29 CHAPTER PARTICLES AND WAVES
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1.
(a) At the higher temperature, the intensity per unit wavelength is greater, and the aximum
occurs at a shorter wavelength (see Section 29.2).
2.
(b) An X-ray photon has a much greater frequency than does a microwave photon (see
Section 24.2). The X-ray photon also has a greater energy E, because E = hf (Equation
29.2), where h is Plancks constant and f is the frequency. The wavelength and frequency
of a photon are related by = c/f (Equation 16.1), where c is the speed of light in a vacuum.
Since the microwave photon has the smaller frequency, it has the greater wavelength.
3.
(c) A green photon has a greater frequency than does a red photon (see Section 24.2).
Therefore, the green photon possesses a greater energy E, because E = hf (Equation 29.2),
where h is Plancks constant and f is the frequency.
4.
(e) As the wavelength of a photon becomes smaller, its frequency and its energy become
larger (see Section 29.3). Since the energy of an incident photon is equal to the maximum
kinetic energy of the photoelectron plus the work function of the metal, increasing the
incident energy increases the maximum kinetic energy of the photoelectron. The work
function is a characteristic of the metal, and does not depend on the photon.
5.
(d) The maximum kinetic energy of the ejected photoelectrons depends on the energy of the
incident photons. Doubling the number of photons per second that strikes the surface does not
change the energy of the incident photons, which depends only on the frequency of the photons.
Thus, the maximum kinetic energy of the ejected photoelectrons does not change. However,
doubling the number of photons per second that strikes the surface means that twice as many
photoelectrons per second are ejected from the surface.
6.
(b) Whether or not electrons are ejected from the surface of the metal depends on the energy of
the incident photons and the work function of the metal The work function depends on the type
of metal (e.g., aluminum or copper) from which the plate is made. (See Section 29.3).
7.
W0 = 3.6 10 19 J
8.
KEmax = 2.7 10 20 J
9.
(d) According to the discussion in Section 29.4, a photon has a momentum whose
magnitude p is related to its wavelength by p = h/.
231 PARTICLES AND WAVES
10. (b) In the Compton effect, an X-ray photon strikes an electron and, like two billiard balls
(particles) colliding on a pool table, the X-ray photon scatters in one direction and the electron
recoils in another direction after the collision.
11. (a) In the Compton effect, some of the energy of the incident photon is given to the recoil
electron. Therefore, the energy of the scattered photon is less than that of the incident
photon. Since the energy of a photon depends inversely on its wavelength (see Section
29.3), the wavelength of the scattered photon is greater than that of the incident photon.
12. = 0.35 nm
13. (c) The de Broglie wavelength depends inversely on the magnitude p of the momentum;
= h/p (Equation 29.8), where h is Plancks constant. Particle A, having the smaller charge,
has the smaller electric potential energy (see Section 19.2). Consequently, after accelerating
through the potential difference, particle A has the smaller kinetic energy, and hence, the
smaller momentum. Thus, particle A has the longer de Broglie wavelength.
14. (e) The de Broglie wavelength depends inversely on the magnitude p of the momentum;
= h/p (Equation 29.8), where h is Plancks constant. Therefore, as the momentum
decreases, the wavelength increases, and vice versa. In A, the proton is moving opposite to
the direction of the electric field, so the proton is slowing down, and its momentum is
decreasing. In B, the proton is accelerating, and its momentum is increasing. In C, the
proton moves parallel to the magnetic field. According to the discussion in Section 21.2, the
proton does not experience a force, so its momentum remains constant. In D the proton is
moving perpendicular to the direction of the magnetic field. According to the discussion in
Section 21.3, such a situation does not change the magnitude of the protons momentum.
15. = 2.1 10 14 m
16. (b) According to the Heisenberg uncertainty principle, the uncertainty y in a particles
position is related to the uncertainty py in its momentum by (see Equation 29.10)
y h / ( 4 p y ) . If p = 0 kg m/s, then y becomes infinitely large.
y
Chapter 29 Problems
232
CHAPTER 29 PARTICLES AND WAVES
PROBLEMS
1.
REASONING AND SOLUTION The energy of a single photon is
34
E = hf = (6.63 10
6
26
J s)(98.1 10 Hz) = 6.50 10
J
The number of photons emitted per second is
Power radiated
5.0 10 4 W
=
=
Energy per photon
6.50 10 26 J
2.
