conditions shown in Figure (b) will exist. The currents and voltages are: I L = 0, VL = 0, IC = 0, IR = 0 V R = 0
Q = C0
IL = 0
+
IR = 0
VC = 0 ,
VC = 0
VR = 0
+ 0
Figure (b)
32.73
When the switch is closed, as shown in Figure (a), the current in the inductor is I : 12.0 7.50I 10.0 = 0 I = 0.267 A When the switch is opened, the initial current in the inductor remains at 0.267 A. IR = V: (0.267 A)R 80.0 V R 300 (a) (b)
Goal Solution To prevent damage from arcing in an electric motor, a discharge resistor is sometimes placed in parallel with the armature. If the motor is suddenly unplugged while running, this resistor limits the voltage that appears across the armature coils. Consider a 12.0-V dc motor with an armature that has a resistance of 7.50 and an inductance of 450 mH. Assume that the back emf in the armature coils is 10.0 V when the motor is running at normal speed. (The equivalent circuit for the armature is shown in Figure P32.73.) Calculate the maximum resistance R that limits the voltage across the armature to 80.0 V when the motor is unplugged.
Chapter 32 Solutions 265
G: We should expect R to be significantly greater than the resistance of the armature coil, for otherwise a large portion of the source current would be diverted through R and much of the total power would be wasted on heating this discharge resistor. O: When the motor is unplugged, the 10-V back emf will still exist for a short while because the motor's inertia will tend to keep it spinning. Now the circuit is reduced to a simple series loop with an emf, inductor, and two resistors. The current that was flowing through the armature coil must now flow through the discharge resistor, which will create a voltage across R that we wish to limit to 80 V. As time passes, the current will be reduced by the opposing back emf, and as the motor slows down, the back emf will be reduced to zero, and the current will stop. A : The steady-state coil current when the switch is closed is found from applying Kirchhoff's loop rule to the outer loop: + 12.0 V - I (7.50 ) - 10.0 V = 0 so We then require that so I= 2.00 V = 0.267 A 7.50
V R = 80.0 V = (0.267 A )R R= V R 80.0 V = = 300 0.267 A I
L : As we expected, this discharge resistance is considerably greater than the coil's resistance. Note that while the motor is running, the discharge resistor turns P = (12 V)2 300 = 0.48 W of power into heat (or wastes 0.48 W). The source delivers power at the rate of about P = IV = [0.267 A + (12 V / 300 )](12 V ) = 3.68 W, so the discharge resistor wastes about 13% of the total power. For a sense of perspective, this 4-W motor could lift a 40-N weight at a rate of 0.1 m/s.
2 2 4 10 -7 T m A (1000) 1.00 10 - 4 m 2 0 N1 A = = 2.51 10 - 4 H = 251 H l1 0.500 m
32.74
(a)
L1 =
(
)
(
)
(b)
M=
N 2 2 N 2 1 N 2BA N 2 0 ( N1 l1 ) I1 A 0 N1N 2 A = = = = l1 I1 I1 I1 I1
-7
[
]
(4 10 M=
(c) dt
T m A (1000)(100) 1.00 10 - 4 m 2 0.500 m or I1R1 = - M dI 2 dt and I1 =
)
(
)
= 2.51 10 - 5 H = 25.1 H
1 = - M dI 2 ,
Q1 = -
dQ1 M dI 2 =- R1 dt dt
M I 2i M tf M M dI 2 = - I 2 f - I 2i = - (0 - I 2i ) = R1 0 R1 R1 R1
-5
(
)
Q1 =
(2.51 10
H (1.00 A )
1000
)
= 2.51 10 -8 C = 25.1 nC
2000 by Harcourt, Inc. All rights reserved.
266 Chapter 32 Solutions 2U is non-zero. I2
32.75
(a)
It has a magnetic field, and it stores energy, so L =
(b)
Every field line goes through the rectangle between the conductors. 1 w-a = I I y=a
(c)
= LI
so
L=
B da
L=
w-a I 0 I 2 0 Ix 2 0 x 1 w-a x dy 0 + = dy = ln y I a 2 2 y 2 ( w -y ) I 2 y a
Thus
L=
0x w - a ln a
32.76
For an RL circuit, R t = 10 - 9 L
I(t) = I max e
-
R t L :
- t R I(t) = 1 - 10 - 9 = e L 1 - t L I max
R
so
Rmax =
(3.14 10 - 8 )(10 - 9 ) = 3.97 10 - 25 (2.50 yr)(3.16 107 s / yr)
(If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm 2 , its resistance would be at least 10 6 ).
32.77
(a) (b)
UB = 2 LI 2 = 2 (50.0 H)(50.0 10 3 A)
1
1
2
= 6.25 1010 J
Two adjacent turns are parallel wires carrying current in the same direction. Since the loops have such large radius, a one-meter section can be regarded as straight. Then one wire creates a field of This causes a force on the next wire of giving B= 0I 2 r
F = IlB sin F = Il
lI 2 0I sin 90 = 0 2 r 2 r
(1.00 m)(50.0 10 3 A) 2 = 2000 N (2)(0.250 m)
Solving for the force,
F = (4 107 N/A2)
Chapter 32 Solutions 267
32.78
P = I ( V ) I= P 1.00 10 9 W = = 5.00 10 3 A V 200 10 3 V B( 2 r ) = 0 I enclosed or B=
From Ampere's law,
0 I enclosed 2 r
(a)
At r = a = 0.0200 m,
I enclosed = 5.00 10 3 A
and
(4 10 B=
(b)
-7
T m A 5.00 10 3 A
2 (0.0200 m )
)(
) = 0.0500 T =
50.0 mT
At r = b = 0.0500 m, B=
I enclosed = I = 5.00 10 3 A
and
(4 10
-7
T m A 5.00 10 3 A
2 (0.0500 m )
)(
) = 0.0200 T =
a
b dr
20.0 mT
(c)
U = u dV =
r=b
r=a
[B(r )]2 (2 rldr ) = 0 I 2l
2 0 4
2
r
=
0 I 2l b ln a 4
3
(4 10 U=
-7
T m A 5.00 10 3 A 4
)(
) (1000 10 m) ln 5.00 cm = 2.29 10
2.00 cm
6
J = 2.29 MJ
(d) The magnetic field created by the inner conductor exerts a force of repulsion on the current i n the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small rectangular section of the outer cylinder of length l and width w . It carries a current of w (5.00 10 A) 2 (0.0500 m)
3
and experiences an outward force
F = IlBsin =
(5.00 10 A)w l(20.0 10 T) sin 90.0 2 (0.0500 m )
3 -3
The pressure on it is
P=
5.00 10 3 A 20.0 10 -3 T F F = = = 318 Pa A wl 2 (0.0500 m )
(
)(
)
2000 by Harcourt, Inc. All rights reserved.
268 Chapter 32 Solutions 4 10 - 7 T m A (1400)( 2.00 A ) 0 NI B= = = 2.93 10 - 3 T (upward) l 1.20 m 2.93 10 - 3 T N B2 J 1 N m u= = = 3.42 3 = 3.42 2 = 3.42 Pa 1J 2 0 2 4 10 -7 T m A m m
*32.79 (a)
(
)
(b)
(
(
)
2
)
(c)
To produce a downward magnetic field, the surface of the super conductor must carry a clockwise current.
(d) The vertical component of the field of the solenoid exerts an inward force on the superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field diverges and has a radially outward horizontal component. This component exerts upward force on the clockwise superconductor current. The total force on the core is upward . You can think of it as a force of repulsion between the solenoid with its north end pointing up, and the core, with its north end pointing down. (e)
2 F = PA = ( 3.42 Pa) 1.10 10 -2 m = 1.30 10 - 3 N
(
)
Note that we have not proven that energy density is pressure. In fact, it is not in some cases; see problem 12 in Chapter 21.
Chapter 33 Solutions
v(t) = Vmax sin( t) = 2 Vrms sin( t) = 200 2 sin[2 (100t)] = (283 V) sin (628 t)
33.1
33.2 (a) (b)
Vrms = P= R=
170 V = 120 V 2
(120 V)2 (Vrms )2 R= = 193 R 75.0 W (120 V)2 = 144 100 W
33.3
Each meter reads the rms value. Vrms = I rms = 100 V = 70.7 V 2
Vrms 70.7 V = = 2.95 A 24.0 R
33.4
(a)
vR = Vmax sin t vR = 0.250 ( Vmax ) , so sin t = 0.250, or t = sin -1 (0.250)
The smallest angle for which this is true is t = 0.253 rad. Thus, if t = 0.010 0 s ,
=
(b)
0.253 rad = 25.3 rad/s 0.010 0 s
The second time when vR = 0.250 ( Vmax ) , t = sin -1 (0.250) again. For this occurrence, t = - 0.253 rad = 2 .89 rad (to understand why this is true, recall the identity sin( - ) = sin from trigonometry). Thus, t= 2.89 rad = 0.114 s 25.3 rad s
33.5
iR = I max sin t Thus, and
becomes
0.600 = sin( 0.00700) (0.00700) = sin1(0.600) = 0.644
= 91.9 rad/s = 2 f
so
f = 14.6 Hz
2000 by Harcourt, Inc. All rights reserved.
Chapter 33 Solutions 271 33.6 P = I rms ( Vrms ) and Vrms = 120 V for each bulb (parallel circuit), so: I1 = I 2 = I3 = Vrms 120 V P1 150 W = = 1.25 A , and R1 = = = 96.0 = R2 1.25 A Vrms 120 V I1
Vrms P3 100 W 120 V = = 0.833 A , and R3 = = = 144 0.833 A Vrms 120 V I3
33.7
Vmax = 15.0 V I max =
and Rtotal = 8.20 + 10.4 = 18.6
Vmax 15.0 V = = 0.806 A = 2 I rms 18.6 Rtotal
2
0.806 A 2 P speaker = I rms Rspeaker = (10.4 ) = 3.38 W 2
33.8
For Imax = 80.0 mA,
Irms =
80.0 mA 2
= 56.6 mA
Vrms 50.0 V (XL)min = Irms = 0.0566 A = 884 XL 884 XL = 2 f L L = 2 f 2 (20.0) 7.03 H
33.9
(a)
XL = L=
Vmax 100 = = 13.3 7.50 I max
XL 13.3 = = 0.0424 H = 42.4 mH 2 (50.0) Vmax 100 = = 40.0 2.50 I max
(b)
XL =
=
XL 40.0 = = 942 rad/s L 42.4 10 - 3
33.10
At 50.0 Hz,
X L 60.0 Hz 50.0 = XL = 2 ( 50.0 Hz)L = 2 ( 50.0 Hz) ( 54.0 ) = 45.0 2 (60.0 Hz) 60.0 2 ( Vrms ) = XL 2 (100 V ) = 3.14 A 45.0
I max =
Vmax = XL
2000 by Harcourt, Inc. All rights reserved.
272 Chapter 33 Solutions
33.11
iL (t ) =
(80.0 V ) sin (65.0 )(0.0155) - 2 Vmax sin ( t - 2) = L (65.0 rad s) 70.0 10-3 H
[
(
)
]
iL (t ) = ( 5.60 A ) sin(1.59 rad) = 5.60 A
33.12
= 2 f = 2 (60.0 / s) = 377 rad / s
XL = L = (377 / s)(0.0200 V s / A) = 7.54 I rms = Vrms 120 V = = 15.9 A 7.54 XL
I max = 2 I rms = 2 (15.9 A) = 22.5 A 2 (60.0) 1 s i(t) = I max sin t = (22.5 A)sin = (22.5 A) sin 120 = 19.5 A s 180 V s 1 1 (19.5 A)2 = 3.80 J U = 2 Li 2 = 2 0.0200 A
33.13
L=
N B where B is the flux through each turn. I 2 VL, rms 2 f
N B, max = LI B,
max
=
XL VL, max XL
(
)
N B, max =
(
)=
120 V s T C m N m J = 0.450 T m2 2 (60.0) N s J V C
33.14
(a)
XC =
1 : 2 f C
1 < 175 2 f (22.0 10 -6 ) f > 41.3 Hz
1 <f 2 (22.0 10 -6 )(175) (b) XC
1 1 , so X(44) = 2 X(22): XC < 87.5 C
33.15 (a) (b)
I max = 2 I rms =
2 ( Vrms ) = 2 ( Vrms ) 2 f C XC
I max = 2 (120 V)2 (60.0 / s)(2.20 10 - 6 C / V) = 141 mA I max = 2 (240 V)2 (50.0 / s)(2.20 10 - 6 F) = 235 mA
Chapter 33 Solutions 273 Qmax = C ( Vmax ) = C
33.16
[
2 ( Vrms ) =
]
2 C ( Vrms )
33.17
I max = ( Vmax ) C = (48.0 V)(2 )(90.0 s -1 )(3.70 10 - 6 F) = 100 mA
33.18
XC =
1 1 = = 2.65 C 2 (60.0 / s)(1.00 10 - 3 C / V)
vC (t) = Vmax sin t , to be zero at t = 0 iC = Vmax sin( t + ) = XC 2 (120 V) 60 s -1 sin 2 + 90.0 = (64.0 A)sin(120 + 90.0) = 32.0 A -1 2.65 180 s
33.19
(a)
XL = L = 2 (50.0)(400 10- 3) = 126 XC = 1 1 = = 719 C 2 (50.0)(4.43 10 -6 )
Z = R 2 + (XL - XC )2 = 500 2 + (126 - 719)2 = 776 Vmax = Imax Z = (250 10- 3)(776) = 194 V (b)
= tan -1
X L - XC 126 - 719 = tan -1 = 49.9 500 R
Thus, the Current leads the voltage.
33.20
L =
1 = C
1 1 = = 1.75 10 4 rad / s -6 LC (57.0 10 )(57.0 10 -6 )
f = 2 = 2.79 kHz
33.21
(a) (b) (c) (d) (e)
XL = L = 2 (50.0 s-1)(250 10-3 H) = 78.5 XC = 1 = 2 (50.0 s -1 )(2.00 10 -6 F) C
[
]
-1
= 1.59 k
Z = R 2 + (XL - XC )2 = 1.52 k I max = Vmax 210 V = = 138 mA Z 1.52 10 3 X L - XC -1 = tan (-10.1) = 84.3 R
= tan -1
2000 by Harcourt, Inc. All rights reserved.
274 Chapter 33 Solutions Z = R 2 + ( XL - XC ) = 68.0 2 + (16.0 - 101) = 109
2
33.22
(a)
XL = L = (100)(0.160) = 16.0 XC = 1 1 = = 101 C (100) 99.0 10 -6
(
)
(b)
I max =
Vmax 40.0 V = = 0.367 A 109 Z XL - XC 16.0 - 101 = = -1.25: 68.0 R
(c)
tan =
= - 0.896 rad = - 51.3
Imax = 0.367 A
= 100 rad/s
= 0.896 rad = 51.3
33.23
XL = 2 f L = 2 (60.0)(0.460) = 173 XC = 1 1 = = 126 2 f C 2 (60.0) 21.0 10 - 6
(
)
(a)
tan =
XL - XC 173 - 126 = = 0.314 150 R
= 0.304 rad = 17.4
(b) Since XL > XC , is positive; so voltage leads the current .
33.24
XC =
1 1 = = 1.33 108 2 f C 2 (60.0 Hz)(20.0 10 -12 F)
Z = (50.0 10 3 )2 + (1.33 108 )2 1.33 108 Irms = Vrms 5000 V = = 3.77 105 A Z 1.33 108 1.88 V
( Vrms )body = Irms Rbody = (3.77 10- 5 A)(50.0 10 3 ) =
Chapter 33 Solutions 275 1 1 = = 49.0 C 2 (50.0)(65.0 10 -6 )
33.25
XC =
XL = L = 2 (50.0)(185 10 -3 ) = 58.1 Z = R 2 + (XL - XC )2 = (40.0)2 + (58.1 - 49.0)2 = 41.0 I max = (a) (b) (c) Vmax 150 = = 3.66 A 41.0 Z
VR = Imax R = (3.66)(40) = 146 V VL = Imax XL = (3.66)(58.1) = 212.5 = 212 V VC = Imax XC = (3.66)(49.0) = 179.1 V = 179 V
(d) VL VC = 212.5 179.1 = 33.4 V
33.26
R = 300 500 -1 XL = L = 2 s (0.200 H) = 200 XC = 1 500 -1 11.0 10 -6 F = 2 s C
2
XL = 200
XL - XC = 109
(
)
-1
{
Z
R = 300
= 90.9 X L - XC = 20.0 R
XC = 90.9
Z = R 2 + ( XL - XC ) = 319
and
= tan -1
33.27
(a)
XL = 2 (100 Hz)( 20.5 H) = 1.29 10 4 Z= Vrms 200 V = = 50.0 4.00 A I rms
(XL - XC )2 = Z2 - R 2 = (50.0 )2 - (35.0 )2
XL - XC = 1.29 10 4 - (b) 1 = 35.7 2 (100 Hz)C C = 123 nF or 124 nF
VL,rms = I rms XL = ( 4.00 A ) 1.29 10 4 = 51.5 kV Notice that this is a very large voltage!
(
)
2000 by Harcourt, Inc. All rights reserved.
276 Chapter 33 Solutions 33.28 XL = L = [(1000 / s)(0.0500 H)] = 50.0 XC = 1/ C = (1000 / s)(50.0 10 - 6 F) Z = R 2 + (XL - XC )2 Z = (40.0)2 + (50.0 - 20.0)2 = 50.0 (a) I rms = ( Vrms ) / Z = 100 V / 50.0 I rms = 2.00 A X - XC = Arctan L R
[
]
-1
= 20.0
= Arctan
30.0 = 36.9 40.0
(b)
P = ( Vrms ) I rms cos = 100 V(2.00 A) cos 36.9 = 160 W
(c)
2 P R = I rms R = (2.00 A)2 40.0 = 160 W
33.29
= 1000 rad/s,
Vmax = 100 V,
R = 400 ,
C = 5.00 10 6 F,
L = 0.500 H
1 L = 500 , = 200 C
2
1 Z = R2 + L - = 400 2 + 300 2 = 500 C I max = Vmax 100 = = 0.200 A 500 Z I2 2 P = I rms R = max R 2
The average power dissipated in the circuit is (0.200 A)2 (400 ) = 8.00 W 2
P=
Chapter 33 Solutions 277 Goal Solution An ac voltage of the form v = (100 V ) sin(1000 t ) is applied to a series RLC circuit. C = 5.00 F, and L = 0.500 H, what is the average power delivered to the circuit? G: Comparing v = (100 V ) sin(1000 t ) with and v = Vmax sin t, we see that
If R = 400 ,
Vmax = 100 V
= 1000 s -1
Only the resistor takes electric energy out of the circuit, but the capacitor and inductor will impede the current flow and therefore reduce the voltage across the resistor. Because of this impedance, the average power dissipated by the resistor must be less than the maximum power from the source: P max =
( Vmax )2 = (100 V)2
2R
2( 400 )
= 12.5 W
2 O: The actual power dissipated by the resistor can be found from P = I rms R, where I rms = Vrms / Z.
A : Vrms =
100 = 70.7 V 2
In order to calculate the impedance, we first need the capacitive and inductive reactances: XC = 1 1 = = 200 C (1000 s -1 )(5.00 10 -6 F) and XL = L = 1000 s -1 (0.500 H) = 500
(
)
Then, I rms =
Z = R 2 + (XL - XC )2 = (400 )2 + (500 - 200 )2 = 500 Vrms 70.7 V = = 0.141 A 500 Z and
2 P = I rms R = (0.141 A ) ( 400 ) = 8.00 W 2
L : The power dissipated by the resistor is less than 12.5 W, so our answer appears to be reasonable. As with other RLC circuits, the power will be maximized at the resonance frequency where XL = XC so that Z = R . Then the average power dissipated will simply be the 12.5 W we calculated first.
33.30
Z = R 2 + ( X L - XC )
2
or
( X L - XC ) =
Z2 - R2
(XL - XC ) = (75.0 )2 - ( 45.0 )2
= tan -1
I rms =
= 60.0
X L - XC 60.0 = tan -1 = 53.1 45.0 R
Vrms 210 V = = 2.80 A 75.0 Z
P = ( Vrms ) I rms cos = ( 210 V )( 2.80 A ) cos( 53.1 ) = 353 W
2000 by Harcourt, Inc. All rights reserved.
278 Chapter 33 Solutions P = I rms (Vrms )cos = (9.00)(180) cos( 37.0) = 1.29 10 3 W
2 P = I rms R
33.31
(a)
so becomes
1.29 10 3 = (9.00)2 R tan( - 37.0) = X L - XC : 16
and so
R = 16.0 XL XC = 12.0
(b)
tan =
X L - XC R
*33.32
XL = L = 2 (60.0 / s)(0.0250 H) = 9.42 Z = R 2 + (XL - XC )2 = (a) (b) (c) I rms =
(20.0)2 + (9.42)2
= 22.1
Vrms 120 V = = 5.43 A Z 22.1 so power factor = cos = 0.905 9.42 = 1 2 (60.0 s -1 )C
= tan -1 (9.42 / 20.0) = 25.2
We require = 0. Thus, XL = XC: and
C = 281 F or
(d)
Pb = Pd
( Vrms )b ( Irms )b cos b =
R( Vrms )b ( I rms )b cos b =
( Vrms )d 2
R
( Vrms )d =
(20.0 )(120 V )( 5.43 A)(0.905) =
109 V
33.33
Consider a two-wire transmission line:
R1
I rms
P 100 106 W = = = 2.00 10 3 A Vrms 50.0 10 3 V
Vrms
RL
2 2 loss = (0.010 0)P = I rms R line = I rms ( 2R1 )
R1
6
Thus, R1
(0.010 0)P = (0.010 0)(100 10 =
2 2 I max
W
2
2 2.00 10 A
3
(
)
) = 0.125
But
R1 =
l or A
4 l = R1
A=
d 2 l = R1 4
Therefore
d=
4 1.70 10 - 8 m 100 10 3 m
(
(0.125 )
)(
) = 0.132 m =
132 mm
Chapter 33 Solutions 279 33.34 Consider a two-wire transmission line: I rms P = Vrms
2
R1
2 I rms R line
and
power loss =
P = 100
Vrms R1
RL
P P Thus, ( 2R1 ) = 100 Vrms R1 =
or
R1 =
( Vrms )2
200 P
d ( Vrms ) = A 200 P
2
or
A=
( 2r )2 200 P d = 4 ( Vrms )2
2r = 800 P d
and the diameter is
( Vrms )
2
33.35
One-half the time, the left side of the generator is positive, the top diode conducts, and the bottom diode switches off. The power supply sees resistance 1 1 2R + 2R
-1
R1 Vrms R1 RL
=R
and the power is
( V rms )2
R
The other half of the time the right side of the generator is positive, the upper diode is an open circuit, and the lower diode has zero resistance. The equivalent resistance is then 1 1 Req = R + + 3R R
-1
=
7R and 4
P=
( V rms )2
Req
=
4( V rms ) 7R
2
The overall time average power is:
[(V
rms
)2
R + 4( V rms ) 7R
2
][
2
]=
11( V rms ) 14 R
2
33.36
At resonance,
1 = 2 f L 2 f C
and
( 2 f )2 L
1
=C
The range of values for C is 46.5 pF to 419 pF
33.37
0 = 2 (99.7 106 ) = 6.26 108 rad / s =
C=
1 LC
1 1 = = 1.82 pF 0 2 L (6.26 108 )2 (1.40 10 - 6 )
2000 by Harcourt, Inc. All rights reserved.
280 Chapter 33 Solutions L = 20.0 mH, C = 1.00 107, R = 20.0 , Vmax = 100 V (a) (b) (c) (d) The resonant frequency for a series RLC circuit is At resonance, From Equation 33.36, I max = Q= Vmax = 5.00 A R f= 1 2 1 = 3.56 kHz LC
33.38
0L = 22.4 R
VL, max = XL I max = 0 L I max = 2.24 kV
33.39
The resonance frequency is 0 = 1 L 2 XL = L = L = 2 LC C
LC . Thus, if = 2 0 , and XC = 1 LC 1 L = = 2 C C 2C Vrms = Z R + 2.25( L C )
2
Z = R 2 + ( XL - XC ) = R 2 + 2.25( L C )
2
so
I rms =
Vrms
and the energy dissipated in one period is Q = P t: Q=
( Vrms )2 R 2 = ( Vrms )2 RC ( R 2 + 2.25( L C ) R 2C + 2.25 L
4 ( 50.0 V ) (10.0 ) 100 10 -6 F
2
LC =
)
4 ( Vrms ) RC LC
2
4R 2C + 9.00 L
With the values specified for this circuit, this gives: Q=
4(10.0 ) 100 10 -6
2
(
(
) (10.0 10 H) F ) + 9.00(10.0 10 H)
32 -3 -3
12
= 242 mJ
33.40
The resonance frequency is 0 = 1 L 2 XL = L = L = 2 LC C
LC .
Thus, if = 2 0 , and XC = 1 LC 1 L = = 2 C C 2C Vrms = Z R + 2.25( L C )
2
Then Z = R 2 + ( XL - XC ) = R 2 + 2.25( L C )
2
so
I rms =
Vrms
and the energy dissipated in one period is Q = P t =
( Vrms )2 R 2 = ( Vrms )2 RC ( R 2 + 2.25( L C ) R 2C + 2.25 L
LC =
)
4 ( Vrms ) RC LC
2
4R 2C + 9.00 L
Chapter 33 Solutions 281 1 = LC 1
-3
*33.41
For the circuit of problem 22,
0 =
(160 10 (
H 99.0 10 -6 F
)(
)
= 251 rad s
Q=
-3 0 L ( 251 rad s) 160 10 H = = 0.591 R 68.0
)
For the circuit of problem 23,
Q=
0L L 1 L 1 460 10 -3 H = = = = 0.987 R R LC R C 150 21.0 10 -6 F
The circuit of problem 23 has a sharper resonance.
33.42
(a) (b)
1 V 2, rms = 13 (120 V ) = 9.23 V
V1, rms I1, rms = V 2, rms I 2, rms (120 V)( 0.350 A) = (9.23 V)I 2, rms I 2, rms = 42.0 W = 4.55 A for a transformer with no energy loss 9.23 V
(c)
P = 42.0 W from (b)
33.43
( Vout )max = N2 ( Vin )max =
N
1
2000 (170 V) = 971 V 350
( Vout )rms =
(971 V) = 687 V 2
33.44
(a)
(V2, rms ) = N21 (V1, rms ) N
I1, rms V1, rms = I 2, rms V 2, rms
N2 =
(2200)(80) = 1600 windings 110 (1.50)(2200) = 30.0 A 110 (1.20)(2200) = 25.3 A 110(0.950)
(b)
(
)
(
) )
I1, rms =
(c)
0.950 I1, rms V1, rms = I 2, rms V 2, rms
(
)
(
I1, rms =
2000 by Harcourt, Inc. All rights reserved.
282 Chapter 33 Solutions 33.45 The rms voltage across the transformer primary is N1 V 2, rms N2 so the source voltage is
(
)
N1 V 2, rms N2
V s, rms = I1, rms Rs +
(
)
N 2 V 2, rms = I1, rms N1 RL
The secondary current is
(V2, rms ) ,
RL
so the primary current is
(
)
Then V s, rms =
N 2 V 2, rms Rs N1RL
(
)
+
N1 V 2, rms N2
(
) ) = 5(50.0 ) 80.0 V - 5(25.0 V) =
2(25.0 V) 2 87.5
and
Rs =
N 2 V 2, rms
(
N1RL
)
N1 V 2, rms V s, rms - N2
(
33.46
(a)
V 2, rms =
N2 V1, rms N1
(
)
N 2 V 2, rms 10.0 10 3 V = = = 83.3 120 V N1 V1, rms
(b)
I 2, rms V 2, rms = 0.900 I1, rms V1, rms I 2, rms 10.0 10 3 V = 0.900 Z2 = V 2, rms I 2, rms =
(
)
(
)
I 2, rms = 54.0 mA
(
)
120 V (120 V ) 24.0
(c)
10.0 10 3 V = 185 k 0.054 A
33.47
(a)
R = (4.50 10 - 4 / m)(6.44 10 5 m) = 290 P loss = I rms R = (10.0 A)2 (290 ) = 29.0 kW
2
and
I rms =
5.00 106 W P = = 10.0 A Vrms 5.00 10 5 V
(b) (c)
P loss 2.90 10 4 = = 5.80 10 - 3 P 5.00 106 It is impossible to transmit so much power at such low voltage. Maximum power transfer occurs when load resistance equals the line resistance of 290 , and is (4.50 10 3 V)2 = 17.5 kW, far below the required 5 000 kW 2 2(290 )
Chapter 33 Solutions 283 Vout = Vin XC = XC
2 R + XC 2
33.48
For the filter circuit,
(a)
At f = 600 Hz ,
1 1 = = 3.32 10 4 2 f C 2 (600 Hz) 8.00 10 -9 F
(
)
and
Vout = Vin
(90.0 )2 + (3.32 10 4
3.32 10 4
)
2
1.00
(b)
At f = 600 kHz,
XC =
1 1 = = 33.2 3 2 f C 2 600 10 Hz 8.00 10 -9 F
(
)(
)
and
Vout = Vin
33.2
(90.0 )2 + (33.2 )2
= 0.346
33.49
For this RC high-pass filter,
Vout = Vin
R
2 R + XC 2
(a) (a) When Vout = 0.500, Vin 0.500
2 (0.500 )2 + XC
then
= 0.500 or XC = 0.866 (b)
If this occurs at f = 300 Hz, the capacitance is C= 1 1 = = 6.13 10 - 4 F = 613 F 2 f XC 2 ( 300 Hz)(0.866 )
(b)
With this capacitance and a frequency of 600 Hz, XC = 2 (600 Hz) 6.13 10 - 4 F R
2 R 2 + XC
(c)
(
1
)
= 0.433 Figures for Goal Solution
Vout = Vin
=
0.500
(0.500 )2 + (0.433 )2
= 0.756
2000 by Harcourt, Inc. All rights reserved.
284 Chapter 33 Solutions Goal Solution The RC high-pass filter shown in Figure 33.22 has a resistance R = 0.500 . (a) What capacitance gives an output signal that has one-half the amplitude of a 300-Hz input signal? (b) What is the gain ( V out / Vin ) for a 600-Hz signal? G: It is difficult to estimate the capacitance required without actually calculating it, but we might expect a typical value in the F to pF range. The nature of a high-pass filter is to yield a larger gain at higher frequencies, so if this circuit is designed to have a gain of 0.5 at 300 Hz, then it should have a higher gain at 600 Hz. We might guess it is near 1.0 based on Figure (b) above. O: The output voltage of this circuit is taken across the resistor, but the input sees the impedance of the resistor and the capacitor. Therefore, the gain will be the ratio of the resistance to the impedance. A: Vout = Vin (a) When solving for C gives R R + (1 C )
2 2
V out / Vin = 0.500 C= 1 Vin R -1 Vout
2
=
1 (2 )(300 Hz)(0.500 ) (2.00)2 - 1
= 613 F
(b)
At 600 Hz, we have so
= ( 2 rad) 600 s -1
Vout = Vin
(
)
0.500 (1200 1 rad / s) (613 F )
2
= 0.756
(0.500 )2 +
L : The capacitance value seems reasonable, but the gain is considerably less than we expected. Based o n our calculation, we can modify the graph in Figure (b) to more transparently represent the characteristics of this high-pass filter, now shown in Figure (c). If this were an audio filter, it would reduce low frequency "humming" sounds while allowing high pitch sounds to pass through. A low pass filter would be needed to reduce high frequency "static" noise.
33.50
V1 = I
2 ( r + R )2 + XL ,
and
2 V 2 = I R 2 + XL 2 2 ( r + R ) 2 + X L = 4( R 2 + X L )
r = 20.0
Thus, when V1 = 2 V 2 or
L = 250 mH V1 V2 R =5.00
(25.0 )
2
2 + XL
= 4( 5.00 ) +
2
2 4XL
which gives
XL = 2 f (0.250 H) =
625 - 100 and f = 8.42 Hz 3
Chapter 33 Solutions 285 Vout = Vin (a) R R + ( X L - XC )
2 2
*33.51
At 200 Hz:
1 = 4
(8.00 )2 (8.00 )2 + 400 L -
2
1 400 C
2
2
At 4000 Hz:
(8.00 )
1 2 + 8000 L - = 4(8.00 ) 8000 C 400 L - 1 = - 13.9 400 C 1 = + 13.9 8000 C and [1] [2]
At the low frequency, XL - XC < 0 . This reduces to For the high frequency half-voltage point, Solving Equations (1) and (2) simultaneously gives (b) (c) When XL = XC , XL = XC requires Vout Vout = = 1.00 Vin Vin max f0 = 1 2 LC = 2
8000 L -
C = 54.6 F
L = 580 H
(5.80 10
1
-4
H 5.46 10 - 5 F
)(
)
= 894 Hz
(d) At 200 Hz,
Vout R 1 = = Z 2 Vin
and
XC > X L ,
XL - XC Z
R
or
Vin
Vout
so the phasor diagram is as shown: R 1 = - cos -1 = - cos -1 Z 2 At f 0 , XL = XC At 4000 Hz, Vout R 1 = = Z 2 Vin 1 = 60.0 2 so so and
Vout leads Vin by 60.0 Vout and Vin have a phase difference of 0 X L - XC > 0
XL - XC Z Vin or
Thus, = cos -1 or
R
Vout
Vout lags Vin by 60.0
(e)
At 200 Hz and at 4 kHz,
(Vout, rms ) = ( 21 Vin, rms ) P=
2
2
R
R
=
1 1 V in, max 2 2
(
)
2
R
=
(10.0 V )2 = 8(8.00 )
1.56 W
At f 0 , P =
1 (Vout,rms )2 = (Vin,rms )2 = 2 (Vin,max )2 = (10.0 V)2 =
R
R
R
2(8.00 )
6.25 W
(f)
We take: Q =
0L
R
=
-4 2 f 0 L 2 (894 Hz) 5.80 10 H = = 0.408 8.00 R
(
)
2000 by Harcourt, Inc. All rights reserved.
286 Chapter 33 Solutions Vout = Vin R 1 R2 + C = R 1 R2 + C R2 1 R2 + C
2 2 2
33.52
For a high-pass filter,
( Vout )1 = ( Vin )1
R 1 R2 + C
2
and
( Vout )2 ( Vin )2
Now
( Vin )2 = ( Vout )1
so
( Vout )2 ( Vin )1
=
=
1 1 1+ RC
2
33.53
Rewrite the circuit in terms of impedance as shown in Fig. (b). Find: Vout = V ab = ZR V ab ZR + ZC [1]
Vin
ZR ZC
a
ZC ZR Vout
From Figure (c),
| ZC | ( ZR + ZC ) Vin ZR + ZC | ( ZR + ZC ) | | (ZR + ZC ) ZR + ZC| (ZR + ZC )
-1
b Figure (a) ZC a
So Eq. [1] becomes Vout =
ZR ZC | ( ZR + ZC ) |
[
[
]
]
Vin
Vab b
ZR
Vout
or
Vout Vin
1 1 ZR + ZC ZR + ZC = -1 1 1 (ZR + ZC )ZR + Z + Z + Z C R C
Figure (b)
ZR Vin ZC ZC ZR
a Vab b
Vout ZR ZC ZR = = Vin ZC ( ZC + ZR ) + ZR ( ZR + 2ZC ) 3ZR + ZC + ( ZR )2 ZC Now, ZR = R and ZC = Vout = Vin Vout = Vin -j where j= -1 C
-^ j
Figure (c) R Z
R 1 where we used = -j. j 1 2 3R - j+ R C j C R 1 - R 2 C j 3R - C = R =
C
(3R)
2
1 + - R 2 C C
2
(
1.00 10 3 3.00 10 3
)
2
+ (1592 - 628)
= 0.317
2
Chapter 33 Solutions 287 33.54 The equation for y = mx + b ) is: v(t ) = v(t ) during the first period (using
2( Vmax ) t - Vmax T =
2 ( Vmax ) 1 T [v(t)] dt = T 0 T 2
[ [
( v)2
] ]
ave
0
T2
T t - 1 dt
2
( v)
2
ave
( Vmax )2 T [2t T - 1] 3 =
T 2 3
t=T
=
t=0
( Vmax )2 (+1)3 - (-1)3 = ( Vmax )2 [ ]
6 3
Vrms =
[
( v)2
]
ave
=
( Vmax )2
3
=
Vmax 3
33.55
0 =
1 1 = = 2000 s -1 -6 LC (0.0500 H)(5.00 10 F)
so the operating frequency of the circuit is =
0 = 1000 s -1 2
Using Equation 33.35,
P=
( Vrms )2 R 2
2 R 2 2 + L2 2 - 0
(
)
2
(Q 12.5)
Figure for Goal Solution
P=
(400)2 (8.00)(1000)2 (8.00)2 (1000)2 + (0.0500)2 (1.00 - 4.00) 106
[
]
2
= 56.7 W
Goal Solution A series RLC circuit consists of an 8.00- resistor, a 5.00- F capacitor, and a 50.0-mH inductor. A variable frequency source applies an emf of 400 V (rms) across the combination. Determine the power delivered to the circuit when the frequency is equal to one half the resonance frequency. G: Maximum power is delivered at the resonance frequency, and the power delivered at other frequencies depends on the quality factor, Q. For the relatively small resistance in this circuit, we could expect a high Q = 0 L R . So at half the resonant frequency, the power should be a small
2 fraction of the maximum power, P av, max = Vrms R = ( 400 V ) 8 = 20 kW. 2
O: We must first calculate the resonance frequency in order to find half this frequency. Then the power delivered by the source must equal the power taken out by the resistor. This power can be found 2 from P av = I rms R where I rms = Vrms / Z.
2000 by Harcourt, Inc. All rights reserved.
288 Chapter 33 Solutions
A : The resonance frequency is
f0 =
1 = 2 LC 2
(0.0500 H)( 5.00 10 -6
1
F
)
= 318 Hz
The operating frequency is f = f 0 / 2 = 159 Hz . We can calculate the impedance at this frequency: XL = 2 f L = 2 (159 Hz)(0.0500 H) = 50.0 and XC = 1 1 = = 200 2 f C 2 (159 Hz) 5.00 10 -6 F
(
)
Z = R 2 + (XL - XC )2 = 8.00 2 + (50.0 - 200)2 = 150 So, I rms = Vrms 400 V = = 2.66 A 150 Z
The power delivered by the source is the power dissipated by the resistor: P av = I rms 2 R = (2.66 A)2 (8.00 ) = 56.7 W L : This power is only about 0.3% of the 20 kW peak power delivered at the resonance frequency. The significant reduction in power for frequencies away from resonance is a consequence of the relatively high Q -factor of about 12.5 for this circuit. A high Q is beneficial if, for example, you want to listen to your favorite radio station that broadcasts at 101.5 MHz, and you do not want to receive the signal from another local station that broadcasts at 101.9 MHz.
33.56
The resistance of the circuit is
R=
V 12.0 V I = 0.630 A = 19.0 Vrms 24.0 V = = 42.1 0.570 A I rms
The impedance of the circuit is Z = Z 2 = R 2 + 2 L2 L=
1 1 Z2 - R2 = (42.1)2 - (19.0)2 = 99.6 mH 377
33.57
(a)
When L is very large, the bottom branch carries negligible current. Also, 1/ C will be negligible compared to 200 and 45.0 V/200 = 225 mA flows in the power supply and the top branch. Now 1/ C and L 0 so the generator and bottom branch carry 450 mA
(b)
Chapter 33 Solutions 289 33.58 (a) With both switches closed, the current goes only through generator and resistor. i(t) = Vmax cos t R
(b)
P=
1 ( Vmax ) 2 R
2
(c)
i(t) =
Vmax R 2 + 2 L2
cos [ t + Arctan( L / R)] 1 L- 0 0C 0 = = Arctan R 1 , so 0 C C= 1 02 L
(d) For
We require 0 L =
(e)
At this resonance frequency,
1 1 U = 2 C ( VC ) = 2 C I 2 XC 2
2
Z= R
(f)
2 1 1 U max = 2 CI max XC 2 = 2 C
( Vmax )2
R2
1 = 0 2C 2
( Vmax )2 L
2R 2
(g)
U max =
1 LI 2 2 max
=
1L 2
( Vmax )2
R2
(h) Now = 2 0 =
2 LC
1 L 1 L L - 2 C - 2 C 3 L C So = Arctan = Arctan = Arctan 2R C R R 1 1 2 C 1 2 LC
(i)
Now L =
=
=
0 2
2000 by Harcourt, Inc. All rights reserved.
290 Chapter 33 Solutions
33.59
(a)
As shown in part (b), and
circuit (a) is a high-pass filter circuit (b) is a low-pass filter .
Vin Vout
(b)
For circuit (a),
Vout = Vin As 0,
2 2 RL + XL 2 RL + ( XL - XC ) 2
=
2 RL + ( L)
2 2
Circuit (a)
2 RL + ( L - 1 C )
Vout RLC 0 Vin Vout 1 Vin XC (high-pass filter) 1 C
2 RL
V out
As , Vout = Vin As 0, As ,
Vin
For circuit (b),
2 RL
+ ( X L - XC ) Vout 1 Vin
2
=
+ ( L - 1 C )
2
Circuit (b)
Vout 1 2 0 Vin LC
(low-pass filter)
33.60
(a) (b)
I R, rms =
Vrms 100 V = = 1.25 A 80.0 R the applied voltage as seen in the
IL I
IR
V
The total current will lag phasor diagram at the right. I L, rms =
Vrms 100 V = = 1.33 A XL 2 60.0 s -1 (0.200 H)
(
)
Thus, the phase angle is:
I L, rms 1.33 A = 46.7 = tan -1 = tan -1 1.25 A I R, rms
*33.61
Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day, or $36 per 30 days at 10 per kWh). Suppose the transmission line is at 20 kV. Then I rms =
(20 000)(500 W) P = Vrms 20 000 V
~10 3 A
If the transmission line had been at 200 kV, the current would be only ~10 2 A .
Chapter 33 Solutions 291 L = 2.00 H, C = 10.0 10 6 F, R = 10.0 , v(t) = (100 sin t) (a) The resonant frequency 0 produces the maximum current and thus the maximum power dissipation in the resistor.
33.62
0 =
P=
1 1 = = 224 rad/s LC (2.00)(10.0 10 - 6 ) = (100)2 = 500 W 2(10.0) Vrms 1 R2 + L - C R
2
(b)
( Vmax )2
2R
(c)
I rms =
Vrms = Z
and
( Irms )max =
Vrms R
2 I rms R =
1 2 I rms 2
( )
max
or
( Vrms )2 R = 1 ( Vrms )2 R
Z2 2 R2 1 R2 + L - = 2R 2 C
2
This occurs where Z 2 = 2R 2:
4 L2C 2 - 2L 2C - R 2 2C 2 + 1 = 0
2 -6 2 4
or
-6
L2C 2 4 - (2LC + R 2C 2 ) 2 + 1 = 0 ) + (10.0)2 (10.0 10 - 6 )2 2 + 1 = 0
[(2.00) (10.0 10 ) ] - [2(2.00)(10.0 10
1 = 48 894 = 221 rad/s
]
Solving this quadratic equation, we find that and
2 = 51 130,
48 894
2 = 51 130 = 226 rad/s
33.63
R = 200 , L = 663 mH, C = 26.5 F, = 377 s -1 , Vmax = 50.0 V
L = 250 ,
(a) I max =
1 C = 100 ,
Z = R 2 + ( XL - XC ) = 250
2
Vmax 50.0 V = = 0.200 A 250 Z X L - XC = 36.8 R (V leads I)
= tan -1
(b) (c) (d)
V R, max = I max R = 40.0 V at = 0 VC, max = I max = 20.0 V at = 90.0 C (I leads V) (V leads I)
VL, max = I max L = 50.0 V at = + 90.0
2000 by Harcourt, Inc. All rights reserved.
292 Chapter 33 Solutions Vrms R, Z
2
*33.64
2 P = I rms R =
so
250 W =
(120 V )2
Z
2
( 40.0 ) :
and
Z = R 2 + ( L - 1 C ) 250 =
2
250 =
(120)2 ( 40.0) ( 40.0)
2
1 + 2 f (0.185) - 2 f 65.0 10 - 6
(
)
2
1600 f 2 + 1.1624 f 2 - 2448.5
(
576 000 f 2
)
2
1=
2304 f 2 1600 f 2 + 1.3511 f 4 - 5692.3 f 2 + 5 995 300 6396.3
so
1.3511 f 4 - 6396.3 f 2 + 5 995 300 = 0
f2=
(6396.3)2 - 4(1.3511)( 5 995 300) 2(1.3511)
= 3446.5 or 1287.4
f = 58.7 Hz or 35.9 Hz
33.65
(a)
From Equation 33.39, Let output impedance so that
N1 V1 = N 2 V 2 Z1 = V1 I1 and the input impedance Z2 = But from Eq. 33.40, N1 = N2 Z1 Z2 V 2 I2 I1 V 2 N 2 = = I 2 V1 N1
N1 Z1I1 = N 2 Z2 I 2
So, combining with the previous result we have N1 = N2 Z1 = Z2 8000 = 31.6 8.00
(b)
33.66
IR =
Vrms ; R
IL =
Vrms ; L
IC =
Vrms ( C)-1
2
(a)
2 I rms = I R + (IC - I L )2 = Vrms
1 1 + C - R2 L
(b)
tan =
1 IC - I L 1 1 = Vrms - XC XL Vrms / R IR
1 1 tan = R - XC X L
Chapter 33 Solutions 293
33.67
(a)
I rms = Vrms
1 1 + C - 2 L R
2
Vrms ( Vrms )max when C = f= f= 1 2 LC 1 2 200 10
-3
1 L
H)(0.150 10 - 6 F)
= 919 Hz
(b)
IR = IL =
Vrms 120 V = = 1.50 A 80.0 R Vrms 120 V = = 1.60 A L (374 s -1 )(0.200 H)
IC = Vrms ( C) = (120 V)(374 s -1 )(0.150 10 - 6 F) = 6.73 mA
(c)
2 I rms = I R + (IC - I L )2 = (1.50)2 + (0.00673 - 1.60)2 = 2.19 A
(d)
I - I 0.00673 - 1.60 = tan -1 C L = tan -1 = 46.7 1.50 IR The current is lagging the voltage .
33.68
(a)
tan =
VL I ( L) L = = V R IR R 200 s (0.500 H) L = 173 = tan( 30.0) tan
-1
L
Thus, R =
(
)
Vin
R
V out
(b)
Vout V R = = cos Vin Vin Vout = ( Vin ) cos = (10.0 V ) cos 30.0 = 8.66 V
V = IZ VL = IXL
VR = IR
2000 by Harcourt, Inc. All rights reserved.
294 Chapter 33 Solutions 33.69 (a) XL = XC = 1884 L= XL 1884 = = 0.150 H 2 f 4000 rad s 1 when and f = 2000 Hz C=
(2 f )XC ( 4000 rad s)(1884 )
( 40.0 )2 + (XL - XC )2
1
=
1
= 42.2 nF
XL = 2 f (0.150 H) XC =
(2 f )( 4.22 10
Z ()
-8
F
)
Z=
(b)
f (Hz) X L () XC ()
Impedence,
300 600 800 1000 1500 2000 3000 4000 6000 10000
283 565 754 942 1410 1880 2830 3770 5650 9420
12600 6280 4710 3770 2510 1880 1260 942 628 377
12300 5720 3960 2830 1100 40 1570 2830 5020 9040
33.70
0 =
1 = 1.00 106 rad s LC
0
0.9990 0.9991 0.9993 0.9995 0.9997 0.9999 1.0000 1.0001 1.0003 1.0005 1.0007 1.0009 1.0010
L ( )
999.0 999.1 999.3 999.5 999.7 999.9 1000 1000.1 1000.3 1000.5 1000.7 1000.9 1001
1 C
( )
Z ( )
2.24 2.06 1.72 1.41 1.17 1.02 1.00 1.02 1.17 1.41 1.72 2.06 2.24
P = I 2 R (W)
0.19984 0.23569 0.33768 0.49987 0.73524 0.96153 1.00000 0.96154 0.73535 0.50012 0.33799 0.23601 0.20016
For each angular frequency, we find Z = R 2 + ( L - 1/ C ) then and I = (1.00 V ) / Z P = I 2 (1.00 )
2
1001.0 1000.9 1000.7 1000.5 1000.3 1000.1 1000.0 999.9 999.7 999.5 999.3 999.1 999.0
The full width at half maximum is: f = (1.0005 - 0.9995) 0 = 2 2 1.00 10 3 s -1 = 159 Hz 2
f = while
1.00 R = = 159 Hz 2 L 2 1.00 10 -3 H
(
)
Chapter 33 Solutions 295 Vout = Vin (a) R R 2 + (1 C )
2
33.71
=
R R 2 + (1 2 f C )
2
C Vin Vout
Vout 1 1 = when =R 3 2 Vin C Hence, f = 1 = = 1.84 kHz 2 2 RC 3
Log Gain versus Log Frequency
0 -1 LogV
out / V in
R
(b)
-2 -3 -4 0 1 2 3 Log f 4 5 6
2000 by Harcourt, Inc. All rights reserved.
Chapter 34 Solutions
34.1 Since the light from this star travels at 3.00 108 m/s, the last bit of light will hit the Earth i n 6.44 1018 m = 2.15 1010 s = 680 years. Therefore, it will disappear from the sky in the year 3.00 108 m / s 1999 + 680 = 2.68 10 3 A.D.
34.2
v=
1 1 = c = 0.750c = 2.25 108 m/s 1.78 k 0 e0
34.3
E B =c
or
220 8 B = 3.00 10 ;
so
B = 7.33 107 T = 733 nT
34.4
Emax Bmax = v is the generalized version of Equation 34.13. Bmax = Emax 7.60 10 -3 V / m N m T C m = 3.80 1011 T = 38.0 pT = v (2 / 3)(3.00 108 m / s) V C N s
34.5
(a) (b) (c)
f = c E B =c k=
or or
f (50.0 m) = 3.00 108 m/s 22.0 8 Bmax = 3.00 10 and
so so
f = 6.00 106 Hz = 6.00 MHz Bmax = (73.3 nT)(k)
2 2 = 50.0 = 0.126 m1 rad/s
= 2 f = 2 (6.00 106 s1) = 3.77 107
B = Bmax cos(kx t) = (73.3 nT) cos (0.126x 3.77 107 t)(k)
34.6
= 2 f = 6.00 109 s1 = 1.88 1010 s-1
k= 2 6.00 109 s1 =c = = 20.0 = 62.8 m1 3.00 108 m/s E 300 V/m Bmax = c = = 1.00 T 3.00 108 m/s B = (1.00 T) cos (62.8x 1.88 1010 t)
V cos 62.8x - 1.88 1010 t E = 300 m
(
)
2000 by Harcourt, Inc. All rights reserved.
Chapter 34 Solutions 299
34.7
(a)
B=
100 V/m E = = 3.33 107 T = 0.333 T c 3.00 10 8 m/s
(b)
2 2 = k = = 0.628 m 1.00 10 7 m 1 f= c = 3.00 108 m/s = 4.77 1014 Hz 6.28 107 m
(c)
34.8
E = Emax cos(kx t)
E = Emax sin(kx t)(k) x 2E = Emax cos(kx t)(k 2) x2
We must show: That is, But this is true, because
E = Emax sin(kx t)() t 2E = Emax cos(kx t)()2 t2
E 2E = 0e0 2 2 x t
- k 2 Emax cos( kx - t ) = - 0e0 ( - ) Emax cos( kx - t )
2
( )
1 k2 1 = = 2 = 0 e0 2 f c
2
The proof for the wave of magnetic field is precisely similar.
*34.9
In the fundamental mode, there is a single loop in the standing wave between the plates. Therefore, the distance between the plates is equal to half a wavelength.
= 2L = 2(2.00 m) = 4.00 m
Thus, f = c 3.00 108 m/s = = 7.50 10 7 Hz = 75.0 MHz 4.00 m
*34.10
dA to A = 6 cm 5% =
2
= 12 cm 5%
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300 Chapter 34 Solutions v = f = (0.12 m 5%) 2.45 10 9 s -1 = 34.11 S=I= U Uc A t = V = uc
(
)
2.9 108 m s 5%
I 1000 W/m2 Energy =u= = = 3.33 J/m3 c Unit Volume 3.00 108 m/s
34.12
Sav =
4.00 10 3 W P = = 7.68 W/m 2 4 (4.00 1609 m)2 4 r 2
Emax = 2 0 cSav = 0.0761 V / m Vmax = Emax L = (76.1 mV/m)(0.650 m) = 49.5 mV (amplitude) or 35.0 mV (rms)
34.13
r = ( 5.00 mi )(1609 m / mi ) = 8.04 10 3 m S= 250 10 3 W P = = 307 W/m 2 4 (8.04 10 3 m) 2 4 r 2
Chapter 34 Solutions 301 Goal Solution What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropically with an average power of 250 kW? G: As the distance from the source is increased, the power per unit area will decrease, so at a distance of 5 miles from the source, the power per unit area will be a small fraction of the Poynting vector near the source. O: The Poynting vector is the power per unit area, where A is the surface area of a sphere with a 5-mile radius. A : The Poynting vector is r = ( 5.00 mi )(1609 m / mi ) = 8045 m 250 10 3 W = 3.07 10 -4 W / m 2 (4 )(8045)2
2
In meters,
and the magnitude is S =
L : The magnitude of the Poynting vector ten meters from the source is 199 W/m , on the order of a million times larger than it is 5 miles away! It is surprising to realize how little power is actually received by a radio (at the 5-mile distance, the signal would only be about 30 nW, assuming a 2 receiving area of about 1 cm ). 34.14 I= 100 W = 7.96 W/m2 4 (1.00 m)2
I u = c = 2.65 10 8 J/m3 = 26.5 n J/m3 (a) (b) (c) uE = 2 u = 13.3 n J/m3 uB = 2 u = 13.3 n J/m3 I = 7.96 W/m2
1 1
34.15
Power output = (power input)(efficiency) Thus, Power input = power out 1.00 106 W = = 3.33 106 W eff 0.300
and
A=
3.33 10 6 W P = = 3.33 10 3 m 2 1.00 10 3 W/m 2 I
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302 Chapter 34 Solutions
2 P Bmax c = 4 r 2 2 0
*34.16
I=
Bmax =
P 2 0 4 r 2 c =
(10.0 10 )(2)(4 10 ) = 4 ( 5.00 10 ) ( 3.00 10 )
3 -7 3 2 8
5.16 1010 T
Since the magnetic field of the Earth is approximately 5 10 5 T, the Earth's field is some 100,000 times stronger.
34.17
(a)
P = I 2R = 150 W; A = 2 rL = 2 (0.900 103 m)(0.0800 m) = 4.52 10 4 m2 S= P = 332 kW/m2 (points radially inward) A
(b)
I 0(1.00) B = 0 2 r = = 222 T 2 (0.900 10 3) E= IR 150 V V = L = 0.0800 m = 1.88 kV/m x
EB Note: S = = 332 kW/m2 0
Chapter 34 Solutions 303 34.18 (a) E B = (80.0 i + 32.0 j 64.0 k)(N/C) (0.200 i + 0.0800 j + 0.290 k)T E B= (16.0 + 2.56 18.56)N 2 s/C 2 m = 0
(b)
S=
1 (80.0 i + 32.0 j 64.0 k)(N/C) (0.200 i + 0.0800 j + 0.290 k)T 0 E B= 4 10 7 T m/A (6.40 k 23.2 j 6.40 k + 9.28 i 12.8 j + 5.12 i)10 6 W/m 2 4 107 = 30.9 W/m2 at 68.2 from the +x axis
S=
S = (11.5 i 28.6 j) W/m 2
34.19
We call the current I rms and the intensity I . The power radiated at this frequency is P = (0.0100)(Vrms)Irms = 0.0100(Vrms)2 = 1.31 W R
If it is isotropic, the intensity one meter away is I= 1.31 W c P 2 = = 0.104 W/m2 = Sav = 2 Bmax 0 4 (1.00 m)2 A 2 0 I = c 2 4 10 -7 T m / A 0.104 W / m 2 3.00 10 m / s
8
Bmax =
(
)(
)=
29.5 n
*34.20 (a)
efficiency =
useful power output 700 W 100% = 50.0% 100% = 1400 W total power input
(b)
Sav =
P 700 W = = 2.69 10 5 W m 2 A (0.0683 m )(0.0381 m )
Sav = 269 kW m 2 toward the oven chamber (c) Sav =
2 Emax 2 0 c
T m m W V 3.00 108 2.69 10 5 Emax = 2 4 10 -7 = 1.42 10 4 = 14.2 kV m 2 A s m m
2000 by Harcourt, Inc. All rights reserved.
304 Chapter 34 Solutions Emax 7.00 105 N/C = = 2.33 mT c 3.00 108 m/s
34.21
(a)
Bmax =
2
(b) (c)
(7.00 10 5)2 Emax = 650 MW/m 2 I = 2 c = 0 2(4 10 7 )(3.00 108) I=
P : P = I A = (6.50 108 W/m2) 4 (1.00 10 3 m) 2 = 510 W A
34.22
Power = SA =
2 Emax (4 r 2 ); solving for r , 2 0 c
r=
P 0 c = 2 Emax 2
(100 W) 0 c = 5.16 m 2 (15.0 V / m)2
34.23
(a)
I=
(10.0 10 - 3 )W = 4.97 kW/m2 (0.800 10 - 3 m)2 I 4.97 10 3 J / m 2 s = = 16.6 J/m3 c 3.00 108 m / s
(b)
uav =
34.24
(a) (b) (c)
E = cB = ( 3.00 10 8 m/s)(1.80 10 - 6 T) = 540 V/m uav = B2 (1.80 10 - 6 )2 = = 2.58 J/m3 0 4 10 - 7
Sav = cuav = (3.00 108 )(2.58 10 - 6 ) = 773 W/m2
(d) This is 77.3% of the flux in Example 34.5 . It may be cloudy, or the Sun may be setting.
34.25
S 25.0 For complete absorption, P = c = = 83.3 nPa 3.00 108
*34.26 (a)
P = (Sav )( A) = 6.00 W / m 2 40.0 10 -4 m 2 = 2.40 10 -2 J / s In one second, the total energy U impinging on the mirror is 2.40 102 J. The momentum p transferred each second for total reflection is 2(2.40 102 J) kg m 2U = 1.60 10 10 (each second) p= c = 8 s 3.00 10 m/s
(
)(
)
(b)
dp 1.60 1010 kg m/s F = dt = = 1.60 1010 N 1s
Chapter 34 Solutions 305 (2)(1340 W / m 2 ) = 8.93 10 -6 N / m 2 3.00 108 m / s 2 A = 6.00 105 m2 gives: F = 5.36 N 5.36 N F a = m = 6000 kg = 8.93 10 4 m/s2 d = 2 at 2 2d = a 2 3.84 108 m
-4
34.27
(a)
The radiation pressure is Multiplying by the total area,
(b) (c)
The acceleration is: It will take a time t where:
1
or
t=
(8.93 10
(
)
2
m/s
)
= 9.27 10 5 s = 10.7 days
34.28
The pressure P upon the mirror is
P=
2Sav c P A 2P 2P A = c A c
where A is the cross-sectional area of the beam and
Sav =
The force on the mirror is then
F = PA =
Therefore,
F=
2(100 10 -3 ) = 6.67 1010 N (3 108 )
34.29
I=
E2 P = max r 2 2 0 c P ( 2 0 c ) = 1.90 kN/C
(a)
Emax =
r2
(b)
15 103 J/s 3.00 108 m/s
(1.00 m) = 50.0 pJ
(c)
U 5 1011 p= c = = 1.67 1019 kg m/s 3.00 108
2000 by Harcourt, Inc. All rights reserved.
306 Chapter 34 Solutions 34.30 (a) If P S is the total power radiated by the Sun, and rE and r M are the radii of the orbits of the planets Earth and Mars, then the intensities of the solar radiation at these planets are: IE = PS 2 4 rE
2
and
IM =
PS 2 4 rM
2
Thus, (b)
1.496 1011 m r 2 I M = IE E = 1340 W m 2 = 577 W/m rM 2.28 1011 m
(
)
Mars intercepts the power falling on its circular face:
2 P M = I M RM = 577 W m 2 3.37 106 m
(
) (
)(
)
2
= 2.06 1016 W SM I M = c c
(c)
If Mars behaves as a perfect absorber, it feels pressure P =
and force F = PA =
IM P 2.06 1016 W 2 RM = M = = 6.87 107 N c c 3.00 108 m s
(
)
(d) The attractive gravitational force exerted on Mars by the Sun is 6.67 10 -11 N m 2 kg 2 1.991 10 30 kg 6.42 10 23 kg GMS M M Fg = = = 1.64 10 21 N 2 2 11 rM 2.28 10 m
(
(
)(
)
)(
)
which is ~1013 times stronger
than the repulsive force of (c).
34.31
(a)
The total energy absorbed by the surface is W 1 1 U = 2 I At = 2 750 2 0.500 1.00 m 2 (60.0 s) = 11.3 kJ m
( )
(
)
(b)
The total energy incident on the surface in this time is 2U = 22.5 kJ, with U = 11.3 kJ being absorbed and U = 11.3 kJ being reflected. The total momentum transferred to the surface is p = (momentum from absorption ) + (momentum from reflection ) p=
3 U 2U 3U 3 11.3 10 J + = = = 1.13 10 -4 kg m s c c c 3.00 108 m s
(
)
34.32
Sav =
2 0 Jmax c 8
or
570 =
2 (4 10 -7 )Jmax (3.00 108 ) 8
so
Jmax = 3.48 A/m2
Chapter 34 Solutions 307 J 2 c P = Sav A = 0 max A 8 4 10 -7 (10.0)2 (3.00 108 ) P= (1.20 0.400) = 2.26 kW 8 (b) Sav =
2 0 Jmax c (4 10 -7 (10.0)2 (3.00 108 ) = = 4.71 kW/m2 8 8
34.33
(a)
*34.34
P=
( V )2
R
or P ( V )
2
y
V = ( - )Ey y = Ey l cos V cos (a) (b) (c) so P cos 2
l
eceiving ntenna
= 15.0: = 45.0: = 90.0:
P = P max cos 2 (15.0) = 0.933P max = 93.3% P = P max cos 2 ( 45.0) = 0.500P max = 50.0% P = P max cos 2 (90.0) = 0
34.35
(a)
Constructive interference occurs when d cos = n for some integer n. cos = n
= n = 2n n = 0, 1, 2, . . . / 2 d
strong signal @ = cos1 0 = 90, 270
(b)
Destructive interference occurs when d cos = 2n + 1 : 2 cos = 2n + 1
weak signal @ = cos1 (1) = 0, 180
2000 by Harcourt, Inc. All rights reserved.
308 Chapter 34 Solutions Goal Solution Two radio-transmitting antennas are separated by half the broadcast wavelength and are driven in phase with each other. In which directions are (a) the strongest and (b) the weakest signals radiated? G: The strength of the radiated signal will be a function of the location around the two antennas and will depend on the interference of the waves. O: A diagram helps to visualize this situation. The two antennas are driven in phase, which means that they both create maximum electric field strength at the same time, as shown in the diagram. The radio EM waves travel radially outwards from the antennas, and the received signal will be the vector sum of the two waves. A : (a) Along the perpendicular bisector of the line joining the antennas, the distance is the same to both transmitting antennas. The transmitters oscillate in phase, so along this line the two signals will be received in phase, constructively interfering to produce a maximum signal strength that is twice the amplitude of one transmitter. (b) Along the extended line joining the sources, the wave from the more distant antenna must travel one-half wavelength farther, so the waves are received 180 out of phase. They interfere destructively to produce the weakest signal with zero amplitude. L : Radio stations may use an antenna array to direct the radiated signal toward a highly-populated region and reduce the signal strength delivered to a sparsely-populated area. 34.36
= =
c = 536 m f c = 188 m f
so so
h= h=
= 134 m 4 = 46.9 m 4
34.37
For the proton:
F = ma qvB sin 90.0 = mv 2 R
The period and frequency of the proton's circular motion are therefore: T= 2 1.67 10 -27 kg 2 R 2 m = = = 1.87 10 -7 s v qB 1.60 10 -19 C (0.350 T )
(
(
)
)
f = 5.34 106 Hz .
The charge will radiate at this same frequency, with
=
3.00 108 m s c = = 56.2 m f 5.34 106 Hz
34.38
For the proton, F = ma yields The period of the proton's circular motion is therefore: The frequency of the proton's motion is The charge will radiate electromagnetic waves at this frequency, with
qvB sin 90.0 = T=
mv 2 R
2 R 2 m = v qB
f = 1/ T
=
2 mc c = cT = f qB
Chapter 34 Solutions 309 *34.39 From the electromagnetic spectrum chart and accompanying text discussion, the following identifications are made: Frequency f 2 Hz = 2 100 Hz 2 kHz = 2 10 3 Hz 2 MHz = 2 106 Hz 2 GHz = 2 10 9 Hz 2 THz = 2 1012 Hz 2 PHz = 2 1015 Hz 2 EHz = 2 1018 Hz 2 ZHz = 2 10 21 Hz 2 YHz = 2 10 24 Hz Wavelength, 2 km = 2 10 3 m 2 m = 2 100 m 2 mm = 2 10 -3 m 2 m = 2 10 -6 m 2 nm = 2 10 -9 m 2 pm = 2 10 -12 m 2 fm = 2 10 -15 m 2 am = 2 10 -18 m Wavelength, =c f 150 Mm 150 km 150 m 15 cm 150 m 150 nm 150 pm 150 fm 150 am Frequency f =c 1.5 10 5 Hz 1.5 108 Hz 1.5 1011 Hz 1.5 1014 Hz 1.5 1017 Hz 1.5 10 20 Hz 1.5 10 23 Hz 1.5 10 26 Hz Classification Radio Radio Radio Microwave Infrared Ultraviolet x-ray Gamma ray Gamma Ray Classification Radio Radio Microwave Infrared Ultraviolet/x-ray x-ray/Gamma ray Gamma ray Gamma ray
*34.40 (a) (b)
f=
c
=
3 108 m/s ~ 108 Hz radio wave 1.7 m
1000 pages, 500 sheets, is about 3 cm thick so one sheet is about 6 10 5 m thick f= 3 108 m/s 6 105 m ~ 1013 Hz infrared
*34.41
f=
3.00 108 m/s c = = 5.45 1014 Hz 5.50 107 m
34.42
(a)
c 3.00 108 m/s = f = = 261 m 1150 103/s c 3.00 108 m/s = f = = 3.06 m 98.1 106/s
so
180 m 261 m = 0.690 wavelengths 180 m 3.06 m = 58.9 wavelengths
(b)
so
2000 by Harcourt, Inc. All rights reserved.
310 Chapter 34 Solutions
34.43
(a) (b)
f = c f = c
gives gives
(5.00 1019 Hz) = 3.00 108 m/s:
= 6.00 10 12 m = 6.00 pm
(4.00 109 Hz) = 3.00 108 m/s: = 0.075 m = 7.50 cm
*34.44
Time to reach object = 2 (total time of flight) = 2 (4.00 10 4 s) = 2.00 10 4 s Thus, d = vt = (3.00 10 8 m/s)(2.00 10 4 s) = 6.00 10 4 m = 60.0 km
1
1
34.45
The time for the radio signal to travel 100 km is:
tr =
100 10 3 m = 3.33 10 4 s 3.00 108 m / s 3.00 m 3 s 343 m/s = 8.75 10
The sound wave to travel 3.00 m across the room in:
ts =
Therefore, listeners 100 km away will receive the news before the people in the newsroom by a total time difference of t = 8.75 10 3 s 3.33 10 4 s = 8.41 10 3 s
*34.46
The wavelength of an ELF wave of frequency 75.0 Hz is
=
c 3.00 108 m s = = 4.00 106 m f 75.0 Hz
The length of a quarter-wavelength antenna would be or
L = 1.00 106 m = 1.00 10 3 km 0.621 mi = 621 mi L = (1000 km ) 1.00 km
Thus, while the project may be theoretically possible, it is not very practical.
34.47
(a)
For the AM band, max =
c f min c f max c f min c f max
= = = =
3.00 108 m / s = 556 m 540 10 3 Hz 3.00 108 m / s = 187 m 1600 10 3 Hz 3.00 108 m / s = 3.41 m 88.0 106 Hz 3.00 108 m / s = 2.78 m 108 106 Hz
min =
(b) For the FM band,
max = min =
Chapter 34 Solutions 311
34.48
CH 4 : fmin = 66 MHz
max = 4.55 m min = 4.17 m
fmax = 72 MHz CH 6 : fmin = 82 MHz
max = 3.66 m min = 3.41 m
fmax = 88 MHz CH 8 : fmin = 180 MHz
max = 1.67 m min = 1.61 m
fmax = 186 MHz
34.49
(a) (b)
P = SA = (1340 W/m2)4 (1.496 1011 m)2 = 3.77 1026 W S=
2 cBmax 2 0 2 Emax 2 0 c
so
Bmax =
2 0S = c
2(4 10 -7 N / A 2 )(1340 W / m 2 ) = 3.35 T 3.00 108 m / s
S=
so
Emax = 2 0 cS = 2 4 10 -7 3.00 108 (1340) = 1.01 kV/m
(
)(
)
*34.50
Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sun is 60. Then the target area you fill in the Sun's field of view is (1.7 m )( 0.3 m )( cos 30 ) = 0.4 m2 Now I = P E = : A At E = IAt = 1340 W (0.6)(0.5)(0.4 m2) 3600 s ~ 106 J m2
34.51
(a)
= - dB = - d (BA cos ) = -A
dt dt
d (Bmax cos t cos ) = ABmax (sin t cos ) dt
(t) = 2 f Bmax A sin 2 f t cos = 2 2r 2 f Bmax cos sin 2 f t
Thus,
max = 2 2r 2 f Bmax cos
, where is the angle between the magnetic field and the
normal to the loop. (b) If E is vertical, then B is horizontal, so the plane of the loop should be vertical and the plane should contain the line of sight to the transmitter .
2000 by Harcourt, Inc. All rights reserved.
312 Chapter 34 Solutions GMs m GMs = ( 4 / 3) r 3 R2 R2
34.52
(a)
F grav =
where M s = mass of Sun, r = radius of particle and R = distance from Sun to particle. Since Frad = S r 2 , c Frad 1 3SR 2 1 = F grav r 4cGMs r r= 3SR 2 4cGM s = 3.78 107 m
(b)
From the result found in part (a), when Fgrav = Frad, we have 3 214 W / m 2 3.75 1011 m
r=
4 6.67 10 -11 N m 2 kg 2 1.991 10 30 kg 1500 kg m 3 3.00 108 m s
(
(
)(
)(
)(
)
2
)(
)
34.53
(a)
Bmax = Sav =
Emax 16 c = 6.67 10 T
(b)
2 Emax = 5.31 1017 W/m 2 2 0 c
(c) (d)
P = Sav A = 1.67 1014 W F = PA = Sav A = 5.56 1023 N ( weight of 3000 H atoms!) c
*34.54 (a)
The electric field between the plates is E = V l, directed downward in the figure. The magnetic field between the plate's edges is B = 0 i 2 r counterclockwise. The Poynting vector is: 1 S= EB= 0
i B S + + + ++ + + E - - - - -+ + + ++ + + - - - - i
( V ) i
2 rl
(radially outward)
(b)
The lateral surface area surrounding the electric field volume is ( V ) i A = 2 rl, so the power output is P = SA = ( 2 rl) = (V)i 2 rl
(c)
As the capacitor charges, the polarity of the plates and hence the direction of the electric field is unchanged. Reversing the current reverses the direction of the magnetic field, and therefore the Poynting vector. The Poynting vector is now directed radially inward.
