Chapter 44
1. The total rest energy of the electron-positron pair is
E = me c 2 + me c 2 = 2me c 2 = 2(0.511 MeV) = 1.022 MeV . With two gamma-ray photons produced in the annihilation process, the wavelength of each photon is (using hc = 1240 eV nm )
=
hc
Chapter 43
1. If MCr is the mass of a 52Cr nucleus and MMg is the mass of a 26Mg nucleus, then the disintegration energy is Q = (MCr 2MMg)c2 = [51.94051 u 2(25.98259 u)](931.5 MeV/u) = 23.0 MeV. 2. Adapting Eq. 42-21, there are
N Pu = M sam 1000 g NA = (6
Chapter 42
1. Kinetic energy (we use the classical formula since v is much less than c) is converted into potential energy (see Eq. 24-43). From Appendix F or G, we find Z = 3 for Lithium and Z = 90 for Thorium; the charges on those nuclei are therefore 3
Chapter 41
1. (a) At absolute temperature T = 0, the probability is zero that any state with energy above the Fermi energy is occupied. (b) The probability that a state with energy E is occupied at temperature T is given by
P( E ) = 1 e
( E E F )/ kT
+1
w
Chapter 40
1. (a) Using Table 40-1, we find l = [ ml ]max = 4. (b) The smallest possible value of n is n = l max +1 l + 1 = 5.
1 (c) As usual, ms = 2 , so two possible values.
2. (a) For l = 3 , the greatest value of ml is ml = 3 .
1 (b) Two states ( ms =
Chapter 39
1. Since En L 2 in Eq. 39-4, we see that if L is doubled, then E1 becomes (2.6 eV)(2) 2 = 0.65 eV. 2. We first note that since h = 6.626 1034 Js and c = 2.998 108 m/s,
6 c.626 10 hc = 1602 c. 10
34
J s 2.998 108 m / s
9
19
h 1240 eV nm. = J / e
Chapter 38
1. Let R be the rate of photon emission (number of photons emitted per unit time) of the Sun and let E be the energy of a single photon. Then the power output of the Sun is given by P = RE. Now E = hf = hc/, where h = 6.626 1034 Js is the Planc
Chapter 37
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
= 1 / 1 2 , and = v/c), we obtain
= 1
Ft I . Gt J HK
2 0
The proper time interval is measured by a clock at rest relative to the muon. Specifically, t0 = 2.200
Chapter 36
1. (a) The condition for a minimum in a single-slit diffraction pattern is given by a sin = m, where a is the slit width, is the wavelength, and m is an integer. For = a and m = 1, the angle is the same as for = b and m = 2. Thus, a = 2b = 2(35
Chapter 35
1. The index of refraction is found from Eq. 35-3: n= c 2.998 108 m s = = 156 . . . v 192 108 m s
2. Note that Snells Law (the law of refraction) leads to 1 = 2 when n1 = n2. The graph indicates that 2 = 30 (which is what the problem gives as t
Chapter 34
1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm. 2. The bird is a distance d2 in front of the mirror; the plane of its image is that same distanc
Chapter 33
1. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm. (b) Similarly, the larger wavelength is approximately 610 nm. (c) From Fig. 33-2 the wavelength at which the eye is most sensitive is about 555 nm. (d) Using t
Chapter 32
1. We use
6 n =1
Bn = 0 to obtain
5
B 6 = Bn = ( 1Wb + 2 Wb 3 Wb + 4 Wb 5 Wb ) = +3 Wb .
n =1
2. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since
Chapter 31
1. (a) The period is T = 4(1.50 s) = 6.00 s. (b) The frequency is the reciprocal of the period:
f= 1 1 = = 167 105 Hz. . T 6.00s
(c) The magnetic energy does not depend on the direction of the current (since UB i2), so this will occur after one
Chapter 30
1. (a) The magnitude of the emf is
=
d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dt
c
h
bg
(b) Appealing to Lenzs law (especially Fig. 30-5(a) we see that the current flow in the loop is clockwise. Thus, the current is to left
Chapter 29
1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must be directed due south. Since B = 0i 2 r ,
i=
2 rB
0
=
2
b
4
m 39 10 6 T T m A
g c
h 16 A. =
(b) The current must be from west to east to produce a field wh
Chapter 28
1. (a) The force on the electron is
r r rr FB = qv B = q vx ? + v y j Bx ? + By j = q ( vx By v y Bx ) k i i = 1.6 1019 C 2.0 106 m s ( 0.15 T ) 3.0 106 m s ( 0.030 T ) 14 = 6.2 10 N k.