7 .7 10 29 photons / s
REASONING The energy of a photon of frequency f is, according to Equation 29.2,
E = hf , where h is Planck's constant. Since the frequency and wavelength are related by
f = c / (see Equation 16.1), the energy of a photon can be written in terms of the
wavelength as E = hc / . These expressions can be solved for both the wavelength and the
frequency.
SOLUTION
a. The wavelength of the photon is
=
hc (6.63 10 34 J s)(3.00 10 8 m / s)
=
= 1.63 10 7 m
18
E
1.22 10
J
b. Using the answer from part (a), we find that the frequency of the photon is
f=
c 3.00 10 8 m / s
=
= 1.84 10 15 Hz
1.63 10 7 m
Alternatively, we could use Equation 29.2 directly to obtain the frequency:
f=
E
1.22 10 18 J
=
= 1.84 10 15 Hz
h 6.63 10 34 J s
c. The wavelength and frequency values shown in Figure 24.9 indicate that this photon
corresponds to electromagnetic radiation in the ultraviolet region of the electromagnetic
spectrum.
233 PARTICLES AND WAVES
3.
REASONING According to Equation 29.3, the work function W0 is related to the photon
energy hf and the maximum kinetic energy KEmax by W0 = hf KE max . This expression
can be used to find the work function of the metal.
SOLUTION KE max is 6.1 eV. The photon energy (in eV) is, according to Equation 29.2,
c
c
h
hf = 6.63 10 34 J s 3.00 10 15 Hz
F eV J
G
h 1.601 10 J I = 12.4 eV
H
K
19
The work function is, therefore,
W0 = hf KE max = 12.4 eV 6.1 ev = 6.3 eV
4.
REASONING According to Equation 29.3, the relation between the photon energy, the
maximum kinetic energy of an ejected electron, and the work function of a metal surface is
hf
K
{ = 1 E max
42 4
3
Photon
energy
Maximum
kinetic energy
of ejected
electron
+ W0
{
Work
function
Equation 16.1 relates the frequency f of a photon to its wavelength via f = c/, where c is
the speed of light in a vacuum. The maximum kinetic energy KE max is related to the mass m
1
and maximum speed vmax of the ejected electron by KEmax = 2 mv 2 (Equation 6.2). With
these substitutions, Equation 29.3 becomes
hf = KE max + W0
or
hc 1 2
= mv
+ W0
2 max
SOLUTION Solving Equation (1) for the wavelength gives
=
hc
1 mv 2
max
2
+ W0
( 6.63 1034 J s ) ( 3.00 108 m/s )
A =
=
2
1 9.11 10 31 kg 7.30 105 m/s + 4.80 1019 J
)(
)
2(
2.75 107 m
(1)
Chapter 29 Problems
( 6.63 1034 J s ) ( 3.00 108 m/s )
B =
=
2
1 9.11 1031 kg 5.00 105 m/s + 4.80 10 19 J
)(
)
2(
5.
234
3.35 107 m
REASONING The energy of the photon is related to its frequency by Equation 29.2,
E = hf . Equation 16.1, v = f , relates the frequency and the wavelength for any wave.
SOLUTION Combining Equations 29.2 and 16.1, and noting that the speed of a photon is
c, the speed of light in a vacuum, we have
=
6.
c
c
hc
(6.63 10 34 J s)(3.0 10 8 m / s)
=
=
=
= 3.1 10 7 m =
19
f
( E / h)
E
6.4 10
J
310 nm
REASONING The photons of this wave must carry at least enough energy to equal the
work function. Then the electrons are ejected with zero kinetic energy. Since the energy of a
photon is E = hf according to Equation 29.2, where f is the frequency of the wave, we have
that W0 = hf. Equation 16.1 relates the frequency to the wavelength according to f = c/,
where c is the speed of light. Thus, it follows that W0 = hc/.
SOLUTION Using Equations 29.2 and 16.1, we find that
c
c
h
h
6.63 10 34 J s 3.00 10 8 m / s
hc
W0 =
=
= 4 .10 10 19 J
485 10 9 m
Since 1 eV = 1.60 1019 J, it follows that
c
W0 = 4 .10 10 19 J
7.
REASONING AND SOLUTION
KE max = hf Wo =
6
c.63 10
=
F eV J
G
h1.601 10 J I =
H
K
c
h
J s 3.00 10 8 m / s
9
215 10 m
2 .56 eV
Equation 29.3 gives
hc
Wo
34
19
h b.68 eV g1.60 10 J I = 3.36 10
F
3
G 1 eV J
H
K
19
19
J
235 PARTICLES AND WAVES
Converting to electron volts
c
KE max = 3.36 10 19 J
8.