Chapter 34 Solutions 313 *34.55 (a) The magnetic field in the enclosed volume is directed upward, di dB = 0n . with magnitude B = 0 ni and increasing at the rate dt dt The changing magnetic field induces an electric field around any circle of radius r , according to Faraday's Law: di E( 2 r ) = - 0n r2 dt or
B
( )
nr di E=- 0 2 dt
E=
E S
0nr di (clockwise) 2 dt
i
Then,
S=
1 1 0nr di EB= ( 0ni) inward, 0 0 2 dt S=
or the Poynting vector is (b)
0n2 r i di (radially inward) 2 dt
The power flowing into the volume is P = SAlat where Alat is the lateral area perpendicular to S . Therefore, n2 r i di 2 2 di P= 0 ( 2 rl) = 0 n r li dt 2 dt Taking Across to be the cross-sectional area perpendicular to B, the induced voltage between the ends of the inductor, which has N = nl turns, is V = and it is observed that
(c)
=N
di dB di A = nl 0n r 2 = 0 n2 r 2l dt cross dt dt
( )
P = ( V ) i
*34.56 (a)
The power incident on the mirror is:
W 2 P I = IA = 1340 2 (100 m ) = 4.21 107 W m
[
]
The power reflected through the atmosphere is P R = 0.746 4.21 107 W = 3.14 107 W (b) (c) S= PR 3.14 107 W = A 4.00 10 3 m
(
)
(
)
2
= 0.625 W/m2
Noon sunshine in Saint Petersburg produces this power-per-area on a horizontal surface: P N = 0.746 1340 W / m 2 sin 7.00 = 122 W / m 2 The radiation intensity received from the mirror is 0.625 W m 2 122 W m 2 100% = 0.513% of that from the noon Sun in January.
(
)
2000 by Harcourt, Inc. All rights reserved.
314 Chapter 34 Solutions
1
34.57
2 u = 2 e0Emax
(Equation 34.21)
Emax =
2u = 95.1 mV/m e0
*34.58
The area over which we model the antenna as radiating is the lateral surface of a cylinder, A = 2 r l = 2 4.00 10 -2 m (0.100 m ) = 2.51 10 -2 m 2 (a) The intensity is then: S= P 0.600 W = = 23.9 W m 2 A 2.51 10 -2 m 2
-3 4 2 W mW mW 1.00 10 W 1.00 10 cm = 0.570 = 5.70 2 cm 2 1.00 mW m cm 2 1.00 m 2
(
)
(b)
The standard is:
0.570
While it is on, the telephone is over the standard by
23.9 W m 2 = 4.19 times 5.70 W m 2
34.59
(a)
Bmax =
Emax 175 V/m 7 c = 3.00 108 m/s = 5.83 10 T
k=
2 2 = (0.0150 m) = 419 rad/m
= kc = 1.26 1011 rad/s
Since S is along x, and E is along y, B must be in the z direction . EmaxBmax = 40.6 W/m2 2 0 (That is S E B.)
(b)
Sav =
(c)
2S Pr = c = 2.71 107 N/m2 a= 2.71 10 -7 N / m 2 0.750 m 2 F PA = = = 4.06 10 -7 m / s 2 m m 0.500 kg
(d)
(
)(
)
Chapter 34 Solutions 315 At steady-state, P in = P out and the power radiated out is P out = eAT 4 . W W AT 4 0.900 1000 2 A = (0.700) 5.67 10 -8 2 m m K4 900 W m 2 T= 0.700 5.67 10 -8 W m 2 K 4
14
*34.60 (a)
Thus,
or
(
)
= 388 K = 115C
(b)
The box of horizontal area A , presents projected area A sin 50.0 perpendicular to the sunlight. Then by the same reasoning, W W AT 4 0.900 1000 2 A sin 50.0 = (0.700) 5.67 10 -8 2 m m K4 900 W m 2 sin 50.0 T= 0.700 5.67 10 -8 W m 2 K 4
or
(
(
)
)
14
= 363 K = 90.0 C
34.61
(a)
I F P= A = c F= IA P 100 J / s = = = 3.33 10 -7 N = (110 kg )a c c 3.00 108 m / s x = 2 at 2
1
a = 3.03 109 m/s2
and
t=
2x = 8.12 10 4 s = 22.6 h a
(b)
0 = (107 kg)v (3.00 kg)(12.0 m/s v) = (107 kg)v 36.0 kg m/s + (3.00 kg)v 36.0 v = 110 = 0.327 m/s t = 30.6 s
2000 by Harcourt, Inc. All rights reserved.
316 Chapter 34 Solutions Goal Solution An astronaut, stranded in space 10.0 m from his spacecraft and at rest relative to it, has a mass (including equipment) of 110 kg. Since he has a 100-W light source that forms a directed beam, he decides to use the beam as a photon rocket to propel himself continuously toward the spacecraft. (a) Calculate how long it takes him to reach the spacecraft by this method. (b) Suppose, instead, he decides to throw the light source away in a direction opposite the spacecraft. If the mass of the light source has a mass of 3.00 kg and, after being thrown, moves at 12.0 m/s relative to the recoiling astronaut , how long does it take for the astronaut to reach the spacecraft? G: Based on our everyday experience, the force exerted by photons is too small to feel, so it may take a very long time (maybe days!) for the astronaut to travel 10 m with his "photon rocket." Using the momentum of the thrown light seems like a better solution, but it will still take a while (maybe a few minutes) for the astronaut to reach the spacecraft because his mass is so much larger than the mass of the light source. O: In part (a), the radiation pressure can be used to find the force that accelerates the astronaut toward the spacecraft. In part (b), the principle of conservation of momentum can be applied to find the time required to travel the 10 m. A : (a) Light exerts on the astronaut a pressure P = F A = S c , and a force of F= SA = c = 100 J / s = 3.33 10 -7 N 3.00 108 m / s
By Newton's 2nd law,
a=
F 3.33 10 -7 N = = 3.03 10 -9 m / s 2 m 110 kg
1 This acceleration is constant, so the distance traveled is x = 2 at 2 , and the amount of time it travels is
t=
2x = a
2(10.0 m ) = 8.12 10 4 s = 22.6 h 3.03 10 -9 m / s 2
(b) Because there are no external forces, the momentum of the astronaut before throwing the light is the same as afterwards when the now 107-kg astronaut is moving at speed v towards the spacecraft and the light is moving away from the spacecraft at (12.0 m / s - v ) . Thus, pi = p f gives 0 = (107 kg )v - ( 3.00 kg )(12.0 m / s - v ) 0 = (107 kg )v - ( 36.0 kg m / s) + ( 3.00 kg )v v= t= 36.0 = 0.327 m / s 110 x 10.0 m = = 30.6 s v 0.327 m / s
L : Throwing the light away is certainly a more expedient way to reach the spacecraft, but there is not much chance of retrieving the lamp unless it has a very long cord. How long would the cord need to be, and does its length depend on how hard the astronaut throws the lamp? (You should verify that the minimum cord length is 367 m, independent of the speed that the lamp is thrown.)
Chapter 34 Solutions 317 2(0.380)S W 2S that is reflected exerts a pressure P1 = r = c c m2 P2 = Sa (0.620)S = c c
34.62
The 38.0% of the intensity S = 1340
The absorbed light exerts pressure Altogether the pressure at the subsolar point on Earth is (a) Ptot = P1 + P2 = 1.38S 1.38(1340 W/m2) = = 6.16 10 6 Pa c 3.00 108 m/s
(b)
1.01 10 5 N / m 2 Pa = = 1.64 1010 times smaller than atmospheric pressure Ptot 6.16 10 -6 N / m 2
34.63
Think of light going up and being absorbed by the bead which presents a face area r b . S I The light pressure is P = c = . c (a) (b) I r b 4 3 Fl = c = mg = 3 r b g
2
2
and
I=
4 gc 3m 3 4
1/3
= 8.32 10 7 W/m 2
P = IA = (8.32 10 7 W/m2) (2.00 103 m)2 = 1.05 kW
34.64
Think of light going up and being absorbed by the bead which presents face area r b . If we take the bead to be perfectly absorbing, the light pressure is P = Fl = F g I= Flc F g c m gc = = A A rb 2 m = V m
4 rb 3 3
2
Sav I Fl = = c A c
(a)
so
From the definition of density, =
so
1 = rb
(
4 3
m
)
1/3
Substituting for rb , I =
2/3 1/3 m gc 4 2/3 4 gc 3m 4 m = gc = 3m 3 3 4
1/3
(b)
P = IA =
r 24gc 3m 1/3 3 4
2000 by Harcourt, Inc. All rights reserved.
318 Chapter 34 Solutions 34.65 The mirror intercepts power P = I1A1 = (1.00 10 3 W/m2) (0.500 m)2 = 785 W In the image, (a) I2 = P 785 W = = 625 kW/m2 A2 (0.0200 m )2
2
(b)
E max I 2 = 2 c 0 Bmax =
so
Emax = (20c I 2 )1/2 = [2(4 107)(3.00 108)(6.25 105)]1/2= 21.7 kN/C
Emax c = 72.4 T
(c)
0.400 P t = mc T J 0.400(785 W)t = (1.00 kg) 4186 kg C (100C 20.0C) t= 3.35 105 J = 1.07 103 s = 17.8 min 314 W
34.66
(a)
=
c 3.00 108 m s = = 1.50 cm f 20.0 10 9 s -1
(b)
J 1.00 10 -9 s = 25.0 10 -6 J = 25.0 J U = P ( t ) = 25.0 10 3 s uav = U U U 25.0 10 -6 J = = = V r 2 l r 2 c( t ) (0.0600 m )2 3.00 108 m s 1.00 10 -9 s
(
)
(c)
( ) ( ) (
(
)(
)
uav = 7.37 10 -3 J m 3 = 7.37 mJ m 3 2uav = e0 2 7.37 10 -3 J m 3 8.85 10
-12 2
(d)
Emax =
C N m2
)
= 4.08 10 4 V m = 40.8 kV/m
Bmax =
Emax 4.08 10 4 V m = = 1.36 10 -4 T = 136 T c 3.00 108 m s J c uav S A= (0.0600 m )2 = 8.33 10 -5 N = 83.3 N A = uav A = 7.37 10 -3 c c m3
(e)
F = PA =
Chapter 34 Solutions 319
34.67
(a)
On the right side of the equation,
(b)
F = ma = qE
or
(C N m )(m s) C)(100 N C) qE (1.60 10 a= = =
2 2 3 -19
C2 m s2
(
)
2
=
N m 2 C2 m 2 s3 N m J = = =W s s C2 s 4 m 3 1.76 1013 m s 2
13 2 3
m
9.11 10 -31 kg
The radiated power is then: v F = mar = m = qvB r The proton accelerates at
2
1.60 10 1.76 10 q2 a2 P= = 3 6 e0 c 6 8.85 10 -12 3.00 108
(
-19 2
(
)( )( )
) )
= 1.75 10 - 27 W
(c)
so
v=
qBr m
1.60 10 -19 (0.350) (0.500) v 2 q 2B2 r a= = = = 5.62 1014 m s 2 -27 2 r m2 1.67 10
2
(
2
(
)
The proton then radiates
1.60 10 5.62 10 q2 a2 P= = 6 e0 c 3 6 8.85 10 -12 3.00 108
(
-19 2
(
)( )(
14 2 3
) )
-24 W = 1.80 10
34.68
P=
S Power P 60.0 W = = = = 6.37 107 Pa c Ac 2 rlc 2 (0.0500 m)(1.00 m)(3.00 108 m \ s)
34.69
F = PA =
SA ( P / A)A P = = , c c c
Pl l = F = , 2 2c
and
=
Therefore,
=
3.00 10 -3 (0.0600) Pl = = 3.00 10 2 deg 2c 2 3.00 108 1.00 10 -11
(
(
)(
)
)
*34.70
We take R to be the planet's distance from its star. projected area r
2
The planet, of radius r , presents a
perpendicular to the starlight.
It radiates over area 4 r 2 .
At steady-state, P in = P out : e I in r 2 = e 4 r 2 T 4 6.00 10 23 W 2 2 4 e r = e 4 r T 4 R2
( ) )
(
)
( )
(
so that
6.00 10 23 W = 16 R 2 T 4
R=
6.00 10 23 W 6.00 10 23 W = = 4.77 10 9 m = 4.77 Gm 4 4 16 T 16 5.67 10 -8 W m 2 K 4 ( 310 K )
(
)
2000 by Harcourt, Inc. All rights reserved.
320 Chapter 34 Solutions E2 2 0c
34.71
The light intensity is
I = Sav =
The light pressure is
P=
S E2 = = 1 e0 E 2 c 2 0c 2 2 and a= e0E 2 A 2m
For the asteroid,
PA = m a
34.72 (a)
f = 90.0 MHz,
Emax = 2.00 10 - 3 V / m = 200 mV/m
=
T=
c = 3.33 m f 1 = 1.11 10 -8 s = 11.1 ns f Emax = 6.67 10 -12 T = 6.67 pT c B = (6.67 pT )k cos 2 t x - 3.33 m 11.1 ns
Bmax =
(b)
x t E = (2.00 mV / m) cos 2 - j 3.33 m 11.1 ns I=
(c)
2 Emax (2.00 10 - 3 )2 = = 5.31 10 - 9 W / m 2 2 0 c 2(4 10 - 7 )(3.00 108 )
(d)
I = cuav P=
so
uav = 1.77 10 -17 J / m 3
(e)
2I (2)(5.31 10 - 9 ) = = 3.54 10 -17 Pa c 3.00 108
Chapter 35 Solutons
35.1
The Moon's radius is 1.74 10 6 m and the Earth's radius is 6.37 10 6 m. The total distance traveled by the light is: d = 2(3.84 10 8 m 1.74 10 6 m 6.37 10 6 m) = 7.52 10 8 m This takes 2.51 s, so v = 7.52 10 8 m = 2.995 10 8 m/s = 299.5 Mm/s 2.51 s
35.2
x = ct ;
2(1.50 10 8 km)(1000 m/km) x c= t = (22.0 min)(60.0 s/min)
= 2.27 10 8 m/s = 227 Mm/s
35.3
The experiment is most convincing if the wheel turns fast enough to pass outgoing light through one notch and returning light through the next: t = 2l c 2l = t = c so
=
c (2.998 10 8 )[2 / (720)] = = 114 rad/s 2l 2(11.45 10 3 )
The returning light would be blocked by a tooth at one-half the angular speed, giving another data point.
35.4
(a)
For the light beam to make it through both slots, the time for the light to travel the distance d must equal the time for the disk to rotate through the angle , if c is the speed of light, d = , so c c= d
(b)
We are given that d = 2.50 m ,
=
1.00 rad = 2 .91 10 - 4 rad, 60.0 180
= 5555
rev 2 rad = 3.49 10 4 rad s s 1.00 rev
4 d ( 2.50 m ) 3.49 10 rad s c= = = 3.00 108 m s = 300 Mm/s -4 2.91 10 rad
(
)
35.5
Using Snell's law, sin 2 =
n1 sin 1 n2
2 = 25.5
2 =
1
n1
= 442 nm
2000 by Harcourt, Inc. All rights reserved.
Chapter 35 Solutions
323
35.6
(a)
f=
3.00 10 8 m/s c = = 4.74 10 14 Hz 6.328 10 - 7 m
(b)
glass =
vglass =
air
n
=
632.8 nm = 422 nm 1.50
(c)
3.00 10 8 m/s cair = = 2.00 10 8 m/s = 200 Mm/s 1.50 n
35.7
n1 sin 1 = n2 sin 2 sin 1 = 1.333 sin 45.0 sin 1 = (1.33)(0.707) = 0.943
1 = 70.5 19.5 above the horizon
Figure for Goal Solution
Goal Solution An underwater scuba diver sees the Sun at an apparent angle of 45.0 from the vertical. What is the actual direction of the Sun? G: The sunlight refracts as it enters the water from the air. Because the water has a higher index of refraction, the light slows down and bends toward the vertical line that is normal to the interface. Therefore, the elevation angle of the Sun above the water will be less than 45 as shown in the diagram to the right, even though it appears to the diver that the sun is 45 above the horizon. We can use Snell's law of refraction to find the precise angle of incidence. Snell's law is: which gives n1 sin 1 = n2 sin 2 sin 1 = 1.333 sin 45.0 sin 1 = (1.333)(0.707) = 0.943 The sunlight is at 1 = 70.5 to the vertical, so the Sun is 19.5 above the horizon. L: The calculated result agrees with our prediction. When applying Snell's law, it is easy to mix up the index values and to confuse angles-with-the-normal and angles-with-the-surface. Making a sketch and a prediction as we did here helps avoid careless mistakes.
O:
A:
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324 *35.8 (a)
Chapter 35 Solutions n1 sin 1 = n2 sin 2 1.00 sin 30.0 = n sin 19.24 n = 1.52 f= 3.00 10 8 m/s c = = 4.74 10 14 Hz in air and in syrup. 6.328 10 - 7 m c 3.00 10 8 m/s = = 1.98 10 8 m/s = 198 Mm/s 1.52 n v 1.98 10 8 m/s = = 417 nm f 4.74 10 14 / s
(c)
(d)
v=
(b)
=
35.9
(a)
Flint Glass: v =
3.00 108 m s c = = 1.81 108 m s = 181 Mm/s 1.66 n 3.00 108 m s c = = 2 .25 108 m s = 225 Mm/s n 1.333 v= 3.00 108 m s c = = 1.36 108 m s = 136 Mm/s n 2.20
(b)
Water:
v=
(c)
Cubic Zirconia:
35.10
n1 sin 1 = n2 sin 2 ; c n2 = 1.90 = v ;
1.333 sin 37.0 = n2 sin 25.0
c v = 1.90 = 1.58 10 8 m/s = 158 Mm/s
35.11
n1 sin 1 = n2 sin 2 ;
n 1 sin 1 2 = sin1 n 2
(1.00)(sin 30) 2 = sin -1 = 19.5 1.50
2 and 3 are alternate interior angles formed by the ray cutting parallel normals. So, 3 = 2 = 19.5 .
1.50 sin 3 = (1.00) sin 4
4 = 30.0
Chapter 35 Solutions
325
35.12
(a)
Water = Glass =
0 436 nm = = 327 nm n 1.333 0 436 nm = = 287 nm n 1.52
(b)
*35.13
sin 1 = nw sin 2 sin 2 = 1 1 sin 1 = sin(90.0 - 28.0) = 0.662 1.333 1.333
2 = sin -1 0.662 = 41.5
h= 3.00 m d = = 3.39 m tan 41.5 tan 2
35.14
(a)
From geometry, 1.25 m = d sin 40.0 so d = 1.94 m
(b)
50.0 above horizontal incident ray
, or parallel to the
*35.15
The incident light reaches the left-hand mirror at distance (1.00 m) tan 5.00 = 0.0875 m above its bottom edge. The reflected light first reaches the right-hand mirror at height 2(0.0875 m) = 0.175 m It bounces between the mirrors with this distance between points of contact with either. 1.00 m Since 0.175 m = 5.72, the light reflects five times from the right-hand mirror and six times from the left.
2000 by Harcourt, Inc. All rights reserved.
326
Chapter 35 Solutions n1 sin 1 = n2 sin 2 1.00 sin 30.0 = 1.50 sin 2
*35.16
At entry,
or
2 = 19.5
The distance h the light travels in the medium is given by cos 2 = (2.00 cm) h or h= (2.00 cm) = 2.12 cm cos 19.5
The angle of deviation upon entry is The offset distance comes from sin = d : h
= 1 - 2 = 30.0 - 19.5 = 10.5
d = (2.21 cm) sin 10.5 = 0.388 cm
*35.17
The distance, h, traveled by the light is h = The speed of light in the material is Therefore, t=
2.00 cm = 2.12 cm cos 19.5 v= c 3.00 108 m/s = = 2.00 108 m/s 1.50 n
h 2.12 10 - 2 m = = 1.06 10 - 10 s = 106 ps v 2.00 108 m/s
*35.18
Applying Snell's law at the air-oil interface, nair sin = noil sin 20.0 yields = 30.4
Applying Snell's law at the oil-water interface nw sin = noil sin 20.0 yields = 22.3
*35.19
time difference = (time light for to travel 6.20 m in ice) (time to travel 6.20 m in air) t = 6.20 m 6.20 m - vice c but v= c n
1.309 1 (6.20 m) = t = (6.20 m) - (0.309) = 6.39 10 -9 s = 6.39 ns c c c
Chapter 35 Solutions *35.20
327
Consider glass with an index of refraction of 1.5, which is 3 mm thick The speed of light i n the glass is 3 108 m/s = 2 108 m/s 1.5 The extra travel time is 3 10 -3 m 3 10 -3 m - ~ 1011 s 2 108 m / s 3 108 m / s 600 nm = 400 nm in glass, 1.5
For light of wavelength 600 nm in vacuum and wavelength
the extra optical path, in wavelengths, is
3 10 - 3 m 3 10 - 3 m - ~ 103 wavelengths 4 10 -7 m 6 10 -7 m
*35.21 (a)
Refraction proceeds according to
(1.00) sin 1 = (1.66) sin 2
v1 cos 1 = v2 cos 2
(1)
For the normal component of velocity to be constant, or We multiply Equations (1) and (2), obtaining: or
(c ) cos 1 = (c 1.66) cos 2
sin 1 cos 1 = sin 2 cos 2 sin 21 = sin 2 2
(2)
The solution 1 = 2 = 0 does not satisfy Equation (2) and must be rejected. solution is 21 = 180 - 2 2 or 2 = 90.0 - 1 . Then Equation (1) becomes:
The physical
sin 1 = 1.66 cos 1 , or tan 1 = 1.66 which yields (b)
1 = 58.9
Light entering the glass slows down and makes a smaller angle with the normal. Both effects reduce the velocity component parallel to the surface of the glass, so that component cannot remain constant, or will remain constant only in the trivial case 1 = 2 = 0
35.22
See the sketch showing the path of the light and are angles of incidence at ray. mirrors 1 and 2. For triangle abca, 2 + 2 + = 180 or
= 180 - 2( + )
(1)
Now for triangle bcdb,
(90.0 - ) + (90.0 - ) + = 180
or
=+
(2)
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328
Chapter 35 Solutions
Substituting Equation (2) into Equation (1) gives
= 180 - 2
Note: From Equation (2), = - . Thus, the ray will follow a path like that shown only if < . For > , is negative and multiple reflections from each mirror will occur before the incident and reflected rays intersect.
35.23
Let n(x) be the index of refraction at distance x below the top of the atmosphere and n( x = h) = n be its value at the planet surface. Then, n( x ) = 1.000 + (a) n - 1.000 x h
The total time required to traverse the atmosphere is t= t=
h dx
0
v
=
h
0
n( x ) 1 h h (n - 1.000) h 2 h n + 1.000 n - 1.000 x dx = + dx = 1.000 + 2 = c c 0 h c 2 c ch
20.0 10 3 m 1.005 + 1.000 = 66.8 s 2 3.00 108 m s Thus, the time in the
(b)
The travel time in the absence of an atmosphere would be h / c . presence of an atmosphere is n + 1.000 = 1.0025 times larger or 0.250% longer . 2
35.24
Let n(x) be the index of refraction at distance x below the top of the atmosphere and n( x = h) = n be its value at the planet surface. Then, n( x ) = 1.000 + (a) n - 1.000 x h
The total time required to traverse the atmosphere is t=
h dx
0
v
=
hn
0
( x) dx = 1
c
1.000 + c 0
h
h (n - 1.000) h 2 h n + 1.000 n - 1.000 x dx = + 2 = c h c 2 ch Thus, the time in the
(b)
The travel time in the absence of an atmosphere would be h / c . presence of an atmosphere is n + 1.000 times larger 2
Chapter 35 Solutions 35.25 From Fig. 35.20 Then nv = 1.470 at 400 nm (1.00)sin = 1.470 sin v sin sin - sin -1 1.458 1.470 and and nr = 1.458 at 700 nm (1.00)sin = 1.458 sin r
329
r - v = r - v = sin -1
= sin -1
sin 30.0 sin 30.0 - sin -1 = 0.171 1.458 1.470
35.26
n1 sin 1 = n2 sin 2
so
n sin 1 2 = sin -1 1 n2
(1.00)(sin 30.0) 2 = sin -1 = 19.5 1.50
3 = ([(90.0 - 19.5) + 60.0] - 180) + 90.0 = 40.5
n3 sin 3 = n4 sin 4 so n sin 3 -1 (1.50)(sin 40.5) 4 = sin -1 3 = 77.1 = sin n4 1.00
35.27
Taking to be the apex angle and min to be the angle of minimum deviation, from Equation 35.9, the index of refraction of the prism material is sin n= Solving for min , + min 2 sin( 2)
min = 2 sin -1 n sin - = 2 sin -1[( 2.20) sin ( 25.0)] - 50.0 = 86.8 2
35.28 (a) (b)
n(700 nm) = 1.458 (1.00) sin 75.0 = 1.458 sin 2; Let 3 + = 90.0, So (c) (d)
2 = 41.5
then + + 60.0 = 180
2 + = 90.0 ;
60.0 - 2 - 3 = 0 60.0 - 41.5 = 3 = 18.5
1.458 sin 18.5 = 1.00 sin 4
4 = 27.6
= (1 - 2 ) + [ - (90.0 - 4 )] = 75.0 - 41.5 + (90.0 - 18.5) - (90.0 - 27.6) = 42.6
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330
Chapter 35 Solutions sin 1 n sin 50.0 = 27.48 1.66
35.29
For the incoming ray, Using the figure to the right,
sin 2 =
( 2 )violet = sin -1 ( 2 )red = sin -1
sin 50.0 = 28.22 1.62 and sin 4 = nsin 3
For the outgoing ray,
3 = 60.0 2
( 4 )violet = sin -1[1.66 sin 32.52] = 63.17 ( 4 )red = sin -1[1.62 sin 31.78] = 58.56 The dispersion is the difference 4 = ( 4 )violet - ( 4 )red = 63.17 58.56 = 4.61
sin 35.30 n=
+ min 2 sin( 2) Then, using the small angle
+ min is also a small angle. 2 approximation ( sin when << 1 rad), we have: For small , min so n
( + min )
2
2
=
+ min
or
min (n - 1)
where is in radians.
35.31
At the first refraction,
(1.00) sin 1 = n sin 2
The critical angle at the second surface is given by n sin 3 = 1.00 , or 3 = sin -1 1.00 = 41.8. 1.50
But, 2 = 60.0 - 3 . Thus, to avoid total internal reflection at the second surface (i.e., have 3 < 41.8), it is necessary that 2 > 18.2. Since sin 1 = n sin 2 , this requirement becomes sin 1 > (1.50) sin (18.2) = 0.468 , or 1 > 27.9
Chapter 35 Solutions 35.32 At the first refraction, (1.00) sin 1 = n sin 2 . The critical angle at the second surface is given by n sin 3 = 1.00 , or
331
3 = sin -1 (1.00 n)
But (90.0 - 2 ) + (90.0 - 3 ) + = 180, which gives 2 = - 3 . necessary that 2 > - sin -1 (1.00 n) . 1.00 sin 1 > n sin - sin -1 n Thus, to have 3 < sin -1 (1.00 n) and avoid total internal reflection at the second surface, it is Since sin 1 = n sin 2 , this requirement becomes or 1.00 1 > sin -1 n sin - sin -1 n
Through the application of trigonometric identities,
1 > sin -1 n2 - 1 sin - cos
35.33
n= tan
sin( + ) sin( / 2) sin ( ) = 1.544 - 5 5 cos
1 2
so and
1.544 sin
( ) = sin(5 + ) = cos( ) sin 5 + sin( ) cos 5
1 2 1 2 1 2 1 2
= 18.1
*35.34
Note for use in every part: so At the first surface is At exit, the deviation is
+ (90.0 - 2 ) + (90.0 - 3 ) = 180
3 = - 2 = 1 - 2 = 4 - 3
The total deviation is therefore = + = 1 + 4 - 2 - 3 = 1 + 4 - (a) At entry: n1 sin 1 = n2 sin 2 or
2 = sin -1
sin 48.6 = 30.0 1.50
Thus, 3 = 60.0 - 30.0 = 30.0 At exit: 1.50 sin 30.0 = 1.00 sin 4 or
4 = sin -1[1.50 sin ( 30.0)] = 48.6
so the path through the prism is symmetric when 1 = 48.6 . (b) (c)
= 48.6 + 48.6 - 60.0 = 37.2
At entry: At exit: sin 2 = sin 45.6 2 = 28.4 1.50
3 = 60.0 - 28.4 = 31.6 = 45.6 + 51.7 - 60.0 = 37.3 3 = 60.0 - 31.5 = 28.5 = 51.6 + 45.7 - 60.0 = 37.3
sin 4 = 1.50 sin ( 31.6) 4 = 51.7 sin 2 =
(d) At entry:
sin 51.6 2 = 31.5 1.50 At exit: sin 4 = 1.50 sin (28.5) 4 = 45.7
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332 35.35 (a) (b) (c)
Chapter 35 Solutions n sin = 1. From Table 35.1,
1 = sin -1 = 24.4 2.419 1 = sin -1 = 37.0 1.66 1 = sin -1 = 49.8 1.309
35.36
sin c =
n2 ; n1
n c = sin -1 2 n1 1.333 c = sin -1 = 33.4 2.419 1.333 c = sin -1 = 53.4 1.66
(a) (b) (c)
Diamond: Flint glass:
Ice: Since n 2 > n 1, there is no critical angle .
35.37
sin c =
n2 n1
(Equation 35.10)
n2 = n1 sin 88.8 = (1.0003)(0.9998) = 1.000 08
*35.38
sin c =
nair 1.00 = = 0.735 npipe 1.36
c = 47.3
Geometry shows that the angle of refraction at the end is
r = 90.0 c = 90.0 47.3 = 42.7
Then, Snell's law at the end, gives 1.00 sin = 1.36 sin 42.7
= 67.2
35.39
For total internal reflection,
n1 sin 1 = n2 sin 90.0 (1.50) sin 1 = (1.33)(1.00) or
1 = 62.4
Chapter 35 Solutions
333
35.40
To avoid internal reflection and come out through the vertical face, light inside the cube must have
3 < sin -1 (1/ n)
So But
2 > 90.0 - sin -1 (1/ n) 1 < 90.0
and n sin 2 < 1
In the critical case, sin -1 (1/ n) = 90.0 - sin -1 (1/ n) 1/n = sin 45.0 n = 1.41
35.41
From Snell's law,
n1 sin 1 = n2 sin 2
At the extreme angle of viewing, 2 = 90.0 (1.59)(sin 1) = (1.00) sin 90.0 So
1 = 39.0
Therefore, the depth of the air bubble is rp rd <d< tan 1 tan 1 or 1.08 cm < d < 1.17 cm
*35.42 (a)
sin 2 v2 = and 2 = 90.0 at the critical angle sin 1 v1 sin 90.0 1850 m s = so c = sin -1 0.185 = 10.7 sin c 343 m s
(b) (c)
Sound can be totally reflected if it is traveling in the medium where it travels slower: air Sound in air falling on the wall from most directions is 100% reflected , so the wall is a good mirror.
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334
Chapter 35 Solutions For plastic with index of refraction n 1.42 surrounded by air, the critical angle for total internal reflection is given by 1 1 = 44.8 c = sin -1 sin -1 n 1.42 In the gasoline gauge, skylight from above travels down the plastic. The rays close to the vertical are totally reflected from both the sides of the slab and from facets at the lower end of the plastic, where it is not immersed in gasoline. This light returns up inside the plastic and makes it look bright. Where the plastic is immersed in gasoline, with index of refraction about 1.50, total internal reflection should not happen. The light passes out of the lower end of the plastic with little reflected, making this part of the gauge look dark. To frustrate total internal reflection in the gasoline, the index of refraction of the plastic should be n < 2.12 , since
*35.43
c = sin -1
( ) = 45.0 .
1.50 2.12
*35.44
Assume the lifeguard's path makes angle 1 with the northsouth normal to the shoreline, and angle 2 with this normal in the water. By Fermat's principle, his path should follow the law of refraction: sin 1 v1 7.00 m s = = = 5.00 sin 2 v2 1.40 m s or
2 = sin -1
sin 1 5
The lifeguard on land travels eastward a distance x = (16.0 m ) tan 1. Then in the water, h e travels 26.0 m - x = (20.0 m ) tan 2 further east. Thus, 26.0 m = (16.0 m ) tan 1 + (20.0 m ) tan 2 or sin 1 26.0 m = (16.0 m ) tan 1 + ( 20.0 m ) tan sin -1 5
We home in on the solution as follows:
1 (deg) right-hand side
50.0 22.2 m
60.0 31.2 m
54.0 25.3 m
54.8 54.81 25.99 m 26.003 m
The lifeguard should start running at 54.8 east of north .
*35.45
Let the air and glass be medium 1 and 2, respectively. By Snell's law, or But the conditions of the problem are such that 1 = 2 2 . We now use the double-angle trig identity suggested. or Thus, 2 = 38.7 and
n2 sin 2 = n1 sin 1 1.56 sin 2 = sin 1 1.56 sin 2 = sin 2 2
1.56 sin 2 = 2 sin 2 cos 2 cos 2 = 1.56 = 0.780 2
1 = 2 2 = 77.5
Chapter 35 Solutions
335
*35.46 (a)
1 = 1 = 30.0
n1 sin 1 = n2 sin 2 (1.00) sin 30.0 = 1.55 sin 2
2 = 18.8
(b)
1 = 1 = 30.0
n sin 1 -1 1.55 sin 30.0 2 = sin -1 1 = 50.8 = sin 1 n2
(c) and (d)
The other entries are computed similarly, and are shown in the table below. (d) glass into air, angles in degrees incidence 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 reflection 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 refraction 0 15.6 32.0 50.8 85.1 none* none* none* none* none*
(c) air into glass, angles in degrees incidence 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 reflection 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 refraction 0 6.43 12.7 18.8 24.5 29.6 34.0 37.3 39.4 40.2
*total internal reflection
35.47
For water, Thus and
sin c =
1 3 = 4/3 4
c = sin -1 (0.750) = 48.6
d = 2 [(1.00 m)tan c ] d = (2.00 m)tan 48.6 = 2.27 m Figure for Goal Solution
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336
Chapter 35 Solutions
Goal Solution A small underwater pool light is 1.00 m below the surface. The light emerging from the water forms a circle on the water's surface. What is the diameter of this circle? G: Only the light that is directed upwards and hits the water's surface at less than the critical angle will be transmitted to the air so that someone outside can see it. The light that hits the surface farther from the center at an angle greater than c will be totally reflected within the water, unable to be seen from the outside. From the diagram above, the diameter of this circle of light appears to be about 2 m. We can apply Snell's law to find the critical angle, and the diameter can then be found from the geometry. The critical angle is found when the refracted ray just grazes the surface (2 = 90). The index of refraction of water is n2 = 1.33, and n1 = 1.00 for air, so n1 sin c = n2 sin 90 gives 1 = sin -1 (0.750) = 48.6 c = sin -1 1.333 tan c = r (1.00 m)
O:
A:
The radius then satisfies So the diameter is
d = 2r = 2(1.00 m ) tan 48.6 = 2.27 m
L:
Only the light rays within a 97.2 cone above the lamp escape the water and can be seen by an outside observer (Note: this angle does not depend on the depth of the light source). The path of a light ray is always reversible, so if a person were located beneath the water, they could see the whole hemisphere above the water surface within this cone; this is a good experiment to try the next time you go swimming!
*35.48
Call 1 the angle of incidence and of reflection on the left face and 2 those angles on the right face. Let represent the complement of 1 and be the complement of 2 . Now = and = because they are pairs of alternate interior angles. We have A = + = + and B = + A + = + + A = 2A
Chapter 35 Solutions *35.49 (a)
337
We see the Sun swinging around a circle in the extended plane of our parallel of latitude. Its angular speed is
=
2 rad = = 7.27 10 -5 rad s t 86 400 s
The direction of sunlight crossing the cell from the window changes at this rate, moving o n the opposite wall at speed v = r = ( 2.37 m ) 7.27 10 - 5 rad s = 1.72 10 - 4 m s = 0.172 mm s (b) The mirror folds into the cell the motion that would occur in a room twice as wide: v = r = 2 (0.174 mm s) = 0.345 mm s
(
)
(c) and (d) As the Sun moves southward and upward at 50.0, we may regard the corner of the window as fixed, and both patches of light move northward and downward at 50.0 .
*35.50
By Snell's law, c With v = n , This is also true for sound. Here,
n1 sin 1 = n2 sin 2 c c sin 1 = sin 2 or v2 v1 sin 12.0 sin 2 = 340 m/s 1510 m/s sin 1 sin 2 = v1 v2
2 = arcsin (4.44 sin 12.0) = 67.4
*35.51 (a)
n=
2.998 108 m s c = = 1.76 107 v km 1.00 h 1.00 10 3 m 61.15 hr 3600 s 1.00 km so
(b)
n1 sin 1 = n2 sin 2
(1.76 10 ) sin
7
1
= (1.00) sin 90.0
1 = 3.25 10 - 6 degree
This problem is misleading. The speed of energy transport is slow, but the speed of the wavefront advance is normally fast. The condensate's index of refraction is not far from unity.
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338
Chapter 35 Solutions
*35.52
Violet light:
(1.00) sin 25.0
= 1.689 sin 2 2 = 14.490
yv = ( 5.00 cm) tan 2 = ( 5.00 cm) tan 14.490 = 1.2622 cm Red Light:
(1.00) sin 25.0 = 1.642 sin 2
2 = 14.915
yR = ( 5.00 cm) tan 14.915 = 1.3318 cm The emergent beams are both at 25.0 from the normal. Thus, w = y cos 25.0 where y = 1.3318 cm - 1.2622 cm = 0.0396 cm w = (0.396 mm ) cos 25.0 = 0.359 mm
35.53
Horizontal light rays from the setting Sun pass above the hiker. The light rays are twice refracted and once reflected, as in Figure (b) below, by just the certain special raindrops at 40.0 to 42.0 from the hiker's shadow, and reach the hiker as the rainbow. The hiker sees a greater percentage of the violet inner edge, so we consider the red outer edge. The radius R of the circle of droplets is R = (8.00 km)(sin 42.0) = 5.35 km Then the angle , between the vertical and the radius where the bow touches the ground, is given by cos = 2.00 km 2.00 km = = 0.374 R 5.35 km or Figure (a)
= 68.1
The angle filled by the visible bow is 360 (2 68.1) = 224, so the visible bow is 224 = 62.2% of a circle 360 Figure (b)
Chapter 35 Solutions
339
35.54
From Snell's law,
(1.00) sin 1 =
4 sin 2 3
x z d R Fish r
x = R sin 2 = r sin 1 so r sin 2 3 = = R sin 1 4
1 2
eye
Fish at depth d Image at depth z
apparent depth z r cos 1 3 cos 1 = = = actual depth d R cos 2 4 1 - sin 2 2 But sin 2 2 = z 3 = d 4 9 3 sin 1 = 1 - cos 2 1 4 16
2
(
)
2
So
cos 1 3 = 4 9 9 1- + cos 2 1 16 16
cos 1 7 + 9 cos 1 16
or
z=
3d cos 1 7 + 9 cos 2 1
35.55
As the beam enters the slab, (1.00) sin 50.0 = (1.48) sin 2 giving 2 = 31.2 . The beam then strikes the top of the slab at x1 = 1.55 mm tan( 31.2) from the left end. Thereafter, the beam strikes a face each time it has traveled a distance of 2 x1 along the length of the slab. Since the slab is 420 mm long, the beam has an additional 420 mm - x1 to travel after the first reflection. The number of additional reflections is 420 mm - x1 420 mm - 1.55 mm tan( 31.2) = = 81.5 2x1 3.10 mm tan( 31.2) or 81 reflections since the answer must be an integer. The total number of reflections made i n the slab is then 82 .
*35.56 (a)
S1 n2 - n1 1.52 - 1.00 = = = 0.0426 S1 n2 + n1 1.52 + 1.00
2
2
(b)
If medium 1 is glass and medium 2 is air, There is no difference
2 S1 n2 - n1 1.00 - 1.52 = = = 0.0426; S1 n2 + n1 1.00 + 1.52
2
(c)
1.76 107 + 1.00 - 2.00 S1 1.76 107 - 1.00 = = 7 7 S1 1.76 10 + 1.00 1.76 10 + 1.00
2
2
2
2 .00 2 .00 S1 1.00 - 2 = 1.00 - = 1.00 - 2.27 10 -7 1.76 107 + 1.00 S1 1.76 107 + 1.00
or
100%
This suggests he appearance would be very shiny, reflecting practically all incident light . See, however, the note concluding the solution to problem 35.51.
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340
Chapter 35 Solutions S1 n - 1 = . S1 n + 1
2
*35.57 (a)
With n1 = 1 and n2 = n, the reflected fractional intensity is The remaining intensity must be transmitted:
2 S2 (n + 1) - (n - 1) = n2 + 2n + 1 - n2 + 2n - 1 = n - 1 = 1- = n + 1 S1 (n + 1)2 (n + 1)2 2 2 2 4( 2.419) S2 n - 1 = 1- = = 0.828 n + 1 S1 (2.419 + 1)2
4n
(n + 1)2
(b)
At entry,
At exit,
S3 = 0.828 S2 S3 S3 S2 2 = = (0.828) = 0.685 or 68.5% S1 S2 S1
Overall,
*35.58 Define T =
4n
(n + 1)2
as the transmission coefficient for one
encounter with an interface. For diamond and air, it is 0.828, as in problem 57. As shown in the figure, the total amount transmitted is
+ . . . + T 2 (1 - T )
T 2 + T 2 (1 - T ) + T 2 (1 - T ) + T 2 (1 - T )
2 4 2n
6
+ ... so the total
1 - T = 1 - 0.828 = 0.172 We have transmission is
(0.828)2 1 + (0.172)2 + (0.172)4 + (0.172)6 + . . .
2
[
]
6
To sum this series, define F = 1 + (0.172) + (0.172) + (0.172) + . . . .
4
Note that (0.172) F = (0.172) + (0.172) + (0.172) + . . ., and
2 2 4 6
1 + (0.172) F = 1 + (0.172) + (0.172) + (0.172) + . . . = F .
2 2 4 6
Then,
1 = F - (0.172) F or F =
2
1 1 - (0.172)
2
The overall transmission is then
(0.828)2 2 1 - (0.172)
= 0.706 or 70.6%
Chapter 35 Solutions 1 = 1.49 sin 42.0 sin 18.0 sin 42.0
341
35.59
n sin 42.0 = sin 90.0
so
n=
sin 1 = n sin 18.0
and
sin 1 =
1 = 27.5
Figure for Goal Solution
Goal Solution The light beam shown in Figure P35.59 strikes surface 2 at the critical angle. Determine the angle of incidence 1 .
G:
From the diagram it appears that the angle of incidence is about 40. We can find 1 by applying Snell's law at the first interface where the light is refracted. At surface 2, knowing that the 42.0 angle of reflection is the critical angle, we can work backwards to find 1 . Define n1 to be the index of refraction of the surrounding medium and n2 to be that for the prism material. We can use the critical angle of 42.0 to find the ratio n2 n1 : n2 sin 42.0 = n1 sin 90.0 So, 1 n2 = = 1.49 n1 sin 42.0
O:
A:
Call the angle of refraction 2 at the surface 1. The ray inside the prism forms a triangle with surfaces 1 and 2, so the sum of the interior angles of this triangle must be 180. Thus,
(90.0 - 2 ) + 60.0 +(90.0 -42.0) = 180
Therefore, Applying Snell's law at surface 1,
2 = 18.0
n1 sin 1 = n2 sin 18.0 sin 1 = (n2 n1 ) sin 2 = (1.49) sin 18.0
1 = 27.5
L:
The result is a bit less than the 40.0 we expected, but this is probably because the figure is not drawn to scale. This problem was a bit tricky because it required four key concepts (refraction, reflection, critical angle, and geometry) in order to find the solution. One practical extension of this problem is to consider what would happen to the exiting light if the angle of incidence were varied slightly. Would all the light still be reflected off surface 2, or would some light be refracted and pass through this second surface?
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342 35.60
Chapter 35 Solutions Light passing the top of the pole makes an angle of incidence 1 = 90.0 - . It falls on the water surface at distance s1 = (L - d) from the pole, tan
and has an angle of refraction 2 from (1.00)sin 1 = n sin 2 . Then s2 = d tan 2 and the whole shadow length is s1 + s2 = L-d sin 1 + d tan sin -1 n tan
s1 + s2 =
L-d 2.00 m cos cos 40.0 + d tan sin -1 = + ( 2.00 m ) tan sin -1 = 3.79 m n tan 40.0 1.33 tan
35.61
(a)
For polystyrene surrounded by air, internal reflection requires 1.00 = 42.2 3 = sin -1 1.49 Then from the geometry, From Snell's law, This has no solution.
2 = 90.0 3 = 47.8
sin 1 = (1.49) sin 47.8 = 1.10
Therefore, total internal reflection
always happens .
(b)
For polystyrene surrounded by water, and From Snell's law,
1.33 = 63.2 3 = sin -1 1.49
2 = 26.8 1 = 30.3
(c)
No internal refraction is possible since the beam is initially traveling in a medium of lower index of refraction.
*35.62
= 1 - 2 = 10.0
Thus,
and
n1 sin 1 = n2 sin 2 with n1 = 1, n2 =
4 3
1 = sin -1(n2 sin 2 ) = sin -1[n2 sin(1 - 10.0)]
Chapter 35 Solutions
343
(You can use a calculator to home in on an approximate solution to this equation, testing different values of 1 until you find that 1 = 36.5 . Alternatively, you can solve for 1 exactly, as shown below.) We are given that sin 1 = 4 sin(1 - 10.0) 3
This is the sine of a difference, so
3 sin 1 = sin 1 cos 10.0 - cos 1 sin 10.0 4 3 sin 10.0 cos 1 = cos 10.0 - sin 1 4 sin 10.0 = tan 1 cos 10.0 - 0.750 and
Rearranging,
1 = tan -1 0.740 = 36.5
35.63
tan 1 =
4.00 cm h
and
2
tan 2 =
2.00 cm h
tan 2 1 = (2.00 tan 2 ) = 4.00 tan 2 2 sin 2 1 sin 2 2 = 4.00 2 (1 - sin 1 ) (1 - sin 2 2 ) Snell's law in this case is: n1 sin 1 = n2 sin 2 sin 1 = 1.333 sin 2 Squaring both sides, Substituting (2) into (1), sin 2 1 = 1.777 sin 2 2 (2) (1)
1.777 sin 2 2 sin 2 2 = 4.00 2 1 - 1.777 sin 2 1 - sin 2 2
Defining x = sin 2 , Solving for x, From x we can solve for 2 : Thus, the height is h=
(1 - 1.777x) (1 - x)
0.444 - 0.444x = 1 - 1.777x and x = 0.417
0.444
=
1
2 = sin -1 0.417 = 40.2
(2.00 cm) (2.00 cm) = = 2.37 cm tan 2 tan( 40.2)
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344 35.64
Chapter 35 Solutions Observe in the sketch that the angle of incidence at point P is , sin = L / R . and using triangle OPQ: Also, Applying Snell's law at point P, Thus, cos = 1 - sin 2 = R 2 - L2 R
(1.00) sin
sin =
= n sin
sin L = nR n n2 R 2 - L2 nR
and
cos = 1 - sin 2 =
From triangle OPS, + ( + 90.0) + (90.0 - ) = 180 or the angle of incidence at point S is = - . Then, applying Snell's law at point S gives (1.00) sin = n sin = n sin( - ) , or L n2 R 2 - L2 R 2 - L2 L sin = n [sin cos - cos sin ] = n - nR nR R R sin = L 2 2 n R - L2 - R 2 - L2 2 R and L = sin -1 2 n2 R 2 - L2 - R 2 - L2 R
35.65
To derive the law of reflection, locate point O so that the time of travel from point A to point B will be minimum. The total light path is L = a sec 1 + b sec 2 1 (a sec 1 + b sec 2 ) v
The time of travel is t =
If point O is displaced by dx, then dt = 1 (a sec 1 tan 1 d1 + b sec 2 tan 2 d 2 ) = 0 v (1)
(since for minimum time dt = 0). Also, so, c + d = a tan 1 + b tan 2 = constant a sec 2 1 d1 + b sec 2 2 d 2 = 0 (2)
Divide equations (1) and (2) to find 1 = 2
Chapter 35 Solutions 35.66 As shown in the sketch, the angle of incidence at point A is: ( d 2) -1 1.00 m = sin -1 = sin = 30.0 2.00 m R If the emerging ray is to be parallel to the incident ray, the path must be symmetric about the center line CB of the cylinder . In the isosceles triangle ABC, = and = 180 - . Therefore, + + = 180 becomes
345
2 + 180 - = 180 or =
= 15.0 2
Then, applying Snell's law at point A , n sin = (1.00) sin or n= sin sin 30.0 = = 1.93 sin sin 15.0
35.67
(a)
At the boundary of the air and glass, the critical angle is given by 1 sin c = n Consider the critical ray PBB : tan c = sin c d4 d or = t cos c 4t
2
Squaring the last equation gives:
sin 2 c sin 2 c d = = 2 2 cos c 1 - sin c 4t 1 d = n2 - 1 4t d= 4t n2 - 1 4(0.600 cm ) = 2.10 cm
2
Since sin c =
1 , this becomes n
or
n = 1 + ( 4t d)
2
(b)
Solving for d,
Thus, if n = 1.52 and t = 0.600 cm ,
d=
(1.52)2 - 1
(c) Since violet light has a larger index of refraction, it will lead to a smaller critical angle and the inner edge of the white halo will be tinged with violet light.
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346 35.68
Chapter 35 Solutions From the sketch, observe that the angle of incidence at point A is the same as the Given that prism angle at point O . = 60.0 , application of Snell's law at point A gives 1.50 sin = 1.00 sin 60.0 or
= 35.3
From triangle AOB , we calculate the angle of incidence (and reflection) at point B.
+ (90.0 - ) + (90.0 - ) = 180
Now, using triangle BCQ :
so
= - = 60.0 - 35.3 = 24.7
(90.0 - ) + (90.0 - ) + (90.0 - ) = 180
= (90.0 - ) - = 30.0 - 24.7 = 5.30
1.00 sin = 1.50 sin 5.30
Thus the angle of incidence at point C is Finally, Snell's law applied at point C gives or
= sin - 1(1.50 sin 5.30) = 7.96
35.69
(a)
Given that 1 = 45.0 and 2 = 76.0 , the first surface gives n sin = (1.00) sin 45.0
Snell's law at
(1)
Observe that the angle of incidence at the second surface is = 90.0 - . Thus, Snell's law at the second surface yields n sin = n sin(90.0 - ) = (1.00) sin 76.0 , or n cos = sin 76.0 Dividing Equation (1) by Equation (2), (2) tan = sin 45.0 = 0.729 or = 36.1 sin 76.0
Then, from Equation (1), (b)
n=
sin 45.0 sin 45.0 = = 1.20 sin sin 36.1
From the sketch, observe that the distance the light travels in the plastic is d = L sin . Also, the speed of light in the plastic is v = c n , so the time required to travel through the plastic is t= d nL (1.20)(0.500 m) = = = 3.40 10 -9 s = 3.40 ns 8 v c sin 3.00 10 m s sin 36.1
(
)
Chapter 35 Solutions sin 1 0.174 0.342 0.500 0.643 0.766 0.866 0.940 0.985 sin 2 0.131 0.261 0.379 0.480 0.576 0.647 0.711 0.740 sin 1 /sin 2 1.3304 1.3129 1.3177 1.3385 1.3289 1.3390 1.3220 1.3315
347
35.70
The straightness of the graph line demonstrates Snell's proportionality. The slope of the line is n = 1.3276 0.01 and n = 1.328 0.8%
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Chapter 36 Solutions
*36.1
I stand 40 cm from my bathroom mirror. I scatter light which travels to the mirror and back to me in time 0.8 m ~ 109 s 3 108 m/s showing me a view of myself as I was at that look-back time. I'm no Dorian Gray!
*36.2
The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.30 m behind the mirror. The image of the choir is 0.800 m + 5.30 m = 6.10 m from the organist. Using similar triangles: 6.10 m h = 0.600 m 0.800 m or h = (0.600 m ) 6.10 m = 4.58 m 0.800 m
View Looking Down South image of choir mirror 0.600 m Organist 0.800 m 5.30 m
h'
36.3
The flatness of the mirror is described by R = , f = , and 1/ f = 0. By our general mirror equation, 1 1 1 + = , p q f or q = -p
Thus, the image is as far behind the mirror as the person is in front. The magnification is then q h' M= p =1= h so h' = h = 70.0"
Figure for Goal Solution
The required height of the mirror is defined by the triangle from the person's eyes to the top and bottom of his image, as shown. From the geometry of the triangle, we see that the mirror height must be: p p h' h ' p q = h' 2p = 2 Thus, the mirror must be at least 35.0" high .
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Chapter 36 Solutions 351 Goal Solution Determine the minimum height of a vertical flat mirror in which a person 5'10" in height can see his or her full image. (A ray diagram would be helpful.) G: A diagram with the optical rays that create the image of the person is shown above. diagram, it appears that the mirror only needs to be about half the height of the person. From this
O:
The required height of the mirror can be found from the mirror equation, where this flat mirror is described by R = , f = , and 1/ f = 0.
A:
The general mirror equation is
1 1 1 + = , p q f
so with f = ,
q = -p
Thus, the image is as far behind the mirror as the person is in front. The magnification is then M= so -q h =1= p h
h = h = 70.0 in.
The required height of the mirror is defined by the triangle from the person's eyes to the top and bottom of the image, as shown. From the geometry of the similar triangles, we see that the length of the mirror must be: p p h 70.0 in L = h = 35.0 in . = h = = 2 p - q 2p 2 Thus, the mirror must be at least 35.0 in high.
L:
Our result agrees with our prediction from the ray diagram. Evidently, a full-length mirror only needs to be a half-height mirror! On a practical note, the vertical positioning of such a mirror is also important for the person to be able to view his or her full image. To allow for some variation i n positioning and viewing by persons of different heights, most full-length mirrors are about 5' i n length.
36.4
A graphical construction produces 5 images, with images I1 and I2 directly into the mirrors from the object O, and (O, I3, I4) and (I1, I2, I5) forming the vertices of equilateral triangles.
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352 Chapter 36 Solutions
*36.5
(1)
The first image in the left mirror is 5.00 ft behind the mirror, or 10.0 ft from the position of the person. The first image in the right mirror is located 10.0 ft behind the right mirror, but this location is 25.0 ft from the left mirror. Thus, the second image in the left mirror is 25.0 ft behind the mirror, or 30.0 ft from the person.
(2)
(3)
The first image in the left mirror forms an image in the right mirror. This first image is 20.0 ft from the right mirror, and, thus, an image 20.0 ft behind the right mirror is formed. This image in the right mirror also forms an image in the left mirror. The distance from this image in the right mirror to the left mirror is 35.0 ft. The third image in the left mirror is, thus, 35.0 ft behind the mirror, or 40.0 ft from the person.
*36.6
For a concave mirror, both R and f are positive. We also know that 1 1 1 1 1 3 q = f p = 10.0 cm 40.0 cm = 40.0 cm , and q 13.3 cm M= p = 40.0 cm = 0.333 The image is 13.3 cm in front of the mirror, is real, and inverted . (b) 1 1 1 1 1 1 = f p = 10.0 cm 20.0 cm = 20.0 cm , and q q 20.0 cm M= p = 20.0 cm = 1.00 The image is 20.0 cm in front of the mirror, is real, and inverted . (c) 1 1 1 1 1 = f p = 10.0 cm 10.0 cm = 0 Thus, q = infinity. q No image is formed. The rays are reflected parallel to each other. q = 20.0 cm
R f = 2 = 10.0 cm
(a)
q = 13.3 cm
*36.7
1 1 1 1 1 q = f p = 0.275 m 10.0 m Thus, the image is virtual . 0.267 q M = p = 10.0 m = 0.0267 Thus, the image is upright (+M)
gives q = 0.267 m
and
diminished ( M < 1
(
)
Chapter 36 Solutions 353
*36.8
With radius 2.50 m, the cylindrical wall is a highly efficient mirror for sound, with focal length f= R 2 = 1.25 m
In a vertical plane the sound disperses as usual, but that radiated in a horizontal plane is concentrated in a sound image at distance q from the back of the niche, where 1 1 1 p + q = f so 1 1 1 + = 2.00 m q 1.25 m q = 3.33 m
36.9
(a)
1 2 1 + q = R p 2 ( 40.0 cm) 1 2 1 1 q = (40.0 cm) (30.0 cm) = 0.0833 cm q (12.0 cm) M= p = (30.0 cm) = 0.400
gives
1 (30.0 cm)
+
1 q
=
so
q = 12.0 cm
(b)
1 1 2 p + q = R 2 (40.0 cm) 1 2 1 1 q = (40.0 cm) (60.0 cm) = 0.0666 cm q (15.0 cm) M= p = (60.0 cm) = 0.250
gives
1 1 (60.0 cm) + q
=
so
q = 15.0 cm
(c)
Since M > 0, the images are upright .
36.10
(a)
q M = - . For a real image, q > 0 so in this case M = - 4.00 p 1 1 2 q = - pM = 120 cm and from + = p q R R= 2pq 2(30.0 cm)(120 cm) = = 48.0 cm (p + q) (150 cm)
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354 Chapter 36 Solutions (b)
36.11
(a)
1 1 2 + = p q R q = 45.0 cm 1 1 2 + = p q R q = 60.0 cm
becomes
2 1 1 = - q (60.0 cm) (90.0 cm ) M= -q (45.0 cm) =- = 0.500 p (90.0 cm) (a)
and
(b)
becomes
2 1 1 = - , q (60.0 cm) (20.0 cm) M=- (- 60.0 cm) q =- = 3.00 p (20.0 cm)
and
(b)
(c)
The image in (a) is real, inverted and diminished. That of (b) is virtual, upright, and enlarged. The ray diagrams are similar to Figures 36.15(a) and 36.15(b), respectively.
Figures for Goal Solution
Chapter 36 Solutions 355 Goal Solution A concave mirror has a radius of curvature of 60.0 cm. Calculate the image position and magnification of an object placed in front of the mirror (a) at a distance of 90.0 cm and (b) at a distance of 20.0 cm. (c) In each case, draw ray diagrams to obtain the image characteristics. G: It is always a good idea to first draw a ray diagram for any optics problem. This gives a qualitative sense of how the image appears relative to the object. From the ray diagrams above, we see that when the object is 90 cm from the mirror, the image will be real, inverted, diminished, and located about 45 cm in front of the mirror, midway between the center of curvature and the focal point. When the object is 20 cm from the mirror, the image is be virtual, upright, magnified, and located about 50 cm behind the mirror. The mirror equation can be used to find precise quantitative values. (a) The mirror equation is applied using the sign conventions listed in the text. 1 1 2 + = p q R M= or 1 2 1 + = 90.0 cm q 60.0 cm so q = 45.0 cm (real, in front of the mirror)
O: A:
-q 45.0 cm =- = -0.500 (inverted) p 90.0 cm or 1 2 1 + = 20.0 cm q 60.0 cm so q = - 60.0 cm (virtual, behind the mirror)
(b)
1 1 2 + = p q R M=-
q -60.0 cm =- = 3.00 (upright) p 20.0 cm
L:
The calculated image characteristics agree well with our predictions. It is easy to miss a minus sign or to make a computational mistake when using the mirror-lens equation, so the qualitative values obtained from the ray diagrams are useful for a check on the reasonableness of the calculated values.
36.12
For a concave mirror, R and f are positive. Also, for an erect image, M is positive. Therefore, q M = - = 4 and q = 4p. p 1 1 1 = p + q f becomes 1 1 1 3 = p 4p = 4p ; 40.0 cm from which, p = 30.0 cm
36.13
(a)
q q = (p + 5.00 m) and, since the image must be real, M = p = 5 or q = 5p. Therefore, p + 5.00 = 5p or p = 1.25 m and q = 6.25 m. From 2pq 2(1.25)(6.25) 1 1 2 + = , R = (q + p) = (6.25 + 1.25) = 2.08 m (concave) p q R in front
(b)
From part (a), p = 1.25 m; the mirror should be 1.25 m of the object.
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356 Chapter 36 Solutions 36.14 (a) The image is the trapezoid abd e as shown in the ray diagram.
a
h
d e
b
qR - q L
F
hL
e
d
hR
C
b
a
pR = 40.0 cm pL = 60.0 cm
(b)
To find the area of the trapezoid, the image distances, qR and qL , along with the heights hR and hL , must be determined. The mirror equation, 1 1 2 + = p q R becomes 1 2 1 + = 40.0 cm qR 20.0 cm or qR = 13.3 cm
-q -13.3 cm hR = hMR = h R = (10.0 cm ) = - 3.33 cm 40.0 cm pR Also 1 2 1 + = 60.0 cm qL 20.0 cm or qL = 12 .0 cm
-12.0 cm hL = hML = (10.0 cm ) = - 2.00 cm 60.0 cm The area of the trapezoid is the sum of the area of a square plus the area of a triangle: At = A1 + A2 = ( qR - qL )hL + 36.15
1 2
(qR - qL )(hR - hL ) =
3.56 cm 2
Assume that the object distance is the same in both cases (i.e., her face is the same distance from the hubcap regardless of which way it is turned). Also realize that the near image (q = 10.0 cm) occurs when using the convex side of the hubcap. Applying the mirror equation to both cases gives: (concave side: R = R , q = - 30.0 cm ) 1 2 1 - = , or p 30.0 R 1 2 1 - =- , or R p 10.0 2 30.0 cm - p = R (30.0 cm)p 2 p - 10.0 cm = R (10.0 cm )p p = 15.0 cm [1]
(convex side: R = - R ,
q = -10.0 cm )
[2]
(a)
Equating Equations (1) and (2) gives: her face is 15.0 cm from the hubcap.
30.0 cm - p = p - 10.0 cm or 3.00
Thus,
(b)
Using the above result ( p = 15.0 cm ) in Equation [1] gives: 30.0 cm - 15.0 cm 2 = R (30.0 cm)(15.0 cm) or 1 2 = , and R 30.0 cm R = 60.0 cm
Chapter 36 Solutions 357
The radius of the hubcap is 60.0 cm .
36.16
1 1 1 + = p q f q= 15.0 cm (behind mirror) 11.0
f=
R = 1.50 cm 2 q 1 = p 11.0
M=
36.17
(a)
The image starts from a point whose height above the mirror vertex is given by 1 1 1 2 + = = p q f R 1 1 1 Therefore, + = 3.00 m q 0.500 m q = 0.600 m
As the ball falls, p decreases and q increases. Ball and image pass when q1 = p1. When this is true, 1 1 1 2 + = = p1 p1 0.500 m p1 or p1 = 1.00 m.
As the ball passes the focal point, the image switches from infinitely far above the mirror to infinitely far below the mirror. As the ball approaches the mirror from above, the virtual image approaches the mirror from below, reaching it together when p2 = q2 = 0. (b) The falling ball passes its real image when it has fallen
1 3.00 m - 1.00 m = 2.00 m = 2 g t 2 , or when t =
2(2.00 m ) = 0.639 s . 9.80 m s 2
The ball reaches its virtual image when it has traversed
1 3.00 m - 0 = 3.00 m = 2 g t 2 , or at t =
2( 3.00 m ) = 0.782 s . 9.80 m s 2
36.18
When R , Equation 36.8 for a spherical surface becomes q = -p(n2 n1 ) . We use this to locate the final images of the two surfaces of the glass plate. First, find the image the glass forms of the bottom of the plate: qB1 = - 1.33 (8.00 cm) = -6.41 cm 1.66
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358 Chapter 36 Solutions This virtual image is 6.41 cm below the top surface of the glass or 18.41 cm below the water surface. Next, use this image as an object and locate the image the water forms of the bottom of the plate. qB2 = - 1.00 (18.41 cm) = -13.84 cm 1.33 or 13.84 cm below the water surface.
Now find image the water forms of the top surface of the glass. q3 = - 1 (12.0 cm) = -9.02 cm, or 9.02 cm below the water surface. 1.33 t = 13.84 cm 9.02 cm = 4.82 cm
Therefore, the apparent thickness of the glass is
36.19
n1 n2 n2 n1 p + q = R q=
=0
and
R
n2 1 n 1 p = 1.309 (50.0 cm) = 38.2 cm of the ice.
Thus, the virtual image of the dust speck is 38.2 cm below the top surface
*36.20
n2 n2 n1 n1 p + q = R 0.0667 They agree.
so
1.00 1.40 1.40 1.00 + 21.0 mm = 6.00 mm
and
0.0667
=
The image is inverted, real and diminished.
Chapter 36 Solutions 359 n2 (n 2 n 1) n1 + = R p q n2 R p q = p(n n ) n R 2 1 1 n1 = 1.50, n2 = 1.00, R = 15.0 cm, and p = 10.0 cm, q= (1.00)(15.0 cm)(10.0 cm) = 8.57 cm (10.0 cm)(1.00 1.50) (1.50)(15.0 cm)
36.21
From Equation 36.8, Solve for q to find In this case, So
Therefore, the apparent depth is 8.57 cm .
36.22
p = and
q = +2R
1.00 n 2 n2 1.00 p + q = R 0+ n2 n2 1.00 R 2R = so n2 = 2.00
36.23 (a)
n1 n2 (n2 - n1 ) + = R p q 1.50 1.00 1 + = q 12.0 cm 20.0 cm
because or
1.00 1.00 1.50 1.50 - 1.00 + = = q 6.00 cm 12.0 cm p q= 1.50 1.00 1.00 - 12.0 cm 20.0 cm = 45.0 cm
(b)
1.50 1.00 1.00 + = q 12.0 cm 10.0 cm
or
q=
1.50 = 90.0 cm 1.00 1.00 - 12.0 cm 10.0 cm 1.50 1.00 1.00 - 12.0 cm 3.00 cm = 6.00 cm
(c)
1.50 1.00 1.00 + = q 12.0 cm 3.00 cm
or
q=
36.24
For a plane surface,
n p n1 n2 n2 - n1 + = becomes q = - 2 . n1 p q R
Thus, the magnitudes of the rate of change in the image and object positions are related by dq n dp = 2 dt n1 dt If the fish swims toward the wall with a speed of 2.00 cm s, the speed of the image is given by vimage = dq 1.00 = (2.00 cm s) = 1.50 cm/s dt 1.33
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360 Chapter 36 Solutions n1 n2 n2 - n1 + = p q R
36.25
n1 = 1.33
n2 = 1.00
p = +10.0 cm
R = -15.0 cm
q = - 9.01 cm , or the fish appears to be 9.01 cm inside the bowl
*36.26
Let R 1 = outer radius and R 2 = inner radius 1 1 1 1 1 0.0500 f = (n 1) R 1 R 2 = (1.50 1) 2.00 m 2.50 cm = cm so f = 20.0 cm
36.27
(a)
1 1 1 1 1 f = (n 1) R 1 R 2 = (0.440) (12.0 cm) (18.0 cm) : f = 16.4 cm 1 1 1 f = (0.440) (18.0 cm) (12.0 cm) : Figure for Goal Solution f = 16.4 cm
(b)
Goal Solution The left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right face has a radius of curvature of magnitude 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate the focal length of the lens. (b) Calculate the focal length if the radii of curvature of the two faces are interchanged. G: Since this is a biconvex lens, the center is thicker than the edges, and the lens will tend to converge incident light rays. Therefore it has a positive focal length. Exchanging the radii of curvature amounts to turning the lens around so the light enters the opposite side first. However, this does not change the fact that the center of the lens is still thicker than the edges, so we should not expect the focal length of the lens to be different (assuming the thin-lens approximation is valid). The lens makers' equation can be used to find the focal length of this lens. The centers of curvature of the lens surfaces are on opposite sides, so the second surface has a negative radius: (a) (b) L: 1 1 1 1 1 = (n - 1) - - = (1.44 - 1.00) f 12.0 cm -18.0 cm R1 R2 1 1 1 = (0.440) - 18.0 cm -12.0 cm f so so f = 16.4 cm f = 16.4 cm
O: A:
As expected, reversing the orientation of the lens does not change what it does to the light, as long as the lens is relatively thin (variations may be noticed with a thick lens). The fact that light rays can be traced forward or backward through an optical system is sometimes referred to as the principle of reversibility. We can see that the focal length of this biconvex lens is about the same magnitude as the average radius of curvature. A few approximations, useful as checks, are that a symmetric biconvex lens with radii of magnitude R will have focal length f R; a plano-convex lens with radius R will have f R / 2; and a symmetric biconcave lens has f -R . These approximations apply when the lens has n 1.5, which is typical of many types of clear glass and plastic. 1 1 1 *36.28 For a converging lens, f is positive. We use p + q = f .
Chapter 36 Solutions 361
(a)
1 1 1 1 1 1 q = f p = 20.0 cm 40.0 cm = 40.0 cm M= q 40.0 p = 40.0 = 1.00
q = 40.0 cm
The image is real, inverted , and located 40.0 cm past the lens. (b) 1 1 1 1 1 = = = 0 q = infinity f p 20.0 cm 20.0 cm q No image is formed. The rays emerging from the lens are parallel to each other. (c) 1 1 1 1 1 1 = f p = 20.0 cm 10.0 cm = 20.0 cm q q 20.0 M= p = 10.0 = 2.00 The image is upright, virtual , and 20.0 cm in front of the lens. q = 20.0 cm
*36.29 (a)
1 1 1 1 1 = f p = 25.0 cm 26.0 cm q
q = 650 cm
The image is real, inverted, and enlarged . (b) 1 1 1 1 1 = f p = 25.0 cm 24.0 cm q q = 600 cm
The image is virtual, upright, and enlarged .
36.30
(a)
1 1 1 1 1 1 + q = f (32.0 cm) + (8.00 cm) = f p q (8.00 cm) M= p = (32.00 cm) = 0.250 Since f > 0, the lens is converging .
so
f = 6.40 cm
(b) (c)
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362 Chapter 36 Solutions 36.31 We are looking at an enlarged, upright, virtual image: h' q M = h =2= p 1 1 1 p + q = f gives so p= q 2.84 cm = = +1.42 cm 2 2
1 1 1 + = 1.42 cm 2.84 cm f
f = 2.84 cm
*36.32
To use the lens as a magnifying glass, we form an upright, virtual image: q M = +2.00 = p We eliminate q = 2.00p:: Solving, or 1 1 1 p + q = f 1 1 1 + 2.00p = 15.0 cm p p = 7.50 cm or 2.00 + 1.00 1 = 15.0 cm 2.00p
36.33
(a)
Note that so which yields a quadratic in p: which has solutions Both solutions are valid.
q = 12.9 cm p 1 1 1 p + 12.9 p = 2.44 p2 + 12.9p = 31.5 p = 9.63 cm or p = 3.27 cm
(b)
For a virtual image,
q = p + 12.9 cm 1 1 1 p 12.9 + p = 2.44
or from which cm.
p2 + 12.9p = 31.8 p = 2.10 cm or p = 15.0
We must have a real object so the 15.0 cm solution must be rejected.
Chapter 36 Solutions 363
36.34
(a)
1 1 1 + = : q f p p = 8.82 cm M=
1 1 1 + = p 30.0 cm 12.5 cm q ( 30.0) p = 8.82 = 3.40, upright
(b)
See the figure to the right.