r r Thus, the magnitude of FB is 6.2 1014 N, and FB points
Chapter 27
1. (a) The energy transferred is U = Pt =
2t
r+R
=
(2.0 V) 2 (2.0 min) (60 s / min) = 80 J . 1.0 + 5.0
(b) The amount of thermal energy generated is
F I F 2.0 V I (5.0 ) (2.0 min) (60 s / min) = 67 J. U = i Rt = G JRt = G H+ R K H + 5.0 J 1.0
Chapter 26
1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is t
Chapter 25
1. (a) The capacitance of the system is
C=
q 70 pC = = 35 pF. . V 20 V
(b) The capacitance is independent of q; it is still 3.5 pF. (c) The potential difference becomes V =
q 200 pC = = 57 V. C 35 pF .
2. Charge flows until the potential differ
Chapter 24
1. If the electric potential is zero at infinity then at the surface of a uniformly charged sphere it is V = q/40R, where q is the charge on the sphere and R is the sphere radius. Thus q = 40RV and the number of electrons is
n= q e = 4 0 R V e
Chapter 23
r r 1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is
rr 2 = E A = EA cos = (1800 N C ) 3.2 103 m cos145 = 1.5 102 N m 2 C.
(
)
rr 2. W
Chapter 22
r r 1. (a) We note that the electric field points leftward at both points. Using F = q0 E , and orienting our x axis rightward (so points right in the figure), we find i
r N F = ( +1.6 1019 C ) 40 ? = (6.4 1018 N) i i C which means the magnitud
Chapter 21
1. Eq. 21-1 gives Coulombs Law, F = k
q1 q2 r2
, which we solve for the distance:
k | q1 | q2 | r= = F
(8.99 10 N m
9
2
C2 ) ( 26.0 106 C ) ( 47.0 106 C ) 5.70N
= 1.39m.
2. (a) With a understood to mean the magnitude of acceleration, Newtons se
Chapter 20
1. An isothermal process is one in which Ti = Tf which implies ln(Tf/Ti) = 0. Therefore, with Vf/Vi = 2.00, Eq. 20-4 leads to
Vf S = nR ln = ( 2.50 mol )( 8.31 J/mol K ) ln ( 2.00 ) = 14.4 J/K. Vi 2. From Eq. 20-2, we obtain
Q = T S = ( 405 K
Chapter 19
1. (a) Eq. 19-3 yields n = Msam/M = 2.5/197 = 0.0127 mol. (b) The number of atoms is found from Eq. 19-2: N = nNA = (0.0127)(6.02 1023) = 7.64 1021. 2. Each atom has a mass of m = M/NA, where M is the molar mass and NA is the Avogadro constant.
Chapter 18
1. Let TL be the temperature and pL be the pressure in the left-hand thermometer. Similarly, let TR be the temperature and pR be the pressure in the right-hand thermometer. According to the problem statement, the pressure is the same in the two
Chapter 17
1. (a) When the speed is constant, we have v = d/t where v = 343 m/s is assumed. Therefore, with t = 15/2 s being the time for sound to travel to the far wall we obtain d = (343 m/s) (15/2 s) which yields a distance of 2.6 km. (b) Just as the
1
Chapter 16
1. (a) The angular wave number is k =
2 2 = = 3.49 m 1. 1.80 m
(b) The speed of the wave is v = f =
(1.80 m ) (110 rad s ) = = 31.5 m s. 2 2
2. The distance d between the beetle and the scorpion is related to the transverse speed vt and longit
Chapter 15
1. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = 2xm, where is the angular frequency ( = 2f since there are 2 radians in one cycle). Therefore, in this circumstance, we obtain
am = 2 x