*36.35
1 1 1 + = : p q f We may differentiate through with respect to p:
p 1 + q1 = constant 1p 2 1q 2 dq =0 dp
dq q2 = 2 = M2 dp p
36.36 (b) (a)
The image is inverted: q + p = 3.00 m = 75.0p + p q = 2.96 m f = 39.0 mm
h' 1.80 m q M = h = 0.0240 m = 75.0 = p p = 39.5 mm 1 1 1 1 1 = p + q = 0.0395 m + 2.96 m f
q = 75.0p
36.37
(a)
1 1 1 + = p q f
1 1 1 + = (20.0 cm) q ( - 32.0 cm) 1 1 q = - + 20.0 32.0
-1
so
= 12.3 cm
The image is 12.3 cm to the left of the lens. (b) M=- (-12.3 cm) q =- = 0.615 p (20.0 cm)
2000 by Harcourt, Inc. All rights reserved.
364 Chapter 36 Solutions (c) See the ray diagram above. 1 1 1 1 1 = (n - 1) - - , or f = 13.3 cm = (1.50 - 1) f R1 R2 15.0 cm ( -12.0 cm ) Ray Diagram:
b a h = 10.0 cm c
10.0 cm
36.38
(a) (b)
qL F
qR
c'
d' h'L
d
pR = 20.0 cm pL = 30.0 cm b'
F
h'R
a'
(c)
To find the area, first find qR and qL , along with the heights hR and hL , using the thin lens equation. 1 1 1 + = p R qR f becomes: 1 1 1 + = 20.0 cm qR 13.3 cm or qR = 40.0 cm
- qR hR = hMR = h = (10.0 cm)(- 2.00) = - 20.0 cm pR 1 1 1 + = : 30.0 cm qL 13.3 cm qL = 24.0 cm hL = hML = (10.0 cm)(- 0.800) = - 8.00 cm Thus, the area of the image is:
1 Area = qR - qL hL + 2 qR - qL hR - hL = 224 cm2
36.39
(a)
The distance from the object to the lens is p, so the image distance is q = 5.00 m - p. Thus, 1 1 1 + = p q f becomes: 1 1 1 + = p 5.00 m - p 0.800 m p 2 - ( 5.00 m )p + ( 4.00 m ) = 0 p = 4.00 m, or p = 1.00 m .
This reduces to a quadratic equation: which yields
Thus, there are two possible object distances, both corresponding to real objects. (b) For p = 4.00 m: q = 5.00 m - 4.00 m = 1.00 m: For p = 1.00 m: q = 5.00 m - 1.00 m = 4.00 m: M=- M=- 1.00 m = 0.250 . 4.00 m 4.00 m = 4.00 . 1.00 m
Chapter 36 Solutions 365
36.40
(a)
Both images are real and inverted , but the magnifications are different, with one being larger than the object and the other smaller. 1 1 1 1 1 1 The image distance is: q = d - p. Thus, + = becomes + = p q f p d-p f This reduces to a quadratic equation: p 2 + ( - d)p + ( f d) = 0 d d2 - 4 f d d = 2 2 d2 -fd 4
which yields:
p=
Since f < d 4 , both solutions are meaningful and the two solutions are not equal to each other. Thus, there are two distinct lens positions that form an image on the screen. (b) The smaller solution for p gives a larger value for q, with a real, enlarged, inverted image . The larger solution for p describes a real, diminished, inverted image .
*36.41
To properly focus the image of a distant object, the lens must be at a distance equal to the focal length from the film ( q1 = 65.0 mm ). For the closer object: 1 1 1 1 1 1 + = becomes + = p2 q2 f 2000 mm q2 65.0 mm and 2000 q2 = (65.0 mm ) 2000 - 65.0
The lens must be moved away from the film by a distance 2000 - 65.0 mm = 2.18 mm D = q2 - q1 = (65.0 mm ) 2000 - 65.0
*36.42 (a)
The focal length of the lens is given by 1 1 1 1 1 = (n - 1) - - = (1.53 - 1.00) f R1 R2 - 32.5 cm 42.5 cm f = - 34.7 cm Note that R1 is negative because the center of curvature of the first surface is on the virtual image side.
R1
R2
2000 by Harcourt, Inc. All rights reserved.
366 Chapter 36 Solutions When p = , the thin lens equation gives q = f . Thus, the violet image of a very distant object is formed at q = - 34.7 cm . image is virtual, upright, and diminished . (b) The same ray diagram and image characteristics apply for red light. Again, q = f , and now 1 1 1 = (1.51 - 1.00) - giving f = - 36.1 cm . f - 32.5 cm 42.5 cm The
I
F
F
36.43
Ray h 1 is undeviated at the plane surface and strikes the second surface at angle of incidence given by 0.500 cm h = 1.43 1 = sin -1 1 = sin -1 20.0 cm R Then, 0.500 (1.00)sin 2 = (1.60)sin 1 = (1.60) 20.0 cm so
2 = 2.29
The angle this emerging ray makes with the horizontal is
2 1 = 0.860
It crosses the axis at a point farther out by f1 where f1 = h1 0.500 cm = tan(0.860) = 33.3 cm tan( 2 1 )
The point of exit for this ray is distant axially from the lens vertex by 20.0 cm (20.0 cm)2 (0.500 cm)2 = 0.00625 cm
so ray h1 crosses the axis at this distance from the vertex: x1 = 33.3 cm 0.00625 cm = 33.3 cm Now we repeat this calculation for ray h2: 12.00 (1.00)sin 2 = (1.60)sin 1 = (1.60) 20.0 f2 = h2 12.0 cm = tan(36.8) = 16.0 cm tan( 2 1 ) 12.0 cm = 36.9 1 = sin -1 20.0 cm
2 = 73.7
x2 = (16.0 cm ) 20.0 cm - (20.0 cm)2 - (12.0 cm)2 = 12.0 cm Now x = 33.3 cm 12.0 cm = 21.3 cm
Chapter 36 Solutions 367
36.44
For starlight going through Nick's glasses,
1 1 1 + = p q f 1 1 1 + = = -1.25 diopters ( - 0.800 m ) f
For a nearby object,
1 1 + = -1.25 m -1 , so p ( - 0.180 m )
p = 23.2 cm
36.45
P=
1 1 1 1 1 = + = - = - 4.00 diopters = - 4.00 diopters, a diverging lens f p q 0.250 m
36.46
Consider an object at infinity, imaged at the person's far point: 1 1 1 + = p q f 1 1 + = -4.00 m -1 q q = -25.0 cm
The person's far point is 25.0 cm + 2.00 cm = 27.0 cm from his eyes. For the contact lenses we want 1 1 1 + = = - 3.70 diopters ( - 0.270 m ) f
36. 47
First, we use the thin lens equation to find the object distance:
1 1 1 + = p ( -25.0 cm ) 10.0 cm
Then, p = 7.14 cm
and
Then, M = -
(- 25.0 cm) = 3.50 q =- p 7.14 cm
36.48
(a)
From the thin lens equation:
1 1 1 or + = p ( - 25.0 cm ) 5.00 cm
p = 4.17 cm
(b)
M=-
q 25.0 cm 25.0 cm =1+ =1+ = 6.00 p f 5.00 cm
36.49
Using Equation 36.20,
L 25.0 cm 23.0 cm 25.0 cm M = - = 575 = - fe 0.400 cm 2.50 cm fo
2000 by Harcourt, Inc. All rights reserved.
368 Chapter 36 Solutions 25.0 cm -12.0 M1 M = M1me = M1 (25.0 cm) = (25.0 cm) = 2.14 cm fe = -140 fe M
36.50
36.51 (a) (b)
f o = 20.0 m
f e = 0.0250 m m=- fo = 800 fe
The angular magnification produced by this telescope is: Since m < 0, the image is inverted .
Chapter 36 Solutions 369 1 1 1 + = p q f q= fp 1 1 = = 1 f -1 p p- f p- f fp M= q f h =- =- h p p- f hf f -p - hf . p
*36.52 (a)
The lensmaker's equation
gives
Then,
gives
h =
(b) (c)
For p >> f ,
f - p -p.
Then,
h =
Suppose the telescope observes the space station at the zenith: h = - hf (108.6 m)( 4.00 m) = =- p 407 10 3 m -1.07 mm
*36.53 (b)
Call the focal length of the objective f o and that of the eyepiece - f e . The distance between the lenses is f o - f e . The objective forms a real diminished inverted image of a very distant object at q1 = f o . This image is a virtual object for the eyepiece at p2 = - f e . For it 1 1 1 + = p q f q2 = becomes 1 1 1 1 + = , =0 - fe q - fe q2
and (a)
The user views the image as virtual . Letting h represent the height of the first image, o = h f o and = h f e . The angular magnification is m= f h f e = = o o h f o fe fo = 3.00. fe
2f 3 o
0
F0
F0
0
I L1
h'
(c)
Here, f o - f e = 10.0 cm and Thus, f fe = o 3.00 f o = 15.0 cm f e = 5.00 cm
Fe L2
Fe O
and
= 10.0 cm.
and
f e = - 5.00 cm
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370 Chapter 36 Solutions *36.54 Let I 0 represent the intensity of the light from the nebula and o its angular diameter. W i t h the first telescope, the image diameter h on the film is given by o = - h f o as h = - o ( 2000 mm ) . The light power captured by the telescope aperture is P1 = I 0 A1 = I 0 ( 200 mm ) 4 , and the
2
light energy focused on the film during the exposure is E1 = P1t1 = I 0 ( 200 mm ) 4 (1.50 min ) .
2
[
[
]
]
Likewise, the
P 2 = I 0 A2 = I 0 (60.0 mm ) 4 and the light energy is E2 = I 0 (60.0 mm ) 4 t2 . Therefore, to
2 2
have the same light energy per unit area, it is necessary that I 0 (60.0 mm ) 4 t2
2
[
light
power
]
captured
by the
aperture
[
of
the
second
]
telescope
is
o (900 mm )
[
[
2
] 4]
=
I 0 ( 200 mm ) 4 (1.50 min )
2
[
o ( 2000 mm )2 4
[
]
]
The required exposure time with the second telescope is t2 =
(200 mm)2 (900 mm)2 1.50 min = ( ) (60.0 mm)2 (2000 mm)2
3.38 min
36.55
Only a diverging lens gives an upright diminished image. The image is virtual and d = p - q = p + q: d p= 1M : f= q M= p so q = Mp and d = p Mp
1 1 1 1 1 M + 1 (1 M)2 + q = f = p + Mp = Mp = Md p
Md (0.500)(20.0 cm) = = 40.0 cm (1 M)2 (1 0.500)2
36.56
If M < 1, the lens is diverging and the image is virtual. q M= p d p= 1M : so q = Mp and d = p Mp
d=p q =p+q
1 1 1 1 1 M + 1 (1 M)2 + q = f = p + Mp = Mp = Md p
f=
Md (1 M)2
If M > 1, the lens is converging and the image is still virtual. Now d = q p. We obtain in this case f= Md (M 1)2
Chapter 36 Solutions 371 *36.57 Start with the first pass through the lens. 1 1 1 1 = = f1 p1 80.0 cm q1 For the mirror, 1 1 1 1 q 2 = f 2 p 2 = 50.0 cm 1 300 cm 1 100 cm q1 = 400 cm to right of lens p2 = 300 cm q2 = 60.0 cm p3 = 160 cm q3 = 160 cm to left of lens q2 60.0 cm 1 M 2 = p = 300 cm = 5 2 M = M 1M 2M 3 = 0.800
For the second pass through the lens, 1 1 1 1 1 q 3 = f 1 p 3 = 80.0 cm 160 cm q1 400 cm M 1 = p = 100 cm = 4.00 1 q3 160 cm M3 = p 3 = 160 cm = 1
Since M < 0 the final image is inverted .
*36.58 (a)
1 1 1 = (n - 1) - f R1 R2 1 1 1 = (1.66 - 1) - - 65.0 cm 50.0 cm R2 1 1 1 = + R2 50.0 cm 42.9 cm so R2 = 23.1 cm
A
D
R1 B C R2
2.00 cm
(b)
The distance along the axis from B to A is
2 R1 - R1 - ( 2.00 cm ) = 50.0 cm - 2
( 50.0 cm)2 - (2.00 cm)
2
= 0.0400 cm
Similarly, the axial distance from C to D is 23.1 cm -
(23.1 cm)2 - (2 .00 cm)
2
= 0.0868 cm
Then, AD = 0.100 cm - 0.0400 cm + 0.0868 cm = 0.147 cm .
2000 by Harcourt, Inc. All rights reserved.
372 Chapter 36 Solutions 1 1 1 1 = = q1 f1 p1 10.0 cm 1 so 12.5 cm
*36.59
q1 = 50.0 cm (to left of mirror)
This serves as an object for the lens (a virtual object), so 1 1 1 1 1 q 2 = f 2 p 2 = 16.7 cm 25.0 cm q2 = 50.3 cm (to right of lens)
Thus, the final image is located 25.3 cm to right of mirror . q1 M1 = p 1 M2 = = 50.0 cm = 4.00 12.5 cm
q2 50.3 cm = = 2.01 p2 25.0 cm
M = M 1M 2 = 8.05 Thus, the final image is virtual, upright , 8.05 times the size of object, and 25.3 cm to right of the mirror.
36.60
We first find the focal length of the mirror. 1 1 1 1 1 9 f = p + q = 10.0 cm + 8.00 cm = 40.0 cm Hence, if p = 20.0 cm, Thus, and f = 4.44 cm
1 1 1 1 1 15.56 q = f p = 4.44 cm 20.0 cm = 88.8 cm
q = 5.71 cm , real
36.61
A hemisphere is too thick to be described as a thin lens. The light is undeviated on entry into the flat face. W e next consider the light's exit from the second surface, for which R = 6.00 cm The incident rays are parallel, so p = . Then, n1 n2 n2 n1 + = p q R becomes 1 (1.00 1.56) 0 + q = 6.00 cm and q = 10.7 cm
Chapter 36 Solutions 373 P 4.50 W = 4 r 2 4 1.60 10 -2 m
*36.62 (a)
I=
(
)
2
= 1.40 kW m 2
(b)
I=
P 4.50 W = = 6.91 mW m 2 4 r 2 4 (7.20 m )2 so q = 0.368 m and
(c)
1 1 1 1 1 1 + = + = 7.20 m q 0.350 m p q f M= q h 0.368 m =- =- 3.20 cm p 7.20 m
h = 0.164 cm P = IA = 6.91 10 - 3 W m 2
(d) The lens intercepts power given by
(
and puts it all onto the image where
6.91 10 -3 W m 2 (15.0 cm ) 4 P I= = = 58.1 W m 2 A (0.164 cm )2 4
2
(
)[
)[
1 4
(0.150 m)2
]
]
36.63
From the thin lens equation,
q1 =
f1 p 1 ( 6.00 cm)(12.0 cm) p 1 f 1 = 12.0 cm ( 6.00 cm) = 4.00 cm
When we require that q2 , the thin lens equation becomes p2 = f2; In this case, Therefore, and p 2 = d ( 4.00 cm) d + 4.00 cm = f2 = 12.0 cm d = 8.00 cm
*36.64 (a)
For the light the mirror intercepts,
2 350 W = 1000 W m 2 Ra
2 P = I 0 A = I 0 Ra
(
)
and we have
Ra = 0.334 m or larger p so q = R . 2 R rad R 0.533 =- (9.30 m rad) 2 180 2
(b)
In
1 1 1 2 + = = p q f R q h =- , h p
M=
so
h = - q( h p) = -
where h p is the angle the Sun subtends. The intensity at the image is then I=
2 4I R 2 4I 0 Ra P = 0 2a = h 2 4 h (R 2)2 9.30 10- 3 rad
(
)
2
120 10 3 W m 2 =
2 16 1000 W m 2 Ra
R
2
( ) (9.30 10 rad)
-3
2
so
Ra = 0.0255 or larger R
2000 by Harcourt, Inc. All rights reserved.
374 Chapter 36 Solutions 36.65 For the mirror, f = R/2 = +1.50 m. In addition, because the distance to the Sun is so much larger than any other figures, we can take p = . The mirror equation, 1 1 1 p + q = f , then gives q = f = 1.50 m . Now, in M = q h' h p = h , the magnification is nearly zero, but we can be more precise: p the angular diameter of the object. Thus, the image diameter is h' = hq = ( 0.533) rad/deg (1.50 m) = 0.140 m = 1.40 cm p 180 1 1 1 = (n - 1) - , becomes: f R1 R2 giving n = 1.99 . is
36.66
(a)
The lens makers' equation,
1 1 1 = (n - 1) - 9.00 cm ( -11.0 cm ) 5.00 cm (b) 1 1 1 + = p1 q1 f
As the light passes through the lens for the first time, the thin lens equation becomes: 1 1 1 + = 8.00 cm q1 5.00 cm M1 = - q1 13.3 cm =- = -1.67 8.00 cm p1
or q1 = 13.3 cm , and
This image becomes the object for the concave mirror with: pm = 20.0 cm - q1 = 20.0 cm - 13.3 cm = 6.67 cm , and f = The mirror equation becomes: giving qm = 10.0 cm and R = + 4.00 cm. 2 1 1 1 + = 6.67 cm qm 4.00 cm M2 = - qm 10.0 cm =- = -1.50 6.67 cm pm
The image formed by the mirror serves as a real object for the lens on the second pass of the light through the lens with: p3 = 20.0 cm - qm = +10.0 cm The thin lens equation yields: or q3 = 10.0 cm, and The final image is a real image located The overall magnification is (c) 1 1 1 + = 10.0 cm q3 5.00 cm M3 = - q3 10.0 cm =- = - 1.00. 10.0 cm p3
10.0 cm to the left of the lens . Mtotal = M1M2 M3 = - 2 .50 .
Since the total magnification is negative, this final image is inverted .
Chapter 36 Solutions 375 36.67 In the original situation, In the final situation, and Our lens equation is p1 + q1 = 1.50 m p2 = p1 + 0.900 m q2 = q1 0.900 m. 1 1 1 1 1 p 1 + q1 = f = p 2 + q2 1 1 1 1 + = + p 1 1.50 m p 1 p1 + 0.900 0.600 p1 1.50 m p 1 + p 1 0.600 p1 + p1 + 0.900 = p 1(1.50 m p 1) (p1 + 0.900)(0.600 p1) p1(1.50 m p1) = (p1 + 0.900)(0.600 p1) p1 = 0.540 1.80 m = 0.300 m
Substituting, we have
Adding the fractions, Simplified, this becomes (a) Thus,
p2 = p1 + 0.900 = 1.20 m (b) 1 1 1 = 0.300 m + 1.50 m 0.300 m f and f = 0.240 m q2 M = p = 0.250
2
(c)
The second image is real, inverted, and diminished , with
36.68
As the light passes through, the lens attempts to form an image at distance q1 where 1 1 1 = - q1 f p1 or q1 = f p1 p1 - f
This image serves as a virtual object for the mirror with p2 = - q1 . The plane mirror then forms an image located at q2 = - p2 = + q1 above the mirror and lens. This second image serves as a virtual object ( p3 = - q2 = - q1 ) for the lens as the light makes a return passage through the lens. The final image formed by the lens is located at distance q3 above the lens where 1 1 1 1 1 1 p1 - f 2p1 - f = - = + = + = f p1 f p1 q3 f p3 f q1 f or q3 = f p1 2p1 - f f p1 = p1 . 2p1 - f
If the final image coincides with the object, it is necessary to require q3 = p1, or This yields the solution p 1 = f or
the object must be located at the focal point of the lens .
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376 Chapter 36 Solutions 1 1 1 + = p q f 1 1 1 + = 3.40 mm q 3.00 mm M1 = - q / p = - me =
36.69
For the objective:
becomes
so
q = 25.5 mm
The objective produces magnification
25.5 mm = - 7.50 3.40 mm
For the eyepiece as a simple magnifier, and overall
25.0 cm 25.0 cm = = 10.0 f 2.50 cm
M = M1me = - 75.0
36.70
(a)
Start with the second lens: This lens must form a virtual image located 19.0 cm to the left of it (i.e., q2 = - 19.0 cm ). The required object distance for this lens is then p2 = q2 f 2 (-19.0 cm)(20.0 cm) = 380 cm = 39.0 q2 - f 2 -19.0 cm - 20.0 cm
The image formed by the first lens serves as the object for the second lens. Therefore, the image distance for the first lens is q1 = 50.0 cm - p2 = 50.0 cm - 380 cm 1570 cm = 39.0 39.0
The distance the original object must be located to the left of the first lens is then given by 1 1 1 39.0 157 - 39.0 118 1 = - = - = = 1570 cm 1570 cm p1 f 1 q1 10.0 cm 1570 cm (b) or p1 = 1570 cm = 13.3 cm 118
q q 1570 cm 118 ( -19.0 cm )( 39.0) M = M1M2 = - 1 - 2 = = - 5.90 380 cm p1 p2 39.0 1570 cm Since M < 0, the final image is inverted .
(c)
36.71
(a)
P=
1 1 1 1 1 = + = + = 44.6 diopters f p q (0.0224 m) 1 1 1 1 1 = + = + = 3.03 diopters f p q (0.330 m)
(b)
P=
Chapter 36 Solutions 377 36.72 The object is located at the focal point of the upper mirror. Thus, the upper mirror creates an image at infinity (i.e., parallel rays leave this mirror). The lower mirror focuses these parallel rays at its focal point, located at the hole in the upper mirror. Thus, the image is real, inverted, and actual size . For the upper mirror: 1 1 1 1 1 1 + = + = : q1 = 7.50 cm q1 7.50 cm p q f For the lower mirror: 1 1 1 + = : q2 7.50 cm q2 = 7.50 cm
Light directed into the hole in the upper mirror reflects as shown, to behave as if it were reflecting from the hole.
36.73
(a)
Lens one: 1 1 1 : + = 40.0 cm q1 30.0 cm M1 = - q1 120 cm =- = - 3.00 40.0 cm p1 q1 = 120 cm
This real image is a virtual object for the second lens, at p2 = 110 cm - 120 cm = -10.0 cm 1 1 1 + = : - 10.0 cm q2 - 20.0 cm M2 = - q2 20.0 cm =- = + 2.00 p2 (-10.0 cm) q2 = 20.0 cm
Moverall = M1M2 = - 6.00 (b) (c) Moverall < 0, so final image is inverted . Lens two converging: 1 1 1 + = - 10.0 cm q2 20.0 cm q2 = 6.67 cm 6.67 cm M2 = - = + 0.667 ( -10.0 cm) Moverall = M1M2 = - 2 .00 Again, Moverall < 0 and the final image is inverted .
2000 by Harcourt, Inc. All rights reserved.
378 Chapter 36 Solutions
Chapter 37 Solutions
37.1
L (632.8 10- 9)(5.00) ybright = d = m = 1.58 cm 2.00 10- 4
37.2
y bright
L = m d
For m = 1,
3.40 10 - 3 m 5.00 10 - 4 m yd = = = 515 nm L 3.30 m
(
)(
)
37.3
Note, with the conditions given, the small angle approximation does not work well. That is, sin , tan , and are significantly different. The approach to be used is outlined below. (a) At the m = 2 maximum, 400 m tan = 1000 m = 0.400
= 21.8 so
(b)
=
d sin (300 m) sin 21.8 = = 55.7 m m 2
The next minimum encountered is the m = 2 minimum; and at that point,
1 5 d sin = m + 2 which becomes d sin = 2
or so
5 5(55.7 m) sin = 2d = 2(300 m) = 0.464 y = (1000 m) tan 27.7 = 524 m
and
= 27.7
Therefore, the car must travel an additional 124 m .
37.4 (a) (b) (c)
v 354 m/s = f = 2000/s = 0.177 m d sin = m d sin = m so so (0.300 m) sin = 1(0.177 m) d sin 36.2 = 1(0.0300 m) so and and
= 36.2
d = 5.08 cm
(1.00 10- 6 m) sin 36.2 = 1 f= 3.00 108 m/s c = = 508 THz 5.90 107 m
= 590 nm
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2
Chapter 37 Solutions
37.5
For the tenth minimum, m = 9. Using Equation 37.3, Also, tan = 9.5 sin y . L
sin =
1 9+ d 2
For small , sin tan . Thus,
d=
=
9.5 L 9.5(589 109 m)(2.00 m) = y 7.26 103 m
= 1.54 103 m = 1.54 mm
Goal Solution Young's double-slit experiment is performed with 589-nm light and a slit-to-screen distance of 2.00 m . The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits. G: For the situation described, the observed interference pattern is very narrow, (the minima are less than 1 mm apart when the screen is 2 m away). In fact, the minima and maxima are so close together that it would probably be difficult to resolve adjacent maxima, so the pattern might look like a solid blur to the naked eye. Since the angular spacing of the pattern is inversely proportional to the slit width, we should expect that for this narrow pattern, the space between the slits will be larger than the typical fraction of a millimeter, and certainly much greater than the wavelength of the light (d >> = 589 nm). Since we are given the location of the tenth minimum for this interference pattern, we should use the equation for destructive interference from a double slit. The figure for Problem 7 shows the critical variables for this problem. d sin = m + sin =
O:
A:
In the equation The first minimum is described by m = 0 and the tenth by m = 9 : Also, tan = y L , but for small , sin tan . Thus, 9.5(589010 -10 m)(2.00 m) = 1.5410 -3 m = 1.54 mm = 1.54 mm 7.2610 -3 m
(
1 2 1 2
) ,
9+ d
( )
d=
9.5 9.5 L = y sin
d=
L:
The spacing between the slits is relatively large, as we expected (about 3 000 times greater than the wavelength of the light). In order to more clearly distinguish between maxima and minima, the pattern could be expanded by increasing the distance to the screen. However, as L is increased, the overall pattern would be less bright as the light expands over a larger area, so that beyond some distance, the light would be too dim to see.
Chapter 37 Solutions
3
*37.6
340 m/s = 2000 Hz = 0.170 m Maxima are at m=0 m=1 m=2 m=3 gives gives gives gives d sin = m :
= 0 0.170 m sin = d = 0.350 m 2 sin = d = 0.971 sin = 1.46 = 29.1 = 76.3
No solution.
1 Minima at d sin = m + 2 :
m=0 m=1 m=2
sin = 2d = 0.243 3 gives sin = 2d = 0.729 gives No solution.
gives
= 14.1 = 46.8
So we have maxima at 0, 29.1, and 76.3 and minima at 14.1 and 46.8.
37.7
(a)
For the bright fringe, y bright = y= m L d where m=1
(546.1 10 -9 m)(1.20 m) = 2.62 10 -3 m = 2.62 mm 0.250 10 -3 m
(b)
For the dark bands, ydark = y2 - y1 =
L 1 m + ; m = 0, 1, 2, 3, . . . d 2
L 1 L (546.1 10 -9 m)(1.20 m) 1 - 0+ = 1+ (1) = 2 2 d d 0.250 10 -3 m
y = 2.62 mm
Figures for Goal Solution
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4
Chapter 37 Solutions
Goal Solution A pair of narrow, parallel slits separated by 0.250 mm is illuminated by green light ( = 546.1 nm). The interference pattern is observed on a screen 1.20 m away from the plane of the slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands. G: The spacing between adjacent maxima and minima should be fairly uniform across the pattern as long as the width of the pattern is much less than the distance to the screen (so that the small angle approximation is valid). The separation between fringes should be at least a millimeter if the pattern can be easily observed with a naked eye. The bright regions are areas of constructive interference and the dark bands are destructive interference, so the corresponding double-slit equations will be used to find the y distances. It can be confusing to keep track of four different symbols for distances. Three are shown in the drawing to the right. Note that: y is the unknown distance from the bright central maximum (m = 0) to another maximum or minimum on either side of the center of the interference pattern.
O:
is the wavelength of the light, determined by the source.
A: (a) For very small and the equation for constructive interference becomes Substituting values, sin tan sin = m d and tan = y / L (Eq. 37.2) (Eq. 37.5)
y bright ( L d)m y bright =
(546 10 -9 m)(1.20 m) (1) = 2.62 mm 0.250 10 -3 m
(b) If you have trouble remembering whether Equation 37.5 or Eq. 37.6 applies to a given situation, you can instead remember that the first bright band is in the center, and dark bands are halfway between bright bands. Thus, Eq. 37.5 describes them all, with m = 0, 1, 2 . . . for bright bands, and with m = 0.5, 1.5, 2.5 . . . for dark bands. The dark band version of Eq. 37.5 is simply Eq. 37.6: ydark =
L m+ d
(
1 2
)
ydark = 1 + L:
(
1 2
1 ) dL - (0 + 2 ) dL = dL = 2.62 mm
This spacing is large enough for easy resolution of adjacent fringes. The distance between minima is the same as the distance between maxima. We expected this equality since the angles are small:
= ( 2.62 mm ) (1.20 m ) = 0.00218 rad = 0.125
When the angular spacing exceeds about 3, then significant figures. sin differs from tan when written to three
Chapter 37 Solutions
5
37.8
Taking m = 0 and y = 0.200 mm in Equation 37.6 gives L 2 dy 2(0.400 10 - 3 m)(0.200 10 - 3 m) = = 0.362 m 442 10 - 9 m
L 36.2 cm Geometric optics incorrectly predicts bright regions opposite the slits and darkness in between. But, as this example shows, interference can produce just the opposite.
37.9
Location of A = central maximum, Location of B = first minimum. So, y = [ ymin - ymax ] = d=
L 1 1 L 0+ -0= = 20.0 m 2 2 d d
Thus,
L (3.00 m)(150 m) = 11.3 m = 2(20.0 m ) 40.0 m
37.10
At 30.0, d sin = m (3.20 10 - 4 m) sin 30.0= m(500 10 - 9 m) so m = 320
There are 320 maxima to the right, 320 to the left, and one for m = 0 straight ahead. There are 641 maxima .
37.11
= =
2 2 y d sin d L 2 (1.20 10 - 4 m) sin 0.500 = 13.2 rad ( 5.00 10 -7 m) 5.00 10 - 3 m 2 (1.20 10 - 4 m) = 6.28 rad -7 1.20 m (5.00 10 m)
(a)
(
)
(b)
=
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6
Chapter 37 Solutions
(c)
If = 0.333 rad =
2 d sin ,
(5.00 10 - 7 m)(0.333 rad) = sin -1 = sin -1 2 (1.20 10 - 4 m) 2 d
= 1.27 10 - 2 deg
(d) If d sin =
, 4
= sin -1
5 10 - 7 m = sin -1 -4 4d 4(1.20 10 m)
= 5.97 10 - 2 deg
37.12
The path difference between rays 1 and 2 is:
= d sin 1 - d sin 2
For constructive interference, this path difference must be equal to an integral number of wavelengths: d sin 1 - d sin 2 = m , or d(sin 1 - sin 2 ) = m
37.13
(a)
The path difference = d sin and when L >> y
yd (1.80 10 - 2 m)(1.50 10 - 4 m) = = 1.93 10 - 6 m = 1.93 m L 1.40 m
(b) (c)
1.93 10 - 6 m = = 3.00 , or = 3.00 6.43 10 - 7 m
Point P will be a maximum wavelength. since the path difference is an integer multiple of the
37.14
(a)
I = cos2 2 Imax Therefore,
(Equation 37.11)
1/2
I = 2 cos1 I max
= 2 cos1 (0.640)1/2 = 1.29 rad
(b)
=
(486 nm)(1.29 rad) = = 99.8 nm 2 2
Chapter 37 Solutions
7
37.15
d sin Iav = Imax cos2 For small , sin =
1/2
y L
and Iav = 0.750 Imax
y=
L Iav cos1 Imax d
y=
(6.00 107)(1.20 m)
(2.50 10
3
m)
0.750 Imax cos1 Imax
1/2
= 48.0 m
37.16
yd I = I max cos 2 L I I max 6.00 10 - 3 m 1.80 10 - 4 m = cos 2 656.3 10 - 9 m (0.800 m )
(
(
)(
)
) =
0.987
37.17
(a)
From Equation 37.8,
=
2 d 2 d sin =
y y + D2
2
-3 -3 2 yd 2 0.850 10 m 2.50 10 m = = 7.95 rad D 600 10 - 9 m ( 2.80 m )
(
(
)( )
)
(b)
I I max
d cos 2 sin cos 2 2 = = cos 2 m 2 d cos sin max = cos 2
I I max
7.95 rad = 0.453 = cos 2 2 2
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8
Chapter 37 Solutions
Goal Solution Two narrow parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screen is 2.80 m away from the slits. (a) What is the phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe? (b) What is the ratio of the intensity at this point to the intensity at the center of a bright fringe? G: It is difficult to accurately predict the relative intensity at the point of interest without actually doing the calculation. The waves from each slit could meet in phase ( = 0 ) to produce a bright spot of constructive interference, out of phase ( = 180) to produce a dark region of destructive interference, or most likely the phase difference will be somewhere between these extremes, 0 < < 180, so that the relative intensity will be 0 < I I max < 1. The phase angle depends on the path difference of the waves according to Equation 37.8. This phase difference is used to find the average intensity at the point of interest. Then the relative intensity is simply this intensity divided by the maximum intensity. (a) Using the variables shown in the diagram for problem 7 we have,
O:
A:
=
2 yd 2 0.85010 -3 m ( 0.00250 m) 2 d 2 d y sin = = = 7.95 rad = 2 + 1.66 rad = 95.5 2 -9 y + L2 L 60010 m ( 2.80 m)
( (
) )
(b)
I I max
d cos 2 sin cos 2 2 95.5 = = = cos 2 = cos 2 = 0.452 2 L 2 l cos 2 ( m ) 2 d cos sin max
L:
It appears that at this point, the waves show partial interference so that the combination is about half the brightness found at the central maximum. We should remember that the equations used in this solution do not account for the diffraction caused by the finite width of each slit. This diffraction effect creates an "envelope" that diminishes in intensity away from the central maximum as shown by the dotted line in Figures 37.13 and P37.60. Therefore, the relative intensity at y = 2.50 mm will actually be slightly less than 0.452.
37.18
(a)
The resultant amplitude is Er = E0 sin t + E0 sin( t + ) + E0 sin( t + 2 ) , where
=
2 d sin .
Er = E0 (sin t + sin t cos + cos t sin + sin t cos 2 + cos t sin 2 ) Er = E0 (sin t) 1 + cos + 2 cos 2 - 1 + E0 (cos t)(sin + 2 sin cos ) Er = E0 (1 + 2 cos )(sin t cos + cos t sin ) = E0 (1 + 2 cos ) sin ( t + )
(
)
Chapter 37 Solutions
9
Then the intensity is where the time average of
2 2 I Er = E0 (1 + 2 cos )2
1 2
sin 2 ( t + ) is 1 2. 1 2 so 2 d sin I = I max 1 + 2 cos
2
2 From one slit alone we would get intensity I max E0
(b)
Look at the N = 3 graph in Figure 37.13. Minimum intensity is zero, attained where cos = -1/2 . One relative maximum occurs at cos = -1.00 , where I = I max . The larger local maximum happens where cos = +1.00, giving I = 9.00 I 0 . The ratio of intensities at primary versus secondary maxima is 9.00 .
*37.19
(a)
We can use sin A + sin B = 2 sin ( A 2 + B 2) cos ( A 2 - B 2) to find the sum of the two sine functions to be E1 + E2 = ( 24.0 kN C) sin (15x - 4.5t + 35.0) cos 35.0 E1 + E2 = (19.7 kN C) sin (15x - 4.5t + 35.0) Thus, the total wave has amplitude 19.7 kN/C and has a constant phase difference of 35.0 from the first wave.
(b)
In units of kN/C, the resultant phasor is E R = E1 + E2 = (12 .0 i) + (12 .0 cos 70.0 i + 12 .0 sin 70.0 j ) = 16.1i + 11.3 j ER =
y ER E2
kx - t
(16.1)2 + (11.3)2
at tan -1 (11.3 16.1) = 19.7 kN/C at 35.0
E1
70.0
x
(c)
E R = 12 .0 cos 70.0 i + 12 .0 sin 70.0 j +15.5cos 80.0i - 15.5sin 80.0 j +17.0 cos 160i + 17.0 sin 160 j
ER E3
y E1 E2
kx - t
x
E R = - 9.18 i + 1.83 j = 9.36 kN/C at 169 The wave function of the total wave is EP = (9.36 kN C) sin (15x - 4.5t + 169)
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10
Chapter 37 Solutions E R = E0 i + (i cos 20.0 + j sin 20.0) +(i cos 40.0 +j sin 40.0)
37.20
(a)
[
]
y ER E3 E2 E1 x
E R = E0 [2 .71i + 0.985 j] = 2 .88 E0 at 20.0 = 2.88E0 at 0.349 rad EP = 2.88E0 sin ( t + 0.349) (b) E R = E0 i + (i cos 60.0 + j sin 60.0) +(i cos120 + j sin 120)
[
]
y ER
E3
E R = E0 [1.00i + 1.73 j] = 2.00E0 at 60.0 = 2.00E0 at 3 rad EP = 2.00E0 sin( t + 3) E R = E0 i + (i cos 120 + j sin 120) +(i cos 240 + j sin 240) E R = E0 [0 i + 0 j] = 0 EP = 0 (d) E R = E0 i + (i cos 3 2 + j sin 3 2) +(i cos 3 + j sin 3 ) E R = E0 [0 i - 1.00 j] = E0 at 270 = E0 at 3 2 rad EP = E0 sin ( t + 3 2)
E1
E2 x
(c)
[
]
y E3 E2 E1 x
[
]
ER
y
E1 E2 E3
x
37.21
E R = 6.00 i + 8.00 j =
(6.00)2 + (8.00)2
at tan -1(8.00 6.00)
y ER 8.00
E R = 10.0 at 53.1 = 10.0 at 0.927 rad EP = 10.0 sin (100 t + 0.927 )
6.00
2
x
37.22
If E1 = E10 sin t and E2 = E20 sin ( t + ) , then by phasor addition, the amplitude of E is E0 =
y
(E10 + E20 cos )2 + (E20 sin )2 =
E10 2 + 2E10E20 cos + E20 2 sin = E20 sin E0
E10
E0
E20
x
and the phase angle is found from
Chapter 37 Solutions E R = 12.0 i + (18.0 cos 60.0 i + 18.0 sin 60.0 j ) E R = 21.0 i + 15.6 j = 26.2 at 36.6 EP = 26.2 sin ( t + 36.6)
11
37.23
y ER
18.0
12.0
60.0
x
37.24
Constructive interference occurs where m = 0, 1, 2, 3, . . ., for 2 (x1 - x2 ) 2 x2 2 x1 = 2 m - 2 f t + - 2 f t + + - - = 2 m 6 8 6 8 1 (x1 - x2 ) 1 + - =m 12 16 1 x1 - x2 = m - 48 m = 0, 1, 2, 3, . . .
37.25
See the figure to the right:
= /2
37.26
2 2 2 ER = E1 + E2 - 2E1E2 cos , where = 180 - .
Since I E 2 , I R = I1 + I 2 + 2 I1I 2 cos
37.27
Take
= 360 N
where N defines the
y The N = 6 case
number of coherent sources. Then, ER =
m=1
E0 sin ( t + m ) = 0
60.0 12.0
N
=
360 N
=
360 6
= 60. 0
In essence, the set of N electric field components complete a full circle and return to zero.
x
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12
Chapter 37 Solutions
*37.28
Light reflecting from the first surface suffers phase reversal. Light reflecting from the second surface does not, but passes twice through the thickness t of the film. So, for constructive interference, we require n 2 + 2t = n
where n = n
is the wavelength in the material. Then
2t =
n = 2n 2
= 4nt = 4 1.33 115 nm = 612 nm
37.29
(a)
The light reflected from the top of the oil film undergoes phase reversal. Since 1.45 > 1.33, the light reflected from the bottom undergoes no reversal. For constructive interference of reflected light, we then have 1 2nt = m + 2
or
m =
2nt m + 1 2
=
2(1.45)(280 nm) m + 1 2
Substituting for m, we have
m = 0: 0 = 1620 nm (infrared) m = 1:
1 = 541 nm (green)
m = 2: 2 = 325 nm (ultraviolet) Both infrared and ultraviolet light are invisible to the human eye, so the dominant color i n the reflected light is green . (b) The dominant wavelengths in the transmitted light are those that produce destructive interference in the reflected light. The condition for destructive interference upon reflection is 2nt = m or
m =
2nt 812 nm = m m
Substituting for m gives: m = 1,
1= 812 nm (near infrared)
m = 2, 2 = 406 nm (violet) m = 3, 3 = 271 nm (ultraviolet)
Of these. the only wavelength visible to the human eye (and hence the dominate wavelength observed in the transmitted light) is 406 nm. Thus, the dominant color in the transmitted light is violet .
Chapter 37 Solutions
13
37.30
Since 1 < 1.25 < 1.33, light reflected both from the top and from the bottom surface of the oil suffers phase reversal. For constructive interference we require 2t = m cons n
and for destructive interference,
m + 1 des 2 2t = n
Then 2 Therefore,
cons 1 640 nm =1+ = = 1.25 2m 512 nm dest
2(640 nm) = 512 nm 2(1.25)
and
m =
t=
37.31
1 Treating the anti-reflectance coating like a camera-lens coating, 2t = m + 2 n Let m = 0:
3.00 cm t = 4n = 4(1.50) = 0.500 cm
This anti-reflectance coating could be easily countered by changing the wavelength of the radar--to 1.50 cm--now creating maximum reflection!
37.32
1 2nt = m + 2 Minimum t =
so
1 t= m+ 2 2n
1 (500 nm) = 96.2 nm 2 2(1.30)
37.33
Since the light undergoes a 180 phase change at each surface of the film, the condition for The film thickness is constructive interference is 2nt = m , or = 2nt m. t = 1.00 10 - 5 cm = 1.00 10 - 7 m = 100 nm . Therefore, the wavelengths intensified in the reflected light are
=
or
2(1.38)(100 nm ) 276 nm = m m
where
m = 1, 2, 3, . . .
1= 276 nm , 2 = 138 nm , . . . . All reflection maxima are in the ultraviolet and beyond.
No visible wavelengths are intensified.
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14
Chapter 37 Solutions
*37.34 (a)
For maximum transmission, we want destructive interference in the light reflected from the front and back surfaces of the film. If the surrounding glass has refractive index greater than 1.378, light reflected from the front surface suffers no phase reversal and light reflected from the back does undergo phase reversal. This effect by itself would produce destructive interference, so we want the distance down and back to be one whole wavelength in the film: 2t = n. t=
656.3 nm = = 238 nm 2n 2(1.378)
(b) (c)
The filter will expand. As t increases in 2nt = , so does increase . Destructive interference for reflected light happens also for in 2nt = 2 , or
= 1.378(238 nm) = 328 nm (near ultraviolet) .
37.35
If the path length = , the transmitted light will be bright. Since = 2d = , dmin =
580 nm = = 290 nm 2 2
37.36
The condition for bright fringes is 2t +
=m 2n n
m = 1, 2, 3, . . .
R
From the sketch, observe that
t
2 r2 2 R r t = R(1 - cos ) R 1 - 1 + = = 2 2 R 2R r
The condition for a bright fringe becomes Thus, for fixed m and , Therefore, nliquidr 2 = nair ri2 and f
1 r2 = m- . 2 n R nr 2 = constant. nliquid = (1.00)
(1.50 cm)2 (1.31 cm)2
= 1.31
Chapter 37 Solutions
15
37.37
For destructive interference in the air, 2t = m . For 30 dark fringes, including the one where the plates meet, t= 29(600 nm) = 8.70 10 6 m 2
Therefore, the radius of the wire is r= d 8.70 m = 4.35 m 2 = 2
Goal Solution An air wedge is formed between two glass plates separated at one edge by a very fine wire as shown i n Figure P37.37. When the wedge is illuminated from above by 600-nm light, 30 dark fringes are observed. Calculate the radius of the wire. G: O: The radius of the wire is probably less than 0.1 mm since it is described as a "very fine wire." Light reflecting from the bottom surface of the top plate undergoes no phase shift, while light reflecting from the top surface of the bottom plate is shifted by , and also has to travel an extra distance 2t, where t is the thickness of the air wedge. For destructive interference, 2t = m
(m = 0,
1, 2, 3, . . .)
The first dark fringe appears where m = 0 at the line of contact between the plates. The 30th dark fringe gives for the diameter of the wire 2t = 29 , and t = 14.5 . A: L: r= t = 7.25 = 7.25 600 10 -9 m = 4.35 m 2
(
)
This wire is not only less than 0.1 mm; it is even thinner than a typical human hair ( ~ 50 m).
37.38
For destructive interference, 2t =
m . n
At the position of the maximum thickness of the air film, m= 2tn 2(4.00 10 - 5 m)(1.00) = = 146.5 5.461 10 - 7 m
The greatest integer value is m = 146.
Therefore, including the dark band at zero thickness, there are
147 dark fringes .
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16
Chapter 37 Solutions
*37.39
For total darkness, we want destructive interference for reflected light for both 400 nm and 600 nm. With phase reversal at just one reflecting surface, the condition for destructive interference is 2nairt = m m = 0, 1, 2, . . .
The least common multiple of these two wavelengths is 1200 nm, so we get no reflected light at 2(1.00)t = 3( 400 nm) = 2(600 nm) = 1200 nm, so t = 600 nm at this second dark fringe. By similar triangles, 600 nm 0.0500 mm = , x 10.0 cm x = 600 10 - 9 m
or the distance from the contact point is
(
0.100 ) 5.00 10 m m =
-5
1.20 mm
37.40
2t = m
m=
2t 2(1.80 10 - 4 m) = = 654 dark fringes 550.5 10 - 9 m
37.41
When the mirror on one arm is displaced by l, the path difference increases by 2 l. A shift resulting in the formation of successive dark (or bright) fringes requires a path length change of one-half wavelength. Therefore, 2 l = m /2, where in this case, m = 250. l = m
-7 ( 250) 6.328 10 m = = 39.6 m 4 4
(
)
37.42
Distance = 2(3.82 10 4 m) = 1700 The light is blue
= 4.49 10 7 m = 449 nm
37.43
Counting light going both directions, the number of wavelengths originally in the cylinder is 2L 2L 2nL m1 = . It changes to m2 = = as the cylinder is filled with gas. If N is the number of n 2L bright fringes passing, N = m2 - m1 = (n - 1) , or the index of refraction of the gas is 35 633 10 - 9 m N n = 1+ = 1+ = 1.000 369 2L 2(0.0300 m )
(
)
Chapter 37 Solutions
17
37.44
Counting light going both directions, the number of wavelengths originally in the cylinder is 2L 2L 2nL m1 = . It changes to m2 = = as the cylinder is filled with gas. If N is the number of n 2L bright fringes passing, N = m2 - m1 = (n - 1) , or the index of refraction of the gas is n = 1+ N 2L
37.45
The wavelength is =
c 3.00 108 m s = = 5.00 m . f 60.0 106 s -1
Along the line AB the two traveling waves going in opposite directions add to give a standing wave. The two transmitters are exactly 2.00 wavelengths apart and the signal from B, when it arrives at A, will always be in phase with transmitter B. Since B is 180 out of phase with A, the two signals always interfere destructively at the position of A. The first antinode (point of constructive interference) is located at distance
5.00 m = = 1.25 m from the node at A. 4 4
*37.46 (a)
My middle finger has width d = 2 cm. Two adjacent directions of constructive interference for 600-nm light are described by d sin = m Thus, and (b) Choose 1 = 20
0 = 0
(2 10 2 m) sin 1 = 1 ( 6 10 7 m)
1 = 2 10 3 degree 1 0 ~ 10 3 degree
2 10 2 m sin 20 = 1
= 7 mm
Millimeter waves are microwaves 3 108 m/s c = ~ 1011 Hz 7 10 3 m
f=
37.47
If the center point on the screen is to be a dark spot rather than bright, passage through the plastic must delay the light by one-half wavelength. Calling the thickness of the plastic t.
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18
Chapter 37 Solutions
t nt t 1 + = = 2 ( n) plastic.
or
t =
2(n - 1)
where n is the index of refraction for the
Chapter 37 Solutions
19
*37.48
No phase shift upon reflection from the upper surface (glass to air) of the film, but there will be a shift of 2 due to the reflection at the lower surface of the film (air to metal). The total phase difference in the two reflected beams is then = 2nt + 2 . For constructive interference, = m , or 2(1.00)t + 2 = m . Thus, the film thickness for the m th order bright fringe is: 1 - , tm = m - =m 2 4 2 2 and the thickness for the m 1 bright fringe is: - . tm-1 = (m - 1) 2 4
Therefore, the change in thickness required to go from one bright fringe to the next is t = tm - tm-1 = 2. To go through 200 bright fringes, the change in thickness of the air film must be: 200( 2) = 100 . Thus, the increase in the length of the rod is L = 100 = 100 5.00 10 - 7 m = 5.00 10 - 5 m, From L = L i ( T ) , we have:
(
)
=
L 5.00 10 - 5 m = = 20.0 10 - 6 C - 1 L i ( T ) (0.100 m )( 25.0C)
*37.49
Since 1 < 1.25 < 1.34, light reflected from top and bottom surfaces of the oil undergoes phase reversal. The path difference is then 2t, which must be equal to mn = m n
for maximum reflection, with m = 1 for the given first-order condition and n = 1.25. So t= m 1(500 nm) = 200 nm 2n = 2(1.25) 1.00 m3 = (200 nm)A
The volume we assume to be constant: A= 1.00 m3 = 5.00 106 m2 = 5.00 km2 200(10 9 m)
37.50
For destructive interference, the path length must differ by m . We may treat this problem as a double slit experiment if we remember the light undergoes a /2-phase shift at the mirror. The second slit is the mirror image of the source, 1.00 cm below the mirror plane. Using Equation 37.5, ydark = m L 1(5.00 10 7 m)(100 m) = = 2.50 mm d (2.00 10 2 m)
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20
Chapter 37 Solutions
37.51
One radio wave reaches the receiver R directly from the distant source at an angle above the horizontal. The other wave undergoes phase reversal as it reflects from the water at P. Constructive interference first occurs for a path difference of (1) d= 2 The angles in the figure are equal because they each form part of a right triangle with a shared angle at R'. It is equally far from P to R as from P to R', the mirror image of the telescope. So the path difference is The wavelength is Substituting for d and in Equation (1), Solving for the angle , 5.00 m sin = 80.0 m d = 2(20.0 m) sin = (40.0 m) sin c 3.00 108 m/s = f = = 5.00 m 60.0 106 Hz (40.0 m) sin = and 5.00 m 2
= 3.58
37.52
2 (15.0 km)2 + h2 = 30.175 km (15.0 km)2 + h2 = 227.63 h = 1.62 km
37.53
From Equation 37.13, Let 2 equal the wavelength for which Then
I Imax
yd = cos2 L
I I2 Imax Imax = 0.640
2= yd 1 I1 L = 1 cos Imax
1/2
yd/L cos 1 (I2/Imax)1/2
But
= (600 nm) cos 1 (0.900) = 271 nm
Substituting this value into the expression for 2, I Note that in this problem, cos 1 Imax
1/2
2 =
271 nm = 421 nm cos 1 (0.6401/2)
must be expressed in radians.
Chapter 37 Solutions
21
37.54
For Young's experiment, use = d sin = m . Then, at the point where the two bright lines coincide, d sin = m1 1= m2 2 so
1 540 m2 6 = = = 2 450 m1 5
sin = 6 2 6(450 nm) = = 0.0180 0.150 mm d
Since sin and L = 1.40 m ,
x = L = (0.0180)(1.40 m ) = 2.52 cm
37.55
For dark fringes, and at the edge of the wedge, When submerged in water, m= 2(1.33)(42)(500 nm) so 500 nm
2nt = m t= 84(500 nm) . 2
2nt = m m + 1 = 113 dark fringes
*37.56
At entrance, 1.00 sin 30.0 = 1.38 sin 2 2 = 21.2 Call t the unknown thickness. Then t cos 21.2 = a c tan 21.2 = t b sin 1 = 2c t a = cos 21.2 c = t tan 21.2 b = 2t tan 21.2 sin 30.0
The net shift for the second ray, including the phase reversal on reflection of the first, is
2an b 2
where the factor n accounts for the shorter wavelength in the film. For constructive interference, we require 2an b 2 = m The minimum thickness will be given by 2an b 2 = 0.
nt = 2an b = 2 cos 21.2 2t(tan 21.2) sin 30.0 2
2 1.38 590 nm = cos 21.2 2 tan 21.2 sin 30.0 t = 2.57t 2 t = 115 nm
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22
Chapter 37 Solutions
37.57
The shift between the two reflected waves is = 2na - b - 2 where a and b are as shown in the ray diagram, n is the index of refraction, and the factor of 2 is due to phase reversal at the top surface. For constructive interference, = m where m has integer values. This condition becomes 1 2na - b = m + 2 From the figure's geometry, a= t , cos 2 c = asin 2 = t sin 2 , cos 2
1
b 2c
1
t
n
2
c
a
(1) b = 2c sin 1 = b= 2nt sin 2 2 cos 2 2t sin 2 sin 1 cos 2
Also, from Snell's law, sin 1 = n sin 2 . Thus,
With these results, the condition for constructive interference given in Equation (1) becomes: t 2nt sin 2 2 2nt 1 2n = 1 - sin 2 2 = m + - cos cos 2 2 cos 2 2
(
)
or
1 2nt cos 2 = m + 2
37.58
(a)
Minimum: 2nt = m 2 1 Maximum: 2nt = m' + 2 1 1 for 1 > 2, m' + 2 < m Then 1 2nt = m 2 = m 2 1 2m 2 = 2m 1 1 so
m = 0, 1, 2, . . . m' = 0, 1, 2, . . .
m' = m 1
so
m=
1 2( 1 2)
(b)
500 m = 2(500 370) = 1.92 2 (wavelengths measured to 5 nm) [Minimum]: 2nt = m 2 2(1.40)t = 2(370 nm) t = 268 nm t = 264 nm
1 [Maximum]: 2nt = m 1 + 2 = 1.5 2(1.40)t = (1.5)500 nm Film thickness = 266 nm
Chapter 37 Solutions
23
37.59
From the sketch, observe that x = h 2 + ( d 2) =
2
d
4h 2 + d 2 2
h
x
x
Including the phase reversal due to reflection from the ground, the total shift between the two waves is = 2x - d - 2. (a)
d/2
For constructive interference, the total shift must be an integral number of wavelengths, or = m where m = 0, 1, 2, 3, . . . . Thus, For the longest wavelength, m = 0, giving 1 2x - d = m + 2 or
=
4x - 2d 2m + 1
= 4 x - 2 d = 2 4h 2 + d 2 - 2d
1 = m - 2 2x - d = m where m = 1, 2, 3, . . . . or 4h 2 + d 2 - d
(b)
For destructive interference, Thus, For the longest wavelength, m = 1 giving
=
2x - d . m
= 2x - d =
37.60
Call t the thickness of the sheet. The central maximum corresponds to zero phase difference. Thus, the added distance r traveled by the light from the lower slit must introduce a phase difference equal to that introduced by the plastic film. As light advances through distance t i n air, the number of cycles it goes through is t / a . t nt = (a /n) a n t t = 2 a a 2 a a r = = (nt t) = (n 1)t a 2 2
The number of cycles in the sheet is
Thus, the sheet introduces phase difference
The corresponding difference in path length is
Note that the wavelength of the light does not appear in this equation. In the figure, the two rays from the slits are essentially parallel, so the angle may be expressed as tan = r d = y L. Substituting for r and solving for y' gives t(n 1)L (5.00 10 5 m)(1.50 1)(1.00 m) L = = 0.0833 m = 8.33 cm y' = r d = d (3.00 10 4 m)
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24
Chapter 37 Solutions
37.61
Call t the thickness of the film. The central maximum corresponds to zero phase difference. Thus, the added distance r traveled by the light from the lower slit must introduce a phase difference equal to that introduced by the plastic film. The phase difference is t = 2 (n 1) a The corresponding difference in path length r is a a t r = = 2 (n 1) = t(n 1) 2 2 a
Note that the wavelength of the light does not appear in this equation. In the figure, the two rays from the slits are essentially parallel. r y' tan = = Thus the angle may be expressed as d L Eliminating r by substitution, y' t(n 1) L = d gives y' = t(n 1)L d
Goal Solution Consider the double-slit arrangement shown in Figure P37.60, where the slit separation is d and the slit to screen distance is L. A sheet of transparent plastic having an index of refraction n and thickness t is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y'. Find y'. G: O: Since the film shifts the pattern upward, we should expect y to be proportional to n, t , and L . The film increases the optical path length of the light passing through the upper slit, so the physical distance of this path must be shorter for the waves to meet in phase ( = 0 ) to produce the central maximum. Thus, the added distance r traveled by the light from the lower slit must introduce a phase difference equal to that introduced by the plastic film. First calculate the additional phase difference due to the plastic. Recall that the relation between phase difference and path difference is = 2 . The presence of plastic affects this by changing the wavelength of the light, so that the phase change of the light in air and plastic, as it travels over the thickness t is
A:
air =
2 t air
and
plastic =
=
2 t air / n
Thus, plastic causes an additional phase change of
2 t (n - 1) air
Next, in order to interfere constructively, we must calculate the additional distance that the light from the bottom slit must travel. r = air = t(n - 1) 2 y = L t( - 1) n dL
In the small angle approximation we can write r = y d L, so L:
As expected, y is proportional to t and L . It increases with increasing n, being proportional to (n - 1). It is also inversely proportional to the slit separation d, which makes sense since slits that are closer together make a wider interference pattern.
Chapter 37 Solutions
25
37.62
=
c 3.00 108 m s = = 200 m f 1.50 106 Hz
For destructive interference, the path difference is one-half wavelength. Thus,
= 100 m = x + x 2 + 2.00 10 4 m 2
(
)
2
- 2.00 10 4 m ,
or Squaring and solving,
2.01 10 4 m - x = x 2 + 2.00 10 4 m x = 99.8 m
(
)
2
37.63
(a)
Constructive interference in the reflected light requires
2t = m +
has m = 0 and the 55th has m = 54, so at the edge of the lens t= 54.5(650 10 - 9 m) = 17.7 m 2
(
1 2
) .
The first bright ring
Now from the geometry in Figure 37.18, the distance from the center of curvature down to the flat side of the lens is R2 -r2 = R-t R= or R 2 - r 2 = R 2 - 2Rt + t 2
r 2 + t 2 (5.00 10 - 2 m)2 + (1.77 10 - 5 m)2 = = 70.6 m 2t 2(1.77 10 - 5 m) so f = 136 m
(b)
1 1 1 1 1 = (n - 1) - = 0.520 - f R2 R2 - 70.6 m
37.64
Bright fringes occur when
2t = 2t =
1 m+ 2 n m n
and dark fringes occur when
The thickness of the film at x is 1 l m+ 2 2hn
t=
h x. l xdark =
Therefore,
xbright =
and
lm 2hn
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26
Chapter 37 Solutions
37.65
7 4 7 4 E R = E1 + E2 + E3 = cos + 3.00 cos + 6.00 cos + 6.00 sin i + sin + 3.00 sin j 6 2 3 6 2 3
E R = -2.13 i - 7.70 j ER =
y 2 E1 x E2 ER E3
(- 2 .13) + (- 7.70)
2
- 7.70 at tan = 7.99 at 4.44 rad - 2 .13
-1
Thus,
EP = 7.99 sin( t + 4.44 rad)
37.66
For bright rings the gap t between surfaces is given by 2t = m + has m = 99.
(
1 2
)
The first bright ring has m = 0 and the hundredth
1 So, t = 2 (99.5) 500 10 - 9 m = 24.9 m .
(
)
Call rb the ring radius. From the geometry of the figure at the right,
2 2 t = r - r 2 - rb - R - R 2 - rb 2 2 2 2 1 rb 1 rb 1 rb 1 rb Since rb << r , we can expand in series: t = r - r 1 - - - R + R 1 - = 2 2 2r 2R 2 r 2 R
2t rb = 1/ r - 1/ R
1/2
2 24.9 10 - 6 m = 1/ 4.00 m - 1/ 12.0 m
(
)
1/2
= 1.73 cm
*37.67
The shift between the waves reflecting from the top and bottom surfaces of the film at the point where the film has thickness t is = 2t nfilm + ( 2) , with the factor of 2 being due to a phase reversal at one of the surfaces. For the dark rings (destructive interference), the total shift should be = m + that t = m 2nfilm .
R
(
1 2
) with m = 0, 1, 2, 3, . . . .
R2
=
nfilm r
This requires
t
To find t in terms of r and R, Since t is much smaller than R,
r 2 + (R - t)2
so r 2 and
=
2Rt + t 2 m r 2 2Rt = 2R . 2nfilm
t 2 << 2Rt m R nfilm
Thus, where m is an integer,
r
Chapter 37 Solutions
27
37.68
(a)
Bright bands are observed when Hence, the first bright band (m = 0) corresponds to Since x1 t1 = , we have x2 t2
1 2nt = m + 2 nt = 4.
t 680 nm = 4.86 cm x2 = x1 2 = x1 2 = ( 3.00 cm) 420 nm t1 1 t2 =
(b)
t1 =
1 420 nm = = 78.9 nm 4n 4(1.33)
t1 78.9 nm = = x1 3.00 cm
2 680 nm = = 128 nm 4n 4(1.33)
(c)
tan =
2.63 10 - 6 rad
37.69
1 2h sin = m + 2 2h y 1 = 2L 2
bright
so
h=
(2.00 m) 606 10- 9 m L = = 0.505 mm 2y 2 1.2 10 - 3 m
(
(
)
)
37.70
Superposing the two vectors, ER = E1 + E2 ER = E1 + E2 = ER = E2 E2 E 2 2 E 2 E0 + 0 cos + 0 sin = E0 + E0 cos + 0 cos 2 + 0 sin 2 3 3 9 9 3
2 2
10 2 2 2 E0 + E0 cos 9 3
Since intensity is proportional to the square of the amplitude, I= 10 2 I max + I max cos 9 3
Using the trigonometric identity cos = 2 cos 2 I=
- 1, this becomes 2
10 2 4 4 I max + I max 2 cos 2 - 1 = I max + I max cos 2 , 9 9 3 3 2 2 4 I max 1 + 3 cos 2 9 2
or
I=
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CHAPTER 38
38.1
sin =
6.328 10 7 = = 2.11 10 3 a 3.00 10 4
y 1.00 m = tan sin = (for small ) 2y = 4.22 mm
38.2
The positions of the first-order minima are y L sin = a. Thus, the spacing between these two minima is y = 2( a)L and the wavelength is
=
-3 -3 y a 4.10 10 m 0.550 10 m = = 547 nm 2 L 2 2.06 m
38.3
m y = sin = a L
y = 3.00 10 3 m
m = 3 1 = 2
and
a=
m L y
a=
(2)(690 10 9 m)(0.500 m) 3.00 10 3 m
= 2.30 10 4 m
*38.4
For destructive interference,
5.00 cm sin = m a = a = 36.0 cm = 0.139
d L = tan d = 91.2 cm gives
and
= 7.98
d = L tan = (6.50 m) tan 7.98 = 0.912 m
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Chapter 38 Solutions 407 *38.5 If the speed of sound is 340 m/s, v 340 m/s = f = 650 /s = 0.523 m Diffraction minima occur at angles described by 1.10 m sin 1 = 1(0.523 m) 1 = 28.4 1.10 m sin 2 = 2(0.523 m) 2 = 72.0 1.10 m sin 3 = 3(0.523 m) 3 nonexistent Maxima appear straight ahead at 0 (1.10 m) sin x = 1.5(0.523 m) and left and right at an angle given approximately by a sin = m
x 46
There is no solution to a sin = 2.5, so our answer is already complete, with three sound maxima.
38.6
(a)
sin =
y m = L a
Therefore, for first minimum, m = 1 and L= 7.50 10 - 4 m 8.50 10 - 4 m ay = 1.09 m = m (1) 587.5 10 - 9 m
(
(
)(
)
)
(b)
w = 2y1
yields y1 = 0.850 mm
w = 2 0.850 10 - 3 m = 1.70 mm
(
)
38.7
sin
y 4.10 10 - 3 m = L 1.20 m
-4 asin 4.00 10 m 4.10 10 - 3 m = = = 7.86 rad 2 1.20 m 546.1 10 - 9 m
(
)
I I max
sin ( / 2) sin (7.86) -2 = = 7.86 = 1.62 10 /2
2 2
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408 Chapter 38 Solutions 38.8 Bright fringes will be located approximately midway between adjacent dark fringes. Therefore, for the second bright fringe, let m = 2.5 and use sin = m a y L . The wavelength will be
ay (0.800 10 - 3 m)(1.40 10 - 3 m) = = 5.60 10 - 7 m = 560 nm mL 2.5(0.800 m)
38.9
Equation 38.1 states that sin = m a , where m = 1, 2, 3, . . . . The requirement for m = 1 is from an analysis of the extra path distance traveled by ray 1 compared to ray 3 in Figure 38.5. This extra distance must be equal to / 2 for destructive interference. When the source rays approach the slit at an angle , there is a distance added to the path Then, for difference (of ray 1 compared to ray 3) of ( a / 2) sin destructive interference, a a sin + sin = 2 2 2 so sin =
- sin . a
sin = sin = sin = 2 - sin a 3 - sin a m - sin a
Dividing the slit into 4 parts leads to the 2nd order minimum: Dividing the slit into 6 parts gives the third order minimum: Generalizing, we obtain the condition for the mth order minimum:
*38.10 (a)
Double-slit interference maxima are at angles given by d sin = m . For m = 0 , For m = 1, (2 .80 m ) sin = 1(0.5015 m ) : Similarly, for m = 2, 3, 4, 5 and 6,
0 = 0 1 = sin - 1(0.179) = 10.3 2 = 21.0 , 3 = 32.5 , 4 = 45.8 , 5 = 63.6 , and 6 = sin -1 (1.07 ) = nonexistent .
Thus, there are (b)
5 + 5 + 1 = 11 directions for interference maxima .
We check for missing orders by looking for single-slit diffraction minima, at a sin = m . For m = 1,
(0.700 m) sin = 1(0.5015 m)
and
1 = 45.8 .
nine bright fringes , at
Thus, there is no bright fringe at this angle.
There are only
= 0, 10.3, 21.0, 32.5, and 63.6 .
Chapter 38 Solutions 409 sin ( asin ) I = I max asin At = 0,
2
(c)
sin 1
and
I I max
1.00
At = 10.3,
asin (0.700 m ) sin 10.3 = = 0.785 rad = 45.0 0.5015 m
I I max sin 45.0 = = 0.811 0.785 and I I max = 0.405
2
Similarly, at = 21.0, At = 32.5, At = 63.6 ,
asin = 1.57 rad = 90.0
and
asin = 2.36 rad = 135 asin = 3.93 rad = 225
I = 0.0901 I max I = 0.0324 I max
and
38.11
5.00 10 7 m = 1.00 10 3 rad sin = a = 5.00 10 4
38.12
y min = L = 1.22 D y= (1.22)(5.00 10 7)(0.0300) 7.00 10 3 = 2.61 m
y = radius of star-image L = length of eye
= 500 nm D = pupil diameter = half angle
38.13
Following Equation 38.9 for diffraction from a circular opening, the beam spreads into a cone of half-angle
(632.8 10 9 m) min = 1.22 D = 1.22 (0.00500 m) = 1.54 10 4 rad
The radius of the beam ten kilometers away is, from the definition of radian measure, r beam = min (1.00 104 m) = 1.544 m and its diameter is d beam = 2r beam = 3.09 m
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410 Chapter 38 Solutions Goal Solution A helium-neon laser emits light that has a wavelength of 632.8 nm. The circular aperture through which the beam emerges has a diameter of 0.500 cm. Estimate the diameter of the beam 10.0 km from the laser. G: A typical laser pointer makes a spot about 5 cm in diameter at 100 m, so the spot size at 10 km would be about 100 times bigger, or about 5 m across. Assuming that this HeNe laser is similar, we could expect a comparable beam diameter. We assume that the light is parallel and not diverging as it passes through and fills the circular aperture. However, as the light passes through the circular aperture, it will spread from diffraction according to Equation 38.9.
O:
A:
The beam spreads into a cone of half-angle
min
632.8 10 -9 m = 1.22 = 1.22 = 1.54 10 -4 rad D (0.00500 m)
(
)
The radius of the beam ten kilometers away is, from the definition of radian measure, r beam = min 1.00 10 4 m = 1.54 m and its diameter is L: dbeam = 2r beam = 3.09 m
(
)
The beam is several meters across as expected, and is about 600 times larger than the laser aperture. Since most HeNe lasers are low power units in the mW range, the beam at this range would be so spread out that it would be too dim to see on a screen.
38.14
d min = 1.22 D = L
d 1 mi 5.80 10 7 m 1.22 = 1.80 mi 1609 m 3 4.00 10 m
d = 0.512 m
The shortening of the wavelength inside the patriot's eye does not change the answer.
38.15
By Rayleigh's criterion, two dots separated center-to-center by 2.00 mm would overlap when
min =
d = 1.22 L D 2.00 10 - 3 m 4.00 10 - 3 m dD = = 13.1 m 1.22 1.22 500 10 - 9 nm
Thus, L =
(
(
)(
)
)
38.16
D = 1.22
min
=
1.22(5.00 10 7) m = 6.10 cm 1.00 10 5
Chapter 38 Solutions 411 wavelength (distance between sources) min = 1.22 = L pupil diameter Thus, 1.22 ( v t ) 1.22 w = , or w = vt d d w 1.22 650 10 - 9 m ( 20.0 m s)(600 s) 5.00 10 - 3 m
38.17
Taillights are red. Take 650 nm:
(
)
= 1.90 m
38.18
wavelength (distance between sources) min = 1.22 = L pupil diameter 1.22 ( v t ) d
so
1.22 w = vt d
w=
where 650 nm is the average wavelength radiated by the red taillights.
38.19
1.22 d = L D d = 1.22
c = f = 0.0200 m
D = 2.10 m
L = 9000 m
(0.0200 m)(9000 m) = 105 m 2.10 m
38.20
Apply Rayleigh's criterion, where
min =
x = 1.22 D d
min = half-angle of light cone, x = radius of spot, d = diameter of telescope, D = distance to Moon.
= wavelength of light,
Then, the diameter of the spot on the Moon is
-9 8 D 2(1.22) 694.3 10 m 3.84 10 m = 2x = 2 1.22 = 241 m d 2.70 m
(
)(
)
38.21
For 0.100 angular resolution,
1.22
(3.00 10
D
-3
m
) = (0.100)
180
D = 2.10 m
38.22 (a) (b)
L = 88.6 10 9 m, 1.22
D = 0.300 m, 2.40 10 - 6 rad
= 590 10 - 9 m
= min = D
d = min L = 213 km
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412 Chapter 38 Solutions 1.00 cm 1.00 10 - 2 m = = 5.00 m 2000 2000
-9 m 1 640 10 m = = 0.128 d 5.00 10 - 6 m
38.23
d=
sin =
(
)
= 7.35
38.24
The principal maxima are defined by d sin = m m = 0, 1, 2, . . . For m = 1,
= d sin
where is the angle between the central (m = 0) and the first order (m = 1) maxima. The value of can be determined from the information given about the distance between maxima and the grating-to-screen distance. From the figure, 0.488 m tan = 1.72 m = 0.284 so
= 15.8
and
sin = 0.273
The distance between grating "slits" equals the reciprocal of the number of grating lines per centimeter d= 1 = 1.88 10 4 cm = 1.88 103 nm 5310 cm 1
The wavelength is = d sin = (1.88 103 nm)(0.273) = 514 nm
38.25
The grating spacing is
d=
(1.00 10 2 m) = 2.22 10 6 m 4500
In the 1st-order spectrum, diffraction angles are given by sin =
: d
sin 1 =
656 10 9 m = 0.295 2.22 10 6 m
Figure for Goal Solution
so that for red and for violet so that
1 = 17.17
sin 2 = 434 10 9 m = 0.195 2.22 106 m
2 = 11.26
Chapter 38 Solutions 413 = 17.17 11.26 = 5.91 2 1 2 2 sin 1 = 13.2 = sin 1 d d 3 1 3 2 = sin 1 d sin 1 d = 26.5
The angular separation is in first-order,
In the second-order spectrum,
Again, in the third order,
Since the red line does not appear in the fourth-order spectrum, the answer is complete.
Goal Solution The hydrogen spectrum has a red line at 656 nm and a violet line at 434 nm. What is the angular separation between two spectral lines obtained with a diffraction grating that has 4500 lines/cm? G: Most diffraction gratings yield several spectral orders within the 180 viewing range, which means that the angle between red and violet lines is probably 10 to 30. The angular separation is the difference between the angles corresponding to the red and violet wavelengths for each visible spectral order according to the diffraction grating equation, d sin = m . The grating spacing is d = 1.00 10 -2 m 4500 lines = 2.22 10 -6 m sin = d:
O:
A:
(
)
In the first-order spectrum (m = 1), the angles of diffraction are given by sin 1r = sin 1v = 656 10 -9 m = 0.295 2.22 10 -6 m 434 10 -9 m = 0.195 2.22 10 -6 m so
1r = 17.17 1v = 11.26
1 = 1r - 1v = 17.17 -11.26 = 5.91 2 = sin -1 3 = sin -1 2 r 2 v - sin -1 = 13.2 d d 3 r 3 v - sin -1 = 26.5 d d
so
The angular separation is In the 2nd-order ( m = 2 )
In the third order ( m = 3 ), In the fourth order, the red line is not visible: L:
4r = sin -1 ( 4 r / d) = sin -1 (1.18) does not exist
The full spectrum is visible in the first 3 orders with this diffraction grating, and the fourth is We can also see that the pattern is dispersed more for higher spectral orders so partially visible. that the angular separation between the red and blue lines increases as m increases. It is also worth noting that the spectral orders can overlap (as is the case for the second and third order spectra above), which makes the pattern look confusing if you do not know what you are looking for.
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414 Chapter 38 Solutions
38.26
sin = 0.350: Line spacing = 1.81 m
d=
632.8 nm = = 1.81 103 nm 0.350 sin
*38.27 (a)
d=
1 4 6 3660 lines/cm = 2.732 10 cm = 2.732 10 m = 2732 nm d sin : m At = 10.09 At = 13.71, At = 14.77,
=
= 478.7 nm = 647.6 nm = 696.6 nm = d sin 2
2 2 sin 2 = d = = 2 sin 1 sin 1
(b)
d=
sin 1
and
so
Therefore, if 1 = 10.09 then sin 2 = 2 sin (10.09) gives 2 = 20.51 Similarly, for 1 = 13.71, 2 = 28.30 and for 1 = 14.77, 2 = 30.66
38.28
1 d = 800/mm = 1.25 10 6 m The blue light goes off at angles m sin m = d : 1 5.00 10 7 m 1 = sin 1 = 23.6 1.25 10 6 m
2 = sin 1 (2 0.400) = 53.1 3 = sin 1 (3 0.400) = nonexistent
The red end of the spectrum is at 1 7.00 10 7 m 1 = sin 1 = 34.1 1.25 10 6 m
2 = sin 1 (2 0.560) = nonexistent
So only the first-order spectrum is complete, and it does not overlap spectrum. the second-order
Chapter 38 Solutions 415 From Equation 38.12, R = Nm where In the 1st order, In the 2nd order, In the 3rd order, (b) From Equation 38.11, In the 3rd order, N = ( 3000 lines cm )( 4.00 cm ) = 1.20 10 4 lines. R = (1)(1.20 10 4 lines) = 1.20 10 4 R = (2)(1.20 10 4 lines) = 2 .40 10 4 R = (3)(1.20 10 4 lines) = 3.60 10 4 R=
38.29
(a)
:
400 nm = = 0.0111 nm = 11.1 pm R 3.60 10 4
=
38.30
sin =
m d
Therefore, taking the ends of the visible spectrum to be v = 400 nm and r = 750 nm , the ends the different order spectra are: End of second order: sin 2r = sin 3v = 2 v d = 2 r d = 1500 nm d
Start of third order:
1200 nm d regardless of the value of the
Thus, it is seen that 2r > 3v and these orders must overlap grating spacing d.
38.31
(a)
Nm =
531.7 nm N(1) = 0.19 nm = 2800
(b)
1.32 10 2 m = 4.72 m 2800
38.32
d sin = m
and, differentiating,
d(cos )d = md d / m2 - 2
2
or
d 1 - sin 2 m
d 1 - m2 2 / d 2 m so
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416 Chapter 38 Solutions 1.00 10 - 3 m mm = 4.00 10 - 6 m = 4000 nm 250 lines mm m= d sin
38.33
d=
d sin = m (a)
The number of times a complete order is seen is the same as the number of orders in which the long wavelength limit is visible. mmax = d sin max ( 4000 nm ) sin 90.0 = = 5.71 700 nm or 5 orders is the maximum .
(b)
The highest order in which the violet end of the spectrum can be seen is: mmax = d sin max ( 4000 nm ) sin 90.0 = = 10.0 400 nm or 10 orders in the short - wavelength region
38.34
d=
1 = 2.38 10 - 6 m = 2380 nm 4200 cm or
d sin = m
= sin -1
m d
and
m y = L tan = L tan sin -1 d
Thus,
-1 m 1 m 2 y = L tan sin -1 - tan sin d d 589.6 589 = 0.554 mm - tan sin -1 y = ( 2.00 m ) tan sin -1 2380 2380 -1 2( 589) 2( 589.6) y = ( 2.00 m ) tan sin -1 - tan sin = 1.54 mm 2380 2380 -1 3( 589) 3( 589.6) y = ( 2.00 m ) tan sin -1 - tan sin = 5.04 mm 2380 2380
For m = 1,
For m = 2 ,
For m = 3 ,
Thus, the observed order must be m = 2 .
38.35
2d sin = m :
=
2d sin 2(0.353 10 9 m) sin (7.60) = = 9.34 10 11 m = 0.0934 nm m (1)
38.36
2d sin = m d =
m (1)(0.129 nm) = 2 sin (8.15) = 0.455 nm 2 sin
Chapter 38 Solutions 417 m 1(0.140 10 9 m) sin = 2d = = 0.249 2(0.281 10 9 m)
38.37
2d sin = m
so
and
= 14.4
38.38
m sin m = 2d : sin 2 = 2 = 2(0.218) 2d so
sin 12.6 =
1 = 0.218 2d
2 = 25.9 3 = sin1 (3 0.218) = 40.9 4 = sin1 (4 0.218) = 60.8 5 = sin1 (5 0.218) = nonexistent
Three other orders appear:
38.39
2d sin = m
m 2 0.166 = sin -1 = 31.9 = sin -1 2 0.314 2d
*38.40
Figure 38.25 of the text shows the situation. 2(2 .80 m ) sin 80.0 = 5.51 m 1 2(2 .80 m ) sin 80.0 = 2.76 m 2 2(2 .80 m ) sin 80.0 = 1.84 m 3
2d sin = m
or
=
2d sin m
m=1
1 = 2 = 3 =
m=2
m=3
*38.41
The average value of the cosine-squared function is 1 one-half, so the first polarizer transmits 2 the light. 3 The second transmits cos2 30.0 = 4 . 1 3 3 If = 2 4 I i = 8 I i
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418 Chapter 38 Solutions
38.42
(a)
1 = 20.0, 2 = 40.0, 3 = 60.0
If = Ii cos2( 1 0) cos2( 2 1) cos2( 3 2) If = (10.0 units) cos2(20.0) cos2(20.0) cos2(20.0) = 6.89 units
(b)
1 = 0, 2 = 30.0, 3 = 60.0
If = (10.0 units) cos2(0) cos2(30.0) cos2(30.0) = 5.63 units
38.43
I = Imax cos2 I 1 Imax = 3.00 I Imax I Imax 1 = 5.00 1 = 10.0
I = cos 1 I max
1/2
1/2
(a)
1 = cos 1 3.00 1 = cos 1 5.00 1 = cos 1 10.0
= 54.7
1/2
(b)
= 63.4
1/2
(c)
= 71.6
38.44
By Brewster's law,
n = tan p = tan(48.0) = 1.11
38.45
sin c =
1 n
or
n=
1 1 = = 1.77 sin c sin 34.4
Also, tan p = n.
Thus, p = tan -1 (n) = tan -1 (1.77 ) = 60.5
38.46
sin c = Thus, sin c =
1 n 1 tan p
and
tan p = n cot p = sin c
or
Chapter 38 Solutions 419 38.47 Complete polarization occurs at Brewster's angle Thus, the Moon is 36.9 above the horizon. tan p = 1.33
p = 53.1
38.48
For incident unpolarized light of intensity I max : After transmitting 1st disk: I= I= I= 1 I 2 max 1 I cos 2 2 max 1 cos 2 cos 2 90 - I ( ) 2 max
After transmitting 2nd disk:
After transmitting 3rd disk:
where the angle between the first and second disk is = t . Using trigonometric identities cos 2 = I= 1 1 (1 + cos 2 ) and cos 2 (90 - ) = sin 2 = (1 - cos 2 ) 2 2
we have
1 1 (1 + cos 2 ) (1 - cos 2 ) 1 1 2 I max = 8 I max (1 - cos 2 ) = 8 I max 2 (1 - cos 4 ) 2 2 2 I= 1 I max (1 - cos 4 t) 16
Since = t , the intensity of the emerging beam is given by
38.49
Let the first sheet have its axis at angle to the original plane of polarization, and let each further sheet have its axis turned by the same angle. The first sheet passes intensity The second sheet passes nd the nth sheet lets through I max cos2 . I max cos4 , I max cos2n 0.90I max 45 cos2 5 5 = 0.885, n= 6 where
= 45 n
45 cos2 6 6 = 0.902,
Try different integers to find (a) So (b)
= 7.50
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420 Chapter 38 Solutions *38.50 Consider vocal sound moving at 340 m s and of frequency 3000 Hz. Its wavelength is
=
v 340 m s = = 0.113 m f 3000 Hz
If your mouth, for horizontal dispersion, behaves similarly to a slit 6.00 cm wide, then a sin = m predicts no diffraction minima. You are a nearly isotropic source of this sound. It spreads out from you nearly equally in all directions. On the other hand, if you use a megaphone with width 60.0 cm at its wide end, then a sin = m predicts the first diffraction minimum at
= sin -1
m 0.113 m = sin -1 = 10.9 0.600 m a
This suggests that the sound is radiated mostly toward the front into a diverging beam of angular diameter only about 20. With less sound energy wasted in other directions, more is available for your intended auditors. We could check that a distant observer to the side or behind you receives less sound when a megaphone is used.
38.51
The first minimum is at This has no solution if or if
a sin = 1.
a >1
a < = 632.8 nm
38.52
x = 1.22
5.00 10 -7 m 3 D = 1.22 250 10 m = 30.5 m -3 d 5.00 10 m
(
)
D = 250 103 m = 5.00 10 7 m d = 5.00 10 3 m
38.53
1 d = 400/mm = 2.50 10 6 m (a) 2 541 10 9 m d sin = m a = sin 1 = 25.6 2.50 10 6 m
(b)
=
541 10 9 m = 4.07 10 7 m 1.33
2 4.07 10 7 m b = sin 1 = 19.0 2.50 10 6 m (c) d sin a = 2 2 d sin b = n n sin b = 1 sin a
Chapter 38 Solutions 421 v 3.00 108 m s = = 0.214 m f 1.40 10 9 s -1
*38.54 (a)
=
min = 1.22
(b)
0.214 m 180 60 60 s = 1.50 arc seconds = 1.22 = 7.26 rad = 7.26 rad 3.60 10 4 m D
d = min L = 7.26 10 - 6 rad ( 26 000 ly ) = 0.189 ly 500 10 - 9 m = 1.22 = 50.8 rad -3 D 12.0 10 m
min =
d : L
(
)
(c)
min = 1.22
(10.5 seconds of arc)
(d)
d = min L = 50.8 10 - 6 rad ( 30.0 m ) = 1.52 10 - 3 m = 1.52 mm
(
)
38.55
min = 1.22
(2.00 m) = 0.244 rad = 14.0 = 1.22 D (10.0 m)
38.56
With a grazing angle of 36.0, the angle of incidence is 54.0 tan p = n = tan 54.0 = 1.38 In the liquid,
n = / n = 750 nm / 1.38 = 545 nm
38.57
(a)
3 500 10 - 9 m m d sin = m , or d = = = 2.83 m sin sin 32.0 Therefore, lines per unit length = 1 1 = d 2 .83 10 - 6 m
(
)
or lines per unit length = 3.53 10 5 m = 3.53 10 3 cm .
-9 m m 500 10 m sin = = = m(0.177 ) d 2.83 10 - 6 m
(b)
(
)
For sin 1.00, we must have Therefore, the highest order observed is Total number primary maxima observed is
m(0.177 ) 1.00 or m 5.65 m=5 2m + 1 = 11
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422 Chapter 38 Solutions Goal Solution Light of wavelength 500 nm is incident normally on a diffraction grating. If the third-order maximum of the diffraction pattern is observed at 32.0, (a) what is the number of rulings per centimeter for the grating? (b) Determine the total number of primary maxima that can be observed in this situation. G: The diffraction pattern described in this problem seems to be similar to previous problems that have diffraction gratings with 2 000 to 5 000 lines/mm. With the third-order maximum at 32, there are probably 5 or 6 maxima on each side of the central bright fringe, for a total of 11 or 13 primary maxima. The diffraction grating equation can be used to find the grating spacing and the angles of the other maxima that should be visible within the 180 viewing range. (a) Use Equation 38.10, d sin = m d= m (3)(5.00 10 -7 m) = = 2.83 10 -6 m sin(32.0) sin
O:
A:
Thus, the grating gauge is
1 = 3.534 10 5 lines / m = 3530 lines / cm d sin = m
-7 m(5.00 10 m) = = m(0.177) d 2.83 10 -6 m
(b)
For sin 1, we require that m(1.77 ) 1 or m 5.65. Since m must be an integer, its maximum value is really 5. Therefore, the total number of maxima is 2m + 1 = 11 L: The results agree with our predictions, and apparently there are 5 maxima on either side of the central maximum. If more maxima were desired, a grating with fewer lines/cm would be required; however, this would reduce the ability to resolve the difference between lines that appear close together.
38.58
For the air-to-water interface, tan p = and nwater 1.33 = 1.00 nair
p = 53.1
2 3
(1.00) sin p = (1.33) sin 2
2 = sin - 1
sin 53.1 = 36.9 1.33 tan p = tan 3 = nglass nwater = 1.50 1.33 so
For the water-to-glass interface, The angle between surfaces is
3 = 48.4
= 3 - 2 = 11.5
Chapter 38 Solutions 423 550 10 - 9 m = 1.22 = 1.22 = 1.34 10 - 4 rad -3 D 5.00 10 m
38.59
The limiting resolution between lines min
( (
) )
Assuming a picture screen with vertical dimension l, the minimum viewing distance for n o visible lines is found from min = (l 485) L . The desired ratio is then 1 1 L = = = 15.4 l 485 min 485 1.34 10 - 4 rad
(
)
38.60
(a)
Applying Snell's law gives n2 sin = n1 sin . From the sketch, we also see that: + + = , or = - ( + ) Using the given identity: which reduces to: sin = sin cos( + ) - cos sin( + ), sin = sin( + ).
Applying the identity again: sin = sin cos + cos sin Snell's law then becomes: or (after dividing by cos ): Solving for tan gives: (b) n2 (sin cos + cos sin ) = n1 sin n2 (tan cos + sin ) = n1 tan . tan = n2 sin n1 - n2 cos
If = 90.0, n1 = 1.00, and n2 = n, the above result becomes: tan = n(1.00) , 1.00 - 0 or n = tan , which is Brewster's law.
38.61
(a)
From Equation 38.1,
= sin -1
m a
In this case m = 1 and
=
c 3.00 108 m/s = = 4.00 10 - 2 m f 7.50 10 9 Hz
Thus,
4.00 10 - 2 m = sin - 1 = 41.8 -2 6.00 10 m I I max sin( 2) = 2
2
(b)
From Equation 38.4,
where =
2 asin
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424 Chapter 38 Solutions
When = 15.0,
=
I
2 (0.0600 m ) sin 15.0 = 2.44 rad 0.0400 m sin (1.22 rad) = = 0.593 1.22 rad
2
and
I max
(c)
sin =
a
so = 41.8:
This is the minimum angle subtended by the two sources at the slit. Let be the half angle between the sources, each a distance l = 0.100 m from the center line and a distance L from the slit plane. Then, L = l cot = ( 0.100 m)cot ( 41.8 2)= 0.262 m
38.62
I I max
=
1 1 (cos 2 45.0)(cos 2 45.0) = 8 2
38.63
(a)
The E and O rays, in phase at the surface of the plate, will have a phase difference
= ( 2 )
after traveling distance d through the plate. Here is the difference in the optical path lengths of these rays. The optical path length between two points is the product of the actual path length d and the index of refraction. Therefore,
= d nO - d nE
The absolute value is used since nO nE may be more or less than unity. Therefore, 2 2 = d nO - d nE = d nO - nE 550 10 - 9 m ( 2) d= = = 1.53 10 - 5 m = 15.3 m 2 nO - nE 2 1.544 - 1.553 For a diffraction grating, the locations of the principal maxima for wavelength are given by sin = m d y L . The grating spacing may be expressed as d = a N where a is the width of the grating and N is the number of slits. Thus, the screen locations of the maxima become
(b) *38.64
(
)
Chapter 38 Solutions 425 y = NLm / a. If two nearly equal wavelengths are present, the difference in the screen locations of corresponding maxima is y = NLm( ) a
For a single slit of width a, the location of the first diffraction minimum is sin = a y L, or y = ( L / a) . If the two wavelengths are to be just resolved by Rayleigh's criterion, y = y from above. Therefore, NLm( ) L = a a or the resolving power of the grating is R"
= Nm .
38.65
The first minimum in the diffraction pattern occurs at sin =
single-slit
ymin a L
Thus, the slit width is given by a=
L ymin
For a minimum located at ymin = 6.36 mm 0.08 mm , the width is a =
(632.810-9 m)(1.00 m)=
6.3610 -3 m
99.5 m 1%
38.66
(a)
From Equation 38.4,
I I max this becomes I I max sin =
2
sin( 2) = ( 2)
2
If we define 2
Therefore, when
I I max
=
1 2
we must have
sin 1 = , or sin = 2 2
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426 Chapter 38 Solutions
(b)
Let y1 = sin and y2 =
. 2
A plot of y1 and y2 in the range 1.00 2 is shown to the right. The solution to the transcendental equation is found to be = 1.39 rad .
(c)
=
2 asin = 2 sin =
gives
= 0.443 . L la a
If
is small, then 0.443 . a a
The pattern is
This gives the half-width, measured away from the maximum at = 0. symmetric, so the full width is given by = 0.443
= - - 0.443 a a
0.886 a
38.67
1 2 1.5 1.4 1.39 1.395 1.392 1.3915 1.39152 1.3916 1.39158 1.39157 1.39156 1.391559 1.391558 1.391557 1.3915574
2 sin 1.19 1.29 1.41 1.394 1.391 1.392 1.3917 1.39154 1.39155 1.391568 1.391563 1.391560 1.391558 1.3915578 1.3915575 1.3915573 1.3915574 bigger than smaller than smaller bigger smaller bigger bigger smaller
We get the answer to seven digits after 17 steps. Clever guessing, like using the value of as the next guess for , could reduce this to around 13 steps.
2 sin
Chapter 38 Solutions 427 sin( 2)2 In I = I maxOE oe 2) ( 2 sin( 2) ( 2) cos( 2)(1 2) - sin( 2)(1 2) dI = I max d ( 2) ( 2)2
*38.68
find
and require that it be zero. The possibility sin( 2) = 0 locates all of the minima and the central maximum, according to
2 = 0, , 2 , . . . ;
=
2 asin = 0, 2 , 4 , . . . ;
asin = 0, , 2 , . . . .
The side maxima are found from
- sin = 0, or tan = . cos L2 l L2 l L2 l 2 2 = 4.4934 , = 7.7253 , and others, giving 2 2
asin = 1.4303 asin = 2.4590
This has solutions
(a) (b)
asin = 4.4934 asin = 7.7253
*38.69 (a)
We require min = 1.22
radius of diffraction disk D = = . L 2L D
Then D 2 = 2.44 L (b) D = 2.44 500 10 - 9 m (0.150 m ) = 428 m
(
)
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Chapter 39 Solutions
39.1
In the rest frame, pi = m1v1i + m2v2i = (2000 kg)(20.0 m/s) + (1500 kg)(0 m/s) = 4.00 104 kg m/s p f = (m 1 + m 2)v f = (2000 kg + 1500 kg)v f Since p i = p f, vf = 4.00 104 kg m/s = 11.429 m/s 2000 kg + 1500 kg
In the moving frame, these velocities are all reduced by +10.0 m/s. v1i = v1i - v = 20.0 m/s (+10.0 m/s) = 10.0 m/s v2i = v2i - v = 0 m/s (+10.0 m/s) = 10.0 m/s v = 11.429 m/s (+10.0 m/s) = 1.429 m/s f Our initial momentum is then pi = m1v1i + m2 v2i = (2000 kg)(10.0 m/s) + (1500 kg)(10.0 m/s) = 5000 kg m/s and our final momentum is p = (2000 kg + 1500 kg) v = (3500 kg)(1.429 m/s) = 5000 kg m/s f f
39.2
(a) (b) (c)
v = v T + v B = 60.0 m/s v = v T v B = 20.0 m/s
2 2 v = vT + vB = 20 2 + 40 2 = 44.7 m/s
39.3
The first observer watches some object accelerate under applied forces. Call the instantaneous velocity of the object v1 . The second observer has constant velocity v21 relative to the first, and measures the object to have velocity v2 = v1 - v21. The second observer measures an acceleration of dv 2 dv 1 a2 = dt = dt
This is the same as that measured by the first observer. In this nonrelativistic case, they measure the same forces as well. Thus, the second observer also confirms that F = ma.
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2
Chapter 39 Solutions
39.4
The laboratory observer notes Newton's second law to hold:
F 1 = ma 1
(where the subscript 1 implies the measurement was made in the laboratory frame of reference). The observer in the accelerating frame measures the acceleration of the mass as a 2 = a 1 a (where the subscript 2 implies the measurement was made in the accelerating frame of reference, and the primed acceleration term is the acceleration of the accelerated frame with respect to the laboratory frame of reference). If Newton's second law held for the accelerating frame, that observer would then find valid the relation F 2 = ma 2 or F 1 = ma 2
(since F 1 = F 2 and the mass is unchanged in each). But, instead, the accelerating frame which is not Newton's second law. observer will find that F 2 = ma 2 ma
*39.5
L = Lp 1 - v 2 c 2 v = c 1 - L Lp
(
)
2
Taking L = Lp / 2 where Lp = 1.00 m gives
Lp 2 1 v = c 1- = c 1 - 4 = 0.866 c Lp
2
39.6
t =
[1 - (v c) ]
tp
2 12
so
tp 2 v = c 1 - t
1/2
12
For
t = 2 tp
t 2 p v = c 1 - 2 tp
1 = c 1 - 4
1/2
= 0.866 c
*39.7
(a)
=
1 1 - (v c)
2
=
1 1 - (0.500)
2
=
2 3
The time interval between pulses as measured by the Earth observer is t = tp = 2 60.0 s = 0.924 s 3 75.0 60.0 s min = 64.9/min 0.924 s
Thus, the Earth observer records a pulse rate of (b)
At a relative speed v = 0.990 c , the relativistic factor increases to 7.09 and the pulse rate recorded by the Earth observer decreases to 10.6 min . That is, the life span of the astronaut (reckoned by the total number of his heartbeats) is much longer as measured by an Earth clock than by a clock aboard the space vehicle.
Chapter 39 Solutions
3
39.8
The observed length of an object moving at speed v is L = L p 1 - v 2 / c 2 with L p as the proper L 2 p = 3L 1p , and v1 = 0.350 c length. For the two ships, we know L 2 = L 1, Thus,
2 L 2 = L1 2
and
v 2 2 9L1p 2 1 - 22 = L1p 2 1 - (0.350) c v2 = 0.950 c
[
]
giving
9-9
v2 2 = 0.878, c2
or
*39.9
t = tp =
tp 1 - v2 / c2
v2 so tp = 1 - v 2 / c 2 t 1 - 2 t 2c 1.00 106 m = 277.8 m s, 3600 s then
and
v2 t - tp = 2 t 2c v = 9.26 10 - 7 c
If and
v = 1000 km h =
(t - tp ) = (4.28 10-13 )(3600 s) = 1.54 10- 9 s =
v2 2 = 1 - (0.950) = 0.312 2 c
1.54 ns
39.10 (a) (b)
-1 = 1 -
astronauts' time: astronauts' distance:
tp = -1 t = (0.312)( 4.42 yr ) = 1.38 yr L = -1 L p = (0.312)( 4.20 ly ) = 1.31 ly
39.11
The spaceship appears length-contracted to the Earth observer as given by L = Lp 1 - v 2 c 2 or L2 = Lp 2 1 - v 2 c 2
(
)
or v2 v2 = (c t)2 2 c2 c or and L2 = v 2t 2 = v2 ( c t )2 c2
Also, the contracted length is related to the time required to pass overhead by: L = vt Equating these two expressions gives Using the given values: this becomes giving Lp 2 - Lp 2
[
Lp 2 + (c t)2
]
v2 = Lp 2 c2
Lp = 300 m (1.41 105 m2)
t = 7.50 10 7 s
v2 = 9.00 104 m2 c2
v = 0.800 c
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4
Chapter 39 Solutions
Goal Solution A spaceship with a proper length of 300 m takes 0.750 s seconds to pass an Earth observer. Determine its speed as measured by the Earth observer. G: We should first determine if the spaceship is traveling at a relativistic speed: classically, 8 v = (300m)/(0.750 s) = 4.00 10 m/s, which is faster than the speed of light (impossible)! Quite clearly, the relativistic correction must be used to find the correct speed of the spaceship, which we can guess will be close to the speed of light. O: We can use the contracted length equation to find the speed of the spaceship in terms of the proper length and the time. The time of 0.750 s is the proper time measured by the Earth observer, because it is the time interval between two events that she sees as happening at the same point in space. The two events are the passage of the front end of the spaceship over her stopwatch, and the passage of the back end of the ship. A : L = Lp / , with L = vt : Squaring both sides, vt = Lp 1 - v 2 / c 2
(
)
1/2
v 2 t 2 = Lp 2 1 - v 2 / c 2
(
)
v 2 c 2 = Lp 2 c 2 / t 2 - v 2 Lp 2 / t 2 Solving for the velocity, v= c Lp / t c 2 + Lp 2 / t 2
So
v=
(3.00 10 ) (300 m) (0.750 10 s) (3.00 10 ) + (300 m) (0.750 10 s)
8 -6 8 2 2 -6
2
= 2.40 108 m / s
L : The spaceship is traveling at 0.8c. We can also verify that the general equation for the speed reduces to the classical relation v = Lp / t when the time is relatively large.
39.12
The spaceship appears to be of length L to Earth observers, v2 L = Lp 1 - 2 c
1/2
where
and
1/2
L = vt
v2 vt = Lp 1 - 2 c
so
v2 v 2t 2 = Lp 2 1 - 2 c
-1/2 v = Lp c 2t 2 + Lp 2 c
Solving for v,
Lp 2 v 2 t 2 + 2 = Lp 2 c
Chapter 39 Solutions
5
*39.13 (a)
For
v = 0.990, = 7.09 c t = 4.60 km 0.990 c
The muon's lifetime as measured in the Earth's rest frame is and the lifetime measured in the muon's rest frame is tp = t 1 4.60 10 3 m = 2.18 s = 7.09 0.990 3.00 108 m s
(
)
(b)
L = Lp 1 - ( v c ) =
2
Lp
=
4.60 10 3 m = 649 m 7.09
39.14
We find Carpenter's speed: GM v= (R + h)
1/2
GMm mv 2 = r r2
1/2
(6.67 10 -11 )(5.98 10 24 ) = 6 6 (6.37 10 + 0.160 10 ) T=
= 7.82 km / s 2 (R + h) 2 (6.53 106 ) = = 5.25 10 3 s v 7.82 10 3
Then the time period of one orbit, (a) The time difference for 22 orbits is
t - tp = ( - 1)tp = 1 - v 2 c 2
2
(
)
-1/2
- 1( 22T )
1 v2 1 7.82 10 3 m / s t - tp 1 + - 1 ( 22T ) = 22 5.25 10 3 s = 39.2 s 2 c2 2 3 108 m / s
(
)
(b)
For one orbit,
t - tp =
39.2 s = 1.78 s . 22
The press report is accurate to one digit .
39.15
For pion to travel 10.0 m in t in our frame, 10.0 m = v t = v( tp ) = Solving for the velocity,
v(26.0 10 - 9 s) 1 - v 2 / c2
(3.85 108 m / s)2 (1 - v 2 / c 2 ) = v 2 1.48 1017 m 2 / s 2 = v 2 (1+ 1.64) v = 2.37 108 m / s = 0.789 c
*39.16
=
1 v2 1- 2 c
= 1.01
so
v = 0.140 c
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6
Chapter 39 Solutions
*39.17 (a)
Since your ship is identical to his, and you are at rest with respect to your own ship, its length is 20.0 m . His ship is in motion relative to you, so you see its length contracted to 19.0 m . We have from which L = Lp 1 - v 2 c 2 v2 L 19.0 m = = 0.950 = 1 - 2 Lp 20.0 m c and v = 0.312 c
(b) (c)
*39.18 (a)
t = tp =
tp 1 - (v c)
2
=
15.0 yr 1 - (0.700)
2
= 21.0 yr
(b) (c)
d = v( t ) = [0.700 c ]( 21.0 yr ) = (0.700)(1.00 ly yr ) ( 21.0 yr ) = 14.7 ly The astronauts see Earth flying out the back window at 0.700 c: d = v tp = [0.700 c ](15.0 yr ) = (0.700)(1.00 ly yr ) (15.0 yr ) = 10.5 ly
[
]
( )
[
]
(d) Mission control gets signals for 21.0 yr while the battery is operating, and then for 14.7 years after the battery stops powering the transmitter, 14.7 ly away: 21.0 yr + 14.7 yr = 35.7 yr
*39.19
The orbital speed of the Earth is as described by F = ma: GmS = r
GmS mE mE v 2 = r r2
4
v=
(6.67 10
-11
N m 2 kg 2 1.99 10 30 kg 1.496 1011 m
)(
) = 2.98 10
ms
The maximum frequency received by the extraterrestrials is f obs = f source 1+ v c = 57.0 106 Hz 1- v c
( (
) 1 - ((2.98 10
1 + 2.98 10 4 m s
4
) (3.00 10 m s) ( 3.00 10 ) (3.00 10 m s) ( 3.00 10
8 8
) = 57.005 66 10 m s)
ms ms
6
Hz
The minimum frequency received is f obs = f source 1- v c = 57.0 106 Hz 1+ v c
) 1 + ((2.98 10
1 - 2.98 10 4 m s
4
8 8
) = 56.994 34 10 m s)
6
Hz
The difference, which lets them figure out the speed of our planet, is
(57.005 66 - 56.994 34) 106 Hz
= 1.13 10 4 Hz
Chapter 39 Solutions
7
39.20
(a)
Let f c be the frequency as seen by the car. Thus, and, if f is the frequency of the reflected wave,
f c = f source f = fc c+v c-v
c+v c-v
Combining gives (b) Using the above result, which gives The beat frequency is then fb =
f = f source
(c + v) (c - v)
f (c - v ) = f source (c + v) ( f - f source )c = ( f + f source )v 2 f source v f b = f - f source = 2 f source v 2v = c
(c)
(2)(30.0 m s)(10.0 10 9 Hz) (2)(30.0 m s) = = 2000 Hz = 2.00 kHz (0.0300 m) 3.00 108 m s c f source fb 2 = 3.00 108 m s = 3.00 cm 10.0 10 9 Hz so v = f b ( 5 Hz)(0.0300 m) = = 0.0750 m s 0.2 mi / h 2 2
=
(d) v=
39.21
(a)
When the source moves away from an observer, the observed frequency is c - vs f obs = f source c + vs When vs << c , the binomial expansion gives c - vs c + vs So,
12 1/2
where
vs = vsource
v = 1 - s c
12
vs 1 + c
-1 2
v v v 1- s 1- s 1- s 2c 2c c
v f obs f source 1 - s c
The observed wavelength is found from c = obs f obs = f source :
obs =
f source f source = f obs f source (1 - vs c ) 1 - vs c
v c 1 1 = obs - = - 1 = - 1 = s 1 - vs c 1 - vs c 1 - vs c Since 1 - vs c 1, (b) vsource c 20.0 nm = 0.050 4 c vsource = c =c 397 nm
2000 by Harcourt, Inc. All rights reserved.
8
Chapter 39 Solutions 0.950 c - 0.750 c ux - v = = 0.696 c 2 1 - 0.950 0.750 1 - ux v / c
39.22
ux =
39.23
ux =
- 0.750 c - 0.750 c ux - v = = 0.960 c 1 - ux v c 2 1 - (- 0.750)(0.750)
*39.24
= 10.0 We are also given: L 1 = 2.00 m, and 1 = 30.0 (both measured in a reference frame moving relative to the rod).
Thus, L1x = L1 cos 1 = (2.00 m)(0.867) = 1.73 m and L1y = L1 sin 1 = (2.00 m)(0.500) = 1.00 m L 2x = a "proper length" is related to L1x by Therefore, L1x = L 2x L 2x = 10.0L1x = 17.3 m and L 2y = L1y = 1.00 m
(Lengths perpendicular to the motion are unchanged). (a) (b) L 2 = (L 2x )2 + (L 2y )2 gives gives L2 = 17.4 m
2 = tan -1
L 2y L 2x
2 = 3.30
39.25
ux = Enterprise velocity v = Klingon velocity From Equation 39.16, ux = 0.900 c - 0.800 c ux - v = = 0.357c ux v 1 - (0.900)(0.800) 1- 2 c
Chapter 39 Solutions x = ( x - v t ) , 0 = 2.00 m - v 8.00 10 - 9 s v=
9
*39.26 (a)
From Equation 39.13,
[
(
)]
= 1.81
2.00 m = 2.50 108 m s -9 8.00 10 s 1 1 - 2.50 108 m s
=
(
) (
2
3.00 108 m s
)
2
(b)
From Equation 39.11,
x = ( x - v t ) = 1.81 3.00 m - 2.50 108 m s 1.00 10 - 9 s = 4.97 m 8 v 1.00 10 - 9 s - 2.50 10 m s ( 3.00 m ) t = t - 2 x = 1.81 2 c 3.00 108 m s t = -1.33 10 - 8 s
[
(
)(
)]
(c)
( (
) )
39.27 (a)
p = mu For an electron moving at 0.0100 c,
=
1 1 - (u c )
2
=
1 1 - (0.0100)2
= 1.00005 1.00
Thus, p = 1.00 9.11 10 - 31 kg (0.0100) 3.00 108 m / s = 2.73 10 - 24 kg m s (b) Following the same steps as used in part (a), we find at 0.500 c
(
)
(
)
= 1.15
(c) At 0.900 c, = 2.29 and
and
p = 1.58 10 - 22 kg m s
p = 5.64 10 - 22 kg m s
*39.28
Using the relativistic form, p =
mu 1 - (u c )
2
= mu, p = mu - mu = ( - 1)mu
we find the difference p from the classical momentum, mu : (a) The difference is 1.00% when ( - 1)mu = 0.0100 mu:
=
(b)
1 1 = 2 0.990 1 - (u c )
1 - (u c ) = (0.990)
2
2
or
u = 0.141 c
The difference is 10.0% when ( - 1)mu = 0.100 mu:
=
1 1 = 2 0.900 1 - (u c )
1 - (u c ) = (0.900)
2
2
or
u = 0.436 c
2000 by Harcourt, Inc. All rights reserved.
10
Chapter 39 Solutions p - mu mu - mu = = -1 mu mu
*39.29
-1=
1 1 - (u c )
2
-1 1+
1 u 1 u -1= c 2 2 c
2
2
2
p - mu 1 90.0 m s = = 4.50 10 -14 mu 2 3.00 108 m s
39.30
p=
mu 1 - (u c )
2
becomes
1-
u2 m 2u2 = c2 p2
which gives: m2c 2 u2 2 + 1 p
m2 1 1 = u2 2 + 2 c p and u= c m c +1 p2
2 2
or
c2
=
*39.31
Relativistic momentum must be conserved: For total momentum to be zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and subscript 1 to the lighter, p2 = p1 or
2 m2u2 = 1m1u1 =
(1.67 10 -27 kg)u2 1 - (u2 c )
2
2.50 10 -28 kg 1 - (0.893)2
(0.893 c)
or
= (4.960 10 -28 kg)c
and
u2 = 0.285 c
Chapter 39 Solutions
11
Goal Solution An unstable particle at rest breaks into two fragments of unequal mass. The rest mass of the lighter fragment is 2.50 10 -28 kg, and that of the heavier fragment is 1.67 10 -27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment? G: The heavier fragment should have a speed less than that of the lighter piece since the momentum of the system must be conserved. However, due to the relativistic factor, the ratio of the speeds will not equal the simple ratio of the particle masses, which would give a speed of 0.134c for the heavier particle. O: Relativistic momentum of the system must be conserved. For the total momentum to be zero after the fission, as it was before, p1 + p2 = 0, where we will refer to the lighter particle with the subscript '1', and to the heavier particle with the subscript '2.' A : 2 m2 v2 + 1m1v1 = 0 so 2.50 10 -28 kg 2 m2 v2 + (0.893c ) = 0 1 - 0.893 2 1.67 10 -27 kg v -28 kg 2 = -4.96 10 2 2 c 1 - v2 c
Rearranging,
Squaring both sides,
(2.79 10 ) vc = (2.46 10 ) 1 - vc
-54 2 2 -55
2 2 2
and
v2 = -0.285c
We choose the negative sign only to mean that the two particles must move in opposite directions. The speed, then, is v2 = 0.285c L : The speed of the heavier particle is less than the lighter particle, as expected. We can also see that for this situation, the relativistic speed of the heavier particle is about twice as great as was predicted by a simple non-relativistic calculation.
39.32 (a)
E = ( 1 - 2 )mc 2 .
For an electron,
mc 2 = 0.511 MeV.
2 1 1 E = - mc = 0.582 MeV (1 - 0.250) (1 - 0.810) 2 1 1 E = - mc = 2.45 MeV 2 1 - 0.810 1 - (0.990)
(b)
39.33
E = mc 2 = 2mc 2 , or = 2 Thus, c 3 3 u 2 , or u = . = 1 - (1 ) = c 2 2
c 3 mc 2 MeV 938.3 MeV 3 = 1.63 10 3 The momentum is then p = mu = 2m = c 3 = c c 2
2000 by Harcourt, Inc. All rights reserved.
12
Chapter 39 Solutions
*39.34
1 The relativistic kinetic energy of an object of mass m and speed u is Kr = - 1 mc 2 1 - u2 / c 2 For u = 0.100 c,
1 Kc = 2 mu2 gives
1 Kr = - 1 mc 2 = 0.005038 mc 2 1 - 0.0100
1 Kc = 2 m(0.100 c)2 = 0.005000 mc 2
The classical equation
different by
0.005038 0.005000 = 0.751% 0.005038
For still smaller speeds the agreement will be still better.
39.35
(a)
ER = mc 2 = (1.67 10 -27 kg)(2.998 108 m / s)2 = 1.50 10 -10 J = 938 MeV 1.50 10 -10 J = 4.81 10 -10 J = 3.00 10 3 MeV [1 - (0.95 c / c)2 ]1/2
(b)
E = mc 2 =
(c)
K = E - mc 2 = 4.81 10 -10 J - 1.50 10 -10 J = 3.31 10 -10 J = 2.07 10 3 MeV
*39.36 (a)
KE = E ER = 5ER E = 6ER = 6(9.11 10 -31 kg)(3.00 108 m / s)2 = 4.92 10 -13 J = 3.07 MeV
(b)
E = mc2 = ER Thus,
=
E 1 =6= Er 1 - u2 c 2
which yields
u = 0.986 c
39.37
The relativistic density is 8.00 g mc 2 m m ER = 2 = = = = 18.4 g/cm3 2 L 1 - u c 2 (1.00 cm )3 1 - (0.900)2 c V c V V Lp Lp p ( )
( )( )
Chapter 39 Solutions
13
*39.38
We must conserve both mass-energy and relativistic momentum. With subscript 1 referring to the 0.868 c particle and subscript 2 to the 0.987c particle,
1 =
1 1 - (0.868)
2
= 2.01
and
2 =
1 1 - (0.987 )
2
= 6.22 which is
Conservation of mass-energy gives or This reduces to:
E1 + E2 = Etotal
1m1c 2 + 2 m2 c 2 = mtotal c 2
2.01m1 + 6.22 m2 = 3.34 10 -27 kg m1 + 3.09 m2 = 1.66 10 -27 kg gives [1]
Since the momentum after must equal zero, p1 = p2 or which becomes Solving [1] and [2] simultaneously,
1m1u1 = 2 m2u2
(2.01)(0.868 c)m1 = (6.22)(0.987 c)m2 m1 = 3.52 m2 m1 = 8.84 10 -28 kg and m2 = 2.51 10 -28 kg [2]
39.39
E = mc 2,
2 2 2
p = mu;
2 2
E 2 = ( mc 2)2;
2 2
p 2 = ( m u ) 2 ; - (mc)2 u2
= (mc 2 )2 1 -
E - p c = ( mc ) - ( mu) c =
2 (mc 2 )2
u2 u2 1- c2 c2
-1
= (mc 2 )2
Q.E.D.
39.40
(a)
K = 50.0 GeV 2 1 mc 2 = 1.67 10 -27 kg 2.998 108 m s = 0.938 GeV -10 J Ge V 1.60 10
(
)(
)
E = K + mc 2 = 50.0 GeV + 0.938 GeV = 50.938 GeV E = p c + mc
2 2 2
( )
2 2
p=
E 2 - mc 2 c
2
( )
2
=
( 50.938 GeV)2 - (0.938 GeV)2
c2
p = 50.9
GeV 50.9 GeV 1.60 10 -10 J -17 = kg m s = 2.72 10 8 c 1 GeV 3.00 10 m s mc 2 1 - (u c )
2
(b)
E = mc 2 =
u = c 1 - mc 2 E
(
)
2
v = 3.00 108 m s 39.41 (a)
(
)
1-
0.938 GeV = 50.938 GeV
2
2 .9995 108 m s
q( V ) = K = ( - 1)me c 2
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14
Chapter 39 Solutions
Thus, =
1 1 - (u c )
2
= 1+
q( V ) from which me c 2
u = 0.302 c
(b)
K = ( - 1)me c 2 = q( V ) = 1.60 10 -19 C 2.50 10 4 J C = 4.00 10 -15 J
(
)(
)
39.42
(a)
E = mc 2 = 20.0 GeV with mc 2 = 0.511 MeV for electrons. Thus, =
20.0 10 9 eV = 3.91 10 4 6 0.511 10 eV
(b)
=
1 1 - (u c )
2
= 3.91 10 4 from which u = 0.999 999 999 7 c
(c)
L = Lp 1 - (u c ) =
2
Lp
=
3.00 10 3 m = 7.67 10 - 2 m = 7.67 cm 3.91 10 4
39.43
Conserving total momentum, pBefore decay = pafter decay = 0 : Conservation of mass-energy gives:
p = p = m u = ( 206 me ) u E + E = E
m c 2 + p c = m c 2 ( 206 me ) +
Substituting from the momentum equation above, 270 u u 1 + = = 1.31 = 0.264 c 206 c p = 270 me c u = 270 me c
( 206 me ) + ( 206 me )
or
1 Then, K = ( - 1)m c 2 = ( - 1)206 me c 2 = - 1 206(0.511 MeV ) = 3.88 MeV 1 - (0.264)2
(
)
Also, E = E - E = m c 2 - m c 2 = ( 270 - 206 )me c 2 206 E = 270 - (0.511 MeV ) = 28.8 MeV 2 1 - (0.264) Let a 0.3-kg flag be run up a flagpole 7 m high. We put into it energy mgh = 0.3 kg(9.8 m/s2) 7 m 20 J
*39.44
Chapter 39 Solutions
15
So we put into it extra mass
m =
E c2
=
20 J (3 108 m/s)2
= 2 1016 kg
for a fractional increase of
2 1016 kg ~1015 0.3 kg
*39.45
E = 2.86 105 J. Also, the mass-energy relation says that E = mc2. Therefore, m= E 2.86 10 5 J = = 3.18 1012 kg c 2 (3.00 108 m / s)2
No, a mass loss of this magnitude (out of a total of 9.00 g) could not be detected .
39.46
(a)
1 K = ( - 1)mc 2 = - 1 mc 2 = 0.25 mc 2 = 2.25 1022 J 1 - u2 / c 2 E = mfuel c2 so mfuel = 2.25 10 22 = 2.50 105 kg 9.00 1016
(b)
39.47
9 7 E P t 0.800 1.00 10 J s ( 3.00 yr ) 3.16 10 s yr = 0.842 kg m = 2 = 2 = 2 c c 3.00 108 m s
(
(
)
( )
)
39.48
Since the total momentum is zero before decay, it is necessary that after the decay pnucleus = pphoton = Also, for the recoiling nucleus, E 2 = p 2 c 2 + mc 2 Thus, E c = 14.0 keV c mc 2 = 8.60 10 - 9 J = 53.8 GeV K 14.0 keV 1+ = +1 2 mc 2 mc
2 2 2
( )
2
2
with or
(mc
1+
2
+K
)
2
= (14.0 keV ) + mc 2
2 2
( )
So
K 1 14.0 keV 14.0 keV = 1+ 1+ mc 2 2 mc 2 mc 2
(Binomial Theorem)
and P=
(14.0 keV)2 K
39.49
2 dE d mc dm = = c2 = 3.77 10 26 W dt dt dt
( )
2 mc 2
(14.0 10 eV) = = 2 ( 53.8 10 eV )
3 2 9
1.82 10 -3 eV
2000 by Harcourt, Inc. All rights reserved.
16
Chapter 39 Solutions
Thus,
3.77 10 26 J s dm = = 4.19 10 9 kg s 2 8 dt 3.00 10 m s
(
)
39.50
2me c 2 = 1.02 MeV :
E 1.02 MeV
39.51
The moving observer sees the charge as stationary, so she says it feels no magnetic force. q(E + v B) = q(E + 0) and E = E + v B
*39.52 (a)
When Ke = K p , In this case, Substituting, but p = 1
1 - up
me c 2 ( e - 1) = mp c 2 p - 1 me c 2 = 0.511 MeV,
(
)
and
mp c 2 = 938 MeV
e = 1 - (0.750)2
[
]
-1/2
= 1.5119
p = 1+
.
me c 2 ( e - 1) (0.511 MeV)(1.5119 - 1) = 1.000279 = 1+ 2 938 MeV mp c Therefore,
-2 up = c 1 - p = 0.0236 c
2 1/2 c
(b)
When pe = pp ,
p mp up = e me ue or p up =
e me ue . mp
Thus, up
p up =
(1.5119)(0.511 MeV
2
c 2 (0.750 c )
2
)
938 MeV c
= 6.1772 10 - 4 c
and
up = 6.1772 10 - 4 1 - c c
which yields
up = 6.18 10 - 4 c
= 185 km s
39.53
(a)
10 13 MeV = ( 1)m pc 2 t = t 10 5 yr = = 10 - 5 yr ~ 10 2 s 1010
so
1010
vp c
(b)
d = ct ~1011 m
Chapter 39 Solutions
17
Goal Solution The cosmic rays of highest energy are protons, which have kinetic energy on the order of 1013 MeV. (a) How long would it take a proton of this energy to travel across the Milky Way galaxy, having a diameter on the order of ~105 light-years, as measured in the proton's frame? (b) From the point of view of the proton, how many kilometers across is the galaxy? G: We can guess that the energetic cosmic rays will be traveling close to the speed of light, so the time it takes a proton to traverse the Milky Way will be much less in the proton's frame than 105 years. The galaxy will also appear smaller to the high-speed protons than the galaxy's proper diameter of 105 light-years. The kinetic energy of the protons can be used to determine the relativistic -factor, which can then be applied to the time dilation and length contraction equations to find the time and distance in the proton's frame of reference. The relativistic kinetic energy of a proton is K = ( - 1)mc 2 = 1013 MeV Its rest energy is
2 m 1 eV mc 2 = 1.67 10 -27 kg 2.998 108 = 938 MeV 1.60 10 -19 kg m 2/ s 2 s
O:
A:
(
)
So
1013 MeV = ( - 1)(938 MeV ) ,
and therefore = 1.07 1010 1 - v2 / c2 :
The proton's speed in the galaxy's reference frame can be found from = 1 1 - v 2 c 2 = 8.80 10 -21 and
v = c 1 - 8.80 10 -21 = 1 - 4.40 10 -21 c 3.00 108 m / s
(
)
The proton's speed is nearly as large as the speed of light. In the galaxy frame, the traversal time is t = x / v = 10 5 light - years / c = 10 5 years (a) This is dilated from the proper time measured in the proton's frame. The proper time is found from t = tp : tp = t / = 10 5 yr 1.07 1010 = 9.38 10 -6 years = 296 s ~ a few hundred seconds (b) The proton sees the galaxy moving by at a speed nearly equal to c, passing in 296 s: Lp = vtp = 3.00 108 ( 296 s) = 8.88 107 km ~ 108 km Lp = 8.88 1010 m 9.46 1015 m / ly = 9.39 10 -6 ly ~ 10 -5 ly
(
(
)(
)
)
L : The results agree with our predictions, although we may not have guessed that the protons would be traveling so close to the speed of light! The calculated results should be rounded to zero significant figures since we were given order of magnitude data. We should also note that the relative speed of motion v and the value of are the same in both the proton and galaxy reference frames.
2000 by Harcourt, Inc. All rights reserved.
18
Chapter 39 Solutions
39.54 (a)
Take the primed frame as: The mother ship: ux = 2(0.500 c) ux + v v+v 2v = = = = 0.800 c 2 2 2 2 2 1+ v c 1+ v c 1 + ux v c 1 + (0.500)2 v+
(b)
The shuttle:
2v 3v + v 3 / c 2 3(0.500 c) + (0.500 c)3 c 2 1 + v2 / c2 ux = = = = 0.929 c v 2v 1 + 3v 2 / c 2 1 + 3(0.500)2 1+ 2 c 1 + v2 / c2
39.55
mc 2 4(938.78 MeV ) - 3728.4 MeV = 100% = 0.712% 4(938.78 MeV ) mc 2
39.56
dearth = vtearth = v tastro 1 - v 2 / c 2 = ( v / c )(1.50 10 - 5 ) 1= v2 (1 + 2.25 10 -10 ) c2
so
2.00 106 yr c = v
1 1 - v2 / c2
30.0 yr
1- so
v2 v2 = (2.25 10 -10 ) c2 c2
v = 1 + 2.25 10 -10 c
(
)
-1/2
1 = 1 - 2 (2.25 10 -10 )
v = 1 - 1.12 10 -10 c
*39.57 (a)
Take the spaceship as the primed frame, moving toward the right at v = + 0.600 c . ux = + 0.800 c , and ux = 0.800 c + 0.600 c ux + v = = 0.946 c 2 1 + (0.800)(0.600) 1 + (ux v ) c = (0.200 ly ) 1 - (0.600) = 0.160 ly
2
Then
(b) (c)
L=
Lp
The aliens observe the 0.160-ly distance closing because the probe nibbles into it from one end at 0.800 c and the Earth reduces it at the other end at 0.600 c . Thus, time = 0.160 ly = 0.114 yr 0.800 c + 0.600 c
(d)
1 1 K= - 1 mc 2 = - 1 4.00 10 5 kg 3.00 108 m s 1 - u2 c 2 1 - (0.946)2
(
)(
)
2
= 7.50 10 22 J
Chapter 39 Solutions
19
39.58
In this case, the proper time is T0 (the time measured by the students on a clock at rest relative to them). The dilated time measured by the professor is: t = T 0 where t = T + t. Here T is the time she waits before sending a signal and t is the time required for the signal to reach the students. Thus, we have: T + t = T0 (1)
To determine the travel time t, realize that the distance the students will have moved beyond the professor before the signal reaches them is: d = v(T + t) The time required for the signal to travel this distance is: t = d v = (T + t) c c
Solving for t gives:
t=
(v c)T 1 - (v c)
T+
Substituting this into equation (1) yields:
(v c)T = T 0 1 - (v c)
-1
or 1 - (v c) 1 - v2 c2 1 - (v / c)
T = (1 - v c ) = T0 = T0 1 - (v c) 1 + (v c)
= T0
Then T = T0
(
)
[1 + (v / c)][1 - (v / c)]
39.59
Look at the situation from the instructor's viewpoint since they are at rest relative to the clock, and hence measure the proper time. The Earth moves with velocity v = 0.280 c relative to the instructors while the students move with a velocity u = 0.600 c relative to Earth. Using the velocity addition equation, the velocity of the students relative to the instructors (and hence the clock) is: u= (a) (- 0.280 c) - (0.600 c) v + u = = - 0.753 c (students relative to clock) 2 1 + v u c 1 + (- 0.280 c)(- 0.600 c) c 2
With a proper time interval of tp = 50.0 min, the time interval measured by the students is: t = tp with
=
1 1 - (0.753c ) / c 2
2
= 1.52
Thus, the students measure the exam to last T = 1.52(50.0 min) = 76.0 minutes (b) The duration of the exam as measured by observers on Earth is: t = tp with
=
1 1 - (0.280c )
2
c2
so
T = 1.04(50.0 min) = 52.1 minutes
2000 by Harcourt, Inc. All rights reserved.
20
Chapter 39 Solutions
*39.60
The energy which arrives in one year is E = P t = 1.79 1017 J / s 3.16 107 s = 5.66 10 24 J Thus, m = E 5.66 10 24 J = 2 c 3.00 108 m / s = 6.28 107 kg
(
)(
)
(
)
2
*39.61
The observer sees the proper length of the tunnel, 50.0 m, but sees the train contracted to length L = Lp 1 - v 2 c 2 = 100 m 1 - (0.950)2 = 31.2 m shorter than the tunnel by 50.0 31.2 = 18.8 m so it is completely within the tunnel.
*39.62
If the energy required to remove a mass m from the surface is equal to its mass energy mc 2 , then G M sm = mc 2 Rg and Rg = GMs (6.67 10 -11 N m 2 / kg 2 )(1.99 10 30 kg) = = 1.47 103 m = 1.47 km c2 (3.00 108 m / s)2
39.63
(a)
At any speed, the momentum of the particle is given by dp dt
p = mu =
mu 1 - (u c )
2
Since F = qE =
qE =
d 2 2 mu 1 - u c dt
(
)
-1/2
qE = m 1 - u2 c 2 qE du 1 - u2 c 2 + u2 c 2 and = m dt 1 - u2 c 2 3/2
(
)
-1/2
du 1 + mu 1 - u2 c 2 dt 2
3/2
(
) (2u c ) du dt
-3/2 2
So
(
)
du qE u2 a= = 1- 2 dt m c a0
(b) (c)
As u c,
0
v
du
3/2 1 - u2 / c 2 t t 0
=
t t=0
qE dt m tdt
so
u=
qEct m 2 c 2 + q 2E 2t 2
x = udt = qEc
0
m c +q E t
2 2
2 2 2
=
c m2 c 2 + q 2E 2t 2 - mc qE
Chapter 39 Solutions
21
*39.64 (a)
f observed = f source
1+ v c 1- v c
implies
c c = +
1+ v c , 1- v c
or
1 - v c + = 1+ v c 1+ 1- v c = 1+ v c
and 1- v c 550 nm - 650 nm = = 0.846 650 nm 1+ v c v v 2 v = 0.716 + 0.716 = (0.846) 1 + c c c
(b)
1+
1-
v = 0.166 c = 4.97 107 m s
39.65
(a)
An observer at rest relative to the mirror sees the light travel a distance D = 2d - x = 2 1.80 1012 m - (0.800 c )t where x = (0.800 c ) t is the distance the ship moves toward the mirror in time t . Since this observer agrees that the speed of light is c , the time for it to travel distance D is: t= D 2(1.80 1012 m) = - 0.800t = 6.67 10 3 s c 3.00 108 m / s
(
)
(b)
The observer in the rocket sees a length-contracted initial distance to the mirror of: L = d 1- v2 = 1.80 1012 m 2 c
(
)
1-
(0.800 c)2 = 1.08 1012 m, 2 c
and the mirror moving toward the ship at speed v = 0.800 c . Thus, he measures the distance the light travels as: D = 2 1.08 1012 m - y
(
)
where y = (0.800 c) (t / 2) is the distance the mirror moves toward the ship before the light reflects off it. This observer also measures the speed of light to be c , so the time for it to travel distance D is: t= D 2 t = 1.08 1012 m - (0.800 c ) , which gives t = 4.00 10 3 s c c 2
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22
Chapter 39 Solutions An observer at rest relative to the mirror sees the light travel a distance D = 2d - x, where x = v t is the distance the ship moves toward the mirror in time t . Since this observer agrees that the speed of light is c , the time for it to travel distance D is t= D 2d - vt 2d = = c c c+v
39.66
(a)
(b)
The observer in the rocket sees a length-contracted initial distance to the mirror of L = d 1- v2 c2
and the mirror moving toward the ship at speed v. Thus, he measures the distance the light travels as D = 2( L - y ) where y = v t 2 is the distance the mirror moves toward the ship before the light reflects off it. This observer also measures the speed of light to be c , so the time for it to travel distance D is: t= D 2 v 2 vt = d 1- 2 - c c 2 c so
(c + v)t =
2d c
(c + v)(c - v)
or
t=
2d c - v c c+v
39.67
(a)
Since Mary is in the same reference frame, S , as Ted, she observes the ball to have the same speed Ted observes, namely ux = 0.800 c . Lp ux 1.80 1012 m 0.800 3.00 108 m s v2 = 1.80 1012 m c2
(b)
t =
=
(
)
= 7.50 10 3 s
(c)
L = Lp 1 -
(
)
1-
(0.600 c)2 = 1.44 1012 m c2
Since v = 0.600 c and ux = - 0.800 c , the velocity Jim measures for the ball is ux =
(- 0.800 c) + (0.600 c) = - 0.385 c ux + v = 2 1 + ( - 0.800)(0.600) 1 + ux v c
(d) Jim observes the ball and Mary to be initially separated by 1.44 1012 m. Mary's motion at 0.600c and the ball's motion at 0.385c nibble into thi distance from both ends. The gap closes at the rate 0.600c + 0.385c = 0.985c, so the ball and catcher meet after a time t = 1.44 1012 m 0.985 3.00 1018 m / s
(
)
= 4.88 10 3 s
Chapter 39 Solutions
23
39.68
(a)
2 2 2 L 0 = L 0x + L 0y
and
L2 = L 2 + L 2 x y Ly = L0y = L0 sin 0 Lx = L0x 1 - ( v c ) = ( L0 cos 0 ) 1 - ( v c )
2 2
The motion is in the x direction:
Thus,
v 2 v 2 2 2 2 2 L2 = L0 2 cos 2 0 1 - + L0 sin 0 = L 0 1 - cos 0 c c L = L0 1 - ( v c ) cos 2 0
2
or
[
]
12
(b)
tan =
Ly Lx
=
L0y L0x 1 - ( v c )
2
= tan 0
39.69
(a)
First, we find the velocity of the stick relative to S using
L = Lp 1 - (ux )
2
c2
Thus
ux = c 1 - L Lp
(
)
2
Selecting the negative sign because the stick moves in the negative x direction in S gives: ux = - c 1 - 0.500 m = - 0.866 c 1.00 m
2
so the speed is
ux = 0.866 c
Now determine the velocity of the stick relative to S, using the measured velocity of the stick relative to S and the velocity of S relative to S. From the velocity addition equation, we have: ux =
(- 0.866 c) + (0.600 c) = - 0.554 c ux + v = 2 1 + (0.600 c )( - 0.866 c ) 1 + vux c
and the speed is
ux = 0.554 c
(b)
Therefore, the contracted length of the stick as measured in S is: L = Lp 1 - (ux c ) = (1.00 m ) 1 - (0.554) = 0.833 m
2 2
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24
Chapter 39 Solutions
39.70
(b)
Consider a hermit who lives on an asteroid halfway between the Sun and Tau Ceti, stationary with respect to both. Just as our spaceship is passing him, he also sees the blast waves from both explosions. Judging both stars to be stationary, this observer concludes that the two stars blew up simultaneously . We in the spaceship moving past the hermit do not calculate the explosions to be simultaneous. We see the distance we have traveled from the Sun as L = Lp 1 - ( v c ) = (6.00 ly ) 1 - (0.800) = 3.60 ly
2 2
(a)
We see the Sun flying away from us at 0.800 c while the light from the Sun approaches at 1.00 c. Thus, the gap between the Sun and its blast wave has opened at 1.80 c, and the time we calculate to have elapsed since the Sun exploded is 3.60 ly 1.80 c = 2.00 yr. We see Tau Ceti as moving toward us at 0.800 c , while its light approaches at 1.00 c, only 0.200 c faster. We see the gap between that star and its blast wave as 3.60 ly and growing at 0.200 c . We calculate that it must have been opening for 3.60 ly 0.200 c = 18.0 yr and conclude that Tau Ceti exploded 16.0 years before the Sun .
*39.71
The unshifted frequency is
f source = f=
c 3.00 108 m s = = 7.61 1014 Hz 394 10 - 9 m
We observe frequency
3.00 108 m s = 6.32 1014 Hz 475 10 - 9 m 1+ v c 1- v c 1+ v c 1- v c
Then
f = f source
gives:
6.32 = 7.61
or
1+ v c 2 = (0.829) 1- v c v = - 0.185 c = 0.185 c (away )
Solving for v yields:
Chapter 39 Solutions
25
39.72
Take m = 1.00 kg. The classical kinetic energy is u u 1 1 Kc = 2 mu2 = 2 mc 2 = 4.50 1016 J c c
2
(
)
2
and the actual kinetic energy is
1 1 Kr = - 1 mc 2 = 9.00 1016 J - 1 1 - (u c )2 1 - (u c )2
(
)
uc 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 0.990
Kc
( J)
Kr
( J)
0.000 0.045 1016 0.180 1016 0.405 1016 0.720 1016 1.13 1016 1.62 1016 2.21 1016 2.88 1016 3.65 1016 4.41 1016
0.000 0.0453 1016 0.186 1016 0.435 1016 0.820 1016 1.39 1016 2.25 1016 3.60 1016 6.00 1016 11.6 1016 54.8 1016
1 2 1 - 1 , yielding u = 0.115 c Kc = 0.990 Kr when 2 (u c ) = 0.990 2 1 - (u c ) Similarly, Kc = 0.950 Kr when u = 0.257 c
and Kc = 0.500 Kr
when u = 0.786 c
39.73
1030 kg / m 3 1.40 10 9 10 3 m ( 4186 J / kgC)(10.0 C) E mc ( T ) Vc ( T ) m = 2 = = = 2 c c2 c2 3.00 108 m / s
3
(
)(
(
)(
)
)
m = 6.71 108 kg
2000 by Harcourt, Inc. All rights reserved.
Chapter 40 Solutions
40.1
T=
2.898 10 - 3 m K = 5.18 10 3 K 560 10 - 9 m
*40.2
(a)
max = max ~
2.898 10 - 3 m K 2.898 10 - 3 m K ~ T 10 4 K 2.898 10 - 3 m K 107 K ~ 10 - 10 m .
~ 10 - 7 m
ultraviolet .
(b)
- ray
40.3
(a)
Using we get max =
maxT = 2.898 10 - 3 m K
2.898 10 - 3 m = 9.99 10 7 m = 999 nm 2900 K
(b)
The peak wavelength is in the infrared region of the electromagnetic spectrum, which is much wider than the visible region of the spectrum.
40.4
Planck's radiation law gives intensity-per-wavelength. Taking E to be the photon energy and n to be the number of photons emitted each second, we multiply by area and wavelength range to have energy-per-time leaving the hole: P= 2 hc 2 ( 2 - 1 ) (d / 2)2
5 1 +2 e 2 2hc ( 1 + 2 )k B T
- 1
= En = nhf
where
E = hf =
2hc 1 +2
P = n= = E ( + )4 e 2hc ( 1 + 2 )kBT - 1 1 2
8 2c d 2 ( 2 - 1 )
(
)
8 2 3.00 108 m s 5.00 10 - 5 m
(
(1001 10
-9
2( 6.626 10 - 34 Js ) ( 3.00 10 8 m s ) (100110 - 9 m )(1.38 10 - 23 J K )(7.50 10 3 K ) 4 m e - 1
)(
) (1.00 10
2
-9
m
)
)
n=
5.90 1016 / s
(e
3.84
-1
)
= 1.30 1015 / s
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2
Chapter 40 Solutions
*40.5
(a)
P = eA T 4 = 1 20.0 10 - 4 m 2 5.67 10 - 8 W m 2 K 4 ( 5000 K ) = 7.09 10 4 W
4
(
)(
)
(b)
maxT = max ( 5000 K ) = 2 .898 10 - 3 m K max = 580 nm
6.626 10 - 34 J s 3.00 108 m s hc We compute: = = 2.88 10 - 6 m kBT 1.38 10 - 23 J K ( 5000 K )
(c)
(
(
)(
)
)
The power per wavelength interval is
P ( ) = A I ( ) =
2 hc 2 A , and 5 exp( hc kBT ) - 1
[
]
2 hc 2 A = 2 6.626 10 - 34 3.00 108 P ( 580 nm ) =
(
)(
) (20.0 10 ) = 7.50 10
2 -4
-19
J m4 s
(580 10
7.50 10 -19 J m 4 s
-9
m
) [exp(2.88 m 0.580 m) - 1]
5
=
1.15 1013 J m s = 7.99 1010 W m e 4.973 - 1
(d) - (i) The other values are computed similarly:
(d) (e) (f) (c) (g) (h) (i) 1.00 nm 5.00 nm 400 nm 580 nm 700 nm 1.00 mm 10.0 cm
hc/k B T 2882.6 576.5 7.21 4.97 4.12 0.00288 2.88 105
e hc/ kB T 1 7.96 101251 2.40 10250 1347 143.5 60.4 0.00289 2.88 105
2 hc 2A/ 5 7.50 1026 2.40 1023 7.32 1013 1.15 1013 4.46 1012 7.50 10 4 7.50 1014
P (), W/m 9.42 101226 1.00 10227 5.44 1010 7.99 1010 7.38 1010 0.260 2.60 109
(j)
We approximate the area under the P ( ) versus curve, between 400 nm and 700 nm, as two trapezoids: 10 W -9 10 W -9 ( 5.44 + 7.99) 10 m ( 580 - 400) 10 m (7.99 + 7.38) 10 m (700 - 580) 10 m P + 2 2 P = 2.13 10 4 W so the power radiated as visible light is approximately 20 kW .
[
]
[
]
Chapter 40 Solutions
3
40.6
(a)
P = eA T 4 , so 3.77 10 26 W = 2 W 1 4 6.96 108 m 5.67 10 - 8 m2 K 4
14
P T= eA
14
(
)
= 5.75 10 3 K
(b)
max =
2 .898 10 - 3 m K 2 .898 10 - 3 m K = = 5.04 10 -7 m = 504 nm T 5.75 10 3 K
40.7
(a)
1.00 eV E = hf = 6.626 10 - 34 J s 620 1012 s -1 = 2.57 eV 1.60 10 -19 J
(
)(
)
(b)
1.00 eV -5 E = hf = 6.626 10 - 34 J s 3.10 10 9 s -1 = 1.28 10 eV 1.60 10 -19 J
(
(
)(
)
(c)
1.00 eV -7 E = hf = 6.626 10 - 34 J s 46.0 106 s -1 = 1.91 10 eV 1.60 10 -19 J
)(
)
(d)
=
c 3.00 108 m s = = 4.84 10 - 7 m = 484 nm, visible light (blue) f 620 1012 Hz c 3.00 108 m s = = 9.68 10 - 2 m = 9.68 cm, radio wave f 3.10 10 9 Hz c 3.00 108 m s = = 6.52 m, radio wave f 46.0 106 Hz
=
=
40.8
6.626 10 - 34 J s 3.00 108 m s hc E = hf = = = 3.37 10 -19 J photon 589.3 10 - 9 m n= 10.0 J s P = = 2 .96 1019 photons s E 3.37 10 -19 J photon
(
)(
)
40.9
Each photon has an energy This implies that there are
E = hf = (6.626 10 34)(99.7 106) = 6.61 10 26 J 150 103 J/s 6.61 10 26 J/photons = 2.27 1030 photons/s
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4
Chapter 40 Solutions
*40.10
Energy of a single 500-nm photon: E = hf = hc = (6.626 10 34 J s)(3.00 108 m/s) = 3.98 10 19 J 500 10 9 m
The energy entering the eye each second E = P t= (IA)t = (4.00 10 11 W/m2)
(8.50 10 3 m) 2(1.00 s) = 2.27 10 15 J 4
The number of photons required to yield this energy E 2.27 10 15 J n=E = = 5.71 103 photons 3.98 10 19 J/photon
40.11
We take = 0.0300 radians. Then the pendulum's total energy is E = mgh = mg(L L cos ) E = (1.00 kg)(9.80 m/s2)(1.00 0.9995) = 4.41 103 J The frequency of oscillation is The energy is quantized, Therefore, f= 1 = 2 2 g L = 0.498 Hz
E = nhf E 4.41 103 J n =hf = = 1.34 1031 (6.626 1034 J s)(0.498 s1)
40.12
The radiation wavelength of = 500 nm that is observed by observers on Earth is not the true wavelength, , emitted by the star because of the Doppler effect. The true wavelength is related to the observed wavelength using: c c = 1 - (v c) 1 + (v c) 1 - (v c) 1 - (0.280) = ( 500 nm ) = 375 nm 1 + (v c) 1 + (0.280)
=
The temperature of the star is given by maxT = 2.898 10 - 3 m K: T= 2.898 10 - 3 m K 2.898 10 - 3 m K = = 7.73 10 3 K max 375 10 - 9
Chapter 40 Solutions
5
40.13
This follows from the fact that at low T or long , the exponential factor in the denominator of Planck's radiation law is large compared to 1, so the factor of 1 in the denominator can be neglected. In this approximation, one arrives at Wien's radiation law.
*40.14
Planck's radiation law is
I ( ,T ) =
5 e hc
(
2 hc 2
kBT
-1
)
]
Using the series expansion Planck's law reduces to
ex = 1 + x + I ( ,T ) =
x2 x3 + +. . . 2! 3!
2 hc 2 2 hc 2 2 ckBT 5 = 5 (1 + hc kBT + . . .) - 1 ( hc kBT ) 4
[
which is the Rayleigh-Jeans law, for very long wavelengths.
40.15
(a)
c =
hc
c
=
(6.626 10 34 J s)(3.00 108 m/s) = 296 nm (4.20 eV)(1.60 1019 J/eV)
fc =
c
=
3.00 108 m/s = 1.01 1015 Hz 296 10 9 m (6.626 10 34)(3.00 108) 180 10 9 V S = 2.71 V = (4.20 eV)(1.60 10 19 J/eV) + (1.60 1019)(VS)
(b)
hc = + e(V S): Therefore,
40.16 (a)
2 1 Kmax = 2 mvmax = 2 (9.11 10 31)(4.60 105) 2 = 9.64 10 20 J = 0.602 eV
1
= E Kmax =
1240 eV nm 0.602 eV = 1.38 eV 625 nm
(b)
fc =
1.60 10 19 J 1.38 eV = 3.34 1014 Hz = h 1 eV 6.626 10 34 J s
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6
Chapter 40 Solutions
40.17
(a)
hc c =
Li:
c
(6.626 10 = (6.626 10
- 34
(2.30 eV)(1.60 10
- 34
J s 3.00 108 m / s
-19
)( )( )(
J / eV
)
) = 540 nm ) = 318 nm ) = 276 nm
Be:
c =
(3.90 eV)(1.60 10
- 34
J s 3.00 108 m / s
-19
J / eV
)
Hg:
c
(6.626 10 =
( 4.50 eV)(1.60 10
J s 3.00 108 m / s
-19
J / eV
)
< c for photo current. Thus, only lithium will exhibit the photoelectric effect.
(b) For lithium, hc = + Kmax
(6.626 10
- 34
J s 3.00 108 m / s
-9
400 10
)(
m
) = (2.30 eV)(1.60 10 ) + K
-19
max
Kmax = 1.29 10 -19 J = 0.808 eV
40.18
From condition (i), hf = e(V S 1) + 1
and
hf = e(V S 2) + 2
(V S 1) = (V S 2) + 1.48 V Then
2 1 = 1.48 eV
h fc 1 = 1 = 0.600hfc 2 = 0.6002
From condition (ii),
2 0.6002 = 1.48 eV 2 = 3.70 eV 1 = 2.22 eV
40.19
(a)
e( VS ) = e( VS ) =
hc -
=
1240 nm eV - 0.376 eV = 1.90 eV 546.1 nm VS = 0.216 V
(b)
hc 1240 nm eV - = - 1.90 eV 587.5 nm
Chapter 40 Solutions
7
Goal Solution Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp ( = 546.1 nm) is used, a retarding potential of 0.376 V reduces the photocurrent to zero. (a) Based on this measurement, what is the work function for this metal? (b) What stopping potential would be observed when using the yellow light from a helium discharge tube ( = 587.5 nm)? G: According to Table 40.1, the work function for most metals is on the order of a few eV, so this metal is probably similar. We can expect the stopping potential for the yellow light to be slightly lower than 0.376 V since the yellow light has a longer wavelength (lower frequency) and therefore less energy than the green light. In this photoelectric experiment, the green light has sufficient energy hf to overcome the work function of the metal so that the ejected electrons have a maximum kinetic energy of 0.376 eV. With this information, we can use the photoelectric effect equation to find the work function, which can then be used to find the stopping potential for the less energetic yellow light. (a) Einstein's photoelectric effect equation is Kmax = hf - , and the energy required to raise an electron through a 1 V potential is 1 eV, so that Kmax = eV s = 0.376 eV. A photon from the mercury lamp has energy: hf = 4.14 10 -15 eV s 3.00 108 m s hc = 546.1 10 -9 m
O:
A:
(
)(
)
E = hf = 2.27 eV Therefore, the work function for this metal is: For the yellow light, = 587.5 nm, and
= hf - Kmax = 2.27 eV - (0.376 eV ) = 1.90 eV
hf = 4.14 10 -15 eV s 3.00 108 m / s hc = 587.5 10 -9 m
(b)
(
)(
)
E = 2.11 eV Therefore, Kmax = hf - = 2.11 eV - 1.90 eV = 0.216 eV , L: so V s = 0.216 V
The work function for this metal is lower than we expected, and does not correspond with any of the values in Table 40.1. Further examination in the CRC Handbook of Chemistry and Physics reveals that all of the metal elements have work functions between 2 and 6 eV. However, a single metal's work function may vary by about 1 eV depending on impurities in the metal, so it is just barely possible that a metal might have a work function of 1.90 eV. The stopping potential for the yellow light is indeed lower than for the green light as we expected. An interesting calculation is to find the wavelength for the lowest energy light that will eject electrons from this metal. That threshold wavelength for K max = 0 is 658 nm, which is red light in the visible portion of the electromagnetic spectrum.)
2000 by Harcourt, Inc. All rights reserved.
8
Chapter 40 Solutions e( VS1 ) = E 1 - e( VS2 ) = E 2 -
40.20
From the photoelectric equation, we have: Since VS2 or and the work function is:
=
and
0.700 ( VS1 ) , then
e( VS2 ) = 0.700(E 1 - ) = E 2 - (1 - 0.700) = E 2 - 0.700E 1
=
E 2 - 0.700E 1 0.300 hc 1240 nm eV = = 3.03 eV 410 eV 1 hc 1240 nm eV = = 2 .79 eV 445 eV 2
The photon energies are:
E 1 =
and
E 2
=
Thus, the work function is and we recognize this as characteristic of
=
2.79 eV - 0.700( 3.03 eV ) = 2.23 eV 0.300
potassium .
*40.21
The energy needed is
E = 1.00 eV = 1.60 10 19 J
The energy absorbed in time t is E = Pt = (IA)t so E t = IA = 1.60 10 19 J = 1.28 107 s = 148 days (500 J/s m2)[ (2.82 10 15 m)2]
The gross failure of the classical theory of the photoelectric effect contrasts with the success of quantum mechanics.
*40.22
Ultraviolet photons will be absorbed to knock electrons out of the sphere with maximum kinetic energy Kmax = hf - , or Kmax =
(6.626 10
- 34
J s 3.00 108 m s 1.00 eV - 4.70 eV = 1.51 eV 1.60 10 -19 J 200 10 - 9 m
)(
)
The sphere is left with positive charge and so with positive potential relative to V = 0 at r = . As its potential approaches 1.51 V, no further electrons will be able to escape, but will fall back onto the sphere. Its charge is then given by V= keQ r Q= 5.00 10 - 2 m (1.51 N m C) rV = = 8.41 10 - 12 C ke 8.99 10 9 N m 2 C 2
or
(
)
Chapter 40 Solutions
9
40.23
(a)
By having the photon source move toward the metal, the incident photons are Doppler shifted to higher frequencies, and hence, higher energy. If v = 0.280 c , Therefore, f = f 1+ v/c = 7.00 1014 1- v/c
(b)
(
)
1.28 = 9.33 1014 Hz 0.720
= 6.626 10 - 34 J s 9.33 1014 Hz = 6.18 10 -19 J = 3.87 eV
f = 3.05 1015 Hz
(
)(
)
(c)
At v = 0.900 c , and
1.00 eV Kmax = hf - = 6.626 10 - 34 J s 3.05 1015 Hz - 3.87 eV = 8.78 eV 1.60 10 -19 J
(
)(
)
*40.24
E=
hc
=
(6.626 10 34 J s)(3.00 108 m/s) 700 10 9 m
= 2.84 10 19 J = 1.78 eV
p=
6.626 10 34 J s h = = 9.47 10 28 kg m/s 700 10 9 m
40.25
(a) (b)
=
6.626 10 3 4 h (1 - cos ) = (1 cos 37.0) (9.11 10 31)(3.00 108) me c
= 4.88 10 13 m
E0 = hc / 0 :
(300 10 )(
3
eV 1.60 10 -19 J / eV = 6.626 10 -34 3.00 108 m / s 0
)(
) (
)(
)
0 = 4.14 10 12 m
E = (c)
and
= 0 + = 4.63 10 -12 m
6.626 10 -34 J s 3.00 108 m / s hc = = 4.30 1014 J = 268 keV 4.63 10 -12 m
(
)
Ke = E0 - E = 300 keV - 268.5 keV = 31.5 keV
40.26
This is Compton scattering through 180: E0 = hc (6.626 10 34 J s)(3.00 108 m/s) = = 11.3 keV 0 (0.110 10 9 m)(1.60 1019 J/eV) h (1 - cos ) = (2.43 10 12 m)(1 cos 180) = 4.86 10 12 m me c E = hc = 10.8 keV and 1 1 pe = h - 0
=
= 0 + = 0.115 nm so
Momentum conservation:
h h i = (i) + pe (i) 0
3.00 108 m / s / c 1 1 pe = 6.626 10 -34 J s + = 22.1 keV/c 1.60 10 -19 J / eV 0.110 10 -9 m 0.115 10 -9 m
(
)(
)
2000 by Harcourt, Inc. All rights reserved.
10
Chapter 40 Solutions
Energy conservation: Check:
11.3 keV = 10.8 keV + K e
2 E 2 = p 2 c 2 + me c 4
so that or
K e = 478 eV (me c 2 + Ke )2 = (pc)2 + (me c 2 )2
(511 keV + 0.478 keV)2 = (22.1 keV)2 + (511 keV)2 2.62 1011 = 2.62 1011
40.27
K e = E 0 E With K e = E , E = E0 E : E0 E = 2
hc hc hc = E = 1 = 2 E = 2 0 0 E 2 0 2 0 = 0 + C (1 cos )
= 0 + C (1 cos )
0.00160 0 = 0.00243 = 70.0 C
1 cos =
40.28
We may write down four equations, not independent, in the three unknowns 0 , , and v using the conservation laws: hc hc = + me c 2 - me c 2 0 (Energy conservation)
h = me v cos 20.0 (momentum in x-direction) 0 0= h - me v sin 20.0 (momentum in y-direction) h (1 - cos 90.0). me c
and Compton's equation - 0 =
It is easiest to ignore the energy equation and, using the two momentum equations, write h/0 h/ =
me v cos 20.0 me v sin 20.0
or 0 = tan 20.0
Then, the Compton equation becomes - tan 20.0 = 0.00243 nm, or
=
0.00243 nm = 0.00382 nm = 3.82 pm 1 - tan 20.0
Chapter 40 Solutions Conservation of momentum in the x direction gives: p = p cos + pe cos or since = , h h cos = pe + 0 [1]
11
40.29
(a)
Conservation of momentum in the y direction gives: 0 = p sin - pe sin , which (neglecting the trivial solution = 0) gives: 2h h = cos , or 0 pe = p = h [2]
Substituting [2] into [1] gives:
= 2 0 cos - 0 =
h (1 - cos ) me c
[3]
Then the Compton equation is
giving
2 0 cos - 0 = 2 cos - 1 =
h (1 - cos ) me c
or
hc 1 (1 - cos ) 0 me c 2
Since E
=
hc , this may be written as: 0
E 2 cos - 1 = (1 - cos ) me c 2 E E cos = 1 + 2 + 2 me c me c 2
which reduces to: me c 2 + E 2me c 2 + E
or
cos =
=
0.511 MeV + 0.880 MeV = 0.732 1.02 MeV + 0.880 MeV
so that
= = 43.0
(b)
Using Equation (3): E =
E hc hc 0.880 MeV = = = = 0.602 MeV = 602 keV 0 ( 2 cos ) 2 cos 2 cos 43.0 p = E c = 0.602 MeV c = 3.21 10 - 22 kg m s
Then,
(c)
From Equation (2), From energy conservation:
pe = p = 0.602 MeV c = 3.21 10 - 22 kg m s Ke = E - E = 0.880 MeV - 0.602 MeV = 0.278 MeV = 278 keV
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12
Chapter 40 Solutions The energy of the incident photon is E0 = p c = hc 0 . (a) Conserving momentum in the x direction gives p = pe cos + p cos , or since = , E0 = pe + p cos c
40.30
(
)
[1]
Conserving momentum in the y direction (with = ) yields 0 = p sin - pe sin , or pe = p = h [2]
Substituting Equation [2] into Equation [1] gives E0 h h cos , or = + c
=
2hc cos E0 h (1 - cos ) , me c
[3] 2hc h 2hc cos - = (1 - cos ) E0 me c E0
By the Compton equation, - 0 = which reduces to
(2m c
e
2
+ E0 cos = me c 2 + E0
)
Thus,
m c2 + E = = cos -1 e 2 0 2me c + E0
(b)
From Equation [3],
=
2hc 2hc me c 2 + E0 cos = E0 E0 2me c 2 + E0 = E0 2me c 2 + E0 , 2 me c 2 + E0
Therefore,
E =
hc hc = 2 ( 2hc E0 ) me c + E0
(
) (2m c
e
2
+ E0
)
and
p =
E c
=
E0 2me c 2 + E0 2c me c 2 + E0 E0 2me c 2 + E0 2 me c 2 + E0
(c)
From conservation of energy,
Ke = E0 - E = E0 -
or
Ke =
2 E0 2me c 2 + 2E0 - 2me c 2 - E0 E0 = 2 me c 2 + E0 2 me c 2 + E0
(
)
Finally, from Equation (2),
pe = p =
E0 2me c 2 + E0 2c me c 2 + E0
Chapter 40 Solutions Thanks to Compton we have four equations in the unknowns , v, and : hc hc = + me c 2 - me c 2 0 h h = cos 2 + me v cos 0 ' 0= h sin 2 - me v sin h (1 - cos 2 ) me c (energy conservation)
13
40.31
(a)
[1]
(momentum in x direction)
[2]
(momentum in y direction) (Compton equation) 2h cos .
[3] [4]
- 0 =
Using sin 2 = 2 sin cos in Equation [3] gives me v =
Substituting this into Equation [2] and using cos 2 = 2 cos 2 - 1 yields h 2h h h = (2 cos 2 - 1) + cos 2 = (4 cos 2 - 1), 0 or
' = 4 0 cos 2 - 0
[5]
Substituting the last result into the Compton equation gives 4 0 cos 2 - 2 0 = h hc 1 - 2 cos 2 - 1 = 2 1 - cos 2 . 2 me c me c
[ (
)]
(
)
With the substitution 0 = hc E0 , this reduces to cos 2 = me c 2 + E0 E 1+ x = where x 0 2 . 2me c 2 + E0 2 + x me c
For x =
0.700 MeV 1+ x = 1.37 , this gives = cos -1 = 33.0 0.511 MeV 2+x
(b)
1+ x 2 + 3x - 1 = 0 . From Equation [5], = 0 4 cos 2 - 1 = 0 4 2+x 2 + x Then, Equation [1] becomes hc hc 2 + x + me c 2 - me c 2 = 0 0 2 + 3x Thus, = 1 + x - x or E0 E 2+x +1= . - 0 me c 2 me c 2 2 + 3x
(
)
2+x , and with x = 1.37 we get = 1.614. 2 + 3x
Therefore,
v = 1 - -2 = 1 - 0.384 = 0.785 or v = 0.785 c . c
2000 by Harcourt, Inc. All rights reserved.
14
Chapter 40 Solutions
40.32
- =
h (1 - cos ) me c h [1 - cos( - )] me c h h h h - cos( - ) + - cos me c me c me c me c
- = - =
Now cos ( ) = cos , so
- = 2
h = 0.00486 nm me c
40.33
(a)
1 1 K = 2 me v 2 = 2 9.11 10 -31 kg 1.40 106 m s
(
)(
)
2
= 8.93 10 -19 J = 5.58 eV
E0 =
hc 1240 eV nm = = 1550 eV 0.800 nm 0 hc 1240 eV nm = = 0.803 nm E 1550 eV - 5.58 eV
E = E0 - K , and =
= - 0 = 0.00288 nm = 2.88 pm (b) = C (1 - cos ) cos = 1 - 0.00288 nm = 1- = - 0.189, so 0.00243 nm C
= 101
*40.34
Maximum energy loss appears as maximum increase in wavelength, which occurs for scattering angle 180. Then = (1 - cos 180)( h / mc ) = 2h / mc where m is the mass of the target particle. The fractional energy loss is 2h mc E0 - E hc 0 - hc - 0 = = = = E0 0 + 0 + 2h mc hc 0 Further, 0 = hc E0 , so (a) 2h mc E0 - E 2E0 = = . E0 hc E0 + 2h mc mc 2 + 2E0
For scattering from a free electron, mc 2 = 0.511 MeV, so 2(0.511 MeV ) E0 - E = = 0.667 E0 0.511 MeV + 2(0.511 MeV )
(b)
For scattering from a free proton, mc 2 = 938 MeV, and 2(0.511 MeV ) E0 - E = = 0.00109 E0 938 MeV + 2(0.511 MeV )
Chapter 40 Solutions
15
40.35
Start with Balmer's equation,
1 1 1 = RH 2 - 2 , 2 n
or =
(4n 2 / RH ) . (n 2 - 4)
Substituting RH = 1.0973732 107 m -1 , we obtain
=
(3.645 10 - 7 m)n 2 364.5n 2 = 2 nm, where n = 3, 4, 5, . . . n2 - 4 n -4
40.36
(a)
Using
1 1 1 = RH 2 - 2 n f ni
, for n f = 2, and ni 3, we get: 4n2
7
=
RH n2 - 4
(
4n2
) (2.00 10
4n2
=
m -1 n2 - 4
)(
)
=
(200.0)n2
n2 - 4
nm
This says that 200 nm 360 nm, which is ultraviolet . Using n 3 , =
(b)
RH n2 - 4
(
) (0.500 10
=
4n2
7
m - 1 n2 - 4
)(
)
=
(800.0)n2 nm n2 - 4
This says that 800 nm 1440 nm , which is in the infrared .
40.37
(a)
Lyman series:
1 1 = R 1- 2 n
n = 2, 3, 4, . . . n=5 n = 4, 5, 6, . . .
1 1 1 = = (1.097 107 ) 1 - 2 94.96 10 - 9 n (b) Paschen series: 1 1 1 =R 2 - 2 3 n
The shortest wavelength for this series corresponds to n = for ionization 1 1 1 = 1.097 107 - 2 9 n This is larger than 94.96 nm, so this wave length Brackett series: 1 1 1 =R 2 - 2 4 n For n = , this gives = 820 nm cannot be associated with the Paschen series n = 5, 6, 7, . . .
1 1 1 = 1.097 107 - 16 n2 Once again this wavelength
n = for ionization min = 1458 nm cannot be associated with the Brackett series
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16
Chapter 40 Solutions
40.38
(a)
min =
hc Emax
Lyman (nf = 1): Balmer (n f = 2): Paschen (nf = 3): Bracket (nf = 4): (b) Emax = hc min
min = min =
hc 1240 eV nm = = 91.2 nm E1 13.6 eV hc 1240 eV nm = 1 = 365 nm E2 13.6 eV 4
(Ultraviolet) (UV) (Infrared)
()
min = . . . = 3 2 (91.2 nm) = 821 nm
min = . . . = 4 2 (91.2 nm) = 1460 nm (IR)
Lyman: Balmer: Paschen: Brackett:
Emax = 13.6 eV Emax = 3.40 eV Emax = 1.51 eV Emax = 0.850 eV
(= E1 ) (= E2 )
(= E3 )
(= E4 )
40.39
Liquid O2 E=
abs = 1269 nm
for each molecule.
hc 1.2398 10 - 6 = = 0.977 eV 1.269 10 - 6
For two molecules,
=
hc = 634 nm, red 2E
By absorbing the red photons, the liquid O2 appears to be blue.
*40.40 (a)
v1 =
ke e 2 2 where r 1 = (1) a 0 = 0.00529 nm = 5.29 10 -11 m me r 1
v1 =
(8.99 10 N m C )(1.60 10 C) (9.11 10 kg)(5.29 10 m)
9 2 2 -19 -31 -11
2
= 2.19 106 m s
(b)
2 1 1 K1 = 2 me v1 = 2 9.11 10 - 31 kg 2.19 106 m s
(
)(
)
2
= 2.18 10 -18 J = 13.6 eV
(c)
8.99 10 9 N m 2 C 2 1.60 10 -19 C ke e 2 U1 = - =- r1 5.29 10 -11 m
(
)(
)
2
= - 4.35 10 -18 J = 27.2 eV
Chapter 40 Solutions
17
40.41
(a)
r 2 = (0.0529 nm )( 2) = 0.212 nm
2 2
(b)
me v2 =
me ke e 2 = r2
(9.11 10
- 31
kg 8.99 10 9 N m 2 C 2 1.60 10 -19 C 0.212 10 - 9 m
)(
)(
)
2
= 9.95 10 - 25 kg m s
(c)
L2 = me v2 r 2 = 9.95 10 - 25 kg m s 0.212 10 - 9 m = 2.11 10 - 34 kg m 2 s
(
)(
)
(d)
K2 =
1 m v2 2 e 2
( m v )2 = e 2
2me
(9.95 10 kg m s) = 2(9.11 10 kg )
- 25 - 31
2
= 5.43 10 -19 J = 3.40 eV
(e)
8.99 10 9 N m 2 C 2 1.60 10 -19 C k e2 U2 = - e = - r2 0.212 10 - 9 m E2 = K2 + U 2 = 3.40 eV - 6.80 eV = 3.40 eV
(
)(
)
2
= -1.09 10 -18 J = 6.80 eV
(f)
40.42
1 1 E = (13.6 eV) 2 - 2 n i nf Where for E > 0 we have absorption and for E < 0 we have emission. (A) for ni = 2 and nf = 5 E = 2.86 eV (absorption) (B) for ni = 5 and nf = 3 E = 0.967 eV (emission) (C) for ni = 7 and nf = 4 E = 0.572 eV (emission) (D) for ni = 4 and nf = 7 E = 0.572 eV (absorption) (a) (b) (c) E= hc so the shortest wavelength is emitted in transition B .
The atom gains most energy in transition A . The atom loses energy in transitions B and C .
40.43
(b)
1 1 1 1 1 = R 2 - 2 = (1.097 107 m -1 ) 2 - 2 2 n 6 f ni E=
so
= 410 nm
(a)
hc (6.626 10 - 34 J s)(3.00 108 m / s) = = 4.85 10 -19 J = 3.03 eV 410 10 - 9 m c 3.00 108 = = 7.32 1014 Hz 410 10 - 9
(c)
f=
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18
Chapter 40 Solutions
*40.44
We use En =
13.6 eV n2
To ionize the atom when the electron is in the n th level, it is necessary to add an amount of energy given by E = En = (a) (b) 13.6 eV n2
Thus, in the ground state where n = 1, we have E = 13.6 eV In the n = 3 level, E = 13.6 eV = 1.51 eV 9
*40.45
k ee 2 k ee 2 1 Starting with 2 m e v 2 = 2r , we have v 2 = m r e and using rn = n2h2 me ke e 2 or vn = ke e 2 nh
gives vn 2 =
ke e 2 n2h2 me me ke e 2
*40.46 (a)
The velocity of the moon in its orbit is
2 r 2 (3.84 108 m) v= T = = 1.02 103 m/s 2.36 106 s
So, (b) We have or (c) We have so
L = mvr = (7.36 1022 kg)(1.02 103 m/s)(3.84 108 m) = 2.89 1034 kg m2/s L = nh n= L 2.89 1034 kg m2/s = = 2.74 1068 h 1.055 10 34 J s
n h = L = mvr = m(GM e /r)1/2 r, r= h2 n 2 = Rn 2 2 m GM e and r (n + 1)2 R n 2R 2n + 1 = = 2 r n R n2 2 = 7.30 10 6 9 n
which is approximately equal to
Chapter 40 Solutions
19
40.47
The batch of excited atoms must make these six transitions to get back to state one: 2 1, and also 3 2 and 3 1, and also 4 3 and 4 2 and 4 1. Thus, the incoming light must have just enough energy to produce the 1 4 transition. It must be the third line of the Lyman series in the absorption spectrum of hydrogen. The absorbing atom changes from energy Ei = - 13.6 eV 13.6 eV = -13.6 eV to E f = - = - 0.850 eV , 12 42
so the incoming photons have wavelength 6.626 10 - 34 J s 3.00 108 m s 1.00 eV hc = = = 9.75 10 - 8 m = 97.5 nm -19 E f - Ei - 0.850 eV - ( -13.6 eV ) 1.60 10 J
(
)(
)
40.48
Each atom gives up its kinetic energy in emitting a photon, so
1 2
mv 2 =
hc (6.626 10 34 J s)(3.00 108 m/s) = (1.216 10 7 m)
= 1.63 1018 J
v = 4.42 104 m/s
40.49
(a)
The energy levels of a hydrogen-like ion whose charge number is Z are given by Z2 En = ( -13.6 eV ) 2 n Thus for He lium ( Z = 2) , the energy levels are En = - 54.4 eV n = 1, 2, 3, . . . n2
(b)
For He + , Z = 2 , so we see that the ionization energy (the energy required to take the electron from the n = 1 to the n = state is E = E - E1 = 0 -
( -13.6 eV)(2)2 (1)2
= 54.4 eV
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20
Chapter 40 Solutions
40.50
r=
n2h2 n 2 h2 = ; Zme ke e 2 Z me ke e 2
n=1
r=
5.29 10 -11 m 1 (1.055 10 -34 J s)2 = Z (9.11 10 - 31 kg)(8.99 10 9 N m 2 / C 2 )(1.602 10 -19 C)2 Z r= r= r= 5.29 10 -11 m = 2.65 10 -11 m = 0.0265 nm 2 5.29 10 -11 m = 1.77 10 -11 m = 0.0177 nm 3 5.29 10 -11 m = 1.32 10 -11 m = 0.0132 nm 4
(a) (b) (c)
For He+, Z = 2 For Li2+, Z = 3 For Be3+, Z = 4
40.51
Since F = qvB =
mv 2 r
we have
qrB = mv, rn = nh qB
or
qr 2B = mvr = nh
so
40.52
(a)
The time for one complete orbit is:
T=
2 r v nh me r
From Bohr's quantization postulate, L = me vr = nh, we see that v = Thus, the orbital period becomes: T=
2 2 me r 2 2 me (a0n 2 )2 2 me a0 3 = = n nh nh h
or
T = t0n3
where
t0 = (b)
2 2 me a0 2 (9.11 10 - 31 kg)(0.0529 10 - 9 m)2 = = 1.52 10 - 16 s h (1.055 10 - 34 J s)
With n = 2, we have T = 8t0 = 8(1.52 10 -16 s) = 1.21 10 -15 s Thus, if the electrons stay in the n = 2 state for 10 s, it will make 10.0 10 - 6 s = 8.23 109 revolutions of the nucleus 1.21 10 -15 s / rev
(c)
Yes, for 8.23 109 "electron years"
Chapter 40 Solutions
21
*40.53
h h 6.626 1034 J s =p = mv = = 3.97 1013 m (1.67 1027 kg)(1.00 106 m/s)
40.54
(a)
p2 = (50.0)(1.60 1019 J) 2m p = 3.81 1024 kg m/s h = p = 0.174 nm p2 = (50.0 103)(1.60 1019 J) 2m p = 1.20 10 22 kg m/s h = p = 5.49 10 12 m The relativistic answer is slightly more precise:
(b)
=
h hc = p (mc 2 + K)2 - m2 c 4
[
]
1/2
= 5.37 10 - 12 m
*40.55 (a)
Electron: so and
h = p p = 2me K
and
p2 m2 v 2 1 K = 2 me v 2 = e = 2me 2me
=
h = 2me K
2 9.11 10 -31 kg ( 3.00) 1.60 10 -19 J
(
6.626 10 -34 J s
)
(
)
= 7.09 10 -10 m = 0.709 nm
(b)
Photon: = c / f and E = hf so f = E h and hc (6.626 10 34 J s)(3.00 108 m/s) = E = = 4.14 107 m = 414 nm (3.00)(1.60 10 19 J)
2000 by Harcourt, Inc. All rights reserved.
22
Chapter 40 Solutions
40.56
From the Bragg condition (Eq. 38.13), m = 2d sin = 2d cos ( 2) But, d = a sin ( 2) where a is the lattice spacing. Thus, with m = 1,
= 2a sin( 2) cos( 2) = a sin =
h = p h = 2me K 6.626 10 - 34 J s
- 31
2 9.11 10
(
kg 54.0 1.60 10
)(
-19
J
)
= 1.67 10 -10 m
Therefore, the lattice spacing is a= 1.67 10 -10 m = = 2.18 10 -10 m = 0.218 nm sin 50.0 sin
*40.57 (a)
~ 10 -14 m or less.
p= h 6.6 10 - 34 J s ~ = 10 -19 kg m s or more. 10 -14 m
The energy of the electron is
2 E = p 2 c 2 + me c 4 ~ 10 -19
(
) (3 10 ) + (9 10 ) (3 10 )
2 8 2 -31 2
12 8 4
~ 10 -11 J~ 108 eV or more,
so that K = E - me c 2 ~ 108 eV - 0.5 106 eV
(
)
~ 108 eV
or more.
(b)
The electric potential energy of the electron would be Ue = 9 10 9 N m 2 C 2 10 -19 C ( -e ) ke q1q2 ~ ~ -10 5 eV r 10 -14 m
(
)(
)
With its kinetic energy much larger than its negative potential energy, the electron would immediately escape the nucleus .
Chapter 40 Solutions
23
Goal Solution The nucleus of an atom is on the order of 10 -14 m in diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be of this order of magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) On the basis of this result, would you expect to find an electron in a nucleus? Explain. The de Broglie wavelength of a normal ground-state orbiting electron is on the order 10 -10 m (the diameter of a hydrogen atom), so with a shorter wavelength, the electron would have more kinetic energy if confined inside the nucleus. If the kinetic energy is much greater than the potential energy from its attraction with the positive nucleus, then the electron will escape from its electrostatic potential well. If we try to calculate the velocity of the electron from the de Broglie wavelength, we find that v= h 6.63 10 -34 J s = = 7.27 1010 m / s -31 -14 me m 9.11 10 kg 10
G:
O:
(
)(
)
which is not possible since it exceeds the speed of light. Therefore, we must use the relativistic energy expression to find the kinetic energy of this fast-moving electron. The relativistic kinetic energy of a particle is K = E - mc 2 , where E 2 = ( pc ) + mc 2 , and the momentum is p = h :
2 2
A:
(a)
( )
p= E=
6.63 10 -34 J s = 6.63 10 -20 N s 10 -14 m
(1.99 10
-11
J
) + (8.19 10
2
-14
J
)
2
= 1.99 10 -11 J
(b)
1.99 10 -11 J - 8.19 10 -14 J = 124 MeV ~ 100 MeV 1.60 10 -19 J / eV The electrostatic potential energy of the electron 10 -14 m away from a positive proton is : K = E - mc 2 = N m2 -19 8.99 10 9 C 1.60 10 C2 2 U = - ke e r = - 10 -14 m
(
)
2
= -2.30 10 -14 J ~ -0.1 MeV
L:
Since the kinetic energy is nearly 1000 times greater than the potential energy, the electron would immediately escape the proton's attraction and would not be confined to the nucleus. It is also interesting to notice in the above calculations that the rest energy of the electron is negligible compared to the momentum contribution to the total energy.
2000 by Harcourt, Inc. All rights reserved.
24
Chapter 40 Solutions
40.58
(a)
From E = m e c2 p E c
=
20.0 103 MeV = 3.91 104 0.511 MeV
(b)
(for m e c 2 << pc) = 1.07 1017 kg m/s
p=
(2.00 104 MeV)(1.60 1013 J/MeV) 3.00 108 m/s
(c)
h 6.626 1034 J s =p = = 6.22 1017 m 1.07 1017 kg m/s Since the size of a nucleus is on the order of 10 14 m, the 20-GeV electrons would be small enough to go through the nucleus.
40.59
(a)
E2 = p 2 c 2 + m 2c4 with E = hf , p= h ,
2
and h2c 2 2 C
mc =
2
h C (Eq. 1)
so (b) For a photon
h 2f
=
h 2c2
2
+
and
1 1 f = 2+ 2 c C
f / c = 1/ .
The third term 1/ C in Equation 1 for electrons and other massive particles shows that they will always have a different frequency from photons of the same wavelength
40.60
(a)
The wavelength of the student is = h p = h mv. If w is the width of the diffraction aperture, then we need w 10.0 = 10.0 ( h mv) , so that v 10.0 6.626 10 -34 J s h - 34 = 10.0 ms = 1.10 10 mw (80.0 kg)(0.750 m) d we get: v t 0.150 m = 1.36 10 33 s 1.10 10 - 34 m / s times the age of the
(b) (c)
Using t =
N o . The minimum time to pass through the door is over 1015 Universe.
Chapter 40 Solutions
25
40.61
The de Broglie wavelength is:
=
h me v h me c it is necessary that
2 2
The Compton wavelength is: Therefore, we see that to have
C =
= C,
v
v = c.
c . 2
This gives:
1- v /c
2
2
= c , or
v v = 1- , yielding v = c c
40.62
VS =
h f- e e
0=
From two points on the graph
h 4.1 1014 Hz - e e h 12 1014 Hz - e e
(
)
and
3.3 V =
(
)
Combining these two expressions we find: (a) (b)
= 1.7 eV
h = 4.2 10 15 V s e At the cutoff wavelength hc h ec = = e c (3.0 108 m/s) = 730 nm (1.7 eV)(1.6 10 19 J/eV)
(c)
c
c = (4.2 10 15 V s)(1.6 10 19 C)
40.63
q2B2R2 (1.60 10 19 C)2(2.00 10 5 T)2(0.200 m)2 Kmax = 2m = = 2.25 10 19 J = 1.40 eV = hf e 2(9.11 10 31 kg)
= hf K max =
hc (4.14 10 15 eV s)(3.00 108 m/s) Kmax= 1.40 eV = 1.36 eV 450 10 9 m
2000 by Harcourt, Inc. All rights reserved.
26
Chapter 40 Solutions
40.64
From the path the electrons follow in the magnetic field, the maximum kinetic energy is seen to be: e 2B2 R 2 Kmax = 2me From the photoelectric equation, Kmax = hf - = hc -
Thus, the work function is
=
hc hc e 2B2 R 2 - Kmax = - 2me
40.65
We want an Einstein plot of K max versus f
, nm f, 1014 Hz Kmax, eV 588 5.10 0.67 505 5.94 0.98 445 6.74 1.35 399 7.52 1.63
(a) (b) slope = 0.402 eV 8% 1014 Hz
e(V S ) = hf h = (0.402) 1.60 10 19 J s = 6.4 10 34 J s 8% 1014
(c)
Kmax = 0 at f 344 1012 Hz
f ( THz)
= hf = 2.32 10 19 J = 1.4 eV
40.66
=
h (6.626 10 - 34 J s) (1 - cos ) = (0.234) = 3.09 10 -16 m mp c (1.67 10 - 27 kg)(3.00 108 m / s) hc (6.626 10 - 34 J s)(3.00 108 m / s) = = 6.20 10 -15 m E0 (200 MeV)(1.60 10 -13 J / MeV)
0 =
= 0 + = 6.51 10 -15 m
(a) (b) E = hc = 191 MeV
K p = 9.20 MeV
Chapter 40 Solutions
27
40.67
M is the mass of the positron which equals m e, the mass of the electron. So
reduced mass =
me M m = e me + M 2 or rpos = 2r Hyd = 1.06 10 -10 m n2
rpos =
n2h2 n2h2 2n2h2 = = Z ke e 2 Z( me / 2)ke e 2 Zme ke e 2
(
)
This is the separation of the two particles. Epos = - m k 2e 4 1 ke 2 e 4 1 ; = - e e2 n2 2h2 n2 4h n = 1, 2, 3, . . . or Epos = EHyd 2 = - 6.80 eV n2
Goal Solution Positronium is a hydrogen-like atom consisting of a positron (a positively charged electron) and an electron revolving around each other. Using the Bohr model, find the allowed radii (relative to the center of mass of the two particles) and the allowed energies of the system. G: Since we are told that positronium is like hydrogen, we might expect the allowed radii and energy levels to be about the same as for hydrogen: r = a0n2 = 5.29 10 -11 m n2 and En = ( -13.6 eV ) / n2 .
(
)
O:
Similar to the textbook calculations for hydrogen, we can use the quantization of angular momentum of positronium to find the allowed radii and energy levels. Let r represent the distance between the electron and the positron. The two move in a circle of radius r/2 around their center of mass with opposite velocities. The total angular momentum is quantized according to Ln = For each particle, F = ma expands to We can eliminate to find ke e 2 mv 2 = r/2 r2
A:
So the separation distances are The orbital radii are r/2 = a0n2, the same as for the electron in hydrogen. The energy can be calculated from Since mv 2 = L: ke e 2 , 2r
1 1 E = K + U = 2 mv 2 + 2 mv 2 -
ke e 2 r
E=
ke e 2 ke e 2 k e 2 -k e 2 6.80 eV - =- e = e 2 =- 2r r 2r n2 4a0n
It appears that the allowed radii for positronium are twice as large as for hydrogen, while the energy levels are half as big. One way to explain this is that in a hydrogen atom, the proton is much more massive than the electron, so the proton remains nearly stationary with essentially no kinetic energy. However, in positronium, the positron and electron have the same mass and therefore both have kinetic energy that separates them from each other and reduces their total energy compared with hydrogen.
2000 by Harcourt, Inc. All rights reserved.
28
Chapter 40 Solutions
40.68
Isolate the terms involving in Equations 40.12 and 40.13. Square and add to eliminate . Solve for b v2 = : 2 c b + c2
1 2 cos 1 2 = 2 me v 2 h2 2 + 2 - 0 0 b= h2 2 me 1 2 cos 1 2 + 2- 0 0
(
)
Substitute into Eq. 40.11:
h 1 1 b 1+ - = = 1- me c 0 b + c2 h2 1 2 cos 2hc 1 1 h2 1 1 1 c + - + 2 - = c2+ 2 2 + 2 - me 0 me 0 0 me 0
2 2
Square each side: From this we get Eq. 40.10:
- 0 = ( h me c )[1 - cos ]
40.69
hf = E =
2 4 2 me ke e 4 1 1 - 2 (n - 1)2 n2 2h
so
f=
2 2 2 me ke e 4 2n - 1 3 (n - 1)2 n2 h
As n approaches infinity, we have f approaching The classical frequency is f = v 1 = 2 r 2 ke e 2 1 me r 3/2 where
2 2 2 me ke e 4 2 n3 h3
r= f=
n2 h 2 4 me ke e 2
2 2 2 me ke e 4 2 n3 h3
Using this equation to eliminate r from the expression for f,
40.70
Show that if all of the energy of a photon is transmitted to an electron, momentum will not be conserved. hc hc hc Energy: = + Ke = me c 2 ( - 1) if =0 (1) 0 Momentum: From (1), h h = + me v = me v if = 0 (2) (3)
2
=
h +1 0 me c
0 me c v = c 1- h + 0 me c
(4)
Substitute (3) and (4) into (2) and show the inconsistency: 0 me c 0 me c + h h(h + 2 0 me c) h h h = 1 + = me c 1 - h + m c = 2 0 0 me c 0 0 (h + 0 me c) 0 e
2
h + 2 0 me c h
40.71
This is impossible, so all of the energy of a photon cannot be transmitted to an electron. h v . Begin with momentum expressions: p = , and p = mv = mc c
Chapter 40 Solutions
29
Equating these expressions,
v h 1 C = = c mc
Thus,
( v c )2 = C 2 2 1 - (v c)
2 2 2 C C C v 2 v = - c = c 1+ C
or
(
(
)
2
) ( C )
2
=
1
2
+1
giving
v=
c 1 + ( / C )2
40.72
(a)
The energy of the ground state is:
E1 = -
hc
series limit
=-
1240 eV nm = 8.16 eV 152.0 nm
From the wavelength of the L line, we see:
E2 - E1 =
hc 1240 nm eV = = 6.12 eV 202.6 nm
E2 = E1 + 6.12 eV = 2.04 eV Using the wavelength of the L line gives: so Next, using the L line gives: and From the L line, so (b) For the Balmer series, For the line, E i = E 3 and so E3 - E1 = 1240 nm eV = 7.26 eV 170.9 nm
E3 = 0 . 9 0 2 e V E4 - E1 = 1240 nm eV = 7.65 eV 162.1 nm
E4 = 0 . 5 0 8 e V E5 - E1 = 1240 nm eV = 7.83 eV 158.3 nm
E5 = 0 . 3 2 5 e V 1240 nm eV hc = E i - E 2 , or = Ei - E 2
=
1240 nm eV = 1090 nm - 0.902 eV ) - (- 2.04 eV) (
Similarly, the wavelengths of the line, line, and the short wavelength limit are found to be: 811 nm , 724 nm , and 609 nm .
2000 by Harcourt, Inc. All rights reserved.
30
Chapter 40 Solutions
(c)
Computing 60.0% of the wavelengths of the spectral lines shown on the energy-level diagram gives: 0.600(202.6 nm) = 122 nm , 0.600(170.9 nm) = 103 nm , 0.600(162.1 nm) = 97.3 nm , 0.600(158.3 nm) = 95.0 nm , and 0.600(152.0 nm) = 91.2 nm .
These are seen to be the wavelengths of the , , , and lines as well as the short wavelength limit for the Lyman series in Hydrogen. (d) The observed wavelengths could be the result of Doppler shift when the source moves away from the Earth. The required speed of the source is found from 1 - (v c) f = = = 0.600 f 1 + (v c) yielding v = 0.471c
40.73
(a)
Starting with Planck's law,
I ( ,T ) =
5 e hc
[
2 hc 2
kBT
-1
] ]
d .
the total power radiated per unit area
0
I ( ,T ) d = hc kBT hc d kBT 2
0
5 e hc kBT - 1
[
2 hc 2
Change variables by letting
x=
and
dx = -
Note that as varies from 0 , x varies from 0 . Then
0
I ( ,T ) d = -
4 2 kBT 4 h3c 2
0
(e
x3
x
-1
)
dx =
4 2 kBT 4 4 h 3 c 2 15
Therefore,
2 k 4 4 0 I ( ,T ) d = 15 h3cB T = T 2
5 4 4 2 5 1.38 10 -23 J K 2 5 kB = = 15 h 3 c 2 15 6.626 10 -34 J s 3 3.00 108 m s
(b)
From part (a),
(
(
)(
)
4
)
2
= 5.67 10 - 8 W m 2 K 4
Chapter 40 Solutions
31
*40.74
Planck's law states I ( ,T ) =
e
5
[
2 hc 2
hc k B T
-1
]
= 2 hc 2 -5 e hc
[
kBT
-1
]
-1
To find the wavelength at which this distribution has a maximum, compute dI = 2 hc 2 -5 - 6 e hc d
[
kBT
-1
]
-1
- -5 e hc
[
kBT
-1
]
-2 hc k T B
e
hc - 2 = 0 kBT
2 hc 2 hc dI e hc kBT = 6 hc k T - 5 + B -1 d e kBT e hc kBT - 1
[
]
[
]
=0
Letting x =
hc xe x , the condition for a maximum becomes x = 5. kBT e -1
We zero in on the solution to this transcendental equation by iterations as shown in the table below. The solution is found to be x 4.00000 4.50000 5.00000 4.90000 4.95000 4.97500 4.96300 4.96900 4.96600 4.96450 4.96550 4.96500 4.96525 4.96513 4.96507 4.96510 4.965115 x= xe x e x - 1 4.0746294 4.5505521 5.0339183 4.9367620 4.9853130 5.0096090 4.9979452 5.0037767 5.0008609 4.9994030 5.0003749 4.9998890 5.0001320 5.0000153 4.9999570 4.9999862 5.0000008 and
(
)
hc = 4.965115 max kBT
- 34
maxT =
hc 4.965115 kB
Thus, max
(6.626075 10 T=
4.965115 1.380658 10
(
J s 2.997925 108 m s
- 23
)(
J K
)
)=
2.897755 10 - 3 m K
This result is very close to Wien's experimental value of maxT = 2.898 10 - 3 m K for this constant.
2000 by Harcourt, Inc. All rights reserved.
32
Chapter 40 Solutions
40.75
=
h (1 - cos ) = - 0 me c
-1
E =
hc h hc = = hc 0 + (1 - cos ) me c 0 + hc hc 1 + (1 - cos ) 2 0 me c 0
-1
E =
hc hc 1 + (1 - cos ) E = 0 me c 2 0
-1
E = E0 1 + 0 2 (1 - cos ) me c
-1
40.76
r1 =
(1)2 h2
Z ke e 2
=
a0 h2 0.0529 nm = = = 3.12 fm 2 82)( 207 ) (82)(207 ) (82)(207me )ke e (
2
E1 =
-13.6 eV 207 82 = 18.9 MeV (1)2 1 1
40.77
This is a case of Compton scattering with a scattering angle of 180. = - 0 = h 2h (1 - cos 180) = m c me c e and = 0 + = 2E0 hc 2h hc + = 1 + E0 me c E0 me c 2
E0 =
hc hc , so 0 = E0 0
The kinetic energy of the recoiling electron is then K = E0 -
2 1 + 2E0 me c 2 - 1 2E0 me c 2 E0 hc = E0 - = E0 = 2 2 1 + 2E0 me c 1 + 2E0 me c 1 + 2E0 me c 2
(
)
Defining a E0 me c 2 , the kinetic energy can be written as K= 2E 0 a 2( hf )a -1 = = 2h f a (1 + 2a) 1 + 2a 1 + 2a
where f is the frequency of the incident photon.
Chapter 40 Solutions Planck's radiation law predicts maximum intensity at a wavelength max we find from d dI 2 -5 (hc/ k B T ) =0= -1 2 hc e d d 0 = 2 hc 2 - 5 (-1) e(hc/ kBT ) - 1
33
40.78
(a)
[
]
-1
[
]
-2
e(hc/ kBT ) - hc / 2 kBT + 2 hc 2 (- 5) - 6 e(hc/ kBT ) - 1 + 5 =0
(
)
[
]
-1
or
- hc e(hc/ kBT )
7 kBT e(hc/ kBT ) - 1
[
]
2
6 e(hc/ kBT ) - 1
[
]
which reduces to Define x = hc kBT .
5 ( kBT hc ) e(hc/ kBT ) - 1 = e(hc/ kBT ) Then we require 5 e x - 5 = xe x . So
[
]
Numerical solution of this transcendental equation gives x = 4.965 to four digits. max = hc 4.965 kBT , in agreement with Wien's law. The intensity radiated over all wavelengths is
0
I ( ,T)d = A + B =
0 5
e(hc/ kBT ) - 1
[
2 hc 2 d
]
Again, define x = hc kBT so = hc xkBT and d = - hc x 2 kBT dx Then, A + B =
0 x = 5 - 2 hc 2 x 5 kBT 5 hc dx
(
)
h 5 c 5 x 2 kBT e x - 1
(
)
=
4 2 kBT 4 h3c 2
0
(e
x 3 dx
x
-1
)
A+B=
4 2 5 kB T 4 15 h 3 c 2
The integral is tabulated as 4 / 15 , so (in agreement with Stefan's law) The intensity radiated over wavelengths shorter than max is
0
max
I( ,T)d = A =
max
0
5 e(hc/ kBT ) - 1
[
2 hc 2 d
]
A=
4 2 kB T 4 h3c 2
With x = hc kBT , this similarly becomes
4.965
x 3 dx ex - 1
So the fraction of power or of intensity radiated at wavelengths shorter than max is
4 4.965 x 3 dx 2 kBT 4 4 15 - 0 h3c 2 e x - 1 4 2 5 kBT 4 3 2
A = A+B
= 1-
15 4
0
4.965
x 3 dx ex - 1
15 h c
2000 by Harcourt, Inc. All rights reserved.
34
Chapter 40 Solutions
(b)
Here are some sample values of the integrand, along with a sketch of the curve:
Approximating the integral by trapezoids gives
15 A 1- 4 ( 4.870) = 0.2501 A+B
40.79
C =
h me c
and
=
h : p
C
p=
=
p h / me c = ; me c h/p E2 - (me c)2 c2
E 2 = c 2 p 2 + (me c 2 )2 :
C
=
1 E2 1 - (me c)2 = me c c 2 (me c)2
E2 E 2 2 - (me c) = -1 me c 2 c
2
40.80
p = mv = 2mE = 2 1.67 10 -27 kg (0.0400 eV ) 1.60 10 -19 J / eV h = m v = 1.43 10 10 m = 0.143 nm
(
)
(
)
This is of the same order of magnitude as the spacing between atoms in a crystal so diffraction should appear.
40.81
Let u represent the final speed of the electron and let eliminate and u from the three conservation equations: hc hc + me c 2 = + me c 2 0 h h + me u - cos = me u cos 0 h sin = me u sin [1]
= 1 - u 2 c 2
(
)
-1/2
.
We must
[2]
[3]
Chapter 40 Solutions
35
Square Equations [2] and [3] and add: h 2 2 h me u 2 h 2 cos 2 h me u cos h2 2 2 + 2 me u2 + 2 + - - = 2 me u 2 2 0 0 0
2 2 h me u 2 h me u cos 2 h 2 cos me u 2 h2 h2 2 + 2 + 2 me u2 + - - = 2 0 0 0 1 - u 2 / c 2
Call the left-hand side b. Then b -
b u 2 2 = me u 2 c2
and u 2 =
b c 2b = 2 2 2 me + b c 2 me c + b
Now square Equation [1] and substitute to eliminate :
2 2 h me c h 2 2 h me c 2 h 2 me c 2 h2 2 2 + 2 me c 2 + 2 + - - = = me c 2 + b 2 0 0 1 - u 2 / c 2
So we have 2h me c 2h me c 2h 2 h2 h2 2 + 2 + 2 me c 2 + - - 0 0 2 0 = me c 2 +
2 Multiply through by 0 me c 2
2h me u 2h me ucos 2h 2 cos h2 h2 2 + 2 + 2 me u2 + - - 0 0 2 0
0 2 +
0 2 u2 2 h u 2 h 0 u cos 2 h 2 cos 2h 2 h 0 2 h2 - - 2 2 = 0 + + - - 2 me c me c me c c2 me c 2 me c 2 me c 2
2 u2 2 h u cos 2 h 2 u 2 h 0 = 0 2 - 1- 2 + 1- 1- + 2 (1 - cos ) c me c me c c me c 2 c The first term is zero. Then 1 - (u cos ) c h -1 1 = 0 + m c 1 - u c (1 - cos ) 1- u c e
Since
-1 = 1 - (u c ) =
2
(1 - u c)(1 + u c)
this result may be written as
1 - (u cos ) c h 1+ u c = 0 + m c 1 - u c (1 - cos ) 1- u c e
2000 by Harcourt, Inc. All rights reserved.
Chapter 41 Solutions
h 6.626 10 -34 J s = = = 9.92 10 -7 m -27 mv (1.67 10 kg)(0.400 m/s)
For destructive interference in a multiple-slit experiment,
41.1
(a)
(b)
1 d sin = m + 2 with m = 0 for the first minimum. Then, = sin -1 = 0.0284 2d y so y = L tan = (10.0 m)(tan 0.0284) = 4.96 mm = tan L
(c) We cannot say the neutron passed through one slit. We can only say it passed through the slits.
41.2
Consider the first bright band away from the center:
d sin = m
.400 (6.00 10-8 m) sin tan -1 0200 = 1 = 1.20 10-10 m
=
1 2
h me v
so
me v =
h
and
K = me v 2 =
me 2 v 2 h2 = = e( V ) 2me 2me 2
6.626 10 -34 J s h2 V = = 2eme 2 2 1.60 10 -19 C 9.11 10 -31 kg 1.20 10 -10 m
(
( )(
) )(
2
)
2
= 105 V
41.3
(a)
The wavelength of a non-relativistic particle of mass kinetic energy energy in joules is
m is given by = h / p = h 2mK where the K is in joules. If the neutron kinetic energy Kn is given in electron volts, its kinetic
K = 1.60 10 -19 J/eV Kn and the equation for the wavelength becomes
(
)
=
h = 2mK
2 1.67 10 -27 kg 1.60 10 -19
(
6.626 10 -34 J s
)(
2.87 10 -11 m = Kn J/eV Kn
)
where (b) If
Kn is expressed in electron volts.
Kn = 1.00 keV = 1000 eV , then
Chapter 41 Solutions
495
41.4
=
h = p
h2 h , so K = 2mK 2m2
If the particles are electrons and
~ 0.1 nm = 10 -10 m , the kinetic energy in electron volts is
2
(6.626 10-34 J s) 1 eV = ~ 102 eV K= 2 -19 2(9.11 10 -31 kg )(10 -10 m) 1.602 10 J
41.5
=
h p
(a)
p=
h 6.626 10 -34 J s = = 6.63 10 -23 kg m/s 1.00 10 -11 m
electrons:
(6.63 10-34 J s) p Ke = = 2me 2(9.11 10 -31 )
2
2
J = 15.1 keV
Ke = p 2 c 2 + me 2 c 4
(b) photons:
(
The relativistic answer is more precisely correct:
)
1/ 2
- me c 2 = 14 /.9 keV
E = pc = 6.63 10 -23 3.00 108 = 124 keV
(
)(
)
41.6
The theoretical limit of the electron microscope is the wavelength of the electrons. If
Ke = 40.0 keV , then E = Ke + me c 2 = 551 keV and
1 2 p= E - me 2 c 4 = c
(551 keV)2 - (511 keV)2 1.60 10 -16
3.00 10 m/s
8
J = 1.10 10 -22 kg m/s 1.00 keV
The electron wavelength, and hence the theoretical limit of the microscope, is then
h 6.626 10 -34 J s = = = 6.03 10 -12 m = 6.03 pm -22 p 1.10 10 kg m/s
41.7
E = K + me c 2 = 1.00 MeV + 0.511 MeV = 1.51 MeV
p 2c 2 = E 2 - me 2 c 4 = p = 1.42 MeV/c
(1.51 MeV)2 - (0.511 MeV)2
so
6.626 10 -34 J s 3.00 108 m/s h hc = = = = 8.74 10 -13 m 6 -19 p 1.42 MeV J 1.42 10 1.60 10
(
(
)(
)(
)
)
d sin = m
Suppose the array is like a flat diffraction grating with openings
0.250 nm apart:
5 8 74 10 -13
Chapter 41 Solutions
496
41.8
(a)
px = mvx h/2 so h 2 J s v = = 0.250 m/s 4 mx 4 (2.00 kg )(1.00 m)
(b)
The duck might move by
(0.25 m/s)(5 s) = 1.25 m .
With original position uncertainty of
1.00 m , we can think of x growing to 1.00 m + 1.25 m = 2.25 m
41.9
For the electron,
p = me v = 9.11 10 -31 kg ( 500 m/s) 1.00 10 -4 = 4.56 10 -32 kg m/s
(
)
(
)
For the bullet,
) p = mv = (0.0200 kg )( 500 m/s)(1.00 10 -4 ) = 1.00 10 -3 kg m/s
x = h = 5.28 10 -32 m 4 p
x =
h 6.626 10 -34 J s = = 1.16 mm 4 p 4 4.56 10 -32 kg m/s
(
Goal Solution An electron ( me objects? G: It seems reasonable that a tiny particle like an electron could be located within a more narrow region than a bigger object like a bullet, but we often find that the realm of the very small does not obey common sense. Heisenberg's uncertainty principle can be used to find the uncertainty in position from the uncertainty in the momentum. A: The uncertainty principle states:
= 9.11 10 -31 kg ) and a bullet (m = 0.0200 kg) each have a speed of 500 m/s , accurate to within 0.0100%. Within what limits could we determine the position of the
O:
xpx h/2 where px = mv and h = h/2 .
v = (0.000100)( 500 m/s) = 0.0500 m/s
Both the electron and bullet have a velocity uncertainty,
For the electron, the minimum uncertainty in position is
x =
h 6.63 10 -34 J s = = 1.16 mm 4 mv 4 9.11 10 -31 kg (0.0500 m/s)
(
)
For the bullet,
h 6.63 10 -34 J s x = = = 5.28 10 -32 m 4 mv 4 (0 0200 kg )(0 0500 m/s)
Chapter 41 Solutions
497
Chapter 41 Solutions
498
41.10
y py = x px
and
dpy h /4
Eliminate
py and solve for x .
2.00 10 m d x = 4px ( y ) = 4 1.00 10 -3 kg (100 m/s) 1.00 10 -2 m = h 6.626 10 -34 J s
(
)
(
)( (
-3
)
)
3.79 10 28 m
This is 190 times greater than the diameter of the Universe!
41.11
px
h 2
so
p = me v
h 4 x
v
h 6.626 10 -34 J s = = 1.16 106 m/s 4 me x 4 9.11 10 -31 kg 5.00 10 -11 m
(
)(
)
41.12
With
x = 2 10 -15 m, the uncertainty principle requires px
The average momentum of the particle bound in a stationary nucleus is zero. The uncertainty in momentum measures the root-mean-square momentum, so we take For an electron, the non-relativistic approximation while v cannot be greater than c . Thus, a better solution would be
h = 2.6 10 -20 kg m/s 2 x
prms = 3 10 -20 kg m/s .
p = me v would predict v = 3 1010 m/s ,
E = me c 2
(
)
2
2 + ( pc)
1/ 2
= 56 MeV = me c 2
so
110 =
v 0.99996c
For a proton,
1 1 - v2 /c2
v = p / m gives v = 1.8 107 m / s, less than one-tenth the speed of light.
41.13
(a)
xi . Thus, the uncertainty principle requires her to release it with typical horizontal momentum px = mvx = h/2 xi . It 1 2 falls to the floor in time given by H = 0 + 2 gt as t = 2 H g , so the total width of the impact
At the top of the ladder, the woman holds a pellet inside a small region points is
h 2H A , x f = xi + ( vx ) t = xi + = xi + xi 2mxi g h 2H A=
where
Chapter 41 Solutions xi = A , and the minimum width of the impact points is
499
so
(x f )min
(b)
A = xi + xi x
i=
A
2h =2 A= m
2(2.00 m) 9.80 m/s 2
1/ 2
2H = g
1/ 4
(x f )min
2 1.0546 10 -34 J s = 5.00 10 -4 kg
(
)
1/ 2
1/ 4
= 5.19 10 -16 m
Chapter 41 Solutions
a
500
41.14
Probability
P=
a
-a
(x) =
2
a
-a
(x 2 + a2 )
a
dx =
x a 1 tan -1 a a -a
P=
1 1 tan -1 1 - tan -1( -1) = - - = 1/ 2 4 4
[
]
41.15
(a)
( x) = A sin
so
2 x = A sin 5.00 1010 x
(
)
2 = 5.00 1010 m -1
=
(5.00 10 )
10
2
= 1.26 10 -10 m
(b)
h 6.626 10 -34 J s p= = = 5.27 10 -24 kg m/s -10 1.26 10 m
(c)
m = 9.11 10 -31 kg
5.27 10 -24 kg m/s p2 K= = 2m 2 9.11 10 -31 kg
(
(
) )
2
= 1.52 10 -17 J =
1.52 10 -17 J = 95.5 eV 1.602 10 -19 J/eV
41.16
For an electron to "fit" into an infinitely deep potential well, an integral number of half-wavelengths must equal the width of the well.
n so = 1.00 10 -9 m 2 2.00 10 -9 h = = n p
h 2 / 2 p2 h2 n2 Since K = = = 2me 2me 2me 2 10 -9
(a)
(
)
(
)
2
= 0.377 n2 eV
(
)
For (b) With
K 6 eV , n=4 ,
n=4 K = 6.03 eV
41.17
(a)
We can draw a diagram that parallels our treatment of standing mechanical waves. In each state, we measure the distance d from one node to another (N to N ), and base our solution upon that: Since
dN to N =
2
and
h
Chapter 41 Solutions
Next,
501
2 -34 Js p2 h2 1 6.626 10 K= = = 2me 8me d d 2 8 9.11 10 -31 kg
(
(
) )
Evaluating,
6.02 10 -38 J m 2 K= d2 3.77 10 -19 eV m 2 K= d2
In state 1,
d = 1.00 10 -10 m K1 = 37.7 eV d = 5.00 10 -11 m K2 = 151 eV
In state 2,
In state 3,
d = 3.33 10 -11 m K3 = 339 eV d = 2.50 10 -11 m K 4 = 603 eV
In state 4,
(b)
When the electron falls from state 2 to state 1, it puts out energy
E = 151 eV - 37.7 eV = 113 eV = hf =
into
hc
8
=
J s 3.00 10 m/s 6.626 10 hc = = 11.0 nm E (113 eV) 1.60 10 -19 J/eV
(
emitting
-34
(
)(
a
)
)
p h o to n
of
wavelength
The wavelengths of the other spectral lines we find similarly: T r a n s i t i o n
E(eV)
(nm)
Chapter 41 Solutions E1 = 2.00 eV = 3.20 10 -19 J
For the ground-state,
502
41.18
E1 =
h2 8me L2
(a)
L=
h = 4.34 10 -10 m = 0.434 nm 8me E1
(b)
h2 h2 E = E2 - E1 = 4 - = 6.00 eV 8me L2 8me L2
41.19
E =
hc h 2 2 2 3h2 = 2 -1 = 8me L2 8me L2
[
]
L=
3h = 7.93 10 -10 m = 0.793 nm 8me c
41.20
E =
hc h 2 2 2 3h2 = 2 -1 = 8me L2 8me L2
[
]
so
L=
3h 8me c
41.21
En =
n2 h 2 8mL2
E = E2 - E1 =
E = hf = hc
so
3( hc) 3h2 = 2 8mL 8mc 2 L2
2
and
-5 6 8mc 2 L2 8 938 10 eV 1.00 10 nm = Hence, = 3hc 3(1240 eV nm)
(
)(
)
2
= 2.02 10 -4 nm (gamma ray)
E= hc = 6.15 MeV
Fi gu re fo r G oa l So lu ti
Chapter 41 Solutions
503
Goal Solution The nuclear potential energy that binds protons and neutrons in a nucleus is often approximated by a square well. Imagine a proton confined in an infinitely high square well of width 10.0 fm , a typical nuclear diameter. Calculate the wavelength and energy associated with the photon emitted when the proton moves from the n = 2 state to the ground state. In what region of the electromagnetic spectrum does this wavelength belong? G: O: Nuclear radiation from nucleon transitions is usually in the form of high energy gamma rays with short wavelengths. The energy of the particle can be obtained from the wavelengths of the standing waves corresponding to each level. The transition between energy levels will result in the emission of a photon with this energy difference. At level 1, the node-to-node distance of the standing wave is twice this distance:
A:
h / p = 2.00 10 -14 m . The proton's kinetic energy is
1.00 10 -14 m , so the wavelength is
K=
1 mv 2 2
6.63 10 -34 J s p2 h2 = = = 2m 2m2 2 1.67 10 -27 kg 2.00 10 -14 m
(
(
)(
)
2
)
2
=
3.29 10 -13 J = 2.06 MeV 1.60 10 -19 J/eV
In the first excited state, level 2, the node-to-node distance is two times smaller than in state 1. The momentum is two times larger and the energy is four times larger: K = 8.23 MeV . The proton has mass, has charge, moves slowly compared to light in a standing-wave state, and stays inside the nucleus. When it falls from level 2 to level 1, its energy change is
2.06 MeV - 8.23 MeV = -6.17 MeV
Therefore, we know that a photon (a traveling wave with no mass and no charge) is emitted at the speed of light, and that it has an energy of +6.17 MeV .
Its frequency is
6.17 106 eV 1.60 10 -19 J/eV E f= = = 1.49 10 21 Hz -34 h 6.63 10 Js
(
)(
)
and its wavelength is
=
c 3.00 108 m/s = = 2.02 10 -13 m f 1.49 10 21 s -1
This is a gamma ray, according to Figure 34.17. L:The radiated photons are energetic gamma rays as we expected for a nuclear transition. In the above calculations, we assumed that the proton was not relativistic ( v < 0.1c ), but we should check this assumption for the highest energy state we examined ( n = 2 ):
v=
2K = m
2 8.23 106 eV 1.60 10 -19 J/eV 1.67 10 -27 kg
(
)(
) = 3.97 107 m/s = 0.133c
(0.133)2 = 1% accuracy.
This appears to be a borderline case where we should probably use relativistic equations, but our classical treatment should give reasonable results, within
Chapter 41 Solutions
504
41.22
= 2D
for the lowest energy state
K=
p2 h2 h2 (6.626 10 -34 J s)2 = = = 2m 2m2 8mD 8 4 1.66 10 -27 kg 1.00 10 -14 m
0.517 MeV
[(
)](
)
2
= 8.27 10 -14 J =
p=
h h 6.626 10 -34 J s = = = 3.31 10 -20 kg m/s 2D 2 1.00 10 -14 m
(
)
41.23
h2 2 En = n 8mL2
(6.626 10-34 J s) E1 = = 8.21 10 -14 J 2 -27 -14 8(1.67 10 kg )(2.00 10 m)
2
E1 = 0.513 MeV
E2 = 4E1 = 2.05 MeV
E3 = 9E1 = 4.62 MeV
41.24
(a)
L 2 4 x 2 L 1 1 2 x < x >= x sin 2 dx = x - cos dx 0 L L L 0 2 2 L
1 x2 < x >= L 2
L
-
0
1 L2 L 16 2
4 x 4 x 4 x L sin L + cos L = L/2 0
0.510 L
L
(b)
Probability
=
0.510 L
0.490 L
4 x 2 1 L 2 x 1 sin 2 sin dx = x - L L L 4 L 0.490 L L
Probability
= 0.20 -
1 (sin 2.04 - sin 1.96 ) = 5.26 10-5 4
0.260 L
(c)
Probability
4 x 1 x = - sin = 3.99 10 -2 L 0.240 L L 4
(d)
In the
n = 2 graph in Figure 41.11 (b), it is more probable to find the particle either near
or
x=
L 4
x=
3L 4
than at the center, where the probability density is zero.
Chapter 41 Solutions
505
x = 0 A
L
2
sin 2
n x L =1 dx = A 2 2 L
or
A=
2 L
41.26
The desired probability is
P=
x = L/ 4
x=0
dx =
2
2 L / 4 2 2 x sin dx L L 0
where Thus,
sin 2 =
1 - cos 2 2
4 x 1 x P= - sin L 4 L 0
L/ 4
=
1 - 0 - 0 + 0 = 0.250 4
41.27
0 x L , the argument 2 x / L of the sine function ranges from 0 to 2 . The probability 2 density ( 2 / L) sin ( 2 x / L) reaches maxima at sin = 1 and sin = -1 at
In
2 x = L 2
and
2 x 3 = 2 L x= L 3L and x = 4 4
The most probable positions of the particle are at
41.28
(a)
The probability is
P=
L/ 3
0
dx =
2
L/ 3
0
2 x 2 2 L/ 3 1 1 x sin 2 dx = dx - cos L L L 0 2 2 L
L/ 3
P=
(b)
2 x 1 x - sin L 2 L 0
=
2 1 3 1 1 sin - = - = 0.196 3 2 3 3 4
The probability density is symmetric about x = L /2 . Thus, the probability of finding the particle between x = 2L / 3 and x = L is the same 0.196 . Therefore, the probability of finding it in the range L / 3 x 2L / 3 is P = 1.00 - 2(0.196) = 0.609 . Classically, the electron moves back and forth with constant speed between the walls, and the probability of finding the electron is the same for all points between the walls. Thus, the classical probability of finding the electron in any range equal to one-third of the available space is
(c)
Pclassical = 1/ 3 .
Chapter 41 Solutions
506
41.29
The ground state energy of a particle (mass
m ) in a 1-dimensional box of width L is E1 =
h2 . 8mL2
(a)
For a proton
(b)
(m = 1.67 10-27 kg) in a 0.200 - nm wide box: (6.626 10 J s) E = = 8.22 10 J = 5.13 10 -3 eV 8(1.67 10 kg )(2.00 10 m) -31 kg ) in the same size box: For an electron (m = 9.11 10 (6.626 10 J s) J = 9.41 eV E = = 1.51 10 kg )(2.00 10 m) 8(9.11 10
-34 2 1 -22 -27 -10 2 -34 2 1 -18 -31 -10 2
(c)
The electron has a much higher energy because it is much less massive.
41.30
(a)
2 x cos L L 2 2 x 2 (x) = sin L L 2 3 x 3 (x) = cos L L
1( x) =
P1( x) = 1( x) P2 ( x) = 2 ( x) P3 ( x) = 3 ( x)
2
=
2
2
2 x cos 2 L L 2 2 x = sin 2 L L 2 3 x = cos L L
41.31
We have
= Ae i( kx - t )
and
2 = - k 2 2 x
Schrdinger's equation:
2 2m = - k 2 = 2 (E - U ) x2 h
k
2
Since
(2 )2 =
2
(2 p)2 = p 2 =
h
2
h
2
and
(E - U ) = p 2 / 2 m
Chapter 41 Solutions
507
*41.32
( x) = A cos kx + B sin kx
2 = - k 2 A cos k x - k 2B sin kx 2 x
2 2m = - (E - U ) x 2 h2
= - kA sin kx + kB cos kx x 2m 2mE - 2 (E - U ) = - 2 ( A cos kx + B sin kx) h h
Therefore the Schrdinger equation is satisfied if
or
2mE - k 2 ( A cos kx + B sin kx) = - 2 ( A cos kx + B sin kx) h
This is true as an identity (functional equality) for all
x if E =
h2 k 2 2m
41.33
Problem 45 in Ch. 16 helps students to understand how to draw conclusions from an identity.
(a)
x2 ( x ) = A 1 - 2 L
Schrdinger's equation becomes
2 Ax d =- 2 dx L
2 2A =- 2 2 x L
2 2m = - 2 (E - U ) 2 x h
x2 -h2 x 2 A 1 - 2 L 2 A 2m x 2 2m - 2 = 2 EA 1 - 2 + 2 2 2 2 L h L h mL L - x
(
) (
)
-
This will be true for all
1 mE mEx 2 x 2 =- 2 + 2 2 - 4 L2 h h L L
and
x if both
1 mE = L2 h2
mE 1 - 4 =0 2 2 h L L
Both of these conditions are satisfied for a particle of energy (b) For normalization,
E=
h2 . L2 m
L x2 2x 2 x 4 1 = A 1 - 2 dx = A 2 1 - 2 + 4 dx -L -L L L L L
2
2
2x 3 x 5 2 L 2 1 = A x - 2 + 4 = A 2 L - L + + L - L + 3 5 3 3L 5L - L
(c)
2
L
L 2 16 L = A 15 5
A=
15 16L
Chapter 41 Solutions
508
P=
47 = 0.580 81
Chapter 41 Solutions
Setting the total energy E equal to zero and rearranging the Schrdinger equation to isolate the potential energy function gives
509
41.34
(a)
h2 1 d 2 U (x) = 2m dx 2
If
( x) = Axe
or
- x 2 / L2
Then
-x /L d 2 3 2 e = 4 Ax - 6 AxL dx 2 L4
(
)
2
2
4 x 2 - 6L2 d 2 = (x) dx 2 L4
and
(
)
U (x) =
h2 4 x 2 - 6 2 2 2mL L
See figure to the right .
41.35
(a) (b)
See figure to the right . The wavelength of the transmitted wave traveling to the left is the same as the original wavelength, which equals 2L .
4
41.37
T = e -2CL
(Use Equation 41.17)
2CL =
2 2 9.11 10 -31 8.00 10 -19 1 055 10
-34
(
)(
)
(2.00 10-10 ) = 4.58
F i
Chapter 41 Solutions
o r G o a l S o l u t i o n
510
Chapter 41 Solutions
511
Goal Solution An electron with kinetic energy E = 5.00 eV is incident on a barrier with thickness L = 0.100 and height U = 10.0 eV (Fig. P41.37). What is the probability that the electron (a) will tunnel through the barrier and (b) will be reflected? G: O: A:
nm
Since the barrier energy is higher than the kinetic energy of the electron, transmission is not likely, but should be possible since the barrier is not infinitely high or thick. The probability of transmission is found from the transmission coefficient equation 41.18. The transmission coefficient is
C=
2m(U - E) = h
2 9.11 10 -31 kg (10.0 eV - 5.00 eV) 1.60 10 -19 J/eV 6.63 10 -34 J s/2
(
(
)
(
) = 1.14 1010 m-1
(a) The probability of transmission is
T = e -2CL = e
-2 1.14 1010 m -1 2.00 10 -10 m
)(
) = e -4.58 = 0.0103
1 - 0.0103 = 0.990
(b) If the electron does not tunnel, it is reflected, with probability
L:Our expectation was correct: there is only a 1% chance that the electron will penetrate the barrier. This tunneling probability would be greater if the barrier were thinner, shorter, or if the kinetic energy of the electron were greater.
41.38
C=
2 9.11 10 -31 ( 5.00 - 4.50) 1.60 10 -19 kg m/s 1.055 10
-34
(
)
(
)
Js
T = e -2CL = exp -2 3.62 10 9 m -1 950 10 -12 m = exp( -6.88) T = 1.03 10 -3
[ (
)(
)]
41.39
From problem 38,
C = 3.62 10 9 m -1
10 -6 = exp -2 3.62 10 9 m -1 L
Taking logarithms, New
[ (
)]
-13.816 = -2 3.62 10 9 m -1 L
(
)
L = 1.91 nm
Chapter 41 Solutions
512
*41.40
With the wave function proportional to are proportional to
2
, to
e - CL .
e - CL , the transmission coefficient and the tunneling current
Then,
I (0.500 nm) e -2(10.0 /nm )(0.500 nm ) = -2(10.0 /nm )(0.515 nm ) = e 20.0(0.015) = 1.35 I (0.515 nm) e
*41.41
With transmission coefficient
e - CL , the fractional change in transmission is
-2 10.0 /nm )( L + 0.002 00 nm ) e -2(10.0 /nm )L - e ( 29.0 0.002 00 ) = 1- e ( = 0.0392 = 3.92% -2(10.0 /nm )L e
41.42
= Be -( m /2h)x
2
so
d m =- x h dx
and
d 2 m 2 m = x+ - 2 h h dx
2
Substituting into Equation 41.19 gives
2mE m 2 m 2 m x+ - = 2 + x h h h h h which is satisfied provided that E = . 2
41.43 Problem 45 in Chapter 16 helps students to understand how to draw conclusions from an identity.
2
2
= Axe - bx
and
2
so
2 2 d = Ae - bx - 2bx 2 Ae - bx dx
2 2 2 d 2 = -2bxAe - bx - 4bxAe - bx + 4b 2 x 3 Ae - bx = -6b + 4b 2 x 2 2 dx
Substituting into Equation 41.19,
-6b + 4b 2 x 2 = -
m 2 2mE + x h h
2
For this to be true as an identity, it must be true for all values of
x.
and
So we must have both
-6b = -
2mE h2 m 2h
4b 2 =
m h
2
(a)
Therefore
b=
E=
(b) (c)
and The wave function is that of the
3bh2 3 = h 2 m
first excited state .
Chapter 41 Solutions
513
*41.44
The longest wavelength corresponds to minimum photon energy, which must be equal to the spacing between energy levels of the oscillator:
hc k = h = h m
so
9.11 10 -31 kg m = 2 c = 2 3.00 108 m/s k 8.99 N/m
(
)
1/ 2
= 600 nm
*41.45
(a)
With
= Be -( m /2h)x , the normalization condition
2
all
2
2
dx = 1
becomes
1 = B2 e -2( m / 2h)x dx = 2B2 e -2( m / 2h)x dx = 2B2
2
-
0
1 2 m /h
where Table B.6 in Appendix B was used to evaluate the integral.
Thus,
1= B
2
h m
and
m B= h
1/ 4
(b)
For small
, the probability of finding the particle in the range - / 2 x / 2 is
2
- / 2
/2
dx = (0)
2
= B e
2 -0
m = h
1/ 2
41.46
(a)
With
(px )2 . Then x h/2px requires
E p 2 kh2 px 2 k h2 + = x + 2 m 2 4 px 2 2 m 8 px 2
< x >= 0 and < px >= 0 , the average value of x 2 is ( x)2 and the average value of px 2 is
(b)
To minimize this as a function of
px 2 , we require
1 1 dE kh2 =0= + (-1) 4 2 2m 8 dpx px
Then
kh2 1 = 4 2m 8 px
px
2
2mkh2 = 8
1/ 2
=
h mk 2
and
E
h mk kh2 3 h k h k + = + . 2(2m) 8h mk 4 m 4 m
Chapter 41 Solutions
514
*41.47
Suppose the marble has mass While it is inside the wall,
20 g . Suppose the wall of the box is 12 cm high and 2 mm thick.
U = mgy = (0.02 kg ) 9.8 m/s 2 (0.12 m) = 0.0235 J
and
(
)
E = K = mv 2 =
C=
1 2
1 2
(0.02 kg)(0.8 m/s)2 = 0.0064 J
2(0.02 kg )(0.0171 J) 1.055 10
-34
Then
2m(U - E) = h
Js
= 2.5 10 32 m -1
and the transmission coefficient is
e -2CL = e
-2 2.5 10 32 2 10 -3
(
)(
) = e -10 10 29 = e -2.30( 4.3 10 29 ) = 10-4.3 10 29 = ~ 10-10 30
*41.48
(a)
= 2L = 2.00 10 -10 m
h 6.626 10 -34 J s p= = = 3.31 10 -24 kg m/s -10 2.00 10 m
p2 E= = 0.172 eV 2m
(b)
(c)
41.49
(a) (b) (c) (d)
See the first figure to the right . See the second figure to the right .
is continuous and
0 as x
Since
is symmetric,
-
or
2
dx = 2 dx = 1
2 0
2 A 2 - 0 2 A 2 e -2 x dx = e -e =1 0 -2
(
)
This gives
A=
Chapter 41 Solutions
(a) Use Schrdinger's equation
515
2 2m = - 2 (E - U ) 2 x h
with solutions
1 = Ae ik1x + Be - ik1x
[region I]
2 = Ce ik 2 x
[region II] Where and
k1 =
2mE h
k2 =
2m(E - U ) h
x = 0:
Then, matching functions and derivatives at and
( 1 )0 = ( 2 )0
A+B=C
d 1 d 2 = dx 0 dx 0
B= C= 2 A 1 + k2 / k1
k1( A - B) = k2C
Then
1 - k2 k1 A 1 + k2 k1
Incident wave
Ae
ikx
reflects
Be
- ikx
, with probability
(k - k ) B2 (1 - k2 / k1 ) R= 2 = = 1 2 2 2 A (1 + k2 / k1 ) (k1 + k2 )
2 2
(b)
With
E = 7.00 eV and U = 5.00 eV ,
k2 = k1
E -U = E
2.00 = 0.535 7.00
= 0.0920
The reflection probability is
(1 - 0.535)2 R= (1 + 0.535)2
The probability of transmission is
T = 1 - R = 0.908
41.51
(k1 - k2 )2 = (1 - k2 / k1 )2 R= (k1 + k2 )2 (1 + k2 / k1 )2
h2 k 2 = E - U for constant U 2m h 2 k12 = E since U = 0 2m
(1)
Chapter 41 Solutions
516
Dividing (2) by (1),
1 1 k2 2 U = 1- = 1- = so 2 2 2 E k1
k2 1 = k1 2
(1 - 1/ 2 ) = ( R= 2 (1 + 1/ 2 ) (
2
and therefore,
) = 0.0294 2 2 + 1)
2 -1
2
Chapter 41 Solutions
41.52 (a) (b) The wave functions and probability densities are the same as those shown in the two lower curves in Figure 41.11 of the textbook.
517
P1 =
0.350 nm 0.150 nm
1
2
2 dx = 1.00 nm
x 2.00 x 1.00 nm 2 x sin 1.00 nm dx = nm 2 - 4 sin 1.00 nm 0.150 nm 0.150 nm
2
0.350 nm
0.350 nm
In the above result we used
sin
2
axdx = ( x / 2) - (1/ 4 a) sin(2 ax)
0.350 nm
P1 =
1.00 1.00 nm 2 x sin x - 1.00 nm 0.150 nm 2 nm
P1 =
1.00 1.00 nm [sin(0.700 ) - sin(0.300 )] = 0.200 0.350 nm - 0.150 nm - nm 2
0.350
(c)
2 0.350 2 2 x x 1.00 4 x P2 = 0.150 sin 1.00 dx = 2.00 2 - 8 sin 1.00 1.00 0.150
1.00 4 x 1.00 P2 = 1.00 x - sin = 1.00(0.350 - 0.150) - [sin(1.40 ) - sin(0.600 )] = 1.00 0.150 4 4 0.351
(d) Using
0.350
En =
n2 h 2 , we find that E1 = 0.377 eV and E2 = 1.51 eV 8mL2
41.53
(a)
mgyi = mv f 2
1 2
v f = 2 gyi = 2 9.80 m/s 2 ( 50.0 m) = 31.3 m/s
(
)
=
h 6.626 10 -34 J s = = 2.82 10 -37 m (not observable) mv (75.0 kg )(31.3 m/s)
so
(b)
Et h/2
E
6.626 10 -34 J s 4 5.00 10
(
-3
s
)
= 1.06 10 -32 J
(c)
E 1.06 10 -32 J = = 2.87 10 -35 % E (75 0 kg ) 9 80 m/s 2 ( 50 0 m)
(
)
Chapter 41 Solutions
Et - h/2 or mc 2 t = h/ 2 . Therefore,
518
41.54
From the uncertainty principle
( )
1 MeV m h h 6.626 10 -34 J s = = = = 2 m 4 c ( t)m 4 ( t)ER 4 8.70 10 -17 s (135 MeV) 1.60 10 -13 J
2.81 10
-8
(
)
Chapter 41 Solutions
1.60 10 -19 J E 180 eV 14 = 1.00 eV = 4.34 10 Hz -34 h 6.626 10 Js
519
41.55
(a)
f=
(b)
=
c 3.00 108 m/s = = 6.91 10 -7 m = 691 nm 14 f 4.34 10 Hz
(c)
Et
h 2
so
E
h h 6.626 10 -34 J s = = = 2.64 10 -29 J = 1.65 10 -10 eV 2 t 4 t 4 2.00 10 -8 s
(
)
41.56
(a)
f=
E h
c hc = f E
(b)
=
(c)
Et
h 2
so
E
h h = 2 t 4 T
41.57
x2 =
-
x 2 dx
2
For a one-dimensional box of width
L, n =
2 n x sin L L
Thus,
x2 =
L2 L2 2 L 2 n x - 2 2 (from integral tables) x sin 2 dx = L 3 2n L 0
41.58
(a)
-
2
dx = 1 becomes
L/ 4 2 x cos 2 A2 dx - L/ 4 L
L x 1 4 x L =A + sin = A2 =1 2 L 2 2 L - L/ 4 4
2
L/ 4
or (b)
A2 =
4 L
and
A=
2 L 0 and L/8 is
The probability of finding the particle between
0
L/8
dx = A 2
2
L/8
0
cos 2
1 1 2 x dx = + = 0.409 L 4 2
Chapter 41 Solutions
520
41.59
For a particle with wave function
(x) =
2 -x/a for x > 0 and 0 for x < 0 e a
(a)
(x) = 0 , x < 0
2
and
2
2 (x) =
0
2 -2 x / a , x>0 e a
(b)
Prob( x < 0) =
0
-
( x) dx =
2
-
(0) dx = 0
2
- (x)
(c)
2
Normalization
dx =
0
-
dx + dx = 1
0
- 0 dx + 0 (2 / a)e
0
-2 x / a
dx = 0 - e -2 x / a
0
= - e - - 1 = 1
(
)
Prob(0 < x < a) =
0
a
2
dx = (2 / a) e -2 x / a dx = e -2 x / a = 1 - e -2 =
0 0
a
a
0.865
41.60
(a)
=
h = p
hc E - me c
2 2 4
=
hc
(mec2 + K ) - (mec2 )
2
2
(6.626 10-34 J s)(3.00 108 m/s) =
(576 keV)2 - (511 keV)2
(b)
1 keV = 4.68 10 -12 m -16 1.60 10 J
50.0 = 2.34 10 -10 m
41.61
(a)
xp h 2 so if x = r , p h 2r
p2 (p) = h2 h = Choosing p = , K = 2me 2me r 2me r 2
2
(b)
U=-
(c)
h2 k e2 ke e 2 - e , so E = K + U = r r 2me r 2 E, ke e 2 h2
To minimize
dE
h
Chapter 41 Solutions
2
521
41.62
(a)
m ke e 2 m k 2e 4 h2 me ke e 2 Then, E = - ke e 2 e 2 = - e e 2 = -13.6 eV 2me h2 h 2h n h nh The requirement that is still valid. = L so p = = 2 2L
E=
(pc)2 + (mc 2 )
2
2
En =
nhc + mc 2 2L
2
2
( )
2
Kn = En - mc 2 =
(b) Taking
nhc + mc 2 2L
( )
- mc 2
L = 1.00 10 -12 m , m = 9.11 10 -31 kg , and n = 1 , we find K1 = 4.69 10 -14 J
6.626 10 -34 J s h2 Nonrelativistic, E1 = = 8mL2 8 9.11 10 -31 kg 1.00 10 -12 m
(
(
)(
)
2
)
2
= 6.02 10 -14 J
Comparing this to
K1 , we see that this value is too large by 28.6% .
41.63
(a)
2 7 ke e 2 e2 1 1 1 ( -7 / 3)e U= -1 + - + -1 + + ( -1) = =- 3d 4 e0 d 2 3 2 4 e0 d
(b)
From Equation 41.9,
K = 2E1 =
2h2
8me 9d 2
( )
=
h2 36me d 2
(c)
E = U + K and
dE = 0 for a minimum: dd
7 ke e 2 h2 - =0 3d 2 18me d 3
d=
(7 )(18ke e 2 me )
3h2
6.626 10 -34 h2 = = 42me ke e 2 ( 42) 9.11 10 -31 8.99 10 9 1.602 10 -19 C
(
(
)(
) )(
2
)
2
=
0.0499 nm
(d) Since the lithium spacing is one atom, we get:
a , where Na3 = V , and the density is Nm /V , where m is the mass of
1/ 3
Vm a= Nm
1/ 3
m = density
1.66 10 -27 kg 7 = 530 kg
1/ 3
m = 2.80 10 -10 m = 0.280 nm
(5.62 times larger than
c ).
Chapter 41 Solutions
522
*41.64
(a)
= Bxe -( m /2h)x
2
2 2 d - m - m = Be -( m / 2h)x + Bx 2 xe -( m / 2h)x = Be ( 2h dx
2h ) x 2
m 2 -( m -B x e h
2h ) x 2
2 d 2 - m -( m / 2h)x 2 m m 2 - m -( m xe x xe = Bx -B 2 xe -( m / 2h)x - B h h h h dx 2
2h ) x 2
d 2 m -( m / 2h)x 2 m 3 -( m / 2h)x 2 = -3B +B xe x e 2 h h dx
Substituting into the Schrdinger Equation (41.19), we have
2 2 2mE m -( m / 2h)x 2 m 3 -( m / 2h)x 2 m 2 -3B xe +B x e = - 2 Bxe -( m / 2h)x + x Bxe -( m / 2h)x h h h h
2
This is true if (b)
-3 = -
3 2E ; it is true if E = h 2 h
x = 0 because = 0 there.
We never find the particle at
(c)
is maximized if
h d m , which is true at x = = 0 = 1 - x2 h m dx
2
(d)
We require
-
dx = 1:
2
1=
-
B2 x 2 e -( m /h)x dx = 2B2 x 2 e -( m /h)x dx = 2B2
2
0
1 4
B2 1/ 2h3 / 2 = (m /h)3 2 (m )3/ 2
Then
21/ 2 m B = 1/ 4 h
3/ 4
4m3 3 = 3 h
1/ 4
(e)
x = 2 h/ m , the potential energy is 2 m 2 x 2 = 2 m 2 ( 4h/ m ) = 2h . This is larger than the total energy 3h / 2 , so there is zero classical probability of finding the particle here.
At
1 1
(f)
Probability Probability
2 2 2 = dx = Bxe -( m / 2h)x = B2 x 2 e -( m /h)x
2
=
2 m 1/ 2 h
3/ 2
4h -( m /h)4(h / m ) 8 m = e h m
1/ 2
e -4
Chapter 41 Solutions
L x 2 x x 2 x A 2 sin 2 sin dx = 1 + 16 sin 2 + 8 sin 0 L L L L
523
41.65
(a)
0
L
dx = 1:
2
L L x 2 x L A2 sin dx = 1 + 16 + 8 sin 0 2 L L 2
x=L 17 L 16L L 17 L x x x A2 + 16 sin 2 cos sin 3 dx = A 2 + =1 0 L L L x=0 3 2 2
A2 =
(b)
2 , so the normalization constant is A = 2 /17 L 17 L
2
- a
a
dx = 1:
-a
A
a
2
cos 2
2 x x x x sin + B sin 2 + 2 A B cos dx = 1 2a a 2a a
The first two terms are
A a and B a . The third term is:
2
2
2A B
-a
a
x x x cos 2 sin 2 a cos 2 a dx = 4 A B 2a
2 2
8a A B x x 3 x cos 2a sin 2a dx = 3 cos 2a -a
a
2
a
=0
-a
so that
a A +B
(
) = 1, giving
A + B = 1/ a .
2
2
*41.66
With one slit open
P1 = 1
2
or
P2 = 2
2
With both slits open, At a maximum, the wave functions are in phase At a minimum, the wave functions are out of phase
P = 1 + 2
Pmax = ( 1 + 2 ) Pmin = ( 1 - 2 )
2
2
2
1 P Now 1 = = 25.0 , so 2 P
2
1 = 5.00
Chapter 41 Solutions
*41.67 (a) (b) (c) (d) The light is unpolarized. It contains both horizontal and vertical field oscillations. The interference pattern appears, but with diminished overall intensity. The results are the same in each case. The interference pattern appears and disappears as the polarizer turns, with alternately increasing and decreasing contrast between the bright and dark fringes. The intensity on the screen is precisely zero at the center of a dark fringe four times in each revolution, when the filter axis has turned by 45, 135, 225, and 315 from the vertical. Looking at the overall light energy arriving at the screen, we see a low-contrast interference pattern. After we sort out the individual photon runs into those for trial 1, those for trial 2, and those for trial 3, we have the original results replicated: The runs for trials 1 and 2 form the two blue graphs in Figure 41.3, and the runs for trial 3 build up the red graph.
524
(e)
Chapter 42 Solutions
42.1
(a)
The point of closest approach is found when E = K +U = 0 + ke q qAu r
9
or
rmin =
ke ( 2e )(79e ) E
rmin (b)
(8.99 10 =
( 4.00 MeV)(1.60 10 -13
N m 2 C 2 158 1.60 10 -19 C J MeV
) (
)
)
2
= 5.68 10 -14 m
The maximum force exerted on the alpha particle is ke q qAu rmin
2
Fmax =
(8.99 10 =
9
N m 2 C 2 158 1.60 10 -19 C
(
) (
5.68 10 -14 m
)
)
2
2
= 11.3 N away from the nucleus
42.2
(a)
The point of closest approach is found when E = K +U = 0 + k ( 2e )( Ze ) ke q qT 2Zke e 2 or rmin = e = r E E
(b)
The maximum force exerted on the alpha particle is Fmax = ke q qT rmin
2
E E2 = 2Zke e 2 = 2 2Zke e 2 2Zke e
2
away from the target nucleus
42.3
(a)
The photon has energy 2.28 eV. And (13.6 eV ) / 2 2 = 3.40 eV is required to ionize a hydrogen atom from state n = 2. So while the photon cannot ionize a hydrogen atom pre-excited to n = 2, it can ionize a hydrogen atom in the n = 3 state, with energy 13.6 eV = 1.51 eV 32
(b)
1 The electron thus freed can have kinetic energy Ke = 2.28 eV 1.51 eV = 0.769 eV = 2 me v 2
v=
2(0.769)(1.60 10 -19 )J = 520 km/s (9.11 10 -31 )kg
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2
Chapter 42 Solutions
*42.4
(a)
Longest wavelength implies lowest frequency and smallest energy: the electron falls from n = 3 to n = 2, losing energy 13.6 eV 13.6 eV + = 1.89 eV 2 22 3
The photon frequency is f = E h and its wavelength is
=
c ch (3.00 108 m / s)(6.626 10 -34 J s) eV = = -19 f E 1.89 eV 1.60 10 J Balmer Series
= 656 nm
(b)
The biggest energy loss is for an electron to fall from ionization, n = , to the n = 2 state. It loses energy 13.6 eV 13.6 eV + = 3.40 eV 22
to emit light of wavelength
3.00 108 m / s 6.626 10 -34 J s hc = = = 365 nm E 3.40 eV 1.60 10 -19 J / eV
(
(
)(
)
)
42.5
(a) (b)
me For positronium, = 2 , so 32 = (656 nm)2 = 1312 nm = 1.31 m For He+, m e, q1 = e, and q2 = 2e, so 32 = (656/4) nm = 164 nm
(infrared region) . (ultraviolet region) .
Goal Solution A general expression for the energy levels of one-electron atoms and ions is k 2 q 2 q 2 En = - e 21 2 2 2h n where ke is the Coulomb constant, q1 and q2 are the charges of the two particles, and is the reduced mass, given by = m1m2 / ( m1 + m2 ) . In Problem 4 we found that the wavelength for the n = 3 to n = 2 transition of the hydrogen atom is 656.3 nm (visible red light). What are the wavelengths for this same transition in (a) positronium, which consists of an electron and a positron, and (b) singly ionized helium? (Note: A positron is a positively charged electron.) G: The reduced mass of positronium is less than hydrogen, so the photon energy will be less for positronium than for hydrogen. This means that the wavelength of the emitted photon will be longer than 656.3 nm. On the other hand, helium has about the same reduced mass but more charge than hydrogen, so its transition energy will be larger, corresponding to a wavelength shorter than 656.3 nm. All the factors in the above equation are constant for this problem except for the reduced mass and the nuclear charge. Therefore, the wavelength corresponding to the energy difference for the transition can be found simply from the ratio of mass and charge variables.
O:
Chapter 42 Solutions
3
A:
For hydrogen, The photon energy is Its wavelength is = 656.3 nm , where (a) For positronium,
=
mp me mp + me
me
E = E3 - E2
= =
c hc = f E me me m = e me + me 2
so the energy of each level is one half as large as in hydrogen, which we could call "protonium." The photon energy is inversely proportional to its wavelength, so for positronium,
32 = 2(656.3 nm) = 1313 nm
(b) For He+,
(in the infrared region)
me , q1 = e , and q2 = 2e ,
so the transition energy is 2 2 = 4 times larger than hydrogen. Then, 656 nm = 164 nm 32 = 4 (in the ultraviolet region)
L : As expected, the wavelengths for positronium and helium are respectively larger and smaller than for hydrogen. Other energy transitions should have wavelength shifts consistent with this pattern. It is important to remember that the reduced mass is not the total mass, but is generally close i n magnitude to the smaller mass of the system (hence the name reduced mass).
*42.6
(a)
For a particular transition from n i to n f ,
EH = -
2 1 hc H ke e 4 1 - 2 = 2 2 2h n f ni H 2 1 hc D ke e 4 1 - 2 = 2 2 2h n f ni D
and me mp me + mp or
ED = -
where
H =
and
D =
me mD me + mD
By division,
EH H D = = ED D H H - D = 1 - H H D
D = H H D
Then,
(b)
H me mp me + mD (1.007 276 u )(0.000 549 u + 2.013 553 u ) = = 0.999728 = D me + mp me mD (0.000 549 u + 1.007 276 u )( 2.013 553 u ) H - D = (1 - 0.999 728)(656.3 nm ) = 0.179 nm
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4
Chapter 42 Solutions
42.7
(a)
In the 3d subshell, n = 3 and l = 2, we have n l ml ms 3 2 +2 +1/2 3 2 +2 -1/2 3 2 +1 +1/2 3 2 +1 -1/2 3 2 0 +1/2 3 2 0 -1/2 3 2 -1 +1/2 3 2 -1 -1/2 3 2 -2 +1/2 3 2 -2 -1/2
(A total of 10 states) (b) In the 3p subshell, n = 3 and l = 1, we have n l ml ms 3 1 +1 +1/2 3 1 +1 -1/2 3 1 +0 +1/2 3 1 +0 -1/2 3 1 -1 +1/2 3 1 -1 -1/2
(A total of 6 states)
42.8
1s (r) =
1
3 a0
e -r/a0
(Eq. 42.3)
P1s (r) =
4r 2 -2r/a0 e 3 a0
(Eq. 42.7)
42.9
(a)
2
dV = 4
0
1 2 r 2 dr = 4 3 r 2 e - 2r a0 0
a0
dr
r=
Using integral tables,
dV = -
2
2 2 a0
- 2r e
a0 2
2 2 a2 a0 = - 2 - 0 = 1 r + a0 r + 2 r = 0 a0 2
so the wave function as given is normalized. (b) Pa0 = 4
3a0 2 a0 2
2 3a0 2
1 3a0 2 2 - 2r 2 r 2 dr = 4 3 r e a0 a0 2
a0
dr
Again, using integral tables, =- 2 2 a0 - 2r e
a0 2 2 a0 r + a0 r + 2 a 3a0 2
Pa0
2 3a0 2
= -
0
2
2 2 a0
2 2 -3 17 a0 -1 5 a0 e - e 4 = 0.497 4
Chapter 42 Solutions
5
42.10
=
1 1 r -r/2a0 e 3/2 a 3 (2a0 ) 0
so
Pr = 4 r 2 2 = 4 r 2
r 2 -r/a0 e 5 24a0
Set
1 4 3 -r/a0 dP = 4r e + r 4 - e -r/a0 = 0 5 dr 24a0 a0
Solving for r, this is a maximum at r = 4 a 0
42.11
=
1
3 a0
e -r/a0
-2 2 2 d = e -r/a0 = 5 ra0 r dr r a 0
d 2 = dr 2
1
7 a0
e -r/a0 =
1 2 a0
-
h2 1 2 e2 2 - ra - 4 e r = E 2me a0 0 0 h 2 4 e0 , me e 2 - e2 = E or 8 e0 a0 E= - ke e 2 2a0
But a0 =
so
This is true, so the Schrdinger equation is satisfied.
42.12
The hydrogen ground-state radial probability density is P(r) = 4 r 2 1s
2
=
2r 4r 2 exp - 3 a0 a0
The number of observations at 2a0 is, by proportion N = 1000 P(2a0 ) (2a0 )2 e -4a0 /a0 = 1000 = 1000(16)e -3 = 797 times P(a0 / 2) (a0 / 2)2 e -a0 /a0
42.13
(a) (b)
For the d state, l = 2, For the f state, l = 3,
L=
6 h = 2.58 10 34 J s 12 h = 3.65 10 34 J s
L = l(l + 1) h =
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6
Chapter 42 Solutions 6.626 10 -34 2 l= 4
*42.14
L = l(l + 1) h
so
-34 2 2
4.714 10 -34 = l(l + 1)
(4.714 10 ) (2 ) l(l + 1) = (6.626 10 )
-34 2
= 1.998 10 1 20 = 4(4 + 1)
so
42.15
The 5th excited state has n = 6, energy
13.6 eV = 0.378 eV 36 (6.626 10 34 J s)(3.00 108 m/s) hc = = 1.14 eV (1090 10 9 m)(1.60 10 19 J/eV) 0.378 eV 1.14 eV = 1.52 eV 13.6 eV = 1.51 eV 33 l(l + 1) h = 6h
The atom loses this much energy: to end up with energy which is the energy in state 3:
While n = 3, l can be as large as 2, giving angular momentum
42.16
For a 3d state, n = 3 and l = 2. Therefore, L = l(l + 1) h = 2(2 + 1) h = ml can have the values 2, 1, 0, 1, and 2, so
6 h = 2.58 10 34 J s
L z can have the values2 h, h, 0, and 2h
Using the relation cos = Lz / L , we find that the possible values of are equal to 145, 114, 90.0, 65.9, and 35.3 .
42.17
(a)
n = 1: For n = 1, l = 0, ml = 0, ms = 2n
2 = 1(1) 2
1 , 2 sets 2
n 1 1 ms 1 2 1 2 1 2 1 2
l 0 0
ml 0 0
ms 1/2 +1/2
= 2 n 2 2 2 2 l 0 1 1 1 ml 0 1 0 1
(b)
For n = 2, we have
yields 8 sets; 2n2 = 2(2) = 8
2
Chapter 42 Solutions
7
Note that the number is twice the number of ml values. Also, for each l there are (2l + 1) different ml values. Finally, l can take on values ranging from 0 to n - 1. So the general expression is s = 2( 2l + 1)
0 n-1
The series is an arithmetic progression: 2 + 6 + 10 + 14, the sum of which is s= (c) (d) (e) n [2a + (n - 1)d] where a = 2, d = 4: 2 s= n [ 4 + (n - 1)4] = 2n2 2
n = 3: 2(1) + 2(3) + 2(5) = 2 + 6 + 10 = 18 n = 4: 2(1) + 2(3) + 2(5) + 2(7) = 32 n = 5: 32 + 2(9) = 32 + 18 = 50
2n2 = 2(3)2 = 18 2n2 = 2(4)2 = 32 2n2 = 2(5)2 = 50
42.18
B =
eh 2me
e = 1.60 10 19 C
h= 1.055 10 34 J s
me = 9.11 10 31 kg
B = 9.27 10 24 J/T = 5.79 10 5 eV/T
42.19
(a)
Density of a proton:
=
1.67 10 -27 kg m = = 3.99 1017 kg/m3 V (4 / 3) (1.00 10 -15 m)3
1/3
(b)
Size of model electron:
3m r= 4
3 9.11 10 - 31 kg m 3 = 17 4 3.99 10 kg
1/3
= 8.17 10 17 m
(c)
Moment of inertia: I =
2 2 2 mr = (9.11 10 -31 kg)(8.17 10 -17 m)2 = 2.43 10 -63 kg m 2 5 5 Lz = I = h Iv = 2 r
Therefore,
v=
6.626 10 -34 J s (8.17 10 -17 m) hr = = 1.77 1012 m/s 2I 2 ( 2) 2.43 10 -63 kg m 2
(
(
)
)
(d) This is 5.91 103 times larger than the speed of light.
2000 by Harcourt, Inc. All rights reserved.
8
Chapter 42 Solutions 2 r r = l(l + 1) h = (l2 + l) h lh T 2 (1.496 1011 m)2 = lh 3.156 107 s so 2.66 1040 = l = 2.52 1074 1.055 10 34 J s
42.20
(a)
L = mvr = m
(5.98 1024 kg)
(b)
E = -U + K = - K =
1 1 mr 2 1 L2 1 l(l + 1)h2 1 l2h2 mv 2 = mv 2 = = 2 2 mr 2 2 mr 2 2 mr 2 2 mr 2
1 (5.98 10 24 2
E dE 1 2lh 2 l = = 2 so l dl 2 mr 2 l E = 5.30 10 33 J = 2.10 10 41 J 2.52 1074
E dE = 2 dl = 2 l
2 1.496 1011 m kg) 1 3.156 107 s 2.52 1074
2
*42.21
n =
eh 2mp
e = 1.60 10 19 C
h = 1.055 10 34 J s
mp = 1.67 10 27 kg
(a) (b)
n = 5.05 10 27 J/T 1 n me B = 1836 = m p
= 31.6 neV/T
Apparently it is harder to "spin up" a nucleus than a electron, because of its greater mass.
42.22
In the N shell, n = 4. For n = 4, l can take on values of 0, 1, 2, and 3. For each value of l, ml can be -l to l in integral steps. Thus, the maximum value for ml is 3. Since Lz = mlh, the maximum value for Lz is Lz = 3h .
42.23
The 3d subshell has l = 2, and n = 3. Also, we have s = 1. Therefore, we can have n = 3; l = 2; ml = 2, 1, 0, 1, 2; s = 1; and ms = 1, 0, 1 , leading to the following table: n l ml s ms 3 2 2 1 1 3 2 2 1 0 3 2 2 1 1 3 2 1 1 1 3 2 1 1 0 3 2 1 1 1 3 2 0 1 1 3 2 0 1 0 3 2 0 1 1 3 2 1 1 1 3 2 1 1 0 3 2 1 1 1 3 2 2 1 1 3 2 2 1 0 3 2 2 1 1
Chapter 42 Solutions
9
42.24
(a) (b)
1s 2 2s 2 2p 4 For the 1s electrons, n = 1, l = 0, ml = 0, ms = + 1/2 and 1/2 For the two 2s electrons, n = 2, l = 0, ml = 0, ms = + 1/2 and 1/2 For the four 2p electrons, n = 2; l = 1; ml = 1, 0, or 1; and ms = + 1/2 or 1/2
42.25
The 4s subshell fills first , for potassium and calcium, before the 3d subshell starts to fill for scandium through zinc. Thus, we would first suppose that [ Ar ]3d 4 4s 2 would have lower energy than [ Ar ]3d 5 4s 1 . But the latter has more unpaired spins, six instead of four, and Hund's rule suggests that this could give the latter configuration lower energy. In fact it must, for [ Ar ]3d 5 4s 1 is the ground state for chromium.
*42.26 (a)
For electron one and also for electron two, n = 3 and l = 1. The possible states are listed here in columns giving the other quantum numbers: electron one electron two electron one electron two ml ms 1
1 2
1
1 2
1
1 2
1
1 2 1 2
1
1 2 -1 2
1 1
1 2
1 0
1 2
1 0
-1 2
1
1
0
1 2
0
1 2
0
1 2
0
1 2 1 2
0
1 2 -1 2
-1 -1 -1 -1 -1 2 2 2 2 2
ml 1 ms - 1 2
0
1 2
0
-1 2
1 1
1 1
1 2 -1 2
1
1 2
1
0
1 1
-1 -1 2 2
ml 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 ms - 1 - 1 - 1 - 1 - 1 1 1 1 1 1 - 1 - 1 - 1 - 1 - 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ml 1 1 0 1 1 1 1 0 0 1 1 1 0 0 1 ms 1 - 1 1 1 - 1 1 - 1 1 - 1 - 1 1 - 1 1 - 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
There are thirty allowed states, since electron one can have any of three possible values for ml for both spin up and spin down, amounting to six states, and the second electron can have any of the other five states. (b) Were it not for the exclusion principle, there would be 36 electron independently. possible states, six for each
42.27
Shell n l ml ms count
K 1 0 0
L 2 0 0
1 1
0
1
M 3 0 0
1 1
0
1
2 2
1
0
1
2
N 4 0 0
12 34 He Be
10 Ne
12 Mg
18 21 Ar
30 Zn
20 Ca
(a) (b) 42.28
zinc or copper 1s2 2s2 2p 6 3s2 3p 6 4s2 3d10 or 1s2 2s2 2p 6 3s2 3p 6 4s1 3d10
Listing subshells in the order of filling, we have for element 110,
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10
Chapter 42 Solutions
1s2 2s2 2p 6 3s2 3p 6 4s2 3d10 4p 6 5s2 4d10 5p 6 6s2 4 f 14 5d10 6p 6 7s2 5 f 14 6d 8 In order of increasing principal quantum number, this is 1s2 2s2 2p 6 3s2 3p 6 3d10 4s2 4p 6 4d10 4 f 14 5s2 5p 6 5d10 5 f 14 6s2 6p 6 6d 8 7s2
42.29
(a)
n+l 1 subshell 1s
2 2s
3 4 5 2p , 3s 3p , 4s 3d , 4p , 5s
6 4d , 5p , 6s
7 4f , 5d , 6p , 7s
(b)
Z = 15:
Filled subshells: Valence subshell: Prediction: Element is phosphorus
1s, 2s, 2p, 3s (12 electrons) 3 electrons in 3p subshell Valance = + 3 or 5 Valence + 3 or 5 (Prediction correct) 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s (38 electrons) 9 electrons in 4d subshell Valence = 1 (Prediction fails) Valence is + 1 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p (86 electrons) Outer subshell is full: inert gas (Prediction correct)
Z = 47:
Filled subshells: Outer subshell: Prediction: Element is silver,
Z = 86:
Filled subshells: Prediction Element is radon, inert
42.30
Electronic configuration: Sodium to Argon [1s2 2s2 2p 6 ] + 3s1 + 3s2 + 3s2 3p1 + 3s2 3p 2 +3s2 3p 3 +3s2 3p 4 +3s2 3p 5 + 3s2 3p 6 [1s2 2s2 2p 6 3s2 3p 6 ]4s1 Na 11 Mg12 Al 13 Si 1 4 P15 S1 6 Cl17 Ar 18 K 19
Chapter 42 Solutions
11
*42.31
n = 3, l = 0, ml = 0
300 corresponds to E 300 =
n = 3, l = 1, ml = 1, 0, 1
Z2 E0 22(13.6) = = 6.05 eV n2 (3)2
31 1, 310, 311 have the same energy since n is the same.
For n = 3, l = 2, ml = 2, 1, 0, 1, 2
32 2, 32 1, 320, 321, 322 have the same energy since n is the same.
All states are degenerate.
42.32
E=
hc
= e(V)
(6.626 10 34 J s)(3.00 108 m/s) 10.0 10 9 m
= (1.60 10 19 )(V)
V = 124 V
*42.33
Ephoton max =
hc
min
= e( V ) = 40.0 keV
min
6.626 10 -34 J s 3.00 108 m s 1.00 eV hc = = = 0.031 0 nm Emax 1.60 10 -19 J 40.0 10 3 eV
(
)(
)
42.34
Some electrons can give all their kinetic energy Ke = e( V ) to the creation of a single photon of x-radiation, with hf = hc = e( V )
6.6261 10 -34 J s 2.9979 108 m s hc 1240 nm V = = = -19 e( V ) V 1.6022 10 C ( V )
(
(
)(
)
)
42.35
Following Example 42.7,
3 E = 4 (42 1) 2(13.6 eV) = 1.71 104 eV = 2.74 10 15 J f = 4.14 1018 Hz and
= 0.072 5 nm
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12
Chapter 42 Solutions
42.36
The K x-rays are emitted when there is a vacancy in the (n = 1) K shell and an electron from the (n = 3) M shell falls down to fill it. Then this electron is shielded by nine electrons originally and by one in its final state. 13.6(Z 9) hc = 32
2
eV +
13.6(Z 1)2 eV 12
Z 2 18Z 81 (6.626 10 -34 J s)(3.00 108 m / s) = 13.6 eV - + - + Z 2 - 2Z + 1 -9 -19 9 9 9 (0.152 10 m)(1.60 10 J / eV) 8Z2 8.17 103 eV = 13.6 eV 9 8 8Z2 8 9
so
601 =
and
Z = 26 Iron
42.37
(a)
Suppose the electron in the M shell is shielded from the nucleus by two K plus seven L electrons. Then its energy is - 13.6 eV(83 - 9)2 = - 8.27 keV 32
Suppose, after it has fallen into the vacancy in the L shell, it is shielded by just two K-shell electrons. Then its energy is - 13.6 eV(83 - 2)2 = - 22.3 keV 22 Thus the electron's energy loss is the photon energy: E = hc (22.3 - 8.27) keV = 14.0 keV
(b)
so
=
6.626 10 - 34 J s (3.00 108 m / s) = 8.85 10 - 11 m 3 -19 J 14.0 10 1.60 10
*42.38
E= for
1240 eV nm 1.240 keV nm hc = =
1 = 0.0185 nm, E = 67.11 keV 2 = 0.0209 nm, E = 59.4 keV 3 = 0.0215 nm, E = 57.7 keV
The ionization energy for K shell = 69.5 keV, so, the ionization energies for the other shells are: L shell = 11.8 keV : M shell = 10.1 keV : N shell = 2.39 keV
Chapter 42 Solutions
13
*42.39 (a)
The outermost electron in sodium has a 3s state for its ground state. The longest wavelength means minimum photon energy and smallest step on the energy level diagram. Since n = 3, n must be 4. With l = 0, l must be 1 , since l must change by 1 in a photon absorption process. 1 1 1 1 = 1.097 107 - -9 2 2 m ( 3 - 1.35) 330 10 m (4 - 1 ) 0.276 = 1
(b)
(1.65)
2
-
(4 - 1 )
1
2
= 0.367 -
( 4 - 1 )2
1
so
( 4 - 1 )2 = 10.98
and
1 = 0.686
42.40
=
c hc (6.626 10 -34 J s)(3.00 108 m / s) = = = 590 nm f hf (2.10 eV)(1.60 10 19 J / eV)
*42.41
We require A = u f B =
16 2h B 3
or
uf =
2 -34 Js 16 2h 16 1.055 10 Js = = 6.21 10 -14 3 3 m3 645 10 -9 m
(
(
)
)
42.42
f=
E = 2.82 1013 s - 1 h c = 10.6 m , f infrared
=
42.43
E = P t = (1.00 106 W)(1.00 10 8 s) = 0.0100 J E = h f = hc (6.626 10 34)(3.00 108) = J = 2.86 10 19 J 694.3 10 9
E 0.0100 N= E = = 3.49 1016 photons 2.86 10 1 9
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14
Chapter 42 Solutions
Goal Solution A ruby laser delivers a 10.0-ns pulse of 1.00 MW average power. If the photons have a wavelength of 694.3 nm, how many are contained in the pulse? G: Lasers generally produce concentrated beams that are bright (except for IR or UV lasers that produce invisible beams). Since our eyes can detect light levels as low as a few photons, there are probably at least 1 000 photons in each pulse. O: From the pulse width and average power, we can find the energy delivered by each pulse. The number of photons can then be found by dividing the pulse energy by the energy of each photon, which is determined from the photon wavelength. A: The energy in each pulse is E = P t = (1.00 106 W)(1.00 10 -8 s) = 1.00 10 -2 J hc (6.626 10 -34 )(3.00 108 ) = J = 2.86 10 -19 J 694.3 10 -9 E 1.00 10 -2 J = = 3.49 1016 photons E 2.86 10 -19 J / photon
The energy of each photon is E = hf = N=
So
L : With 1016 photons/pulse, this laser beam should produce a bright red spot when the light reflects from a surface, even though the time between pulses is generally much longer than the width of each pulse. For comparison, this laser produces more photons in a single ten-nanosecond pulse than a typical 5 mW helium-neon laser produces over a full second (about 1.6 1016 photons/second).
*42.44
In
n -n L G = e ( u l)
we require
1.05 = e
(1.00 10 -18 m 2 )(nu - nl )(0.500 m )
ln(1.05) = 9.76 1016 m - 3 5.00 10 -19 m 3
Thus, ln(1.05) = 5.00 10 -19 m 3 (nu - nl )
(
)
so
nu - nl =
42.45
(a)
- E3 ( k B 300 K ) N3 N g e -(E - E ) k 300 K ) - hc = =e 3 2 ( B =e N 2 N g e - E2 ( kB 300 K )
( k B 300 K )
where is the wavelength of light radiated in the 3 2 transition:
- ( 6.626 10 - 34 N3 =e N2 J s 3.00 10 8 m s
)(
) (632.8 10 - 9 m )(1.38 10 - 23 J K )( 300 K ) = e -75.8 =
1.22 10 - 33
(b)
- ( 6.626 10 - 34 N3 = e hc/ kBT = e N2
J s 3.00 10 8 m s
)(
) (694.3 10 - 9 m )(1.38 10 - 23 J K )( 4.00 K ) = e -5187
To avoid overflowing your calculator, note that 10 = e ln 10 . Take N3 = e ln 10 (- 5187/ln 10) = 10 - 2253 N2
Chapter 42 Solutions
15
*42.46
N u Nl = e ( u l ) B where the subscript u refers to an upper energy state and the subscript l to a lower energy state.
- E -E k T
(a)
Since Eu - El = Ephoton = hc , Thus, we require 1.02 = e - hc
kBT
N u Nl = e - hc or
kBT
ln(1.02) = - T= -
(694.3 10
(6.626 10
-34
J s 3.00 108 m s
-9
m 1.38 10
)(
)(
- 23
J K T
)
)
2.07 10 4 K = - 1.05 106 K ln(1.02)
A negative-temperature state is not achieved by cooling the system below 0 K, but by heating it above T = , for as T the populations of upper and lower states approach equality. (b) Because Eu - El > 0, and in any real equilibrium state T > 0, e - ( Eu -El )
kBT
<1
and N u < Nl.
Thus, a population inversion cannot happen in thermal equilibrium.
42.47
(a)
I=
(3.00 10 -3 J) = 4.24 1015 W/m 2 (1.00 10 -9 s) (15.0 10 -6 m)2 (0.600 10 -9 m)2 = 1.20 10 12 J = 7.50 MeV (30.0 10 -6 m)2
(b)
(3.00 10 -3 J)
*42.48 (a)
The energy difference between these two states is equal to the energy that is absorbed. Thus, E = E2 E1 = 3 We have E = 2 kBT, ( 13.6 eV) ( 13.6 eV) = 10.2 eV = 1.63 10 18 J 4 1 2 2(1.63 10 18 J) or T = 3k E = = 7.88 104 K B 3(1.38 10 23 J/K)
(b)
42.49
4r 3 rav = rP(r)dr = 3 (e -2r/a0 )dr 0 0 a 0
Make a change of variables with
3 -x
2r = x and a0
dr =
a0 dx. 2
0
Then rav =
a0 4
0
x e
dx =
a0 - x 3 e -x + 3 - x 2 e -x + 2e -x (- x - 1) 4
[
(
)]
=
3 a0 2
2000 by Harcourt, Inc. All rights reserved.
16
Chapter 42 Solutions
*42.50
4r 2 - 2r 1 = e 0 a3 r 0
a0
4 1 dr = 3 r a0
0
4 1 1 - 2a r r e ( 0 ) dr = 3 = 2 a0 a0 ( 2 a0 )
1 2 1 = = , and find that the average reciprocal value is NOT the 3a0 2 3a0 r reciprocal of the average value. We compare this to
42.51 (a)
The wave equation for the 2s state is given by Eq. 42.7: Taking r = a0 = 0.529 10 -10 m , we find
1 1 2s (r ) = 4 2 a0
32
r -r 2 - e a0
2a0
2s ( a0 ) =
(b) (c)
2
1 1 4 2 0.529 10 -10 m
32
[2 - 1] e - 1 2 =
1.57 1014 m - 3 2
2s ( a0 ) = 1.57 1014 m -3 2
(
)
2
= 2 .47 10 28 m - 3
2 P2s ( a0 ) = 4 a0 2s ( a0 ) = 8.69 108 m -1 2
Using Equation 42.5 and the results to (b) gives
*42.52
We define the reduced mass to be , and the ground state energy to be E1: 207 9.11 10 -31 kg 1.67 10 -27 kg 207me mp m1m2 = = = = 1.69 10 -28 kg m1 + m2 207me + mp 207 9.11 10 -31 kg + 1.67 10 -27 kg
(
(
) (
)(
)
)
2 1.69 10 -28 kg 8.99 10 9 N m 2 C 2 1.60 10 -19 C k 2 q 2 q2 E1 = - e2 1 2 = - 2 2h (1) 2 1.055 10 -34 J s
(
)(
(
)
)(
2
)
3
e
= - 2.52 10 3 eV
To ionize the muonium "atom" one must supply energy + 2.52 keV .
42.53
(a) (b)
(3.00 108 m/s)(14.0 10 12 s) = 4.20 mm E= hc = 2.86 10 19 J N= 3.00 J = 1.05 1019 photons 2.86 10 19 J 1.05 1019 = 8.82 1016 mm 3 119
(c)
V = (4.20 mm) (3.00 mm)2 = 119 mm3
n=
Chapter 42 Solutions
17
42.54
(a) (b)
The length of the pulse is L = ct The energy of each photon is d2 V = L 4 E = hc
so
N=
E E = E hc
(c)
n=
4 E N = 2 V ct d hc
42.55
We use
1 r 3 2s (r) = (2 a0 )-1/2 2 - e -r/2a0 4 a0 P(r) = 4 r 2 2 = r -r/a0 1 r2 3 2 - a e 8 a0 0
2
By Equation 42.5,
(a)
2 2 r r2 r 2r 2 1 r 1 dP(r) 1 2r = 3 2 - - 3 2 - - 3 2 - e -r/a0 = 0 a0 a0 8 a0 a0 a0 a0 dr a0 a0
or
r r 2r r r -r/a0 1 r =0 3 2 - a 2 2 - a - a - a 2 - a e 8 a0 0 0 0 0 0 dP = 0 at r = 0, r = 2a0 , and r = to be minima with P(r ) = 0. dr
2
Therefore we require the roots of
[ . . . . . ] = 4 - (6r / a0 ) + (r / a0 ) = 0
with solutions r = 3 5 a0 .
(
)
We substitute the last two roots into P(r) to determine the most probable value: When r = 3 - 5 a0 = 0.7639 a0 , then When then
( ) r = ( 3 + 5 )a0 = 5.236a0 ,
2
P(r ) = 0.0519 / a0 P(r ) = 0.191/ a0
Therefore, the most probable value of r is
(3 + 5 )a0 =
Let
5.236a0
(b)
0 0
P(r)dr = P(r)dr =
0
r -r/a0 1 r2 dr 3 2 - a e 8 a0 0
r u= a , 0
dr = a 0 du,
0
1 4 1 1 2 (u - 4u3 + 4u2 )e -u du = - (u4 + 4u2 + 8u + 8)e -u = 1 u (4 - 4u + u2 )e -u dr = 0 8 8 8 0
This is as desired .
2 at 2 1 Fz 2 z ( dBz dz) x and = t = v 2 m0 2 2m0
*42.56
z =
z =
eh 2me
dBz 2m0 (z)v 2 2me 2(108)(1.66 10 - 27 kg )(10 4 m 2 / s 2 )2(9.11 10 - 31 kg )(10 - 3 m) = = = 0.389 T/m dz x 2 eh (1.00 m 2 )(1.60 10 -19 C)(1.05 10 - 34 J s)
2000 by Harcourt, Inc. All rights reserved.
18
Chapter 42 Solutions
42.57
With one vacancy in the K shell, excess energy
1 1 E -(Z - 1)2 (13.6 eV) 2 - 2 = 5.40 keV 2 1
We suppose the outermost 4s electron is shielded by 20 electrons inside its orbit: Eionization Note the experimental ionization energy is 6.76 eV. 2 2 (13.6 eV) = 3.40 eV 42
K = E - Eionization 5.39 keV
*42.58
E=
1240 eV nm hc = = E so E 1 = 4.00 eV E 2 = 3.10 eV E 3 = 0.900 eV
1 = 310 nm, 2 = 400 nm, 3 = 1378 nm,
and the ionization energy = 4.10 eV The energy level diagram having the fewest levels and consistent with these energies is shown at the right.
42.59
(a)
One molecule's share of volume Al: V=
3
1.00 10 -6 m 3 mass per molecule 27.0 g mol - 29 = m3 = 1.66 10 23 2.70 g density 6.02 10 molecules mole
V = 2.55 10 -10 m ~ 10 -1 nm
-6 3 238 g 1.00 10 m -29 m3 = 2.09 10 6.02 10 23 molecules 18.9 g
U:
V=
3
V = 2.76 10 -10 m ~ 10 -1 nm
(b)
The outermost electron in any atom sees the nuclear charge screened by all the electrons below it. If we can visualize a single outermost electron, it moves in the electric field of net charge, +Ze - (Z - 1)e = +e , the charge of a single proton, as felt by the electron in hydrogen. So the Bohr radius sets the scale for the outside diameter of every atom. An innermost electron, on the other hand, sees the nuclear charge unscreened, and the scale size of its (K-shell) orbit is a0 Z.
Chapter 42 Solutions
19
42.60
(a)
No orbital magnetic moment to consider: moments, for antiparallel spins
higher energy for
N S
N S
parallel magnetic
of the electron and proton.
(b)
E=
hc = 9.42 10 -25 J = 5.89 eV h 2 so E 1.00 eV = 1.04 10 -30 eV -19 J 2 10 yr 3.16 10 s yr 1.60 10
(c)
E t
(
1.055 10 -34 J s
7
)(
7
)
42.61
P=
2.50 a0
4r 2 -2r e a0 3
a0
1 dr = 2
5.00
5.00
z 2 e -z dz where z
2r a0
1 P = - (z 2 + 2z + 2)e -z 2
1 1 37 = - [0] + ( 25.0 + 10.0 + 2.00)e -5 = (0.00674) = 0.125 2 2 2
Goal Solution For hydrogen in the 1s state, what is the probability of finding the electron farther than 2.50 ao from the nucleus? G: From the graph shown in Figure 42.8, it appears that the probability of finding the electron beyond 2.5 a0 is about 20%. The precise probability can be found by integrating the 1s radial probability distribution function from r = 2.50 ao to . The general radial probability distribution function is With P( r ) = 4 r 2
2
O:
A:
1s = a0 3
(
)
-1/2 -r/a 0
e
it is
P(r ) = 4r 2 a0 -3 e -2r/a0 P=
The required probability is then
2.50a0
P(r )dr =
2.50a0
4r 2 -2r/a0 e dr a0 3
Let z = 2r a0 and dz = 2dr a0 : Performing this integration by parts, P=- L:
1 2 1 (0) + 2 (25.0 + 10.0 + 2.00)e -5.00 = 37 2
P=
1 2
5.00
z
2 -z
e dz + 2z + 2 e -z
P=-
1 2 z 2
]
5.00
( )(0.00674) = 0.125
The probability of 12.5% is less than the 20% we estimated, but close enough to be a reasonable result. In comparing the 1s probability density function with the others in Figure 42.8, it appears that the ground state is the most narrow, indicating that a 1s electron will probably be found in the narrow range of 0 to 4 Bohr radii, and most likely at r = a0.
2000 by Harcourt, Inc. All rights reserved.
20
Chapter 42 Solutions
42.62
The probability, P, of finding the electron within the Bohr radius is P=
a0 r=0
P1s (r ) dr =
4r 2 -2r e r=0 a 3 0
a0
a0
dr
Defining z 2r a0 , this becomes 1 P = - (z 2 + 2z + 2)e -z 2
2 0
= -
1 1 10 ( 4 + 4 + 2) e -2 - (0 + 0 + 2) e 0 = 2 - 2 = 0.323 2 2 e
[
]
The electron is likely to be within the Bohr radius about one-third of the time. The Bohr model indicates none of the time.
42.63
(a)
For a classical atom, the centripetal acceleration is a= v2 1 e2 = 4 e0 r 2 me r e2 m v2 e2 + e = - 4 e0 r 2 8 e0 r so e 2 dr -1 e 2 a2 -e 2 dE e2 = = = 2 3 3 2 dt 8 e0 r dt 6 e0 c 6 e0 c 4 e0 r me e4 dr = - dt 12 2e 0 2 r 2 me 2 c 3
-10
2
E= -
Therefore, -
r=0 m
(b)
r= 2.00 10
12 2e0 2 r 2 me 2 c 3 dr = e 4
2.00 10 -10
T
t=0
dt
12 2e0 2 me 2 c 3 r 3 3 e4
= T = 8.46 10 10 s
0
Since atoms last a lot longer than 0.8 ns, the classical laws (fortunately!) do not hold for systems of atomic size.
42.64
(a) (b)
+3e 0.85e 0.85e = 1.30e The valence electron is in an n = 2 state, with energy 13.6 eV(1.30)2 13.6 eV Zeff = = 5.75 eV n2 22 To ionize the atom you must put in + 5.75 eV This differs from the experimental value by 6%, so we could say the effective value of Z is accurate within 3%.
2
Chapter 42 Solutions 2 9.27 10 - 24 J / T (0.350 T ) = 6.626 10 - 34 J s f
21
42.65
E = 2 BB = h f
so
(
)
(
)
and f = 9.79 10 9 Hz
42.66
The photon energy is E4 - E3 = 20.66 - 18.70 eV = 1.96 eV =
hc
=
(6.626 10 - 34 J s)(3.00 108 m / s) = 633 nm 1.96 1.60 10 -19 J
42.67
(a)
6.626 10 -34 3.00 108 hc 1 = = ke e 2 2 8.99 10 9 1.60 10 -19
(
(
)(
)(
) )
2
= 137
(b)
C
re
=
h mc 2 hc 2 = = mc ke e 2 ke 2
(c)
1 a0 h2 mc 1 hc 137 = = = = 2 2 2 2 C mke e h 2 ke e 1 RH 4 ch3 mke e 2 1 hc 4 = = = 4 = 2 4 2 2 RH a0 mke e a0 ke e h
(d)
42.68
=
1 (2 )-1/2 1 4 a
0
3/2
r -r/2a0 r = A 2 - e -r/2a0 2 - e a0 a0
r 2 Ae -r/2a0 3 = 2 - 4a 2 2 r a0 0
Substituting into Schrdinger's equation and dividing by , 2m r r 1 1 - = - 2 [E - U] 2 - 2 a0 2 4a0 h a0 Now
2 2 h2 1 ke 4a0 m h 1 h2 - E -U = = - 2 2 4 2m a0 2m a0 4 m h2
(
(
)(
)
)
and
1 1 1 r r 2 2 - 4a = 2 2 - a 4a0 a0 0 0
is a solution.
2000 by Harcourt, Inc. All rights reserved.
22
Chapter 42 Solutions
*42.69
The beam intensity is reduced by absorption of photons into atoms in the lower state. The number of transitions per time and per area is -BNl I ( x ) n dx c. The beam intensity is increased by stimulating emission in atoms in the upper state, with transition rate +BN u I ( x ) n dx c. The net rate of change in photon numbers per area is then -B( Nl - N u ) I ( x ) n dx c. Each photon has energy h f , so the net change in intensity is dI ( x ) = - h f B( Nl - N u ) I ( x ) n dx c = - h f B N I ( x ) n dx c Then, dI ( x ) h f B N n = - dx c I (x) so
I
I (L)
0
L dI ( x ) h f B N n = - dx x=0 I (x) c
I ( L) h f B N n ln[ I ( L)] - ln [ I 0 ] = ln ( L - 0) = - I0 c I ( L) = I 0 e - h f B N n L c = I 0 e - L This result is also expressed in problem 42.44 as I ( L) = G = e - ( nl -nu )L = e + ( nu -nl )L I0
*42.70 (a)
Suppose the atoms move in the +x direction. The absorption of a photon by an atom is a completely inelastic collision, described by mvi i + h ( -i) = mv f i so v f - vi = - h m
This happens promptly every time an atom has fallen back into the ground state, so it happens every 10 -8 s = t . Then, a= v f - vi t = - h 6.626 10 -34 J s ~- m t 10 -25 kg 500 10 -9 m 10 -8 s
(
)(
)(
)
ms
- 106 m s 2
(b)
With constant average acceleration,
v 2 = vi2 + 2a( x ) f
0 ~ 10 m s
(
3
)
2
+ 2 -10 m s x
6 2
(
)
so
(10 x ~
3
)
2
106 m s 2
~1m
Chapter 43 Solutions
F= U= q2 (1.60 10 -19 )2 (8.99 10 9 ) = N = 0.921 10 - 9 N (5.00 10 -10 )2 4 e0 r 2 - q2 (1.60 10 -19 )2 (8.99 10 9 ) = - J 2.88 eV 4 e0 r 5.00 10 -10
43.1
(a)
toward the other ion.
(b)
*43.2
We are told and or By substitution,
K + C l + 0.7 eV K + + C l - C l + e - C l - + 3.6 eV C l - C l + e - - 3.6 eV K + C l + 0.7 eV K + + C l + e - - 3.6 eV K + 4.3 eV K + + e -
or the ionization energy of potassium is 4.3 eV
43.3
(a)
Minimum energy of the molecule is found from dU = -12Ar -13 + 6Br -7 = 0, yielding dr 2A r0 = B
16
(b)
E=U
r=
-U
A B 1 - = - - r = r0= 0 - 2 2 2A B 4 4A B
1 B2 B2 = 2 A 4A
This is also the equal to the binding energy, the amount of energy given up by the two atoms as they come together to form a molecule. (c) 2 0.124 10 -120 eV m12 r0 = 1.488 10 -60 eV m 6
(
)
16
= 7.42 10 -11 m = 74.2 pm
(1.488 10 E= 4(0.124 10
*43.4
-60
eV m 6
-120
)
2
eV m12
)
= 4.46 eV
At the boiling or condensation temperature,
kBT 10 -3 eV = 10 -3 1.6 10 -19 J T 1.6 10 -22 J 1.38 10 -23 J K ~10 K
(
)
2000 by Harcourt, Inc. All rights reserved.
546 Chapter 43 Solutions
43.5
=
m1m2 132.9(126.9) = 1.66 10 - 27 kg = 1.08 10 - 25 kg m1 + m2 132.9 + 126.9
(
)
I = r 2 = (1.08 10 - 25 kg)(0.127 10 - 9 m)2 = 1.74 10 - 45 kg m 2 (a) (I )2 J(J + 1)h2 1 E = 2 I2 = = 2I 2I J = 0 gives E = 0 J = 1 gives E = h2 (6.626 10 - 34 J s)2 = = 6.41 10 - 24 J = 40.0 eV I 4 2 (1.74 10 - 45 kg m 2 ) to f = 9.66 10 9 Hz If r is 10% too small, f is 20% too large.
h f = 6.41 10 - 24 J - 0 (b) f= E1 h2 h = = r- 2 h h I 4 2 r 2
43.6
h f = E =
2 2 h2 [2(2 + 1)] - h [1(1 + 1)] = h ( 4) 2I 2I 2I 2
4( h 2 ) h 6.626 10 -34 J s I= = = = 1.46 10 - 46 kg m 2 2h f 2 2 f 2 2 2.30 1011 Hz
(
)
43.7
For the HCl molecule in the J = 1 rotational energy level, we are given r 0 = 0.1275 nm. Erot = h2 J(J + 1) 2I h2 1 2 = 2 I I or
Taking J = 1, we have Erot =
=
2h2 h = 2 2 I I mm 2 2 I = r0 = 1 2 r0 m1 + m2
The moment of inertia of the molecule is given by Equation 43.3.
(1 u)(35 u) 2 I= r 0 = (0.972 u)(1.66 10 -27 kg / u)(1.275 10 -10 m)2 = 2.62 10 -47 kg m 2 1 u + 35 u Therefore,
= 2
h 1.055 10 -34 J s = 2 = 5.69 1012 rad / s I 2.62 10 -47 kg m 2
Chapter 43 Solutions 547
Goal Solution An HCl molecule is excited to its first rotational-energy level, corresponding to J = 1. If the distance between its nuclei is 0.127 5 nm, what is the angular speed of the molecule about its center of mass? G: For a system as small as a molecule, we can expect the angular speed to be much faster than the few rad/s typical of everyday objects we encounter. O: The rotational energy is given by the angular momentum quantum number, J . The angular speed can be calculated from this kinetic rotational energy and the moment of inertia of this onedimensional molecule. A : For the HCl molecule in the J = 1 rotational energy level, we are given r0 = 0.1275 nm. Erot = h J ( J + 1) 2I so with J = 1, Erot = h2 1 2 = I 2 I and
=
2h2 h 2 = I I2
The moment of inertia of the molecule is given by:
mm (1 u)(35 u) 2 I = r0 2 = 1 2 r0 2 = r0 m1 + m2 1 u + 35 u
I = (0.972 u)(1.66 10 -27 kg / u)(1.275 10 -10 m)2 = 2.62 10 -47 kg m 2 Therefore, 1.055 10 -34 J s 2 = 5.69 1012 rad / s -47 2 2.62 10 kg m
L : This angular speed is more than a billion times faster than the spin rate of a music CD, which rotates at 200 to 500 revolutions per minute, or = 20 rad / s to 50 rad / s. 43.8 I = m1r 1 + m2 r 2 Then r1 = r2 =
2 2
where so
m1r 1 = m2 r 2 m2 r 2 m1 r1 + = + r2 = r =r =
and and
r1 + r 2 = r r2 = r1 = m1r m1 + m2 m2 r m1 + m2
m2 r 2 m1 m1r 1 m2 +
Also, I = m1
Thus,
2 m2 m1 r 2
m1r 1 m2
and m1m2 r 2 = r2 m1 + m2
(m1 + m2 )2 (m1 + m2 )2
2 m2 r 2
m1m2 r 2 ( m2 + m1 )
(m1 + m2 )2
43.9
(a)
=
22.99( 35.45) 1.66 10 -27 kg = 2.32 10 -26 kg (22.99 + 35.45)
(
)
I = r 2 = 2.32 10 -26 kg 0.280 10 -9 m (b)
(
)(
)
2
=
1.81 10 - 45 kg m 2
h2 3h2 h2 2h2 2h 2 hc h2 = 2( 2 + 1) - 1(1 + 1) = - = = 2I 2I I I I 4 2 I 3.00 108 m / s 4 2 1.81 10 - 45 kg m 2 c 4 2 I = = = 1.62 cm 2h 2 6.626 10 -34 J s
(
(
)
(
)
)
2000 by Harcourt, Inc. All rights reserved.
548 Chapter 43 Solutions
43.10
The energy of a rotational transition is E = h2 I J where J is the rotational quantum number of the higher energy state (see Equation 43.7). We do not know J from the data. However, 6.626 10 -34 J s 3.00 108 m s hc 1 eV E = = -19 1.60 10 J For each observed wavelength,
( )
(
)(
)
(mm ) 0.1204 0.0964 0.0804 0.0690 0.0604
E (eV ) 0.01032 0.01288 0.01544 0.01800 0.02056
The Es consistently increase by 0.00256 eV. E1 = h2 I = 0.00256 eV 1.055 10 -34 J s h2 I= = E1 (0.00256 eV)
and
(
)
2
1 eV = 2.72 10 -47 kg m 2 -19 1.60 10 J r= I = 2.72 10 -47 m = 0.130 nm 1.62 10 -27
For the HCl molecule, the internuclear radius is
43.11
=
m1m2 35 = 1.66 10 -27 kg = 1.61 10 -27 kg m1 + m2 36 k = 1.055 10 -34
E vib = h
(
)
480 = 5.74 10 -20 J = 0.358 eV 1.61 10 -27
43.12
(a)
Minimum amplitude of vibration of HI is
1 kA 2 2 1 = 2 hf : A =
hf (6.626 10 -34 J s)(6.69 1013 / s) = = 1.18 10 -11 m = 0.0118 nm 320 N / m k A= (6.626 10 -34 J s)(8.72 1013 / s) = 7.72 10 -12 m = 0.00772 nm 970 N / m
(b)
For HF,
Since HI has the smaller k , it is more weakly bound.
Chapter 43 Solutions 549
43.13
(a)
The reduced mass of the O 2 is The moment of inertia is then
=
(16 u)(16 u) = 8 u = 8 1.66 10 -27 ( (16 u) + (16 u)
kg = 1.33 10 -26 kg
)
I = r 2 = 1.33 10 -26 kg 1.20 10 -10 m
(
)(
)
2
= 1.91 10 -46 kg m 2
The rotational energies are Thus And for J = 0, 1, 2, (b)
E rot
1.055 10 -34 J s h2 = J ( J + 1) = J ( J + 1) 2I 2 1.91 10 -46 kg m 2
E rot = 2.91 10 -23
(
( ( J) J(J + 1)
)
2
)
E rot = 0, 3.64 10 -4 eV, 1.09 10 -3 eV
k 1 1 1.055 10 -34 J s E vib = v + h = v+ 2 2 E vib = v +
(
)
1177 N / m 8(1.66 10 -27 kg) 1 (0.196 eV) 2
1 1 eV = v+ 3.14 10 -20 J -19 2 1.60 10 J
(
)
For v = 0, 1, 2,
E vib = 0.0982 eV, 0.295 eV, 0.491 eV
43.14
In Benzene, the carbon atoms are each 0.110 nm from the axis and each hydrogen atom is (0.110 + 0.100 nm) = 0.210 nm from the axis. Thus, I = mr 2 : I = 6 1.99 10 -26 kg 0.110 10 -9 m
(
)( ( (
)
2
+ 6 1.67 10 -27 kg 0.210 10 -9 m = 1.89 10 -45 kg m 2
(
)(
)
2
The allowed rotational energies are then Erot 1.055 10 -34 J s h2 = J ( J + 1) = J ( J + 1) = 2.95 10 -24 J J ( J + 1) = 18.4 10 -6 eV J ( J + 1) 2I 2 1.89 10 -45 kg m 2
)
2
)
(
)
(
)
Erot =
(18.4 eV) J ( J + 1)
where J = 0, 1, 2, 3, . . . Erot = 0, 36.9 eV, 111 eV, 221 eV, and 369 eV
The first five of these allowed energies are:
43.15
hf =
h2 J 4 2 I
where the rotational transition is from J 1 to J , and I = 1.46 10 - 46 kg m 2 from Example 43.1.
where f = 6.42 1013 Hz J=
4 2 If 4 2 (1.46 10 - 46 kg m 2 )(6.42 1013 / s) = = 558 h 6.626 10 - 34 J s
2000 by Harcourt, Inc. All rights reserved.
550 Chapter 43 Solutions
*43.16
The emission energies are the same as the absorption energies, but the final state must be below ( v = 1, J = 0) . The transition must satisfy J = 1, so it must end with J = 1. To be lower in energy, it must be ( v = 0, J = 1) . The emitted photon energy is therefore hf photon = E vib + E rot v=1 hf photon = h f vib - h f rot Thus, f photon = f vib - f rot = 6.42 1013 Hz - 1.15 1011 Hz = 6.41 1013 Hz
J = 0
+ E rot - E vib v=0
J = 1
= E vib - E vib - E rot v=1 v = 0
J =1
- E rot
J = 0
*43.17
The moment of inertia about the molecular axis is
4 I x = 2 mr 2 + 2 mr 2 = 5 m 2.00 10 -15 m 5 5 2 2
(
)
2
The moment of inertia about a perpendicular axis is I y = m
m R R +m = 2.00 10 -10 m 2 2 2
(
)
2
The allowed rotational energies are E rot = h2 2I J ( J + 1) , so the energy of the first excited state is E1 = h I . The ratio is therefore
2
(
)
E1, x E1, y
(h I ) = I = (h I ) I
2 x 2 y
y x
=
2 1 m 2.00 10 -10 m 2 4 m 2.00 10 -15 m 5
( (
) = (10 ) )
5 8
5 2
= 6.25 10 9
*43.18
Consider a cubical salt crystal of edge length 0.1 mm. The number of atoms is 10 -4 m -9 0.261 10 m
3
~ 1017
3
This number of salt crystals would have volume If it is cubic, it has edge length 40 m.
(10
-4
10 -4 m m -9 0.261 10 m
)
3
~ 10 5 m 3
43.19
U= -
ke e 2 1 (1.60 10 -19 )2 1 = - 1.25 10 -18 J = 7.84 eV 1- = -(1.7476)(8.99 10 9 ) 1- -9 m 8 r0 (0.281 10 )
43.20
Visualize a K+ ion at the center of each shaded cube, a C l ion at the center of each white one. The distance ab is Distance ac is Distance ad is 2 ( 0.314 nm) = 0.444 nm 2(0.314 nm) = 0.628 nm 2 2 + ( 2)2 (0.314 nm) = 0.769 nm
Chapter 43 Solutions 551
43.21
U= -
ke e 2 ke e 2 ke e 2 ke e 2 ke e 2 ke e 2 ke e 2 ke e 2 2k e 2 1 1 1 - + + - - + + - ... = - e 1- + - + . . . 2 3 4 r r 2r 2r 3r 3r 4r 4r r ln(1 + x ) = 1 - x2 x3 x4 + - + ... 2 3 4 or U = - ke e2 r where = 2 ln 2
But,
so,
U= -
2ke e 2 ln 2, r
43.22
EF =
h 2 3ne 2m 8
23
(6.625 10 -34 J s)2 23 23 = (3 / 8 ) n -31 -19 2(9.11 10 kg)(1.60 10 J / eV)
EF = (3.65 10 -19 )n2 3 eV with n measured in electrons/m3
43.23
The density of conduction electrons n is given by 8 2 mEF ne = 3 h2
3/2
EF =
h 2 3ne 2m 8
23
or
- 31 kg)(5.48)(1.60 10 -19 J) 8 2(9.11 10 = 3 (6.626 10 - 34 J s)3
[
]
3/2
= 5.80 10 28 m -3
The number-density of silver atoms is 1 atom 1u nAg = 10.6 10 3 kg / m 3 = 5.91 10 28 m -3 - 27 kg 108 u 1.66 10
(
)
So an average atom contributes
5.80 = 0.981 electron to the conduction band 5.91
43.24
(a)
1 mv 2 2
= 7.05 eV
v=
2(7.05 eV)(1.60 10 -19 J / eV) = 1.57 106 m/s - 31 kg 9.11 10 However, the energy of an electron at
(b)
Larger than 10 4 m/s by ten orders of magnitude. room temperature is typically kBT =
1 40
eV .
2000 by Harcourt, Inc. All rights reserved.
552 Chapter 43 Solutions
43.25
For sodium, M = 23.0 g mol and = 0.971 g cm 3 . ne = 6.02 10 23 electrons mol 0.971 g cm 3 NA = M 23.0 g mol
(a)
(
)(
)
ne = 2.54 10 22 electrons cm 3 = 2.54 10 28 electrons m 3
(b)
h 2 3ne 2 EF = 2m 8
3
(6.626 10 = 2 (9.11 10
-31
-34 -31
Js
) 3 (2.54 10 8 kg )
2
28
m -3
)
2 3
= 5.05 10 -19 J = 3.15 eV
(c)
vF =
2E F m
=
2 5.05 10 -19 J 9.11 10 kg
(
)=
1.05 106 m s
*43.26
The melting point of silver is 1234 K. Its Fermi energy at 300 K is 5.48 eV. The approximate fraction of electrons excited is 1.38 10 -23 J K (1234 K ) kBT = 2% EF ( 5.48 eV) 1.60 10 -19 J eV
(
(
)
)
43.27
Taking EF = 5.48 eV for sodium at 800 K,
E - EF ) f = e(
[
kBT
+1
]
-1
= 0.950
e(
E - EF ) k B T
= (1/ 0.950) - 1 = 0.0526
E - EF = ln(0.0526) = - 2.94 kBT E - EF = - 2.94 (1.38 10 -23 )(800) J = - 0.203 eV 1.60 10 -19 J / eV or E = 5.28 eV
Chapter 43 Solutions 553
Goal Solution Calculate the energy of a conduction electron in silver at 800 K if the probability of finding an electron i n that state is 0.950. The Fermi energy is 5.48 eV at this temperature. G: Since there is a 95% probability of finding the electron in this state, its energy should be slightly less than the Fermi energy, as indicated by the graph in Figure 43.21. The electron energy can be found from the Fermi-Dirac distribution function. Taking EF = 5.48 eV for silver at 800 K, and given f (E) = 0.950, we find f (E) = 1 = 0.950 ( E-EF ) kBT + 1 e or so e( E-EF )
k BT
O: A:
=
1 - 1 = 0.05263 0.950
E - EF = ln(0.05263) = -2.944 kBT
E - EF = -2.944kBT = -2.944 1.38 10 -23 J / K (800 K )
(
)
E = EF - 3.25 10 -20 J = 5.48 eV - 0.203 eV = 5.28 eV L: As expected, the energy of the electron is slightly less than the Fermi energy, which is about 5 eV for most metals. There is very little probability of finding an electron significantly above the Fermi energy in a metal. d = 1.00 mm , so V = 1.00 10 -3 m g(E) = CE 1 g (E) =
2
43.28
(
)
3
= 1.00 10 -9 m 3
2
The density of states is
=
8 2 m3 h3
E1
2
or
8 2 9.11 10 -31 kg
(6.626 10 )
(
-34
Js
)
)
3 2
3
( 4.00 eV)(1.60 10 -19
J eV
)
g(E) = 8.50 10 46 m -3 J -1 = 1.36 10 28 m -3 eV -1 So, the total number of electrons is N = [ g(E)]( E)V = 1.36 10 28 m -3 eV -1 (0.0250 eV ) 1.00 10 -9 m 3 = 3.40 1017 electrons
(
(
)
43.29
Eav = 1 ne
0
EN (E) dE N (E) = 0 for E > E F ; N(E) = CE1/2
At T = 0, Since f (E) = 1 for E < E F and f (E) = 0 for E > E F , we can take Eav = 1 ne
0
EF
CE 3/2 dE =
C ne
0
EF
E 3 2 dE =
2C 5 E 5ne
2
But from Equation 43.24,
C 3 -3/2 = EF , so that ne 2
Eav =
3 2 3 -3/2 5/2 E EF E = 5 2 F F 5
2000 by Harcourt, Inc. All rights reserved.
554 Chapter 43 Solutions
43.30
Consider first the wave function in x. At x = 0 and x = L , = 0. Therefore, Similarly, sin kxL = 0 sin kyL = 0 sin kzL = 0 and and and k xL = , 2 , 3 , . . . k yL = , 2 , 3 , . . . kzL = , 2 , 3 , . . .
= A sin
From
ny y nx x nz z sin sin L L L we have inside the box, where U = 0, h2 2 2 2 2 (nx + ny + nz ) 2me L2
d 2 d 2 d 2 2me + + 2 = 2 (U - E) , dx 2 dy 2 dz h
2 n 2 2 ny 2 n 2 2 2m - x 2 - 2 - z 2 = 2 e (-E) L L L h
E=
nx , ny , nz = 1, 2, 3, . . .
Outside the box we require = 0. The minimum energy state inside the box is nx = ny = nz = 1, with E = 3h2 2 2me L2
43.31
(a)
The density of states at energy E is Hence, the required ratio is
g(E) = CE 1
2 1 2 2
g(8.50 eV ) C (8.50) = g(7.00 eV ) C (7.00) 1
= 1.10 CE 1
2
(b)
From Eq. 43.22, the number of occupied states having energy E is N (E) = N (8.50 eV ) (8.50) = N (7.00 eV ) (7.00) 1 N (8.50 eV ) (8.50) = N (7.00 eV ) (7.00) 1
e
( E-EF ) kBT
+1
kBT kBT
1 2 2
Hence, the required ratio is
e(7.00 - 7.00 ) 8.50 - 7.00 ) e(
+ 1 + 1
At T = 300 K , kBT = 4.14 10 -21 J = 0.0259 eV ,
1 2 2
2.00 (1.50 ) 0.0259 + 1 e
And
N (8.50 eV ) = 1.55 10 - 25 N (7.00 eV )
Comparing this result with that from part (a), we conclude that very few states with E > EF are occupied.
43.32
(a)
Eg = 1.14 eV for Si h f = 1.14 eV = (1.14 eV)(1.60 10 19 J/eV) = 1.82 10 19 J so f 2.75 1014 Hz
(b)
c = f;
=
c 3.00 108 m / s = = 1.09 10 -6 m = 1.09 m f 2.75 1014 Hz
(in the infrared region)
Chapter 43 Solutions 555
43.33
Photons of energy greater than 2.42 eV will be absorbed. This means wavelength shorter than
=
hc (6.626 10 - 34 J s)(3.00 108 m / s) = = 514 nm E 2.42 1.60 10 -19 J
All the hydrogen Balmer lines except for the red line at 656 nm will be absorbed.
43.34
Eg =
hc (6.626 10 - 34 J s)(3.00 108 m / s) = J 1.91 eV 650 10 - 9 m
43.35
If 1.00 10 -6 m , then photons of sunlight have energy E hc
max
(6.626 10 =
J s 3.00 108 m s 1 eV = 1.24 eV -6 -19 1.60 10 1.00 10 m J Since Si has an Therefore,
-34
)(
)
Thus, the energy gap for the collector material should be Eg 1.24 eV . Si is acceptable as a material for a solar collector.
energy gap Eg 1.14 eV , it will absorb radiation of this energy and greater.
Goal Solution Most solar radiation has a wavelength of 1 m or less. What energy gap should the material in a solar cell have in order to absorb this radiation? Is silicon appropriate (see Table 43.5)? G: Since most photovoltaic solar cells are made of silicon, this semiconductor seems to be an appropriate material for these devices. To absorb the longest-wavelength photons, the energy gap should be no larger than the photon energy. The minimum photon energy is 6.63 10 -34 J s 3.00 108 m / s hc 1 eV = 1.24 eV hf = = -6 -19 1.60 10 J 10 m Therefore, the energy gap in the absorbing material should be smaller than 1.24 eV. L: So silicon, with gap of 1.14 eV < 1.24 eV, is an appropriate material for absorbing solar radiation.
O:
A:
(
)(
)
*43.36
If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here,
(h f )max =
hc
min
= 5.50 eV :
min
6.626 10 -34 J s 3.00 108 m s hc = = 5.5 eV ( 5.5 eV) 1.60 10 -19 J eV
(
(
)(
)
)
min = 2.26 10 -7 m = 226 nm
2000 by Harcourt, Inc. All rights reserved.
556 Chapter 43 Solutions
43.37
I = I 0 e e ( V ) and
(
kBT
- 1 Thus, e e ( V ) V =
)
kBT
= 1 + I I0
kBT ln(1 + I I 0 ) e
At T = 300 K , I = 9.00I 0 ,
V =
(1.38 10
-23
J K ( 300 K )
1.60 10 -19 C
)
I I ln 1 + = ( 25.9 mV ) ln 1 + I0 I0
(a) (b)
If If
V = ( 25.9 mV ) ln (10.0) = 59.5 mV
I = - 0.900I 0 , V = ( 25.9 mV ) ln (0.100) = 59.5 mV
The basic idea behind a semiconductor device is that a large current or charge can be controlled by a small control voltage.
43.38
The voltage across the diode is about 0.6 V. The voltage drop across the resistor is (0.025 A)(150 ) = 3.75 V. Thus, E - 0.6 V - 3.8 V = 0 and E = 4.4 V
*43.39
First, we evaluate I 0 in I = I 0 e e ( V ) T = 300 K .
(
kBT
- 1 , given that I = 200 mA when V = 100 mV and
)
1.60 10 -19 C (0.100 V ) e( V ) = = 3.86 so kBT 1.38 10 -23 J K ( 300 K )
( (
)
)
I0 =
e
e ( V ) k B T
I
-1
=
200 mA = 4.28 mA e 3.86 - 1
If V = -100 mV, I = I 0 e e ( V )
e( V ) = - 3.86 ; and the current will be kBT
(
kBT
- 1 = ( 4.28 mA ) e - 3.86 - 1 = 4.19 mA
)
(
)
43.40
(a) (b)
See the figure at right. For a surface current around the outside of the cylinder as shown, B= N 0I l or NI = Bl (0.540 T)(2.50 10 -2 m) = = 10.7 kA 0 (4 10 -7 ) T m / A
Chapter 43 Solutions 557
43.41
By Faraday's law (Equation 32.1),
I B B =L =A . t t t I = A( B) (0.0100 m ) (0.0200 T ) = = 203 A L 3.10 10 -8 H
2
Thus,
The direction of the induced current is such as to maintain the B field through the ring.
Goal Solution Determine the current generated in a superconducting ring of niobium metal 2.00 cm in diameter if a 0.0200-T magnetic field in a direction perpendicular to the ring is suddenly decreased to zero. The inductance of the ring is 3.10 10-8 H. G: The resistance of a superconductor is zero, so the current is limited only by the change in magnetic flux and self-inductance. Therefore, unusually large currents (greater than 100 A) are possible. O: The change in magnetic field through the ring will induce an emf according to Faraday's law of induction. Since we do not know how fast the magnetic field is changing, we must use the ring's inductance and the geometry of the ring to calculate the magnetic flux, which can then be used to find the current. A : From Faraday's law (Eq. 31.1), we have
=
I B B =A =L t t t
or
I =
AB (0.0100 m ) (0.0200 T ) = = 203 A L 3.10 10 -8 H
2
The current is directed so as to produce its own magnetic field in the direction of the original field. L : This induced current should remain constant as long as the ring is superconducting. If the ring failed to be a superconductor (e.g. if it warmed above the critical temperature), the metal would have a nonzero resistance, and the current would quickly drop to zero. It is interesting to note that we were able to calculate the current in the ring without knowing the emf. In order to calculate the emf, we would need to know how quickly the magnetic field goes to zero.
43.42
(a)
V = IR If R = 0, then V = 0, even when I 0.
(b)
The graph shows a direct proportionality. Slope = 1 I (155 - 57.8) mA = 43.1 -1 = = R V ( 3.61 - 1.356) mV
R = 0.0232 (c) Expulsion of magnetic flux and therefore fewer current-carrying paths could explain the decrease in current.
2000 by Harcourt, Inc. All rights reserved.
558 Chapter 43 Solutions
*43.43 (a)
Since the interatomic potential is the same for both molecules, the spring constant is the same. Then f = 1 2 k where
12 =
(12 u)(16 u) = 6.86 u
12 u + 16 u
and 14 =
(14 u)(16 u) = 7.47 u
14 u + 16 u
Therefore, f 14 = (b) 1 2 1 k = 14 2
12 k 12 13 = f 12 = 6.42 10 Hz 12 14 14
(
)
6.86 u = 6.15 1013 Hz 7.47 u
The equilibrium distance is the same for both molecules. 14 14 2 I14 = 14 r 2 = 12 r = I12 12 12 I14 = 7.47 u 1.46 10 -46 kg m 2 = 1.59 10 -46 kg m 2 6.86 u
(
)
(c)
The molecule can move to the ( v = 1, J = 9) state or to the ( v = 1, J = 11) state. The energy it can absorb is either E = hc h2 h2 1 1 = 1 + 2 h f 14 + 9(9 + 1) - 0 + 2 h f 14 + 10(10 + 1) , 2I14 2I14
( (
) )
(
)
or
E =
hc h2 h2 1 1 = 1 + 2 h f 14 + 11(11 + 1) - 0 + 2 h f 14 + 10(10 + 1) . 2I14 2I14
(
)
The wavelengths it can absorb are then
=
c c or = f 14 - 10h ( 2 I14 ) f 14 + 11h ( 2 I14 ) 3.00 108 m s 6.15 1013 Hz - 10 1.055 10 -34 J s
These are:
=
2 1.59 10 -46 kg m 2
(
(
)
= 4.96 m
)
= 4.79 m
and
=
3.00 108 m s 6.15 1013 Hz + 11 1.055 10 -34 J s
2 1.59 10 -46 kg m 2
(
(
)
)
Chapter 43 Solutions 559
43.44
For the N2 molecule,
k = 2297 N / m , m = 2.32 10 - 26 kg, r = 1.20 10 -10 m , = m / 2 I = r 2 = (1.16 10 - 26 kg)(1.20 10 -10 m)2 = 1.67 10 - 46 kg m 2
= k = 4.45 1014 rad / s,
For a rotational state sufficient to allow a transition to the first exited vibrational state, h2 J(J + 1) = h 2I Thus J = 37 so J(J + 1) = 2I 2(1.67 10 -46 )(4.45 1014 ) = = 1410 h 1.055 10 -34
43.45
1 Emax = 4.5 eV = v + h 2 8.25 > 7.5 v=7
so
(4.5 eV)(1.6 10 -19 J / eV) 1 v+ -34 14 -1 2 J s)(8.28 10 s ) (1.055 10
43.46
With 4 van der Waal bonds per atom pair or 2 electrons per atom, the total energy of the solid is 6.02 10 23 atoms E = 2(1.74 10 -23 J / atom) = 5.23 J/g 4.00 g
43.47
The total potential energy is given by Equation 43.16:
U total = -
ke e 2 B + m r r
The total potential energy has its minimum value U 0 at the equilibrium spacing, r = r 0 . At this point, or dU dr
r = r0
= 0, dU dr = ke e 2 d B - r + r m dr =
r = r0
ke e 2
2 r0
r = r0
-
mB r0
m+1
=0
Thus,
B=
ke e 2 m - 1 r m 0 ke e 2 k e2 m - 1 1 1 k e2 + e r0 m = - e 1- r m r0 m r0 0
Substituting this value of B into U total ,
U0 = -
*43.48
3 1 Suppose it is a harmonic-oscillator potential well. Then, 2 h f + 4.48 eV = 2 h f + 3.96 eV is the depth of the well below the dissociation point. We see h f = 0.520 eV , so the depth of the well is 1 h f + 4.48 eV = 1 (0.520 eV ) + 4.48 eV = 4.74 eV 2 2
2000 by Harcourt, Inc. All rights reserved.
560 Chapter 43 Solutions
*43.49 (a)
For equilibrium,
dU = 0: dx
d Ax -3 - Bx -1 = - 3Ax -4 + Bx -2 = 0 dx
-2
(
)
x describes one equilibrium position, but the stable equilibrium position is at 3Ax0 = B. x0 = (b) 3A = B 3 0.150 eV nm 3 3.68 eV nm
(
)=
0.350 nm U0 = U x = x =
0
The depth of the well is given by 2B 3 2 3 2 1 3 A
A B AB 3 2 - = 3 2 3 3 x0 x0 3 A = 7.02 eV
2
-
BB 1 2 31 2A1
2
U0 = U x = x = -
0
2
= -
2 ( 3.68 eV nm ) 33
2
3 2
(
0.150 eV nm 3
)
1 2
(c)
Fx = -
dU = 3Ax -4 - Bx -2 dx dFx dx
1 2 x = xm
To find the maximum force, we determine finite xm such that
=0
Thus, -12Ax -5 + 2Bx -3 Fmax = 3A
[
]
x = x0 2
=0
so that
xm =
6A B
Then
B2 (3.68 eV nm) B B = - -B = - 6A 6A 12A 12 0.150 eV nm 3
2
(
)
or
Fmax = - 7.52
eV 1.60 10 -19 J 1 nm -9 - 9 = -1.20 10 N = -1.20 nN nm 1 eV 10 m
43.50
(a)
For equilibrium,
dU = 0: dx
d Ax -3 - Bx -1 = -3Ax -4 + Bx -2 = 0 dx
(
)
x describes one equilibrium position, but the stable equilibrium position is at 3Ax0 = B
-2
or
x0 = 3A B A B AB 3 2 - = 3 2 3 3 x0 x0 3 A BB 1 2 3 A1
1 2
(b)
The depth of the well is given by dU = 3Ax -4 - Bx -2 dx
U0 = U x = x =
0
2
-
2
= -2
B3 27A
(c)
Fx = -
To find the maximum force, we determine finite xm such that dFx dx = -12Ax -5 + 2Bx -3
x = xm
[
]
x = x0
=0
then
Fmax = 3A
B2 B B = - -B 6A 6A 12A
2
Chapter 43 Solutions 561
*43.51 (a)
At equilibrium separation, r = re ,
dU dr
r = re
-a( r e - r 0 ) -a( r e - r 0 ) = - 2aB e - 1 e =0 e
-a r e -r 0
We have neutral equilibrium as re and stable equilibrium at or
(
) = 1,
re = r 0
(b)
At r = r 0 , U = 0 . As r , U B . The depth of the well is B .
(c)
We expand the potential in a Taylor series about the equilibrium point: U (r ) U r 0 +
( )
dU dr
r = r0
1 (r - r 0 ) + 2 d U dr 2
2
r = r0
(r - r 0 )
2
-2( r - r 0 ) - ( r - r 0 ) -2( r - r 0 ) 1 U (r ) 0 + 0 + 2 ( - 2Ba)-ae - ae - 1 r - r0 e r = r 0
(
)
2
Ba2 r - r 0
(
)
2
This is of the form for a simple harmonic oscillator with Then the molecule vibrates with frequency
1 kx 2 2
1 = 2 k r - r0
(
)
2
k = 2Ba2 f= 1 2 a k = 2 2B a = B 2
(d) The zero-point energy is
1 h 2
ha B 1 = 2 hf = 8 B- ha B 8
Therefore, to dissociate the molecule in its ground state requires energy
2000 by Harcourt, Inc. All rights reserved.
562 Chapter 43 Solutions
43.52
T=0 E / EF
0 0.500 0.600 0.700 0.800 0.900 1.00 1.10 1.20 1.30 1.40 1.50
T = 0.1T F f (E)
1.00 1.00 1.00 1.00 1.00 1.00 0.500 0.00 0.00 0.00 0.00 0.00
T = 0.2T F f (E)
1.000 0.993 0.982 0.953 0.881 0.731 0.500 0.269 0.119 0.0474 0.0180 0.00669
T = 0.5T F f (E)
0.993 0.924 0.881 0.818 0.731 0.622 0.500 0.378 0.269 0.182 0.119 0.0759
e
E TF -1 EF T
e e e e e e e e e e e
- - - - - - 0 + + + +
e
E TF -1 EF T
e e e e e e e e e e e
-10.0 -5.00 -4.00 -3.00 -2.00 -1.00 0 1.00 2.00 3.00 4.00
e
E TF -1 EF T
e e e e e e e e e e e
-5.00 -2.50 -2.00 -1.50 -1.00 -0.500 0 0.500 1.00 1.50 2.00
e
E TF -1 EF T
f (E)
0.881 0.731 0.690 0.646 0.599 0.550 0.500 0.450 0.401 0.354 0.310 0.269
e e e e e e e e e e e
-2.00 -1.00 -0.800 -0.600 -0.400 -0.200 0 0.200 0.400 0.600 0.800
e +
e 5.00
e 2.50
e1.00
Chapter 43 Solutions 563
43.53
(a)
There are 6 C l - ions at distance r = r 0 . The contribution of these ions to the electrostatic potential energy is - 6 ke e 2 r 0 . There are 12 Na+ ions at distance r = 2 r 0 . Their contribution to the electrostatic potential energy is +12 ke e 2 2 r 0 . Next, there are 8 Cl - ions at distance r = 3 r 0 . These contribute a term of - 8 ke e 2 the electrostatic potential energy. 3 r 0 to
To three terms, the electrostatic potential energy is: 12 8 ke e 2 k e2 U = -6 + - = - 2.13 e r0 2 3 r0 (b) U = - ke e 2 with = 2.13 r0
or
The fourth term consists of 6 Na+ at distance r = 2r0 . Thus, to four terms, U = ( - 2.13 + 3) k e2 ke e 2 = 0.866 e r0 r0
So we see that the electrostatic potential energy is not even attractive to 4 terms, and that the infinite series does not converge rapidly when groups of atoms corresponding to nearest neighbors, next-nearest neighbors, etc. are added together.
2000 by Harcourt, Inc. All rights reserved.
Chapter 44 Solutions
*44.1
An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water molecule there are eight neutrons and ten protons. So protons and neutrons are nearly equally numerous in your body, each contributing mass (say) 35 kg: 1 nucleon 28 28 35 kg ~ 10 protons and ~ 10 neutrons 27 1.67 10 kg The electron number is precisely equal to the proton number, ~ 1028 electrons
44.2
1 mv 2 2
= q( V )
and
mv 2 = qvB r
2m( V ) = qr 2B2
12
r=
2m( V ) 2(1000 V ) = 1.60 10 -19 C (0.200 T )2 qB2
(
)
m
r = 5.59 1011 (a) For
12
m m kg
C:
m = 12 u and r = 5.59 1011 m 12 1.66 10 - 27 kg = 0.0789 m = 7.89 cm kg
(
)
For
13
C:
r = 5.59 1011 r1 = 2m1( V ) qB2
m 13 1.66 10 - 27 kg = 0.0821 m = 8.21 cm kg
(
)
(b)
With
and
r2 = r1 r2
2m2 ( V ) , qB2 = m1 . m2
the ratio gives
r1 r2
=
7.89 cm = 0.961 8.21 cm
and
12 u m1 = = 0.961 13 u m2
so they do agree.
2000 by Harcourt, Inc. All rights reserved.
566 Chapter 44 Solutions Q 1Q 2 (2)(6)(1.60 10 19 C)2 = (8.99 109 N m2/C 2) = 27.6 N r2 (1.00 1014 m)2 away from the nucleus..
*44.3
(a)
F = ke
(b)
a=
F 27.6 N 27 2 m = 6.64 10 27 kg = 4.17 10 m/s Q 1Q 2 r = (8.99 109 N m2/C 2)
(c)
U = ke
(2)(6)(1.60 10 19 C)2 (1.00 1014 m)
= 2.76 10 13 J = 1.73 MeV
44.4 (a) (b)
E = 7.70 MeV dmin = 4 ke Ze 2 2 ke Ze 2 2 (8.99 10 9 )(79)(1.60 10 -19 )2 = 29.5 10 15 m = 29.5 fm = = E mv 2 7.70(1.60 10 -13 )
The de Broglie wavelength of the is
=
h = m v
h = 2m E
6.626 10 -34 2(6.64 10 -27 )7.70(1.60 10 -13 )
= 5.18 10 15 m = 5.18 fm
(c)
is much less than the distance of closest approach , the may be considered a particle.
Since
44.5
(a)
The initial kinetic energy of the alpha particle must equal the electrostatic potential energy at the distance of closest approach. Ki = U f = ke qQ r min
r min
8.99 10 9 N m 2 C 2 ( 2)(79) 1.60 10 -19 C k qQ = e = Ki (0.500 MeV) 1.60 10 -13 J MeV
(
(
)
(
)
)
2
= 4.55 10 -13 m
(b)
k qQ 1 Since Ki = 2 m vi2 = e , r min 2ke qQ = m r min 2 8.99 10 9 N m 2 C 2 ( 2)(79) 1.60 10 -19 C
vi =
(
( 4.00 u)(1.66 10 -27
kg u 3.00 10 -13 m
)
)(
(
)
)
2
= 6.04 106 m s
Chapter 44 Solutions 567 Goal Solution (a) Use energy methods to calculate the distance of closest approach for a head-on collision between an alpha particle having an initial energy of 0.500 MeV and a gold nucleus (197Au) at rest. (Assume the gold nucleus remains at rest during the collision.) (b) What minimum initial speed must the alpha particle have in order to get as close as 300 fm? G: The positively charged alpha particle (q = +2e) will be repelled by the positive gold nucleus (Q = +79e), so that the particles probably will not touch each other in this electrostatic "collision." Therefore, the closest the alpha particle can get to the gold nucleus would be if the two nuclei did touch, in which case the distance between their centers would be about 6 fm (using r = r0 A1/3 for the radius of each nucleus). To get this close, or even within 300 fm, the alpha particle must be traveling very fast, probably close to the speed of light (but of course v must be less than c ). At the distance of closest approach, rmin , the initial kinetic energy will equal the electrostatic potential energy between the alpha particle and gold nucleus. qQ (8.99 10 9 N m 2 / C 2 )(2)(79)(1.60 10 -19 C)2 = = 455 fm K (0.500 MeV)(1.60 10 -13 J / MeV)
O:
A:
(a)
K = U = k e
qQ rmin
and qQ rmin
rmin = ke
(b)
Since
1 K = 2 mv 2 = ke
v=
2ke qQ = mrmin
2(8.99 10 9 N m 2/ C 2 )(2)(79)(1.60 10 -19 C)2 = 6.04 106 m / s 4(1.66 10 -27 kg)(3.00 10 -13 m)
L:
The minimum distance in part (a) is about 100 times greater than the combined radii of the particles. For part (b), the alpha particle must have more than 0.5 MeV of energy since it gets closer to the nucleus than the 455 fm found in part (a). Even so, the speed of the alpha particle in part (b) is only about 2% of the speed of light, so we are justified in not using a relativistic approach. In solving this problem, we ignored the effect of the electrons around the gold nucleus that tend to "screen" the nucleus so that the alpha particle sees a reduced positive charge. If this screening effect were considered, the potential energy would be slightly reduced and the alpha particle could get closer to the gold nucleus for the same initial energy.
*44.6
It must start with kinetic energy equal to K i = U f = ke qQ r f . the radii of the r f = r 0 A1
1 3 4 2 He 1 3
Here r f stands for the sum of
and
197 79 Au
nuclei, computed as
+ r 0 A2
= 1.20 10 -15 m 4 1 3 + 197 1 3 = 8.89 10 -15 m
(
)(
)
Thus,
Ki = Uf
(8.99 10 =
9
N m 2 C 2 (2)(79) 1.60 10 -19 C 8.89 10 -15 m
)
(
)
2
= 4.09 10 -12 J = 25.6 MeV
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568 Chapter 44 Solutions
44.7
(a) (b)
r = r0 A1/ 3 = (1.20 10 15 m)(4)1/3 = 1.90 10 15 m r = r0 A1/ 3 = (1.20 10 15 m)(238)1/3 = 7.44 10 15 m
*44.8
From r = r0 A1/ 3 , the radius of uranium is r U = r 0(238)1/3. Thus, if r = 2 rU
1
then A = 30
r 0 A 1/3 = 2 r 0(238)1/3
1
from which
44.9
The number of nucleons in a star of two solar masses is A= 2 1.99 10 30 kg 1.67 10
-27
(
) )(
kg nucleon
= 2.38 10 57 nucleons
Therefore
r = r 0 A1/3 = 1.20 10 -15 m 2.38 10 57
(
)
13
= 16.0 km
*44.10
V = 4 r 3 = 4.16 10 -5 m 3 3 m = V = (2.31 1017 kg / m 3 )(4.16 10 -5 m 3 ) = 9.61 1012 kg F=G and
(9.61 1012 kg)2 m1m2 = (6.67 10 -11 Nm 2 / kg 2 ) = 6.16 1015 N toward the other ball. r2 (1.00 m)2
44.11
The stable nuclei that correspond to magic numbers are: Z magic:
2
He
8
O
20
Ca
28
Ni
50
Sn
82
Pb 126
N magic:
3 4 37 39 40 51 52 T 1 , He 2 , N 15 , O 16 , C l 17 , K 19 , Ca 20 , V 23 , Cr 24 , Sr 88 , Y 89 , 7 8 38 39 208 209 210 Zr 90 , Xe 136 , Ba 138 , La 139 , Ce 140 , Pr 141 , Nd 142 , Pb 82 , Bi 83 , Po 84 40 54 56 57 58 59 60
Chapter 44 Solutions 569 44.12 (a) (b) (c) (d) Of the 102 stable nuclei listed in Table A.3, Even Z , Even N Even Z , Odd N Odd Z , Even N Odd Z , Odd N 48 6 44 4
44.13
(a)
(b)
44.14
(a)
fn =
2 B 2(1.9135)(5.05 10 -27 J / T)(1.00 T) = = 29.2 MHz h 6.626 10 -34 J s 2(2.7928)(5.05 10 -27 J / T)(1.00 T) = 42.6 MHz 6.626 10 -34 J s fp = 2(2.7928)(5.05 10 -27 )(50.0 10 -6 ) = 2.13 kHz 6.626 10 -34
(b)
fp =
(c)
In the Earth's magnetic field,
44.15 (a)
Using atomic masses as given in Table A.3, For 2H 1, 2.014 102 + 1(1.008 665) + 1(1.007 825) 2
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570 Chapter 44 Solutions Eb = (0.00119 4 u) 931.5 MeV = 1.11 MeV/nucleon u
Chapter 44 Solutions 571 2 (1.008 665) + 2 (1.007 825) 4.002 602 4
(b)
For 4He,
Eb = 0.00759 u = 7.07 MeV/nucleon (c) For
56
Fe 26, 30(1.008 665) + 26(1.007 825) 55.934 940 = 0.528 u
0.528 Eb = 56 = 0.00944 u = 8.79 MeV/nucleon (d) For
238
U 92, 146(1.008 665) + 92(1.007 825) 238.050 784 = 1.934 2 u
1.934 2 = 0.00813 u = 7.57 MeV/nucleon Eb = 238
44.16 Nuclei
55 M n 56 Fe 59Co
M = Zm H + Nm n M Z 25 26 27 N 30 30 32 M in u 54.938048 55.934940 58.933198
BE M(931.5) = A A M in u 0.517527 0.528460 0.555357 BE/A in MeV 8.765 8.786 8.768
56Fe has a greater BE/A than its neighbors. This tells us finer detail than is shown in Figure 44.8.
44.17
(a) (b) (c)
The neutron-to-proton ratio, ( A - Z) / Z is greatest for
139 139
139 55 Cs
and is equal to 1.53.
La has the largest binding energy per nucleon of 8.378 MeV.
C s with a mass of 138.913 u. We locate the nuclei carefully on Figure 44.3, the neutronproton plot of stable nuclei. Cesium appears to be farther from the center of the zone of stability. Its instability means extra energy and extra mass.
44.18
Use Equation 44.4. The
23 11 Na, 23 12 Mg ,
Eb = 8.11 MeV/nucleon A Eb A = 7.90 MeV/nucleon
23 11 Na
and for
The binding energy per nucleon is greater for repulsion in Na23.)
by 0.210 MeV . (There is less proton
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572 Chapter 44 Solutions
44.19
The binding energy of a nucleus is For For
15 8 O: 15 7 N:
Eb ( MeV ) = Z M(H) + Nmn - M
[
( X)](931.494 MeV u)
A Z
[ ] Eb = [7 (1.007 825 u ) + 8 (1.008 665 u ) - 15.000 108 u](931.494 MeV u ) = 115.49 MeV
15 7N
Eb = 8 (1.007 825 u ) + 7 (1.008 665 u ) - 15.003 065 u (931.494 MeV u ) = 111.96 MeV
Therefore, the binding energy of
is larger by 3.54 MeV .
44.20
(a)
The radius of the
40
Ca nucleus is:
R = r 0 A1 3 = 1.20 10 -15 m ( 40)
(
)
13
= 4.10 10 -15 m
The energy required to overcome electrostatic repulsion is U= 3keQ = 5R
2
3 8.99 10 9 N m 2 C 2 20 1.60 10 -19 C 5 4.10 10 -15 m
40 20 Ca
(
(
)[ (
)
)]
2
= 1.35 10 -11 J = 84.1 MeV
(b)
The binding energy of
is
Eb = 20 (1.007 825 u ) + 20 (1.008 665 u ) - 39.962 591 u (931.5 MeV u ) = 342 MeV (c) The nuclear force is so strong that the binding energy greatly exceeds the minimum energy needed to overcome electrostatic repulsion.
[
]
44.21
43 Removal of a neutron from 20 Ca would result in the residual nucleus, separation energy is Sn, the overall process can be described by
42 20 Ca.
If the required
mass
(
43 20 Ca
)+S
n
= mass
(
42 20 Ca
) + mass(n)
Sn = (41.958 618 + 1.008 665 42.958 767) u = (0.008 516 u)(931.5 MeV/u) = 7.93 MeV
44.22
(a)
The first term overstates the importance of volume and the second term subtracts this overstatement. For spherical volume
4 R3 3
(b)
4 R
2
=
R 3
For cubical volume
R R3 = 6 2 6R
The maximum binding energy or lowest state of energy is achieved by building "nearly" spherical nuclei.
Chapter 44 Solutions 573
44.23 For A = 200, so For A 100 , so
Eb = Eb f - Ebi Eb = 7.4 MeV A Ebi = 200(7.4 MeV ) = 1480 MeV Eb A 8.4 MeV Eb f = 2(100)(8.4 MeV ) = 1680 MeV Eb = Eb f - Ebi Eb = 1680 MeV - 1480 MeV = 200 MeV
44.24
(a)
"Volume" term: "Surface" term: "Coulomb" term:
E 1 = C 1A = (15.7 MeV)(56) = 879 MeV E 2 = C 2 A 2/3 = (17.8 MeV)(56)2/3 = 260 MeV E3 = C 3 Z(Z 1) (26)(25) = (0.71 MeV) = 121 MeV 1/3 A (56)1/3
"Asymmetry" term: Eb = 491 MeV (b) E1 = 179%; Eb
E4 = C 4
(56 52)2 (A 2Z)2 = (23.6 MeV) = 6.74 MeV A 56
E2 = - 53.0%, Eb
E3 = - 24.6%; Eb
E4 = - 1.37% Eb
44.25
dN = N dt T1/ 2 =
so
=
1 dN - = (1.00 10 -15 )(6.00 1011 ) = 6.00 10 - 4 s -1 N dt
ln 2 = 1.16 10 3 s (= 19.3 min)
*44.26
R = R0 e
- t
= (6.40 mCi)e
ln 2 - ( 40.2 d ) 8.04 d
= (6.40 mCi) e - ln 2
(
)
5
= (6.40 mCi )
1 = 0.200 mC i 25
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574 Chapter 44 Solutions From R = R 0 e - t ,
44.27
(a)
=
1 R 0 1 10.0 ln = 5.58 10 -2 h -1 = 1.55 10 - 5 s -1 ln = t R 4.00 h 8.00 R0
T1/ 2 =
ln 2 = 12.4 h
(b) (c)
N0 =
=
10.0 10 - 3 Ci 3.70 1010 / s 13 = 2.39 10 atoms -5 1 Ci 1.55 10 / s
R = R 0 e - t = (10.0 mCi) exp ( - 5.58 10 - 2 30.0) = 1.87 mC i
Goal Solution A freshly prepared sample of a certain radioactive isotope has an activity of 10.0 mCi. After 4.00 h, its activity is 8.00 mCi. (a) Find the decay constant and half-life. (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30.0 h after it is prepared? G: Over the course of 4 hours, this isotope lost 20% of its activity, so its half-life appears to be around 10 hours, which means that its activity after 30 hours (~3 half-lives) will be about 1 mCi. The decay constant and number of atoms are not so easy to estimate. From the rate equation, R = R0 e - t , we can find the decay constant , which can then be used to find the half life, the original number of atoms, and the activity at any other time, t . (a) 10.0 mCi 1 1 R = 1.55 10 -5 s -1 = ln o = ln t R ( 4.00 h )(60.0 s / h ) 8.00 mCi T1/2 = (b) ln 2 0.693 = = 12.4 h 0.0558 h -1
O:
A:
The number of original atoms can be found if we convert the initial activity from curies into becquerels (decays per second): 1 Ci 3.7 1010 Bq R0 = 10.0 mCi = 10.0 10 -3 Ci 3.70 1010 Bq / Ci = 3.70 108 Bq Since R0 = N 0 , N0 = R0 3.70 108 decays / s = = 2.39 1013 atoms 1.55 10 -5 s
-2
(
)(
)
(c) L:
R = Ro e - t = (10.0 mCi)e -(5.58 10
h -1 )(30.0 h)
= 1.87 mCi
Our estimate of the half life was about 20% short because we did not account for the non-linearity of the decay rate. Consequently, our estimate of the final activity also fell short, but both of these calculated results are close enough to be reasonable. The number of atoms is much less than one mole, so this appears to be a very small sample. To get a sense of how small, we can assume that the molar mass is about 100 g/mol, so the sample has a mass of only m 2.4 1013 atoms (100 g / mol ) 6.02 10 23 atoms / mol 0.004 g
(
)
(
)
This sample is so small it cannot be measured by a commercial mass balance! The problem states that this sample was "freshly prepared," from which we assumed that all the atoms within the sample are initially radioactive. Generally this is not true, so that N 0 only accounts for the formerly radioactive atoms, and does not include additional atoms in the sample that were not radioactive. Realistically then, the sample mass should be significantly greater than our above estimate.
Chapter 44 Solutions 575 ln 2 = 26.0 h = 0.0266/h ln (0.100) = t
44.28
R = R0 e- t R = 0.100 = e - t R0 2.30 = 0.0266 t h
where so
t = 86.4 h
44.29
The number of nuclei which decay during the interval will be First we find :
N1 - N 2 = N 0 e - t 1 -e - t 2
=
ln 2 0.693 = = 0.0107 h -1 = 2.97 10 -6 s -1 T 1/2 64.8 h R0 = (40.0 Ci)(3.70 10 4 cps / Ci) = 4.98 1011 nuclei 2.97 10 -6 s -1
and
N0 =
Substituting these values,
-1 -1 N1 - N 2 = 4.98 1011 e -(0.0107 h )(10.0 h) - e -(0.0107 h )(12.0 h)
(
)
Hence, the number of nuclei decaying during the interval is
N1 - N 2 = 9.47 109 nuclei
44.30
The number of nuclei which decay during the interval will be First we find :
N1 - N 2 = N 0 e - t 1 - e - t 2
=
ln 2 T1/ 2 and N0 = -e
- t 2
so
e - t = e ln 2(-t/T1/2 ) = 2 -t/T1/2 R 0 T1/2 ln 2
R0
=
R 0 T1/2 ln 2
-t 1 /T 1/2
Substituting in these values
N1 - N 2 =
(e
- t 1
)=
R 0 T1/2 ln 2
(2
- 2 -t2 /T1/2
)
44.31
ln 2 1.00 g 23 R = N = 59.93 g / mol 6.02 10 5.27 yr
(
)
1 yr decays R = 1.32 10 21 = 4.18 1013 Bq 7 yr 3.16 10 s
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576 Chapter 44 Solutions
65 * 28 Ni 211 82 Pb 55 27 Co 0 -1 e 1 1H
44.32
(a) (b) (c) (d) (e)
(or p)
44.33
Q = M238
(
U
- M234
Th
- M4
He
)(931.5 MeV u)
Q = ( 238.050 784 - 234.043 593 - 4.002 602) u (931.5 MeV u ) = 4.27 MeV
44.34
0.0210 g NC = 6.02 10 23 molecules mol 12.0 g mol
(
) )
(N
( N0 ) 14 C = 1.37 109 ,
C
= 1.05 10 21 carbon atoms of which 1 in 7.70 1011 is a 14 C atom ln 2 = 1.21 10 -4 yr -1 = 3.83 10 -12 s -1 5730 yr
14 C =
R = N = N0 e - t At t = 0, At time t , 7(86400 s) 3 decays R 0 = N 0 = 3.83 10 -12 s -1 1.37 10 9 = 3.17 10 week 1 week
(
)(
)
R=
837 = 951 decays week 0.88 so t= -1 R ln R0
Taking logarithms,
ln t=
R = - t R0
-1 951 ln = 1.21 10 -4 yr -1 3.17 10 3
9.96 10 3 yr
44.35
3 In the decay 3 H 2 He + 1
E = ( m)c 2 = M 3 H - M 3 He c 2 since 1 2 the antineutrino is massless and the mass of the electron is accounted for in the masses of 3 3 1 H and 2 He .
0 - 1e +
, the energy released is
Thus,
E = [ 3.016 049 u - 3.016 029 u] (931.5 MeV u ) = 0.0186 MeV = 18.6 keV
Chapter 44 Solutions 577
44.36
(a)
For e + decay, Q = ( MX - MY - 2me ) c 2 = [ 39.962 591 u - 39.964 000 u - 2 (0.0000 549 u )](931.5 MeV u ) Q = -2.34 MeV Since Q < 0, the decay cannot occur spontaneously.
(b)
For alpha decay, Q = ( MX - M - MY ) c 2 = [97.905 287 u - 4.002 602 u - 93.905 085 u](931.5 MeV u ) Q = -2.24 MeV Since Q < 0, the decay cannot occur spontaneously.
(c)
For alpha decay, Q = ( MX - M - MY ) c 2 = [143.910 082 u - 4.002 602 u - 139.905 434 u](931.5 MeV u ) Q = 1.91 MeV Since Q > 0, the decay can occur spontaneously.
44.37
(a) (b)
e- + p n + For nuclei, Add seven electrons to both sides to obtain
15
O + e - 15 N + . atom 15 N atom + . 7
15 8O
(c)
From Table A.3, m = 15.003 065 u 15.000 108 u = 0.002 957 u Q = (931.5 MeV/u)(0.002 957 u) = 2.75 MeV
m
( O) = m( N) + cQ
15 15
2
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578 Chapter 44 Solutions 44.38 (a) Let N be the number of
238U
nuclei and N' be 206Pb nuclei. so N = ( N + N ) e - t or e t = 1 + N N
Then N = N 0 e - t and N 0 = N + N Taking logarithms,
N t = ln 1 + N
where = (ln 2) / T1/2 .
Thus, N = 1.164 N
T1 2 N t= ln 1 + N ln 2 for the
238
If
U 206 Pb chain with T1
2
= 4.47 10 9 yr , the age is:
4.47 10 9 yr 1 9 t= ln 1 + 1.164 = 4.00 10 yr ln 2 (b) From above, e t = 1 + N . N
2
Solving for
N N
gives
235
e- t N = N 1 - e- t
With t = 4.00 10 9 yr and T1
= 7.04 108 yr for the
U 207 Pb chain,
ln 2 (ln 2) 4.00 109 yr t = = 3.938 and t = 7.04 108 yr T1 2 With t = 4.00 10 9 yr and T1
2
(
)
N = 0.0199 N
= 1.41 1010 yr for the N = 4.60 N
232
Th 208 Pb chain,
t =
(ln 2)( 4.00 109 yr)
1.41 1010 yr
= 0.1966 and
44.39
Chapter 44 Solutions 579 4.00 10 -12 Ci 3.70 1010 Bq 1.00 10 3 L 3 4.00 pCi L = = 148 Bq m 1L 1 Ci 1 m3 N= T1 2 Bq 3.82 d 86 400 s R = R = 148 3 = 7.05 10 7 atoms m 3 ln 2 1 d ln 2 m
*44.40 (a)
(b)
(c)
g atoms 1 mol 222 g mass = 7.05 10 7 = 2.60 10 -14 m 3 6.02 10 23 atoms 1 mol m3 Since air has a density of 1.20 kg m 3 , the fraction consisting of radon is fraction = 2.60 10 -14 g m 3 = 2 .17 10 -17 1.20 kg m 3
*44.41
Number remaining: Fraction remaining: (a) (b) (c) With T1 = 3.82 d and t = 7.00 d,
N = N0 e
- ( ln 2 )t T 1 2
N - ( ln 2 )t T 1 2 = e- t = e N0 N - ln 2 7.00 3.82 = e ( )( ) ( ) = 0.281 N0 N - ln 2 365.25 ) ( 3.82 ) = e ( )( = 1.65 10 - 29 N0 as one daughter in the series of decays starting from the
2
When t = 1.00 yr = 365.25 d, Radon is continuously created long-lived isotope
238
U.
44.42
Q = M 27 Al + M - M 30 P - mn c 2 Q = [ 26.981 538 + 4.002 602 - 29.978 307 - 1.008 665] u (931.5 MeV u ) = - 2.64 MeV
[
]
44.43
(a) (b)
197 79 Au
1 + 0 n 197 Au * 198 Hg + 79 80
0 - 1e + 197 79 Au 1 atom + 0 n 198 80 Hg
Consider adding 79 electrons: Q = M 197 Au + mn - M 198 Hg c 2
atom + + Q
Q = [196.966 543 + 1.008 665 - 197.966 743] u (931.5 MeV u ) = 7.89 MeV
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580 Chapter 44 Solutions
*44.44 (a) (b) (c)
For X , A = 24 + 1 - 4 = 21 and Z = 12 + 0 - 2 = 10, so X is
21 10 Ne 144 54 Xe
A = 235 + 1 - 90 - 2 = 144 and Z = 92 + 0 - 38 - 0 = 54, so X is A = 2 - 2 = 0 and Z = 2 - 1 = + 1, so X must be a positron. As it is ejected, so is a neutrino: X = 0 e+ 1 and
0 X = 0
*44.45
Neglect recoil of product nucleus, (i.e., do not require momentum conservation). The energy balance gives Kemerging = Kincident + Q. To find Q : Q = ( MH + MAl ) - ( MSi + mn ) c 2 Q = (1.007 825 + 26.981 528) - ( 26.986 721 + 1.008 665) u (931.5 MeV u ) = - 5.61 MeV Thus, Kemerging
=
[
]
[
]
6.61 MeV - 5.61 MeV = 1.00 MeV
*44.46 (a)
10 5B
+ 4 He 13 C + 1H 2 6 1
13 6C
The product nucleus is (b)
13 6C
+ 1H 10 B + 4 He 1 5 2
10 5B
The product nucleus is
44.47
9 4 Be + 1.666
8 1 MeV 4 Be + 0 n , so M 8 Be = M 9 Be -
4 4
Q - mn c2 8.005 3 u Q c2
M 8 Be = 9.012 174 u -
4
(-1.666 MeV) - 1.008 665 u =
931.5 MeV u
9 4 Be
1 + 0n
10 4 Be
+ 6.810 MeV , so M 10 Be = M 9 Be + mn -
4 4
M 10 Be = 9.012 174 u + 1.008 665 u -
4
6.810 MeV = 10.013 5 u 931.5 MeV u
Chapter 44 Solutions 581 Goal Solution Using the Q values of appropriate reactions and from Table 44.5, calculate the masses of 8 Be and atomic mass units to four decimal places. G:
10
Be i n
The mass of each isotope in atomic mass units will be approximately the number of nucleons (8 or 10), also called the mass number. The electrons are much less massive and contribute only about 0.03% to the total mass. In addition to summing the mass of the subatomic particles, the net mass of the isotopes must account for the binding energy that holds the atom together. Table 44.5 includes the energy released for each nuclear reaction. Precise atomic masses values are found in Table A.3. The notation means Therefore
9 9
O:
A:
Be ( , n ) 8 Be
with Q = -1.666 MeV
Be + 8 Be + n - 1.666 MeV
m 8 Be = m 9Be - mn +
( ) ( ) ( )
10
1.666 MeV 931.5 MeV / u
m 8 Be = 9.012174 - 1.008665 + 0.001789 = 8.0053 u The notation means
9 9
Be (n, ) Be + n
Be
with Q = 6.810 MeV
10
Be + + 6.810 MeV 6.810 MeV 931.5 MeV / u
m m L:
(
10
Be = m 9Be + mn +
) ( )
(
10
Be = 9.012174 + 1.008665 - 0.001789 = 10.0135 u
)
As expected, both isotopes have masses slightly greater than their mass numbers. We were asked to calculate the masses to four decimal places, but with the available data, the results could be reported accurately to as many as six decimal places.
44.48
236 90 92 U 37 Rb
+
143 55 Cs
1 + 30 n ,
so
Q = M 236 U - M 90 Rb - M 143 Cs - 3mn c 2 37 55 92
From Table A.3, Q = 236.045 562 - 89.914 811 - 142.927 220 - 3(1.008 665) u (931.5 MeV u ) = 165 MeV
[
]
44.49
N 0 - N 0 e - T h /2 1 - 2 -1/ 2 1 - e -ln 2/2 N1 = = -ln 2/2 = -1/ 2 = -ln 2 - T h /2 - Th N2 N0 e e -e 2 - 2 -1 - N0 e
2
2000 by Harcourt, Inc. All rights reserved.
582 Chapter 44 Solutions
1 1H 1 + 7 Li 7 Be + 0 n 3 4
*44.50
Q = [(MH + MLi ) - (MBe + Mn )](931.5 MeV / u) Q = [(1.007 825 u + 7.016 003 u) - (7.016 928 u + 1.008 665 u)](931.5 MeV / u) Q = (-1.765 10 -3 u)(931.5 MeV / u) = -1.644 MeV Thus, m incident projectile 1.007 825 KE min = 1 + Q = 1 + 7.016 003 (1.644 MeV) = 1.88 MeV m target nucleus
44.51
(a)
N0 =
1.00 kg mass = = 2.52 10 24 mass per atom ( 239.05 u ) 1.66 10 -27 kg u
(
)
(b)
=
ln 2 ln 2 = = 9.106 10 -13 s -1 4 T1/ 2 2.412 10 yr 3.156 107 s yr
(
)(
)
R0 = N 0 = 9.106 10 -13 s -1 2.52 10 24 = 2.29 1012 Bq (c) R = R 0 e - t , so t = -1 R 1 R 0 ln = ln R0 R
(
)(
)
t=
2.29 1012 Bq 1 yr 1 ln = 3.38 1013 s = 1.07 106 yr -13 -1 3.156 107 s 0.100 Bq 9.106 10 s
44.52
(a)
57 57 27 Co 26 Fe
+
0 +1e
0 + 0
The Q - value for this positron emission is
Q = M 57 Co - M 57 Fe - 2me c 2
[
]
Q = 56.936 294 - 56.935 396 - 2 (0.000 549) u (931.5 MeV u) = - 0.186 MeV Since Q < 0, this reaction cannot spontaneously occur . (b)
14 14 6 C 7N
[
]
+
0 - 1e
0 + 0
The Q - value for this e - decay is Q = M 14 C - M 14 N c 2 . Q = [14.003 242 - 14.003 074] u (931.5 MeV u ) = 0.156 MeV = 156 keV Since Q > 0, the decay can spontaneously occur . (c) The energy released in the reaction of (b) is shared by the electron and neutrino. K e can range from zero to 156 keV . Thus,
[
]
Chapter 44 Solutions 583 r = r 0 A1/3 = 1.20 10 -15 A1/3 m . F=
44.53
(a)
When A = 12,
r = 2.75 10 15 m
(b)
ke (Z - 1)e 2 (8.99 10 9 N m 2 / C 2 )(Z - 1)(1.60 10 -19 C)2 = r2 r2
When Z = 6 and r = 2.75 10 15 m, F = 152 N U= ke q1q2 ke (Z - 1)e 2 (8.99 10 9 )(1.6 10 -19 )2 (Z - 1) = = r r r
(c)
When Z = 6 and r = 2.75 10 15 m, U = 4.19 10 13 J = 2.62 MeV (d) A = 238; and Z = 92, r = 7.44 10 15 m F = 379 N
U = 2.82 10 12 J = 17.6 MeV
44.54
(a)
9.00
ln(counts/min) vs time
6.00
ln(cpm)
3.00
0.00 0.00
4.00
8.00
12.00
t (h)
A least-square fit to the graph yields: and (b)
= - slope = - - 0.250 h -1 = 0.250 h -1,
ln (cpm) t = 0 = intercept = 8.30
(
)
1h = 4.17 10 - 3 min -1 = 0.250 h -1 60.0 min T1
2
=
ln 2 ln 2 = = 166 min = 2.77 h 4.17 10 -3 min -1
(c)
From (a), intercept = ln (cpm)0 = 8.30. Thus,
(cpm)0 = e 8.30
R0
counts min = 4.02 10 3 counts min
(d)
N0 =
=
4.02 10 3 counts min 1 (cpm )0 = = 9.65 106 atoms -3 -1 Eff 4.17 10 min (0.100)
(
)
2000 by Harcourt, Inc. All rights reserved.
584 Chapter 44 Solutions Because the reaction p n + e+ + would violate the law of conservation of energy , mp = 1.007 276 u (b) (c) mn = 1.008 665 u me + = 5.49 10 -4 u Note that mn + me + > mp
44.55
(a)
The required energy can come from the electrostatic repulsion of protons in the nucleus. Add seven electrons to both sides of the reaction for nuclei to obtain the reaction for neutral atoms Q = c2 m
13 13 e+ e- 13 7N 13 13 7 N 6 C
+ e+ +
atom 13 C atom + e + + e - + 6
[ ( N) - m( C) - m - m - m ] Q = (931.5 MeV / u)[13.005 738 - 13.003 355 - 2 ( 5.49 10 ) - 0] u
-4
Q = (931.5 MeV/u)(1.285 10 3 u) = 1.20 MeV
44.56
(a)
If we assume all the t=
87
Sr came from
87
Rb, then N = N 0 e - t yields N = N 87 Rr and N 0 = N 87 Sr + N 87 Rb .
-1 N T1 2 N 0 , ln ln = N 0 ln 2 N
where
10
t= (b)
(4.75 10
10
yr
ln 2
) ln 1.82 10
+ 1.07 10 9 9 = 3.91 10 yr 1.82 1010
87
It could be no longer . The rock could be younger if some
Sr were originally present.
44.57
(a)
Let us assume that the parent nucleus (mass M p) is initially at rest, and let us denote the masses of the daughter nucleus and alpha particle by M d and M , respectively. Applying the equations of conservation of momentum and energy for the alpha decay process gives Md vd = M v
2 2 1 1 Mp c 2 = Md c 2 + M c 2 + 2 M v + 2 Md vd
(1) (2)
2 2 1 1 Q = ( Mp - Md - M )c 2 = 2 M v + 2 Md vd
The disintegration energy Q is given by
(3)
Eliminating vd from Equations (1) and (3) gives
2 M M M 1 M 2 1 2 2 2 1 1 1 Q = 2 Md v + 2 M v = v + 2 M v = 2 M v 1 + = K 1 + 2 Md Md Md Md 2
(b)
K =
Q 4.87 MeV = = 4.78 MeV 1 + ( M Md ) 1 + ( 4 / 222)
Chapter 44 Solutions 585
145 141 61 Pm 59 Pr
44.58
(a) (b) (c)
The reaction is
+
Q = (MPm - M - MPr )931.5 = (144.912 745 - 4.002 602 - 140.907 647)931.5 = 2.32 MeV The alpha and daughter have equal and opposite momenta E =
2 p 2m
p = pd
Ed =
2 pd 2md
1 md 141 E 2m E = = 97.2% or 2.26 MeV = = 2 = = 2 1 1 E tot E + Ed md + m 141 + 4 p p + + 2ma 2md 2m 2md This is carried away by the alpha
2 p 2m
44.59
(a)
If E is the energy difference between the excited and ground states of the nucleus of mass M, and h f is the energy of the emitted photon, conservation of energy gives E = h f + E r Where Er is the recoil energy of the nucleus, which can be expressed as Er
=
(1)
Mv 2 (Mv)2 = 2 2M
(2)
Since momentum must also be conserved, we have hf Mv = c Hence, Er can be expressed as Er = (h f )2 . 2Mc 2 Er (E)2 2Mc 2 (3)
When h f << Mc 2, we can make the approximation that h f E, so (E)2 2Mc 2
(b)
Er =
where E = 0.0144 MeV and
Mc 2 = (57 u)(931.5 MeV/u) = 5.31 10 4 MeV
Therefore,
Er =
(1.44 10 -2 MeV)2 = 1.94 10 3 eV (2)(5.31 10 4 MeV)
2000 by Harcourt, Inc. All rights reserved.
586 Chapter 44 Solutions
40
*44.60 (a)
One liter of milk contains this many
K nuclei:
6.02 10 23 nuclei mol 0.0117 18 N = ( 2.00 g ) 100 = 3.60 10 nuclei 39.1 g mol
=
1 yr ln 2 ln 2 = = 1.72 10 -17 s -1 T1 2 1.28 10 9 yr 3.156 107 s
R = N = 1.72 10 -17 s -1 3.60 1018 = 61.8 Bq (b) For the iodine, R = R 0 e - t with = t= ln 2 . 8.04 d
(
)(
)
1 R 0 8.04 d 2000 = 40.3 d ln = ln 61.8 ln 2 R
*44.61 (a)
For cobalt-56, =
ln 2 ln 2 365.25 d = = 3.28 yr -1 . T1 2 77.1 d 1 yr
- ( 3.28 yr -1 ) ( 946 yr ) R - ln 10 )1349 = e- t = e = e -3106 = e ( = ~ 10 -1349 R0
The elapsed time from July 1054 to July 2000 is 946 yr. R = R 0 e - t implies (b) For carbon-14,
- (1.21 10 - 4 R = e- t = e R0
=
yr -1 ( 946 yr )
ln 2 = 1.21 10 -4 yr -1 5730 yr
)
= e -0.114 = 0.892
*44.62
We have N 235 = N 0, 235 e
- 235 t
and
N 238 = N 0, 238 e
- 238 t T h, 235 + ( ln 2 )t T h, 238
N 235 ( -(ln 2)t = 0.00725 = e N 238 Taking logarithms,
)
ln 2 ln 2 - 4.93 = - + t 9 9 0.704 10 yr 4.47 10 yr 1 1 - 4.93 = - + (ln 2)t 9 9 0.704 10 yr 4.47 10 yr t=
or
(- 1.20 10
- 4.93
-9
yr
-1
) ln 2
= 5.94 10 9 yr
Chapter 44 Solutions 587 44.63 (a) Add two electrons to both sides of the 4 1H atom 4 He atom + Q 1 2 Q = mc 2 = 4 M 1H - M 4 He c 2
1 2
reaction
to have
it in
energy
terms:
[
]
1.60 10 -13 J = 4.28 10 -12 J Q = 4 (1.007 825 u ) - 4.002 602 u (931.5 MeV u ) 1 MeV
[
]
(b)
N=
1.99 10 30 kg = 1.67 10 -27 kg atom
1.19 10 57 atoms = 1.19 10 57 protons
(c)
The energy that could be created by this many protons in this reaction is:
(1.19 10
P= E t
57
4.28 10 -12 J 45 protons = 1.27 10 J 4 protons
)
so
t=
E 1.27 10 45 J = = 3.38 1018 s = 107 billion years P 3.77 10 26 W
44.64
(a)
Q = M 9 Be + M 4 He - M 12 C - mn c 2 Q = [9.012 174 u + 4.002 602 u - 12.000 000 u - 1.008 665 u] (931.5 MeV u ) = 5.69 MeV
[
]
(b)
Q = 2 M 2 H - M 3 He - mn
[
]
]
Q = 2 (2.014 102) - 3.016 029 - 1.008 665 u (931.5 MeV u) = 3.27 MeV (exothermic)
[
44.65
E = B so the energies are E 1 = +B and E 2 = B = 2.7928n and n = 5.05 10 27 J/T
E = 2B = 2 (2.7928)(5.05 10 27 J/T)(12.5 T) = 3.53 10 25 J = 2.20 10 6 eV
2000 by Harcourt, Inc. All rights reserved.
588 Chapter 44 Solutions 1 yr ln 2 ln 2 = = 4.17 10 -9 s -1 T1 2 5.27 yr 3.156 107 s s 7 = 7.89 10 s
44.66
(a)
=
3.156 107 t = 30.0 months = ( 2.50 yr ) 1 yr R = R 0 e - t = ( N 0 )e - t
10 R t (10.0 Ci ) 3.70 10 Bq Ci ( 4.17 10 - 9 e e = so N 0 = 4.17 10 -9 s -1
(
)
s -1 7.89 10 7 s
)(
)
N 0 = 1.23 10 20 nuclei 59.93 g mol -2 Mass = 1.23 10 20 atoms = 1.23 10 g = 12.3 mg 23 6.02 10 atoms mol
(
)
(b)
We suppose that each decaying nucleus promptly puts out both a beta particle and two gamma rays, for Q = (0.310 + 1.17 + 1.33) MeV = 2.81 Mev P = QR = ( 2.81 Mev ) 1.6 10 -13 J MeV 3.70 1011 s -1 = 0.166 W
(
)(
)
44.67
Ze For an electric charge density = 4 R3 3 Using Gauss's Law inside the sphere,
r Ze E 4 r 2 = 3 : 4 R3 e0 3
4
3
E=
1 Zer 4 e0 R 3 1 Ze 4 e0 r 2
(r R)
E=
(r R)
We now find the electrostatic energy: 1 1e R 2 0 0 4 e
2
U=
r=0 2 2
1e
0E
2
4 r 2 dr
U=
Z2 e 2r 2 Z2e 2 R 5 1 1 Z2e 2 3 Z2e 2 1 4 r 2 dr + 2 e0 4 r 2 dr = 6 + = 6 4 R 4 e R 20 e0 R 8 e0 5R R r 0 0
Chapter 44 Solutions 589
93 43Tc
44.68
(a)
For the electron capture, The disintegration energy is
+
0 93 -1e 42 Mo +
Q = M 93 Tc - M 93 Mo c 2 . Q = [92.910 2 - 92.906 8] u (931.5 MeV u) = 3.17 MeV > 2.44 MeV
[
]
Electron capture is allowed For positron emission, The disintegration energy is
to all specified excited states
93 93 43Tc 42 Mo
in
93 42 Mo.
+
0 +1e +
Q = M 93 Tc - M 93 Mo - 2 me c 2 .
[
]
Q = 92.910 2 - 92.906 8 - 2 (0.000 549) u (931.5 MeV u) = 2.14 MeV
[
]
Positron emission can reach the 1.35, 1.48, and 2.03 MeV states but there is insufficient energy to reach the 2.44 MeV state. (b) The daughter nucleus in both forms of decay is
93 42 Mo
.
44.69
1 K = 2 mv 2 ,
so
v=
2K = m
2 (0.0400 eV ) 1.60 10 -19 J eV 1.67 10
-27
(
kg
) = 2.77 10
3
ms
The time for the trip is
t=
x 1.00 10 4 m = = 3.61 s v 2.77 10 3 m s
The number of neutrons finishing the trip is given by N = N 0 e - t . The fraction decaying is 1 - N - ( ln 2 )t = 1- e N0
T1
2
- ln 2 3.61 s 624 s ) = 1 - e ( )( = 0.004 00 = 0.400%
2000 by Harcourt, Inc. All rights reserved.
590 Chapter 44 Solutions 44.70 (a) At threshold, the particles have no kinetic energy relative to each other. That is, they move like two particles which have suffered a perfectly inelastic collision. Therefore, in order to calculate the reaction threshold energy, we can use the results of a perfectly inelastic collision. Initially, the projectile Ma moves with velocity va while the target MX is at rest. We have M a v a = ( M a + M X ) vc from momentum conservation: The initial energy is: The final kinetic energy is:
2 1 Ei = 2 Ma va
Ma va Ma 2 1 1 E f = 2 ( M a + M X ) vc = 2 ( M a + M X ) = M + M Ei X ) M a + MX ( a
2
From this, we see that E f is always less than E i and the loss in energy, E i - E f , is given by MX Ma Ei = E i - E f = 1 - Ei M a + MX M a + MX In this problem, the energy loss is the disintegration energy -Q and the initial energy is the threshold energy E th . Therefore, MX -Q = E th M a + MX (b) M + Ma Ma or E th = -Q X = -Q 1 + M X MX
First, calculate the Q - value for the reaction: Q = M 14 N + M 4 He - M 17 O - M 1H c 2 Q = [14.003 074 + 4.002 602 - 16.999 132 - 1.007 825] u (931.5 MeV u ) = -1.19 MeV Then, M + Ma 4.002 602 = - ( -1.19 MeV ) 1 + Eth = - Q X = 1.53 MeV MX 14.003 074
[
]
44.71
R= R 0 exp ( t) ln R = ln R 0 t slope = The logarithmic plot shown in Figure P44.71 is fitted by ln R = 8.44 0.262t. If t is measured in minutes, then the decay constant is 0.262 per minute. The half-life is T1
2
(the equation of a straight line)
=
ln 2 ln 2 = = 2.64 min 0.262 /min The
The reported half-life of 137Ba is 2.55 min. difference reflects experimental uncertainties.
Chapter 45 Solutions
*45.1 m = (m n + M U ) (M Zr + M Te + 3m n ) m = (1.008 665 u + 235.043 924 u) (97.912 0 u + 134.908 7 u + 3(1.008 665 u)) m = 0.205 89 u = 3.418 1028 kg so Q = mc2 = 3.076 1011 J = 192 MeV
45.2
Three different fission reactions are possible:
1 0n
1 0n 1 0n
+ +
235 90 92 U 38 Sr 235 90 92 U 38 Sr
+ +
144 54 Xe 142 54 Xe
+ 2 1n 0
1 + 4 0n
144 54 Xe 142 54 Xe
+
235 90 92 U 38 Sr
+
143 54 Xe
1 + 3 0n
143 54 Xe
45.3
1 0n
+
232 233 233 90Th 90Th 91Pa
+ e- +
233 233 91Pa 92 U
+ e- +
45.4
1 0n
+
238 239 239 92 U 92 U 93 Np
+ e- +
239 239 93 Np 94 Pu
+ e- +
45.5
(a)
Q = ( m)c 2 = [mn + MU235 - MBa141 - MKr92 - 3mn ] c 2 m = (1.008 665 + 235.043 924) - (140.913 9 + 91.897 3 + 3 1.008 665) u = 0.215 39 u Q = (0.215 39 u ) (931.5 MeV u ) = 201 MeV
[
]
(b)
f=
0.215 39 u m = = 9.13 10 -4 = 0.0913% mi 236.052 59 u
45.6
If the electrical power output of 1000 MW is 40.0% of the power derived from fission reactions, the power output of the fission process is 1000 MW J 8.64 10 4 14 = 2.50 10 9 = 2.16 10 J/d 0.400 s d The number of fissions per day is This also is the number of
235
J 1 fission 1 eV 24 -1 2.16 1014 = 6.74 10 d d 200 106 eV 1.60 10 -19 J
235
U nuclei used, so the mass of
U used per day is
235 g/mol nuclei 6.74 10 24 = 2.63 10 3 g/d = 2.63 kg/d 6.02 10 23 nuclei/mol d In contrast, a coal-burning steam plant producing the same electrical power uses more than 6 106 kg/d of coal.
2000 by Harcourt, Inc. All rights reserved.
Chapter 45 Solutions 593 45.7 The available energy to do work is 0.200 times the energy content of the fuel.
(1.00 kg fuel)
12
0.0340
U 1000 g 1 mol 6.02 10 23 (208)(1.60 10 -13 J) 1 kg 235 g fuel mol fission
235 11
(2.90 10 J)(0.200) = 5.80 10
d = 5.80 106 m = 5.80 Mm
J = 1.00 10 5 N d
(
)
Goal Solution Suppose enriched uranium containing 3.40% of the fissionable isotope
5
235 92 U
is used as fuel for a ship. The
water exerts an average frictional drag of 1.00 10 N on the ship. How far can the ship travel per kilogram of fuel? Assume that the energy released per fission event is 208 MeV and that the ship's engine has an efficiency of 20.0%. G: Nuclear fission is much more efficient for converting mass to energy than burning fossil fuels. However, without knowing the rate of diesel fuel consumption for a comparable ship, it is difficult to estimate the nuclear fuel rate. It seems plausible that a ship could cross the Atlantic ocean with only a few kilograms of nuclear fuel, so a reasonable range of uranium fuel consumption might be 10 km / kg to 10 000 km/kg. The fuel consumption rate can be found from the energy released by the nuclear fuel and the work required to push the ship through the water. One kg of enriched uranium contains 3.40%
235 92 U
O:
A:
so
m235 = (1000 g )(0.0340) = 34.0 g
In terms of number of nuclei, this is equivalent to 1 23 22 N 235 = ( 34.0 g ) 6.02 10 atoms / mol = 8.71 10 nuclei 235 g / mol
(
)
If all these nuclei fission, the thermal energy released is equal to
(8.71 10
22
MeV 1.602 10 -19 J eV = 2.90 1012 J nuclei 208 nucleus efficiency = work output heat input or e= fd cos Qh
)
(
)
Now, for the engine,
So the distance the ship can travel per kilogram of uranium fuel is d= L: 0.200 2.90 1012 J eQh = = 5.80 106 m f cos(0) 1.00 10 5 N
(
)
The ship can travel 5 800 km/kg of uranium fuel, which is on the high end of our prediction range. The distance between New York and Paris is 5 851 km, so this ship could cross the Atlantic ocean on just one kilogram of uranium fuel.
2000 by Harcourt, Inc. All rights reserved.
594 Chapter 45 Solutions
1 3
45.8
(a)
For a sphere:
V = 4 r 3 3
and
r=
3V 4
3
so
4 r 2 A = = 4.84V 1/3 V (4 / 3) r 3 A 6l2 = = 6V 1/3 V l3 2a2 + 8a2 A = = 6.30V 1/3 V 2a3
(b)
For a cube:
V = l3
and
l = V1
so
(c)
For a parallelepiped:
V = 2a
3
and
V a= 2
1/3
so
(
)
(d) Therefore, the sphere has the least leakage and the parallelepiped has the greatest leakage for a given volume.
45.9
mass of
235
106 g 12 U available (0.007 ) 10 9 metric tons = 7 10 g 1 metric ton
(
)
7 1012 g nuclei = 1.8 10 34 nuclei number of nuclei ~ 6.02 10 23 mol 235 g mol The energy available from fission (at 208 MeV/event) is E ~ 1.8 10 34 events ( 208 MeV / event ) 1.60 10 -13 J / MeV = 6.0 10 23 J This would last for a time of t= 1 yr E 6.0 10 23 J ~ = 8.6 1010 s ~ 3000 yr 3.16 107 s P 7.0 1012 J s
(
)
(
)
(
)
45.10
60.0 s In one minute there are 1.20 ms = 5.00 104 fissions. So the rate increases by a factor of (1.000 25)50000 = 2.68 105
45.11
P = 10.0 MW = 1.00 107 J/s If each decay delivers 1.00 MeV = 1.60 1013 J, then the number of decays/s = 6.25 1019 Bq
Chapter 45 Solutions 595 45.12 (a) The Q value for the D-T reaction is 17.59 MeV. Heat content in fuel for D-T reaction: r DT = (b) (17.59 MeV)(1.60 10 -13 J/MeV) = 3.39 1014 J/kg ( 5 u)(1.66 10 -27 kg/u)
(3.00 10 9 J / s)(3600 s / hr) = 31.9 g/h burning of D and T (3.39 1014 J / kg)(10 -3 kg / g)
1
Heat content in fuel for D-D reaction: Q = 2 (3.27 + 4.03) = 3.65 MeV average of two Q values ( 3.65 MeV)(1.60 10 -13 J/MeV) = 8.80 1013 J/kg -27 ( 4 u)(1.66 10 kg/u) r DD = (3.00 10 9 J / s)(3600 s / hr) = 122 g/h burning of D (8.80 1012 J / kg)(10 -3 kg / g)
45.13
(a)
At closest approach, the electrostatic potential energy equals the total energy E. k ( Z e )( Z2 e ) Uf = e 1 = E: rmin
(8.99 10 E=
9
N m 2 C 2 1.6 10 -19 C Z1Z2 1.00 10 -14 m
)(
)
2
=
(2.30 10
-14
J Z1Z2
)
(b)
For both the D-D and the D-T reactions, Z1 = Z2 = 1. Thus, the minimum energy required i n both cases is 1 MeV E = 2.30 10 -14 J = 0.144 MeV 1.60 10 -13 J
(
)
45.14
(a)
r f = r D + r T = 1.20 10 -15 m
(
)[(2)
13
+ ( 3)
13
]=
3.24 10 -15 m
(b)
8.99 10 9 N m 2 C 2 1.60 10 -19 C k e2 Uf = e = rf 3.24 10 -15 m
(
)(
)
2
= 7.10 10 -14 J= 444 keV
(c)
mD 2 Conserving momentum, mD vi = ( mD + mT )v f , or v f = vi = 5 vi mD + mT Ki + U i = K f + U f : mD 2 1 1 Ki + 0 = 2 ( mD + mT )v 2 + U f = 2 ( mD + mT ) vi + U f f mD + mT mD 1 mD 2 Ki + 0 = 2 mD vi + U f = Ki + U f mD + mT mD + mT mD 1 - Ki = U f : mD + mT m + mT 5 Ki = U f D = ( 444 keV ) = 740 keV mT 3
2
(d)
(
)
(e)
Possibly by tunneling.
2000 by Harcourt, Inc. All rights reserved.
596 Chapter 45 Solutions
3 1 Average KE per particle is 2 kBT = 2 mv 2 .
45.15
(a)
Therefore,
vrms =
3kBT = m
3 1.38 10 -23 J / K 4.00 108 K 2 1.67 10
(
(
-27
)(
kg
)
)=
2.23 106 m/s
(b)
t=
x 0.1 m ~ 6 v 10 m / s
~ 107 s
45.16
(a)
V = 317 106 mi 3
(
) 1609 m 1 mi )(
3
= 1.32 1018 m 3
mwater = V = 10 3 kg m 3 1.32 1018 m 3 = 1.32 10 21 kg MH 2.016 2 mH 2 O = mH 2 = 1.32 10 21 kg = 1.48 10 20 kg 18.015 MH 2 O
(
)
(
)
mDeuterium = (0.0300%) mH 2 = 0.0300 10 -2 1.48 10 20 kg = 4.43 1016 kg The number of deuterium nuclei in this mass is N= 4.43 1016 kg mDeuterium = = 1.33 10 43 -27 mDeuteron (2.014 u) 1.66 10 kg u
(
)(
)
(
)
Since two deuterium nuclei are used per fusion, N 2 = 6.63 10 42 . The energy released per event is
2 1H
+ 2 H 4 He + Q , the number of events is 1 2
Q = M 2 H + M 2 H - M 4 He c 2 = 2 ( 2.014 102) - 4.002 602 u (931.5 MeV u ) = 23.8 MeV The total energy available is then E= (b) 1.60 10 -13 N Q = 6.63 10 42 ( 23.8 MeV ) 2 1 MeV
[
] [
]
(
)
J 31 = 2 .52 10 J
The time this energy could possibly meet world requirements is t= 1 yr 2.52 10 31 J E = = 3.61 1016 s = 1.14 10 9 yr ~ 1 billion years. 12 3.16 107 s P 100 7.00 10 J s
(
) (
)
Chapter 45 Solutions 597 45.17 (a) Including both ions and electrons, the number of particles in the plasma is N = 2nV where n is the ion density and V is the volume of the container. Application of Equation 21.6 gives the total energy as 3 E = 2 NkBT = 3nVkBT = 3 2.0 1013 cm -3 50 m 3
(
)(
cm ) 101 m (1.38 10
6 3
3
-23
J K 4.0 108 K
)(
)
E = 1.7 107 J (b) From Table 20.2, the heat of vaporization of water is L v = 2.26 106 J kg . The mass of water that could be boiled away is m= E 1.7 107 J = = 7.3 kg L v 2.26 106 J kg
45.18
(a)
Lawson's criterion for the D-T reaction is n 1014 s cm 3 .
For a confinement time of
-3
= 1.00 s , this requires a minimum ion density of n = 10
(b)
14
cm
At the ignition temperature of T = 4.5 107 K and the ion density found above, the plasma pressure is 106 cm 3 -23 P = 2nkBT = 2 1014 cm -3 J K 4.5 107 K = 1.24 10 5 J m 3 1.38 10 1 m3
(
)
(
)(
)
(c)
The required magnetic energy density is then uB = B2 10 P = 10 1.24 10 5 J m 3 = 1.24 106 J m 3 , 2 0
(
)
B 2 4 10 -7 N A 2 1.24 106 J m 3 = 1.77 T
(
)(
)
45.19
Let the number of 6 Li atoms, each having mass 6.015 u, be N 6 while the number of atoms, each with mass 7.016 u, is N7 . Then, N 6 = 7.50% of N total = 0.0750( N 6 + N7 ) , or N7 = 0.925 N 0.0750 6
7
Li
Also, total mass = [ N 6 (6.015 u ) + N7 (7.016 u )] 1.66 10 -27 kg u = 2.00 kg , or This yields 0.925 N 6 (6.015 u ) + (7.016 u) 1.66 10 -27 kg u = 2.00 kg . 0.0750 N 6 = 1.30 10 25 N7 = as the number of 6 Li atoms and
(
)
(
)
0.925 1.30 10 25 = 1.61 10 26 0.0750
(
)
as the number of 7 Li atoms.
2000 by Harcourt, Inc. All rights reserved.
598 Chapter 45 Solutions 45.20 The number of nuclei in 1.00 metric ton of trash is N = 1000 kg (1000 g / kg ) 6.02 10 23 nuclei / mol At an average charge of 26.0 e/nucleus, Therefore
(
) (56.0 g / mol) = 1.08 10 nuclei q = (1.08 10 )( 26.0)(1.60 10 ) = 4.47 10
28 28 -19
10
C
t=
q 4.47 1010 = = 4.47 10 4 s = 12.4 h I 1.00 106
45.21
N0 =
mass present 5.00 kg = = 3.35 10 25 nuclei mass of nucleus (89.9077 u ) 1.66 10 -27 kg u
(
)
=
ln 2 ln 2 = = 2.38 10 -2 yr -1 = 4.52 10 -8 min -1 T1/2 29.1 yr
R0 = N 0 = 4.52 10 -8 min -1 3.35 10 25 = 1.52 1018 counts min 10.0 counts min R = e- t = = 6.60 10 -18 R0 1.52 1018 counts min giving t= and
(
)(
)
t = - ln 6.60 10 -18 = 39.6
(
)
39.6 39.6 = = 1.66 10 3 yr -2 -1 2.38 10 yr
45.22 (a)
Source: 100 mrad of 2-MeV - rays/h at a 1.00-m distance. For - rays, dose in rem = dose in rad. Thus a person would have to stand 10.0 hours source. (b) to receive 1.00 rem from a 100-mrad/h
If the - radiation is emitted isotropically, the dosage rate falls off as 1/ r 2 . Thus a dosage 10.0 mrad/h would be received at a distance r = 10.0 m = 3.16 m .
45.23
(a)
The number of x-rays taken per year is n = (8 x - ray d)( 5 d wk )( 50 wk yr ) = 2.0 10 3 x - ray yr The average dose per photograph is 5.0 rem yr = 2.5 10 -3 rem x - ray 3 2.0 10 x - ray yr
(b)
The technician receives low-level background radiation at a rate of 0.13 rem yr . The dose of 5.0 rem yr received as a result of the job is
Chapter 45 Solutions 599 5.0 rem yr = 38 times background levels 0.13 rem yr
2000 by Harcourt, Inc. All rights reserved.
600 Chapter 45 Solutions 1 I0 ln I 1 ln ( 2) = 0.436 cm 1.59 cm -1 1 ln 1.00 10 4 = 5.79 cm 1.59 cm -1
45.24
(a)
I = I0e - x ,
so
x= x= x=
With = 1.59 cm -1 , the thickness when I = I 0 2 is (b) When I0 = 1.00 10 4 , I
(
)
45.25
1 rad = 10 -2 J / kg t=
Q = mc T P t = mc T
mc T m( 4186 J / kgC)( 50.0 C) = = 2.09 106 s 24 days! P (10) 10 -2 J / kg s (m)
(
)
Note that power is the product of dose rate and mass.
45.26
10 -2 J / kg Q absorbed energy = = (1000 rad) = 10.0 J / kg m 1 rad unit mass The rise in body temperature is calculated from Q = mc T where c = 4186 J/kg C for water and the human body T = Q 1 = (10.0 J / kg) = 2.39 10 - 3 C mc 4186 J / kgC (Negligible)
45.27
If half of the 0.140-MeV gamma rays are absorbed by the patient, the total energy absorbed is E=
(0.140 MeV) 1.00 10 -8
2
g 6.02 10 23 nuclei 12 = 4.26 10 MeV 1 mol 98.9 g mol
E = 4.26 1012 MeV 1.60 10 -13 J MeV = 0.682 J Dose = 0.682 J 1 rad = 1.14 rad 60.0 kg 10 -2 J kg
(
)(
)
Thus, the dose received is
Chapter 45 Solutions 601 6.02 10 23 nuclei mol 12 The nuclei initially absorbed are N 0 = 1.00 10 -9 g = 6.70 10 89.9 g mol
45.28
(
)
The number of decays in time t is
N = N 0 - N = N 0 1 - e - t = N 0 1 - e 1.00 yr t = = 0.0344 T1 2 29.1 yr
(
)
- ( ln 2 ) t T 1
2
At the end of 1 year,
and The energy deposited is
N = N 0 - N = 6.70 1012 1 - e - 0.0238 = 1.58 1011 E = 1.58 1011 (1.10 MeV ) 1.60 10 -13 J MeV = 0.0277 J 0.0277 J -4 Dose = = 3.96 10 J kg 70.0 kg = 0.0396 rad
(
)(
)
(
)
(
)
Thus, the dose received is
45.29
(a)
1 5.00 10 -12 F 1.00 10 3 V 1 C(V)2 E 2 2 = = = 3.12 107 E 0.500 MeV (0.500 MeV ) 1.60 10 -13 J / MeV
(
(
)(
)
2
)
(b)
5.00 10 -12 F 1.00 10 3 V Q C(V) N= = = = 3.12 1010 electrons e e 1.60 10 -19 C
(
)(
)
45.30
(a)
amplification =
1 C( V ) energy discharged 2 C( V ) = = 2E E E
2
2
(b)
N=
C( V ) charge released = e charge of electron
45.31
(a)
EI = 10.0 eV is the energy required to liberate an electron from a dynode. Let ni be the number of electrons incident upon a dynode, each having gained energy e( V ) as it was accelerated to this dynode. The number of electrons that will be freed from this dynode is N i = ni e ( V ) EI : At the first dynode, ni = 1 and N1 = (1)e(100 V ) = 101 electrons 10.0 eV
2000 by Harcourt, Inc. All rights reserved.
602 Chapter 45 Solutions
(b)
For the second dynode,
ni = N1 = 101, ni = N 2 = 10 2
so
N2 = N3 =
(101 )e(100 V ) = 10 2 . 10.0 eV (10 2 )e(100 V ) = 10 3 . 10.0 eV
At the third dynode,
and
Observing the developing pattern, we see that the number of electrons incident on the seventh and last dynode is n7 = N 6 = 106 .
(c)
The number of electrons incident on the last dynode is n7 = 106 . electrons deliver to that dynode is given by E = ni e( V ) = 106 e (700 V - 600 V ) = 10 8 eV
The total energy these
*45.32 (a)
The average time between slams is 60 min 38 = 1.6 min . Sometimes, the actual interval is nearly zero. Perhaps about equally as often, it is 2 1.6 min . Perhaps about half as often, it is 4 1.6 min . Somewhere around 5 1.6 min = 8.0 min , the chances of randomness producing so long a wait get slim, so such a long wait might likely be due to mischief. R = R0 e - t .
(b)
The midpoints of the time intervals are separated by 5.00 minutes. Subtracting the background counts, 337 - 5 (15) = [372 - 5 (15)] e or ln
- ln 2 T1
We use
(
2
)( 5.00 min )
2
262 = ln (0.882) = - 3.47 min T1 297
which yields T1
2
= 27.6 min .
(c)
As in the random events in part (a), we imagine a 5 count counting uncertainty. smallest likely value for the half-life is then given by ln 262 - 5 = - 3.47 min T1 2 , or T1 297 + 5
The
(
2
)min = 21.1 min ( )max = 38.8 min
The largest credible value is found from ln = 262 + 5 = - 3.47 min T1 2 , yielding T1 297 - 5
2
Thus, T1
2
38.8 + 21.1 38.8 - 21.1 min = ( 30 9) min = 30 min 30% 2 2
Chapter 45 Solutions 603 45.33 The initial specific activity of (R / m)0 =
59Fe
in the steel,
20.0 Ci 100 Ci 3.70 10 4 Bq 6 = = 3.70 10 Bq / kg 0.200 kg kg 1 Ci
- 4 -1 R = (R / m)0 e - t = (3.70 106 Bq / kg)e -(6.40 10 h )(1000 h) = 1.95 106 Bq / kg m
After 1000 h,
The activity of the oil, Therefore,
Roil =
800 Bq / liter (6.50 liters) = 86.7 Bq 60.0 86.7 Bq Roil = = 4.45 10 -5 kg (R / m) 1.95 106 Bq / kg
min oil =
So that wear rate is
4.45 10 - 5 kg = 4.45 10 - 8 kg/h 1000 h
*45.34
The half-life of
14
14
O is 70.6 s, so the decay constant is =
ln 2 ln 2 = = 0.009 82 s -1 T1 2 70.6 s
The
O nuclei remaining after five min is
N = N 0 e - t = 1010 e
( )
- 0.009 82 s -1 ( 300 s )
(
)
= 5.26 108
The number of these in one cubic centimeter of blood is 1.00 cm 3 1.00 cm 3 5 = 5.26 108 N = N = 2.63 10 total vol. of blood 2000 cm 3
(
)
and their activity is R = N = 0.009 82 s -1 2.63 10 5 = 2.58 10 3 Bq ~ 10 3 Bq
(
)(
)
*45.35 (a)
The number of photons is 10 4 MeV 1.04 MeV = 9.62 10 3 . Since only 50% of the photons are detected, the number of 65 Cu nuclei decaying is twice this value, or 1.92 10 4 . In two half3 lives, three-fourths of the original nuclei decay, so 4 N 0 = 1.92 10 4 and N 0 = 2.56 10 4 . This is 1% of the
65
Cu , so the number of
65
Cu is 2.56 106 ~ 106 .
(b)
Natural copper is 69.17% 63 Cu and 30.83% 65 Cu . Thus, if the sample contains N Cu copper atoms, the number of atoms of each isotope is N 63 = 0.6917 N Cu and N 65 = 0.3083 N Cu . Therefore, N 63 0.6917 0.6917 0.6917 N = 2.56 106 = 5.75 106 = or N 63 = 0.3083 65 0.3083 N 65 0.3083 mCu = (62.93 u )N 63 + (64.93 u )N 65 :
(
)
The total mass of copper present is then
mCu = (62.93) 5.75 106 + (64.93) 2.56 106 u 1.66 10 -24 g u = 8.77 10 -16 g
[
(
)
(
)] (
)
~ 10 -15 g
45.36
(a)
Starting with N = 0 radioactive atoms at t = 0, the rate of increase is (production decay)
2000 by Harcourt, Inc. All rights reserved.
604 Chapter 45 Solutions
dN = R N dt The variables are separable.
so
dN = (R N)dt
N = 0
-
N
t dN = dt t=0 R-N
1 R - N ln =t R
so
R - N ln = -t R 1-
R - N - t =e R Therefore,
and
N = e- t R
R 1 - e- t
N=
(
)
(b)
The maximum number of radioactive nuclei would be
R/
45.37
(a)
At 6 108 K, each carbon nucleus has thermal energy of
3k T 2 B
= (1.5)(8.62 10 -5 eV / K)(6 108 K) = 8 104 eV
(b)
The energy released is
E = 2m C12 - m Ne 20 - m He 4 c 2 E = (24.000 000 19.992 435 4.002 602)(931.5) MeV = 4.62 MeV
[ ( ) (
) (
)]
In the second reaction,
E = 2m C12 - m Mg 24
[ ( ) (
)](931.5)MeV / u
E = (24.000 000 23.985 042)(931.5)MeV = 13.9 MeV
(c)
The energy released is the energy of reaction of the # of carbon nuclei in a 2.00-kg sample, which corresponds to 6.02 10 23 atoms / mol 4.62 MeV / fusion event 1 kWh E = 2.00 10 3 g 2 nuclei / fusion event 2.25 1019 MeV 12.0 g / mol
(
)
E =
(1.00 10 )(4.62) kWh = 2 ( 2.25 10 )
26 19
1.03 107 kWh
Chapter 45 Solutions 605 Suppose each 235 U fission releases 208 MeV of energy. Then, the number of nuclei that must have undergone fission is N= total release 5 1013 J = = 1.5 10 24 nuclei energy per nuclei ( 208 MeV ) 1.60 10 -13 J MeV
45.38
(a)
(
)
(b)
1.5 10 24 nuclei mass = (235 g mol) 0.6 kg 23 6.02 10 nuclei mol
45.39
For a typical loss is
235
U, Q = 208 MeV ; and the initial mass is 235 u. Thus, the fractional energy 208 MeV Q = = 9.50 10 -4 = 0.0950% 2 235 u ) (931.5 MeV u ) ( mc
For the D-T fusion reaction, The initial mass is The fractional loss in this reaction is 0.375% = 3.95 0.0950% .
Q = 17.6 MeV m = (2.014 u) + ( 3.016 u) = 5.03 u 17.6 MeV Q = = 3.75 10 -3 = 0.375% mc 2 ( 5.03 u ) (931.5 MeV u )
or
the fractional loss in D - T fusion is about 4 times that in 235 U fission
45.40
To conserve momentum, the two fragments must move in opposite directions with speeds v1 and v2 such that m m1v1 = m2 v2 or v2 = 1 v1 m2 The kinetic energies after the break-up are then
2 1 K1 = 2 m1v1
and
m 2 m 2 1 1 K2 = 2 m2 v2 = 2 m2 1 v1 = 1 K1 m2 m2 K1 K1 m2 = = K1 + K2 K1 + ( m1 m2 )K1 m1 + m2 1- m2 m1 = m1 + m2 m1 + m2
2
The fraction of the total kinetic energy carried off by m1 is
and the fraction carried off by m2 is
2000 by Harcourt, Inc. All rights reserved.
606 Chapter 45 Solutions ln 2 ln 2 = = 1.78 10 -9 s -1 T1 2 (12.3 yr ) 3.16 107 s yr
45.41
The decay constant is
=
(
)
The tritium in the plasma decays at a rate of 2.00 1014 106 cm 3 50.0 m 3 R = N = 1.78 10 -9 s -1 3 3 cm 1 m
(
)
(
)
1 Ci R = 1.78 1013 Bq = 1.78 1013 Bq = 482 Ci 10 3.70 10 Bq
(
)
The fission inventory is
4 1010 Ci ~ 108 times greater 482 Ci
than this amount.
Goal Solution The half-life of tritium is 12.3 yr. If the TFTR fusion reactor contained 50.0 m 3 of tritium at a density equal to 2.00 1014 ions / cm 3 , how many curies of tritium were in the plasma? Compare this value with a fission inventory (the estimated supply of fissionable material) of 4 1010 Ci . G: It is difficult to estimate the activity of the tritium in the fusion reactor without actually calculating it; however, we might expect it to be a small fraction of the fission (not fusion) inventory.
3 The decay rate (activity) can be found by multiplying the decay constant by the number of 1 H particles. The decay constant can be found from the half-life of tritium, and the number of particles from the density and volume of the plasma.
O:
A:
The number of Hydrogen-3 nuclei is
3 particles cm N = 50.0 m 3 2.00 1014 = 1.00 10 22 particles 100 m m3
(
)
=
1 yr ln 2 0.693 = = 1.78 10 -9 s -1 3.16 107 s T1/2 12.3 yr
The activity is then 1 Ci R = N = 1.78 10 -9 s -1 1.00 10 22 nuclei = 1.78 1013 Bq = 1.78 1013 Bq = 482 Ci 10 3.70 10 Bq
(
)(
)
(
)
L:
Even though 482 Ci is a large amount of radioactivity, it is smaller than 4.00 1010 Ci by about a hundred million. Therefore, loss of containment is a smaller hazard for a fusion power reactor than for a fission reactor.
Chapter 45 Solutions 607 45.42 Momentum conservation: Thus, KLi = 0 = mLi v Li + m v , or, mLi vLi = m v
2 (m v ) m2 2 1 1 ( mLi vLi ) 2 mLi vLi = = = v 2 2 mLi 2mLi 2mLi 2
KLi
( 4.002 6 u ) 2 2 2 6 6 = 9.30 10 m s = (1.14 u ) 9.30 10 m s 2 (7.016 9 u )
KLi = 1.14 1.66 10 -27 kg 9.30 106 m s = 1.64 10 -13 J = 1.02 MeV
(
)
2
45.43
The complete fissioning of 1.00 gram of U235 releases Q = (1.00 g) MeV atoms J 200 1.60 10 -13 = 8.20 1010 J 6.02 10 23 fission 235 grams / mol mol MeV
If all this energy could be utilized to convert m kilograms of 20.0C water to 400C steam (see Chapter 20 of text for values), then Q = mcw T + mL v + mcs T Q = m ( 4186 J / kgC)(80.0 C) + 2.26 106 J / kg + ( 2010 J / kgC)( 300 C) Therefore m= 8.20 1010 J = 2.56 104 kg 3.20 106 J / kg
[
]
45.44
When mass m of 235 U undergoes complete fission, releasing 200 MeV per fission event, the total energy released is: m Q= N A ( 200 MeV ) 235 g mol where N A is Avogadro's number.
If all this energy could be utilized to convert a mass mw of liquid water at Tc into steam at T h , then, Q = m w c w (100C - Tc ) + L v + c s (T h - 100C)
[
]
m N A ( 200 MeV )
where c w is the specific heat of liquid water, Lv is the latent heat of vaporization, and c s is the specific heat of steam. Solving for the mass of water converted gives mw =
[c
w
(100C - Tc ) + L v + c s (T h - 100C)]
Q
=
(235 g
mol ) c w (100C - Tc ) + L v + c s (T h - 100C)
[
]
2000 by Harcourt, Inc. All rights reserved.
608 Chapter 45 Solutions 45.45 (a) The number of molecules in 1.00 liter of water (mass = 1000 g) is 1.00 10 3 g 23 25 N = 6.02 10 molecules mol = 3.34 10 molecules 18.0 g mol
(
)
The number of deuterium nuclei contained in these molecules is N = 3.34 10 25 molecules
(
1 ) 3300deuteron = 1.01 10 molecules
22
deuterons
Since 2 deuterons are consumed per fusion event, the number of events possible is N 2 = 5.07 10 21 reactions, and the energy released is Efusion = 5.07 10 21 reactions ( 3.27 MeV reaction) = 1.66 10 22 MeV Efusion = 1.66 10 22 MeV 1.60 10 -13 J MeV = 2 .65 10 9 J (b) In comparison to burning 1.00 liter of gasoline, the energy from the fusion of deuterium is Efusion 2.65 10 9 J = = 78.0 times larger . Egasoline 3.40 107 J
( (
)
)(
)
45.46
The number of nuclei in 0.155 kg of
210
Po is
155 g 23 23 N0 = 6.02 10 nuclei g = 4.44 10 nuclei 209.98 g mol
(
)
The half-life of
210
Po is 138.38 days, so the decay constant is given by
=
ln 2 ln 2 = = 5.80 10 -8 s -1 T1 2 (138.38 d) 8.64 10 4 s d
(
)
The initial activity is
R0 = N 0 = 5.80 10 -8 s -1 4.44 10 23 nuclei = 2.58 1016 Bq
210 206 84 Po 82 Pb
(
)(
)
The energy released in each
+ 4 He reaction is 2
Q = M 210 Po - M 206 Pb - M 4 He c 2 : 84 82 2
MeV = 5.41 MeV Q = [ 209.982 848 - 205.974 440 - 4.002 602] u 931.5 u Thus, assuming a conversion efficiency of 1.00%, the initial power output of the battery is decays J MeV -13 = 223 W P = (0.0100)R0Q = (0.0100) 2.58 1016 5.41 decay 1.60 10 s MeV
Chapter 45 Solutions 609 45.47 (a) The thermal power transferred to the water is P w = 0.970 ( waste heat )
P w = 0.970 ( 3065 - 1000) MW = 2.00 10 9 J s rw is the mass of heated per hour: rw = 2.00 10 9 J s ( 3600 s h ) Pw = = 4.91 108 kg h c( T ) ( 4186 J kgC)( 3.50 C)
(
)
The volume used per hour is
4.91 108 kg h = 4.91 10 5 m 3 h 1.00 10 3 kg m 3 3065 106 J s 1 kg 3600 s = 0.141 kg/h rf = 10 7.80 10 J g 1000 g 1 h
(b)
The
235
U fuel is consumed at a rate
*45.48 (a) (b)
V = 4 r 2 ( r ) = 4 14.0 10 3 m
(
) (0.05 m) = 1.23 10
2
8
m3
~ 108 m 3
The force on the next layer is determined by atmospheric pressure. N 1.23 108 m 3 = 1.25 1013 J W = P( V ) = 1.013 10 5 m2
(
)
~ 1013 J
(c)
1 1.25 1013 J = 10 ( yield) , so yield = 1.25 1014 J
~ 1014 J
(d)
1.25 1014 J = 2.97 10 4 ton TNT ~ 10 4 ton TNT 4.2 10 9 J ton TNT
or
~ 10 kilotons
*45.49 (a) (b)
V = l3 =
m m , so l =
1 3
70.0 kg = 3 3 18.7 10 kg m
13
= 0.155 m Then it becomes
Add 92 electrons to both sides of the given nuclear reaction. 238 4 206 92 U atom 8 2 He atom + 82 Pb atom + Qnet .
Qnet = M 238 U - 8 M 4 He - M 206 Pb c 2 = 238.050 784 - 8 ( 4.002 602) - 205.974 440 u (931.5 MeV u ) 2 82 92 Qnet = 51.7 MeV (c) If there is a single step of decay, the number of decays per time is the decay rate R and the energy released in each decay is Q . Then the energy released per time is P = QR . If there is a series of decays in steady state, the equation is still true, with Q representing the net decay energy.
[
]
2000 by Harcourt, Inc. All rights reserved.
610 Chapter 45 Solutions (d) The decay rate for all steps in the radioactive series in steady state is set by the parent uranium: 7.00 10 4 g 23 26 N = 6.02 10 nuclei mol = 1.77 10 nuclei 238 g mol
(
)
=
ln 2 ln 2 1 = = 1.55 10 -10 9 T1 2 4.47 10 yr yr
1 R = N = 1.55 10 -10 1.77 10 26 nuclei = 2.75 1016 decays yr, yr
(
)
so (e)
1 P = QR = ( 51.7 MeV ) 2.75 1016 1.60 10 -13 J MeV = 2 .27 10 5 J yr yr
(
)
dose in rem = dose in rad x RBE 5.00 rem rad rad rad = dose in 1.10 , giving dose in yr = 4.55 yr yr yr
The allowed whole-body dose is then
(70.0 kg) 4.55
rad 10 -2 J kg = 3.18 J/yr yr 1 rad
45.50
ET E(thermal) = kBT = 0.039 eV
3 2
ET =
()E
1 2 n
where n number of collisions, n = 25.6 = 26 collisions
and
0.039 =
( ) (2.0 10 )
1 2 n 6
Therefore,
45.51
From conservation of energy:
K + K n = Q m v = mn vn
1 m mn 2 m
2
or
1 m v 2 2
+ mn vn 2 = 17.6 MeV
1 2
Conservation of momentum:
or
m v = n vn . m
The energy equation becomes:
m + m 1 2 2 2 1 vn + 2 mn vn = n mn vn = 17.6 MeV m 2
(
)
Thus,
m 4.002 602 Kn = (17.6 MeV ) = = 14.1 MeV mn + m 1.008 665 + 4.002 602
Chapter 45 Solutions 611 Goal Solution Assuming that a deuteron and a triton are at rest when they fuse according to
2
H + 3 H 4 He + n + 17.6 MeV ,
determine the kinetic energy acquired by the neutron. G: The products of this nuclear reaction are an alpha particle and a neutron, with total kinetic energy of 17.6 MeV. In order to conserve momentum, the lighter neutron will have a larger velocity than the more massive alpha particle (which consists of two protons and two neutrons). Since the kinetic energy of the particles is proportional to the square of their velocities but only linearly proportional to their mass, the neutron should have the larger kinetic energy, somewhere between 8.8 and 17.6 MeV. Conservation of linear momentum and energy can be applied to find the kinetic energy of the neutron. We first suppose the particles are moving nonrelativistically. The momentum of the alpha particle and that of the neutron must add to zero, so their velocities must be in opposite directions with magnitudes related by mn v n + m v = 0 or
O:
A:
(1.0087 u)vn = ( 4.0026 u)v
At the same time, their kinetic energies must add to 17.6 MeV E = mn vn 2 + m v 2 =
1 2 1 2 1 2 1 (1.0087 u )vn2 + 2 ( 4.0026 u )v 2 = 17.6 MeV
Substitute v = 0.2520vn :
1u E = (0.50435 u )vn 2 + (0.12710 u )vn 2 = 17.6 MeV 2 931.494 MeV / c vn = 0.0189c 2 = 0.173c = 5.19 107 m / s 0.63145
Since this speed is not too much greater than 0.1c, we can get a reasonable estimate of the kinetic energy of the neutron from the classical equation, K = mv 2 =
1 2 1 2
(1.0087 u)(0.173c)
931.494 MeV / c 2 = 14.1 MeV u
L:
The kinetic energy of the neutron is within the range we predicted. For a more accurate calculation of the kinetic energy, we should use relativistic expressions. Conservation of momentum gives
nmn v n + m v = 0
yielding Then and vn = 0.171c, implying that
1.0087
vn 1 - vn / c
2 2
= 4.0026
v 1 - v 2/ c 2
vn 2 v 2 = 2 2 c 15.746c - 14.746vn 2
( n - 1)mnc 2 + ( - 1)m c 2 = 17.6 MeV ( n - 1)mnc 2 = 14.0 MeV
2000 by Harcourt, Inc. All rights reserved.
612 Chapter 45 Solutions
45.52
From Table A.3, the half-life of
32
P is 14.26 d. Thus, the decay constant is
=
ln 2 ln 2 = = 0.0486 d -1 = 5.63 10 -7 s -1 . T1 2 14.26 d R0 5.22 106 decay s = = 9.28 1012 nuclei 5.63 10 -7 s -1
N0 =
At t = 10.0 days, the number remaining is N = N 0 e - t = 9.28 1012 nuclei e
(
)
- 0.0486 d -1 (10.0 d )
(
)
= 5.71 1012 nuclei
so the number of decays has been N 0 - N = 3.57 1012 and the energy released is J = 0.400 J E = 3.57 1012 (700 keV ) 1.60 10 -16 keV If this energy is absorbed by 100 g of tissue, the absorbed dose is 0.400 J 1 rad Dose = = 400 rad 0.100 kg 10 -2 J kg
(
)
45.53
(a)
The number of Pu nuclei in 1.00 kg =
6.02 1023 nuclei/mol (1000 g) 239.05 g/mol
The total energy = (25.2 1023 nuclei)(200 MeV) = 5.04 1026 MeV E = (5.04 1026 MeV)(4.44 10 20 kWh/MeV) = 2.24 107 kWh (b) or 22 million kWh
E = mc2 = (3.016 049 u + 2.014 102 u 4.002 602 u 1.008 665 u) (931.5 MeV/u) E = 17.6 MeV for each D-T fusion
(c)
En = (Total number of D nuclei)(17.6)(4.44 1020) En = (6.02 1023)(1000/2.014)(17.6)(4.44 1020) = 2.34 108 kWh
(d) En = the number of C atoms in 1.00 kg 4.20 eV En = (6.02 1026/12.0)(4.20 10 6 MeV)(4.44 10 20) = 9.36 kWh (e) Coal is cheap at this moment in human history. We hope that safety and waste disposal problems can be solved so that nuclear energy can be affordable before scarcity drives up the price of fossil fuels.
Chapter 45 Solutions 613
4 4 1H atom 2 He atom + Q 1
*45.54
Add two electrons to both sides of the given reaction. Then where or
Q = ( m)c 2 = 4 (1.007 825) - 4.002 602 u (931.5 MeV u ) = 26.7 MeV Q = ( 26.7 MeV ) 1.60 10 -13 J MeV = 4.28 10 -12 J
[
]
(
)
The proton fusion rate is then rate = power output 3.77 10 26 J s = = 3.53 10 38 protons s -12 energy per proton 4.28 10 J ( 4 protons)
(
)
*45.55 (a)
QI = [ MA + MB - MC - ME ] c 2 , and QII = [ MC + MD - MF - MG ] c 2 Qnet = QI + QII = [ MA + MB - MC - ME + MC + MD - MF - MG ] c 2 Qnet = QI + QII = [ MA + MB + MD - ME - MF - MG ] c 2 Thus, reactions may be added. Any product like C used in a subsequent reaction does not contribute to the energy balance.
(b)
Adding all five reactions gives or Adding two electrons to each side
0 1 1 1H + 1H + -1e 0
+ 1H + 1H + 1 1
4 2 He + 2 4 2 He
0 -1e
4 2 He + 2
+ Qnet
4 1H + 2 -1e 1
+ Qnet
Thus, Qnet = 4 M 1H - M 4 He c 2 = 4 (1.007 825) - 4.002 602 u (931.5 MeV u ) = 26.7 MeV
1 2
[
]
4 1H atom 1
atom + Qnet
[
]
45.56
(a)
The mass of the pellet is
3 g 4 1.50 10 -2 cm -7 m = V = 0.200 = 3.53 10 g 2 cm 3 3 2
The pellet consists of equal numbers of 2.50 and the total number of atoms is
H and
3
H atoms, so the average atomic weight is
3.53 10 -7 g 23 16 N = 6.02 10 atoms mol = 8.51 10 atoms 2.50 g mol
(
)
When the pellet is vaporized, the plasma will consist of 2N particles ( N nuclei and N electrons). The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ. The 3 k T as temperature of the plasma is found from E = ( 2N ) 2 B
(
)
T= (b)
2.00 10 3 J E = = 5.68 108 K 3NkB 3 8.51 1016 1.38 10 -23 J K
(
)(
)
Each fusion event uses 2 nuclei, so N 2 events will occur. The energy released will be E= 8.51 1016 J N -13 Q= = 1.20 10 5 J = 120 kJ (17.59 MeV ) 1.60 10 2 MeV 2
2000 by Harcourt, Inc. All rights reserved.
614 Chapter 45 Solutions *45.57 (a) The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome the 4 Coulomb-repulsion barrier to 1 H + 3 He 2 He + e + + , estimated as ke ( e )( 2e ) r . The Coulomb 1 2 barrier to Bethe's fifth and eight reactions is like ke ( e )(7e ) r , temperature should be like 7 15 106 K 5 107 K . 2 (b) For
12
(
)
larger by 2
7
times, so the
C + 1H 13 N + Q , Q1 = (12.000 000 + 1.007 825 - 13.005 738)(931.5 MeV ) = 1.94 MeV
For the second step, add 13 13 - + N atom C atom + e + e + Q.
seven
electrons
to
both
sides
to
have:
Q2 = 13.005 738 - 13.003 355 - 2 (0.000 549) (931.5 MeV ) = 1.20 MeV Q3 = Q7 = 2 (0.000 549)(931.5 MeV ) = 1.02 MeV Q4 = [13.003 355 + 1.007 825 - 14.003 074](931.5 MeV ) = 7.55 MeV Q5 = [14.003 074 + 1.007 825 - 15.003 065](931.5 MeV ) = 7.30 MeV Q6 = 15.003 065 - 15.000 108 - 2 (0.000 549) (931.5 MeV ) = 1.73 MeV Q8 = [15.000 108 + 1.007 825 - 12 - 4.002 602](931.5 MeV ) = 4.97 MeV The sum is 26.7 MeV , the same as for the proton-proton cycle. (c) Not all of the energy released heats the star. When a neutrino is created, it will likely fly directly out of the star without interacting with any other particle.
[
]
[
]
45.58
(a)
I 2 I0e - 2 x - - x = = e ( 2 1) I1 I 0 e - 1 x I 50 - 5.40 - 41.0 )( 0.100 ) =e ( = e 3.56 = 35.2 I100 I 50 - 5.40 - 41.0 )(1.00 ) =e ( = e 35.6 = 2.89 1015 I100 Thus, a 1.00-cm aluminum plate has essentially removed the long-wavelength x-rays from the beam.
(b)
(c)
Chapter 46 Solutions
46.1
Assuming that the proton and antiproton are left nearly at rest after they are produced, the energy of the photon E, must be E = 2E0 = 2(938.3 MeV) = 1876.6 MeV = 3.00 10 10 J Thus, E = hf = 3.00 10 10 J f= 3.00 10 10 J = 4.53 10 23 Hz 6.626 10 34 J s
c 3.00 10 8 m/s = f = = 6.62 10 16 m 4.53 10 23 Hz
46.2
The minimum energy is released, and hence the minimum frequency photons are produced, when the proton and antiproton are at rest when they annihilate. That is, E = E 0 and K = 0. To conserve momentum, each photon must carry away one-half the energy. Thus, Emin = hfmin = (2E0) 2 = E0 = 938.3 MeV = 2.27 10 23 Hz
Thus, fmin =
(938.3 MeV)(1.60 10 13 J/MeV) 6.626 10 34 J s
c 3.00 10 8 m/s = f = = 1.32 10 15 m min 2.27 10 23 Hz
*46.3
In p+ + p,
we start with energy we end with energy
2.09 GeV 938.3 MeV + 938.3 MeV + 95.0 MeV + K 2
where K 2 is the kinetic energy of the second proton. Conservation of energy gives K 2 = 118 MeV
2000 by Harcourt, Inc. All rights reserved.
2
Chapter 46 Solutions
Goal Solution A photon with an energy E = 2.09 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 95.0 MeV. What is the kinetic energy of the antiproton? ( mp c 2 = 938.3 MeV). G: An antiproton has the same mass as a proton, so it seems reasonable to expect that both particles will have similar kinetic energies. The total energy of each particle is the sum of its rest energy and its kinetic energy. Conservation of energy requires that the total energy before this pair production event equal the total energy after. E = ERp + K p + ERp + K p
O:
A:
(
) (
)
The energy of the photon is given as E = 2.09 GeV = 2.09 10 3 MeV . From Table 46.2, we see that the rest energy of both the proton and the antiproton is ERp = ERp = mp c 2 = 938.3 MeV If the kinetic energy of the proton is observed to be 95.0 MeV, the kinetic energy of the antiproton is K p = E - ERp - ERp - K p = 2.09 10 3 MeV - 2(938.5 MeV ) - 95.0 MeV = 118 MeV
L:
The kinetic energy of the antiproton is slightly (~20%) greater than the proton. The two particles most likely have different shares in momentum of the gamma ray, and therefore will not have equal energies, either.
*46.4
The reaction is muon-lepton number before reaction: electron-lepon number before reaction:
+ + e + (1) + (0) = 1 (0) + (1) = 1
Therefore, after the reaction, the muon-lepton number must be 1. Thus, one of the neutrinos must be the anti-neutrino associated with muons, and one of the neutrinos must be the neutrino associated with electrons: and e Then + + e + e
46.5
The creation of a virtual Z 0 boson is an energy fluctuation E = 93 10 9 eV. It can last n o longer than t = h 2E and move no farther than c ( t ) = 6.626 10 -34 J s 3.00 108 m s hc 1 eV -18 = m = ~ 10 -18 m = 1.06 10 9 4 E 1.60 10 -19 J 4 93 10 eV
(
(
)(
)
)
Chapter 46 Solutions
3
46.6
(a) From Table A-3, (b)
E = (m n m p m e )c 2 E = (1.008 665 1.007 825)931.5 = 0.782 MeV pp = pe
Assuming the neutron at rest, momentum is conserved, relativistic energy is conserved, Since p p = p e , Solving the algebra If pe c = me ve c = 1.19 MeV, then Solving,
(m c )
p
2 2
2 + pp c 2 +
(m c )
e
2 2
2 + pe c 2 = mnc 2
(938.3)2 + ( pc)2
pc = 1.19 MeV
+
(0.511)2 + ( pc)2
= 939.6 MeV
1.19 MeV ve = 0.511 MeV = c x 2 = (1 x 2)5.43 ve = 0.919c and
x 1x
2
ve = 2.33 where x = c
x = ve/c = 0.919
Then
m pv p = em ev e : vp = (1.19 MeV)(1.60 1013 J/MeV) e me ve c = = 3.80 10 5 m/s = 380 km/s (1.67 10 2 7 )(3.00 10 8) mp c
(c)
The electron is relativistic, the proton is not.
*46.7
The time for a particle traveling with the speed of light to travel a distance of 3 1015 m is d 3 1015 m t = v = = 3 108 m/s ~ 1023 s
*46.8
With energy 938.3 MeV, the time that a virtual proton could last is at most t in E t ~ h. The distance it could move is at most c t ~ hc (1.055 1034 J s)(3 108 m/s) = = ~ 1016 m E (938.3)(1.6 1013 J)
2000 by Harcourt, Inc. All rights reserved.
4
Chapter 46 Solutions
46.9
By Table 46.2, Therefore,
M 0 = 135 MeV/c 2 E = 67.5 MeV for each photon E = 1.63 1022 Hz h
p=
E MeV = 67.5 c c
and
f=
*46.10
In ? + p+ n + +, charge conservation requires the unknown particle to be neutral. Baryon number conservation requires baryon number = 0. The muon-lepton number of ? must be 1. So the unknown particle must be .
46.11
+ + K +
0
KS + + -
0
(or 0 + 0 )
p + +
0
n p + e + + e
46.12
(a) (b) (c)
p + p + + e
Le
0+00+1 1 + 1 +1 + 1
and
L
0 + 0 1 + 0
+pp++
p+pp++
charge
baryon number baryon number charge
1+11+0 1+11+1+1
(d) p + p p + p + n (e)
+ p n + 0
0+10+0
*46.13 (a)
Baryon number and charge are conserved, with values of 0 + 1 = 0 + 1 and 1 + 1 = 1 + 1 i n both reactions. Strangeness is not conserved in the second reaction.
(b)
Chapter 46 Solutions
5
2000 by Harcourt, Inc. All rights reserved.
6
Chapter 46 Solutions
46.14
Baryon number conservation allows the first and forbids the second .
46.15
(a) (b) (c) (d) (e) (f)
- - +
K + + +
L : 0 1 - 1 L : 0 -1 + 1 Le : - 1+ 0 0 - 1 Le : 1+ 0 0 + 1 L : 1+ 0 0 + 1 L : 1 0 + 0 + 1 and Le : 0 1 - 1 + 0
e + p+ n + e + e + n p+ + e - + n p+ + - - e - + e +
*46.16
Momentum conservation requires the pions to have equal speeds. The total energy of each is so E 2 = p 2 c 2 + (mc 2 )2 gives Solving, 497.7 MeV 2 (248.8 MeV)2 = (pc)2 + (139.6 MeV)2 pc = 206 MeV = mvc = mc 2 1 - (v c)
2
v c
pc 206 MeV 1 v = 1.48 = = 2 2 139.6 MeV mc 1 - (v c) c
(v c) = 1.48
2
1 - (v c)
2
and so
(v c)2 = 2.18 [1 - (v c)2 ] = 2.18 - 2.18 (v c)2
v = c 2.18 = 0.828 3.18 and v = 0.828 c
3.18( v c ) = 2.18
46.17
(a) (b) (c)
p+ + + 0
Baryon number is violated: 1 0 + 0 This reaction can occur . Baryon number is violated: 1 + 1 1 + 0 This reaction can occur .
e
p+ + p+ p+ + p+ + 0 p+ + p+ p+ + +
(d) + + + (e) (f) n 0 p+ + e +
This reaction can occur . Violates baryon number : 0 0 + 1
+ + + n
Chapter 46 Solutions
7
Violates muon-lepton number : 0 1 + 0
2000 by Harcourt, Inc. All rights reserved.
8
Chapter 46 Solutions
46.18
(a) (b)
p e+ +
Baryon number: +1 0 + 0
B 0 , so baryon number is violated. pe = p
2 Ee = p c
From conservation of momentum:
2 2 Then, for the positron, Ee = ( pe c ) + E0, e becomes 2
( )
2
2 2 2 + E0, e = E + E0, e
From conservation of energy: E0, p = Ee + E so Equating this to the result from above gives or Thus, Also, (c) E =
2 2 E0, p - E0, e
or
Ee = E0, p - E
2 2 Ee = E0, p - 2E0, pE 2 + E . 2 + E ,
2 2 2 E + E0, e = E0, p - 2E0,pE
2E0, p
=
(938.3 MeV)2 - (0.511 MeV)2 2 (938.3 MeV )
= 469 MeV
Ee = E0, p - E = 938.3 MeV 469 MeV = 469 MeV p = E c = 469 MeV/c and pe = p = 469 MeV/c
The total energy of the positron is Ee = 469 MeV . But, which yields: Ee = E0, e = E0, e 1 - (v / c)2 so 1 - (v c) =
2
E0, e Ee
=
0.511 MeV = 1.09 10 -3 469 MeV
v = 0.999 999 4 c
*46.19 (a) (b) (c) (d)
The relevant conservation laws are:
Le = 0, L = 0, and L = 0.
+ 0 + e+ + ?
? + p - + p + + 0 p + - + ?
Le : 0 0 - 1 + Le
Le = 1
and we have a e L = 1 and we have a and we have a and we have a and we have a
L : L + 0 + 1 + 0 + 0 L : 0 0 + 1 + L L : 0 - 1 + L L : + 1 0 + L
L = - 1
+ + + ? + ?
L = 1 L = 1
Conclusion for (d):
L = 1 for one particle, and L = 1 for the other particle. We have and .
46.20
The 0 + + - decay must occur via the strong interaction. The K 0 + + - decay must occur via the weak interaction. S
Chapter 46 Solutions
9
46.21
(a) (b) (c)
0 p +
Strangeness: 1 0 + 0
(strangeness is not conserved )
+ p 0 + K 0
p + p 0 + 0
Strangeness: 0 + 0 1 + 1 (0 = 0 and strangeness is conserved ) Strangeness: 0 + 0 +1 1 (0 = 0 and strangeness is conserved ) Strangeness: 0 + 0 0 1 Strangeness: 2 1 + 0 Strangeness: 2 0 + 0 (0 1: strangeness is not conserved ) (2 1 so strangeness is not conserved (2 0 so strangeness is not conserved )
(d) + p + + (e) 0 + ) (f) 0 p +
46.22
(a) (b) (c)
e + n p + e + e 0 p + 0
Le : 0 1 + 0, Le : 0 0 + 1 + 1
and
L : 1 0
Strangeness: 1 0 + 0,
and charge: 0 +1 + 0
(d) p e+ + 0 (e) 0 n + 0
Baryon number: +1 0 + 0 Strangeness: 2 0 + 0
*46.23 (a) (b)
- + p 2 violates conservation of baryon number as 0 + 1 0. not allowed
K - + n 0 + - Baryon number = 0 + 1 1 + 0 Charge = 1 + 0 0 1 Strangeness, 1 + 0 1 + 0 Lepton number, 0 0 The interaction may occur via the strong interaction since all are conserved. K- - + 0 Strangeness, 1 0 + 0 Baryon number, 0 0 Lepton number, 0 0 Charge, 1 1 + 0 Strangeness is violated by one unit, but everything else is conserved. Thus, the reaction can occur via the weak interaction , but not the strong or electromagnetic interaction. - - + 0 Baryon number, 1 1 + 0 Lepton number, 0 0 Charge, 1 1 + 0 Strangeness, 3 2 + 0 May occur by weak interaction , but not by strong or electromagnetic.
(c)
(d)
(e)
2 Baryon number, 0 0 Lepton number, 0 0 Charge, 0 0 Strangeness, 0 0
2000 by Harcourt, Inc. All rights reserved.
10
Chapter 46 Solutions
No conservation laws are violated, but photons are the mediators of the electromagnetic interaction. Also, the lifetime of the is consistent with the electromagnetic interaction .
*46.24 (a)
- 0 + - + Baryon number: + 1 + 1 + 0 + 0 Le : 0 0 + 0 + 0 L : 0 0 + 0 + 0 Conserved quantities are: K 0 2 0 S Baryon number: 0 0 Le : 0 0 L : 0 0 Conserved quantities are: K- + p 0 + n Baryon number: 0 + 1 1 + 1 Le : 0 + 0 0 + 0 L : 0 + 0 0 + 0 Conserved quantities are: 0 0 + Baryon number: + 1 1 + 0 Le : 0 0 + 0 L : 0 0 + 0 Conserved quantities are: e+ + e- + + - Baryon number: 0 + 0 0 + 0 Le : 1 + 1 0 + 0 L : 0 + 0 0 + 0 Conserved quantities are: p + n 0 + - Baryon number: 1 + 1 1 + 1 Le : 0 + 0 0 + 0 L : 0 + 0 0 + 0 Conserved quantities are: Charge: 1 + 0 0 1 L : 0 + 0 0 + 0 Strangeness: 0 + 0 +1 1 B, S, charge , Le , L , and L Charge: 0 0 L : 0 0 Strangeness: +1 0 B, charge , Le , L , and L Charge: 1 0 1 + 0 L : 0 0 + 1 + 1 Strangeness: 2 1 + 0 + 0 B, charge , Le , and L
(b)
(c)
Charge: 1 + 1 0 + 0 L : 0 + 0 0 + 0 Strangeness: 1 + 0 1 + 0 S, charge , Le , L , and L
(d)
Charge: 0 0 L : 0 0 + 0 Strangeness: 1 1 + 0 B, S, charge , Le , L , and L
(e)
Charge: +1 1 +1 1 L : 0 + 0 + 1 1 Strangeness: 0 + 0 0 + 0 B, S, charge , Le , L , and L
(f)
Chapter 46 Solutions
11
*46.25 (a)
K+ + p ? + p The strong interaction conserves everything. 0 +1 B +1 so Baryon number, +1 + 1 Q + 1 so Charge, so Lepton numbers, 0 + 0 L + 0 +1 + 0 S + 0 so Strangeness,
B=0 Q = +1 Le = L = L = 0 S=1
The conclusion is that the particle must be positively charged, a non-baryon, with strangeness of +1. Of particles in Table 46.2, it can only be the K + . Thus, this is an elastic scattering process. The weak interaction conserves all but strangeness, and S = 1. (b) - ? + - Baryon number, Charge, Lepton numbers, Strangeness, +1 B + 0 -1 Q - 1 0L+0 -3 S + 0 B=1 Q=0 Le = L = L = 0 S = -2
so so so so S = 1:
The particle must be a neutral baryon with strangeness of 2. Thus, it is the 0 . (c) K+ ? + + + : Baryon number, 0B+0+0 +1 Q + 1 + 0 Charge, Lepton Numbers Le , 0 Le + 0 + 0 L , 0 L - 1 + 1 L , 0 L + 0 + 0 1 S + 0 + 0 Strangeness:
so so so so so so
B=0 Q=0 Le = 0 L = 0 L = 0 S = 1 (for weak interaction): S = 0
The particle must be a neutral meson with strangeness = 0 0 .
*46.26 (a) strangeness baryon number charge proton 0 1 e u 0 1/3 2e/3 u 0 1/3 2e/3 d 0 1/3 e/3 total 0 1 e
(b) strangeness baryon number charge neutron 0 1 0 u 0 1/3 2e/3 d 0 1/3 e/3 d 0 1/3 e/3 total 0 1 0
*46.27 (a)
The number of protons
6.02 10 23 molecules 10 protons 26 N p = 1000 g molecule = 3.34 10 18.0 g
protons
2000 by Harcourt, Inc. All rights reserved.
12
Chapter 46 Solutions
and there are
6.02 10 23 molecules 8 neutrons 26 N n = (1000 g ) molecule = 2.68 10 18.0 g 3.34 1026 electrons
neutrons
So there are for electric neutrality The up quarks have number and there are (b)
2 3.34 1026 + 2.68 1026 = 9.36 1026 up quarks 2 2.68 1026 + 3.34 1026 = 8.70 1026 down quarks
Model yourself as 65 kg of water. Then you contain 65 3.34 1026 ~ 1028 electrons 65 9.36 1026 ~ 1029 up quarks 65 8.70 1026 ~ 1029 down quarks Only these fundamental particles form your body. You have no strangeness, charm, topness or bottomness.
46.28
(a) strangeness baryon number charge (b) strangeness baryon number charge
K0 1 0 0 0 1 1 0
d 0 1/3 e/3 u 0 1/3 2e/3
s 1 1/3 e/3 d 0 1/3 e/3
total 1 0 0 s 1 1/3 e/3 total 1 1 0
46.29
Quark composition of proton = uud and of neutron = udd. Thus, if we neglect binding energies, we may write and Solving simultaneously, we find m u = 3 (2m p m n) = 3 2(938.3 MeV / c 2 ) - 939.6 MeV / c 2 = 312 MeV/c2 and from either (1) or (2), m d = 314 MeV/c 2
1 1
m p = 2m u + m d m n = m u + 2m d
(1) (2)
[
]
*46.30
In the first reaction, - + p K 0 + 0 , the quarks in the particles are: ud + uud ds + uds. There is a net of 1 up quark both before and after the reaction, a net of 2 down quarks both
Chapter 46 Solutions
13
before and after, and a net of zero strange quarks both before and after. Thus, the reaction conserves the net number of each type of quark. In the second reaction, - + p K 0 + n , the quarks in the particles are: ud + uud ds + udd. In this case, there is a net of 1 up and 2 down quarks before the reaction but a net of 1 up, 3 down, and 1 anti-strange quark after the reaction. Thus, the reaction does not conserve the net number of each type of quark.
46.31
(a)
- + p 0 + 0
In terms of constituent quarks: up quarks: down quarks: strange quarks: ud + uud ds + uds . 1 + 2 0 + 1, or 1 1 1 + 1 1 + 1, or 2 2 0 + 0 1 + 1, or 0 0
(b)
+ + p + + +
up quarks: down quarks: strange quarks:
ud + uud us + uus 1 + 2 1 + 2, or 3 3 1 + 1 0 + 0, or 0 0 0 + 0 1 + 1, or 0 0
(c)
- + p + + 0 + - up quarks: down quarks: strange quarks:
us + uud us + ds + sss 1 + 2 1 + 0 + 0, or 1 1 0 + 1 0 + 1 + 0, or 1 1 1 + 0 1 1 + 3, or 1 1
(d)
p + p K0 + p + + + ?
uud + uud ds + uud + ud + ?
The quark combination of ? must be such as to balance the last equation for up, down, and strange quarks. up quarks: down quarks: strange quarks: quark composite = uds = 0 or 0 2+2=0+2+1+? 1+1=1+11+? 0 + 0 = 1 + 0 + 0 + ? (has 1 u quark) (has 1 d quark) (has 1 s quark)
46.32
0 + p + + + X
dds + uud uds + 0 + ?
The left side has a net 3d, 2u and 1s. The right-hand side has 1d, 1u, and 1s leaving 2d and 1u missing. The unknown particle is a neutron, udd. Baryon and strangeness numbers are conserved.
2000 by Harcourt, Inc. All rights reserved.
14
Chapter 46 Solutions
*46.33 (a) (b) (c) (d)
Compare the given quark states to the entries in Tables 46.4 and 46.5. suu = + ud = - sd = 0 ssd = -
*46.34 (a) (b)
1 u ud : charge = - 2 e + - 2 e + 3 e = -e . This is the antiproton . 3 3 1 1 u d d : charge = - 2 e + 3 e + 3 e = 0 . This is the antineutron . 3
(
) (
) ( )
(
) ( ) ( )
*46.35
Section 39.4 says
f observer = f source
1 + va c 1 - va c
The velocity of approach, va , is the negative of the velocity of mutual recession: va = -v. Then, c c 1- v c = 1 + v c and
=
1+ v c 1- v c
46.36
v = HR (Equation 46.7) v (2.00 106 ly) = 3.4 104 m/s v (2.00 108 ly) = 3.4 106 m/s v (2.00 109 ly) = 3.4 107 m/s
H=
(1.7 102 m/s) ly 1 + v/c = 590(1.0001133) = 590.07 nm 1 - v/c 1 + 0.01133 = 597 nm 1 - 0.01133 1 + 0.1133 = 661 nm 1 - 0.1133
(a) (b) (c)
' =
' = 590 ' = 590
46.37
(a)
' 650 nm = 434 nm = 1.50 =
v = 0.383c,
1 + v/c 1 - v/c
1 + v/c 1 v/c = 2.24 38.3% the speed of light v (0.383)(3.00 108) R= H = = 6.76 109 light years (1.7 102)
(b)
Equation 46.7, v = HR
Chapter 46 Solutions
15
Goal Solution A distant quasar is moving away from Earth at such high speed that the blue 434-nm hydrogen line is observed at 650 nm, in the red portion of the spectrum. (a) How fast is the quasar receding? You may use the result of Problem 35. (b) Using Hubble's law, determine the distance from Earth to this quasar. G: The problem states that the quasar is moving very fast, and since there is a significant red shift of the light, the quasar must be moving away from Earth at a relativistic speed (v > 0.1c). Quasars are very distant astronomical objects, and since our universe is estimated to be about 9 15 billion years old, we should expect this quasar to be ~10 light-years away. As suggested, we can use the equation in Problem 35 to find the speed of the quasar from the Doppler red shift, and this speed can then be used to find the distance using Hubble's law. 1+ v c 650 nm = = 1.498 = 434 nm 1- v c Therefore, v = 0.383c or 1+ v c = 2.243 1- v c
O:
A:
(a)
or squared,
38.3% the speed of light
(b) Hubble's law asserts that the universe is expanding at a constant rate so that the speeds of galaxies are proportional to their distance R from Earth, v = HR
8 v (0.383) 3.00 10 m / s R= = = 6.76 10 9 ly H 1.70 10 -2 m / s ly
so,
(
(
)
)
L:
The speed and distance of this quasar are consistent with our predictions. It appears that this quasar is quite far from Earth but not the most distant object in the visible universe.
*46.38 (a)
= n n
1+ v/c = (Z + 1) n 1- v/c
1 + v/c 2 1 v/c = ( Z + 1)
v v 1 + c = ( Z + 1)2 c ( Z + 1) 2 v ( Z 2 + 2 Z + 2) = Z 2 + 2 Z c Z2 + 2 Z v = c 2 Z + 2 Z + 2 v c Z2 + 2 Z R=H = H Z 2 + 2 Z + 2
(b)
2000 by Harcourt, Inc. All rights reserved.
16
Chapter 46 Solutions
*46.39
The density of the Universe is = 1.20c = 1.20 3H 2 8 G . Consider a remote galaxy at distance r . The mass interior to the sphere below it is 4 r 3 3H 2 4 r 3 0.600H 2 r 3 M = = 1.20 8 G 3 = G 3 both now and in the future when it has slowed to rest from its current speed v = Hr. The energy of this galaxy is constant as it moves to apogee distance R:
1 mv 2 2
(
)
-
GmM GmM =0- r R r R
so
1 mH 2 r 2 2
-
Gm 0.600H 2 r 3 Gm 0.600H 2 r 3 =0- R r G G
- 0.100 = - 0.600
so
R = 6.00 r from its current dimensions.
The Universe will expand by a factor of 6.00
*46.40 (a)
kBT 2mp c 2
so
T
2 mp c 2 kB
=
( (
2 (938.3 MeV ) 1.60 10 -13 1.38 10 -23 J K 1 MeV
)
J 13 ~ 10 K J ~ 1010 K
(b)
kBT 2me c 2
T
2 (0.511 MeV ) 1.60 10 -13 2me c 2 = kB 1.38 10 -23 J K 1 MeV
)
*46.41 (a)
Wien's law: Thus,
maxT = 2.898 10 - 3 m K max =
2.898 10 - 3 m K 2.898 10 - 3 m K = = 1.06 10 - 3 m = 1.06 mm T 2.73 K
(b)
This is a microwave .
*46.42 (a)
hG L= = c3
(1.055 10
-34
J s 6.67 10 -11 N m 2 kg 2
8
(3.00 10
)(
ms
)
3
)=
1.61 10 - 35 m
(b)
This time is given as T =
L 1.61 10 -35 m = = 5.38 10 - 44 s , c 3.00 108 m s
which is approximately equal to the ultra-hot epoch.
Chapter 46 Solutions
17
46.43
(a)
E t h, and
t
r 1.4 10 -15 m = = 4.7 10 -24 s c 3.0 108 m s 1 MeV 2 J = 1.4 10 MeV 1.60 10 -13 J
E
h 1.055 10 -34 J s = = 2.3 10 -11 t 4.7 10 -24 s
m=
E 1.4 10 2 MeV c 2 ~ 10 2 MeV c 2 c2
(b)
From Table 46.2, m c 2 = 139.6 MeV , a pi-meson
*46.44 (a) (b) (c)
+ p + + 0 is forbidden by charge conservation
+ e is forbidden by energy conservation p + + + + is forbidden by baryon number conservation
46.45
The total energy in neutrinos emitted per second by the Sun is: (0.4)(4)(1.5 1011)2 = 1.1 1023 W Over 109 years, the Sun emits 3.6 1039 J in neutrinos. This represents an annihilated mass mc 2 = 3.6 1039 J m = 4.0 1022 kg About 1 part in 50,000,000 of the Sun's mass, over 109 years, has been lost to neutrinos.
2000 by Harcourt, Inc. All rights reserved.
18
Chapter 46 Solutions
Goal Solution The energy flux carried by neutrinos from the Sun is estimated to be on the order of 0.4 W/m 2 at Earth's surface. Estimate the fractional mass loss of the Sun over 109 years due to the radiation of neutrinos. (The mass of the Sun is 2 1030 kg. The Earth-Sun distance is 1.5 1011 m.) G: Our Sun is estimated to have a life span of about 10 billion years, so in this problem, we are examining the radiation of neutrinos over a considerable fraction of the Sun's life. However, the mass carried away by the neutrinos is a very small fraction of the total mass involved in the Sun's nuclear fusion process, so even over this long time, the mass of the Sun may not change significantly (probably less than 1%). The change in mass of the Sun can be found from the energy flux received by the Earth and Einstein's famous equation, E = mc 2 . Since the neutrino flux from the Sun reaching the Earth is 0.4 W/m 2, the total energy emitted per second by the Sun in neutrinos in all directions is
O:
A:
(0.4 W / m )(4 r ) = (0.4 W / m )(4 )(1.5 10
2 2 2
11
m
)
2
= 1.13 10 23 W
In a period of 10 9 yr, the Sun emits a total energy of
(1.13 10
23
J / s 10 9 yr 3.156 107 s / yr = 3.57 10 39 J
)(
)(
)
in the form of neutrinos. This energy corresponds to an annihilated mass of E = m c 2 = 3.57 10 39 J so m = 3.97 10 22 kg
Since the Sun has a mass of about 2 10 3 kg, this corresponds to a loss of only about 1 part i n 50 000 000 of the Sun's mass over 10 9 yr in the form of neutrinos.
L:
It appears that the neutrino flux changes the mass of the Sun by so little that it would be difficult to measure the difference in mass, even over its lifetime! p + p p ++ + X We suppose the protons each have 70.4 MeV of kinetic energy. From conservation of momentum, particle X has zero momentum and thus zero kinetic energy. Conservation of energy then requires Mp c 2 + M c 2 + M X c 2 = Mp c 2 + K p + Mp c 2 + K p
46.46
(
) (
)
MX c 2 = Mp c 2 + 2Kp - M c 2 = 938.3 MeV + 2 (70.4 MeV ) - 139.6 MeV = 939.5 MeV X must be a neutral baryon of rest energy 939.5 MeV . Thus X is a neutron .
Chapter 46 Solutions
19
*46.47
We find the number N of neutrinos: 1046 J = N(6 MeV) = N(6 1.6 1013 J) N = 1.0 1058 neutrinos The intensity at our location is 1 ly N 1.0 10 58 N 14 2 = = = 3.1 10 / m 2 5 2 8 7 A 4 r 4 (1.7 10 ly) (3.0 10 m / s)(3.16 10 s) The number passing through a body presenting 5000 cm2 = 0.50 m2 is then 1 3.1 1014 2 0.50 m 2 = 1.5 1014 or m
2
(
)
~ 1014
*46.48
By relativistic energy conservation,
E + me c 2 = E c X= =
3me c 2 1 - v2 / c2
(1)
By relativistic momentum conservation,
3me v 1 - v2 / c2 E
2
(2) v c 3me c 2 X 1 - X2
Dividing (2) by (1),
E + me c
=
Subtracting (2) from (1), Solving, 1= 3 - 3X 1- X
2
me c 2 = and X=
3me c 2 1 - X2
-
4 so 5
E = 4me c 2 = 2.04 MeV
46.49
m c2 = 1115.6 MeV 0 p + m p c 2 = 938.3 MeV m c 2 = 139.6 MeV
The difference between starting mass-energy and final mass-energy is the kinetic energy of the products. and p p = p = p K p + K = 37.7 MeV Applying conservation of relativistic energy, (938.3)2 + p 2 c 2 - 938.3 + (139.6)2 + p 2 c 2 - 139.6 = 37.7 MeV Solving the algebra yields Then, p c = ppc = 100.4 MeV Kp = K = (m p c 2)2 + (100.4)2 (139.6)2 + (100.4)2 m pc 2 = 5.35 MeV 139.6 = 32.3 MeV
2000 by Harcourt, Inc. All rights reserved.
20
Chapter 46 Solutions qBr = 1.60 10 -19 C (0.250 kg / C s)(1.33 m ) cpp = 1.60 1011 pp = 99.8 MeV c kg m2 = 1.60 1011 J = 99.8 MeV s2
46.50
Momentum of proton is pp = 5.32 1020 Therefore, The total energy of the proton is kg m s
(
)
2 Ep = E0 + (cp)2 = (938.3)2 + (99.8)2 = 944 MeV
For pion, the momentum qBr is the same (as it must be from conservation of momentum i n a 2-particle decay). p = 99.8 MeV c E0 = 139.6 MeV
2 E = E0 + (cp)2 = (139.6)2 + (99.8)2 = 172 MeV
Thus,
ETotal after = ETotal before = Rest Energy (This is a 0 particle!)
Rest Energy of unknown particle = 944 MeV + 172 MeV = 1116 MeV Mass = 1116 MeV/c 2
46.51
0 0 + From Table 46.2, m = 1192.5 MeV/c 2 and m = 1115.6 MeV c 2
Conservation of energy requires E0 , = E0 , + K + E , or p2 1192.5 MeV = 1115.6 MeV + + E 2m
(
)
Momentum conservation gives p = p , so the last result may be written as
2 p 1192.5 MeV = 1115.6 MeV + + E 2m 2 p c 2 1192.5 MeV = 1115.6 MeV + + E 2m c 2
or Recognizing that we now have Solving this quadratic equation,
m c 2 = 1115.6 MeV
and
p c = E ,
2 E
1192.5 MeV = 1115.6 MeV + E 74.4 MeV
2 (1115.6 MeV )
+ E
Chapter 46 Solutions
21
46.52
p + p p + n ++ The total momentum is zero before the reaction. Thus, all three particles present after the reaction may be at rest and still conserve momentum. This will be the case when the incident protons have minimum kinetic energy. Under these conditions, conservation of energy gives 2 mp c 2 + K p = mp c 2 + mnc 2 + m c 2 so the kinetic energy of each of the incident protons is Kp = mnc 2 + m c 2 - mp c 2 2 =
(
)
(939.6 + 139.6 - 938.3) MeV =
2
70.4 MeV
46.53
Time-dilated lifetime:
T = T0 =
0.900 1010 s 1 v2/c2
=
0.900 1010 s 1 (0.960)2
= 3.214 1010 s
distance = (0.960)(3.00 108 m/s)(3.214 1010 s) = 9.26 cm
46.54
- - +
From the conservation laws, and Thus, or Since and m c 2 = 139.5 MeV = E + E p = p , E = p c
2 E = p c
[1]
( )
2
+ (105.7 MeV ) = ( p c) + (105.7 MeV )
2 2 2
2
2 2 E - E = (105.7 MeV )
[2] [1] [2]
E + E = 139.5 MeV
(E + E )(E - E ) = (105.7 MeV)2
E - E =
then Subtracting [3] from [1],
(105.7 MeV)2
139.5 MeV
= 80.1 E = 29.7 MeV
[3]
2E = 59.4 MeV
and
2000 by Harcourt, Inc. All rights reserved.
22
Chapter 46 Solutions
*46.55
The expression e -E kBT dE gives the fraction of the photons that have energy between E and E + dE. The fraction that have energy between E and infinity is
-E k T dE e -E k T ( - dE kBT ) e E e E = E = = e -E k T -E k T -E k T -E k T dE e (- dE kBT ) e 0 e 0 0
B B B B B B
-E k B T
We require T when this fraction has a value of 0.0100 (i.e., 1.00%) and E = 1.00 eV = 1.60 10 -19 J
- 1.60 10 -19 J
Thus, 0.0100 = e or
(
) (1.38 10 -23 J K )T
ln (0.0100) = -
(
1.60 10 -19 J
1.38 10 -23 J K T
)
= -
1.16 10 4 K T
giving
T = 2.52 10 3 K
46.56
(a)
This diagram represents the annihilation of an electron and an antielectron. From charge and leptonnumber conservation at either vertex, the exchanged particle must be an electron, e .
(b)
This is the tough one. A neutrino collides with a neutron, changing it into a proton with release of a muon. This is a weak interaction. The exchanged particle has charge +1e and is a W + .
(a)
(b)
46.57
(a) (b)
The mediator of this weak interaction is a Z0 boson . The Feynman diagram shows a down quark and its antiparticle annihilating each other. They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge. In this case the product (a) (b) particle is a photon . For conservation of both energy and momentum, we would expect two photons; but momentum need not be strictly conserved, according to the uncertainty principle, if the photon travels a sufficiently short distance before producing another matter-antimatter pair of particles, as shown in Figure P46.57. Depending on the color charges of the d and d quarks,
Chapter 46 Solutions
23
the ephemeral particle could also be a gluon , as suggested in the discussion of Figure 46.13(b).
2000 by Harcourt, Inc. All rights reserved.