F
h1.601eV J I =
G 10 J
H
K
19
2 .10 eV
REASONING The intensity S = 680 W/m2 of the photons is equal to the total amount of
energy delivered by the photons per second per square meter of surface area. Therefore, the
intensity of the photons is equal to the energy E of one photon multiplied by the number N
of photons that reach the surface of the earth per second per square meter:
S = NE
(1)
The energy E of each photon is given by E = hf (Equation 29.2), where h = 6.631034 Js
c
is Plancks constant and f is the frequency of the photon. We will use f =
(Equation 16.1) to determine the frequency f of the photons from their wavelength and the
speed c of light in a vacuum.
SOLUTION Solving Equation (1) for N, we obtain N = S/E. Substituting E = hf
(Equation 29.2) into this result yields
N=
S
S
=
E hf
(2)
c
(Equation 16.1) into Equation (2), we find that
680 W/m 2 730 10 9 m
S
S
S
N=
=
=
=
hf
c hc
6.63 1034 J 3.00 108 m/s
s
h
Substituting f =
(
(
)(
)(
)
)
= 2.5 1021 photons/ ( s 2 )
m
9.
REASONING AND SOLUTION The number of photons per second, N, entering the owl's
eye is N = SA / E , where S is the intensity of the beam, A is the area of the owl's pupil, and
E is the energy of a single photon. Assuming that the owl's pupil is circular,
A = r2 =
cd h, where d is the diameter of the owl's pupil.
1
2
2
Combining Equations 29.2
and 16.1, we have E = hf = hc / . Therefore,
c
h
2
1
(5.0 10 13 W / m 2 ) 2 8.5 10 3 m (510 10 9 m)
SA
N=
=
=
hc
(6.63 10 34 J s)(3.0 10 8 m / s)
73 photons / s
Chapter 29 Problems
236
10. REASONING The wavelength of the light shining on the surface is related to the
maximum kinetic energy KEmax of the electrons ejected from the surface by
c
hf = KE max + W0 (Equation 29.3), where h is Plancks constant, f = (Equation 16.1) is
the frequency of the light, and W0 is the work function of the surface. Substituting Equation
16.1 into Equation 29.3, we see that
hf =
hc
= KE max + W0
(1)
The work function W0 is a property of the metal surface itself, so it remains the same for any
wavelength of incident light. When the wavelength of the incident light is 1 = 221 nm, the
maximum kinetic energy of the ejected electrons is KE max,1 = 3.281019 J, and when the
wavelength is 2, the maximum kinetic energy is twice as great: KE max,2 = 2KEmax,1. We
will use this information, with Equation (1), to determine the unknown wavelength 2.
SOLUTION Solving Equation (1) for the work function W0 yields
W0 = KE max +
hc
(2)
Because the work function W0 isnt affected by changing the wavelength of the incident
light from 1 to 2, we have that
W0 = KE max, 2 +
hc
hc
= KE max,1 +
2
1
KE max, 2 KE max, 1 +
or
hc hc
=
1 2
(3)
Solving Equation (3) for 2 and substituting KEmax,2 = 2 KEmax,1, we obtain
2 =
hc
hc
KE max, 2 KE max, 1 +
1
=
hc
hc
2KE max, 1 KE max, 1 +
1
=
1
KE max, 1
hc
+
1
1
(4)
In the last step of Equation (4), we have divided both the numerator and the denominator by
the product hc. Substituting the given values the into Equation (4), we find that
237 PARTICLES AND WAVES
2 =
1
( 3.28 10 J )
1
+
34
8
( 6.63 10 J s ) ( 3.00 10 m/s ) 221 109 m
19
= 1.62 107 m = 162 nm
11. REASONING The wavelength of the photon is related to its frequency f by = c/f
(Equation 16.1), where c is the speed of light. The frequency of the photon is proportional to
its energy E via f = E/h (Equation 29.2), where h is Plancks constant. Thus, = ch/E. The
photon energy is equal to the sum of the maximum kinetic energy KE max of the ejected
electron and the work function W0 of the metal; E = KE max + W0 (Equation 29.3).
Substituting this expression for E into = ch/E gives
=
ch
KE max + W0
(1)
2
The maximum kinetic energy is related to the maximum speed vmax by KE max = 1 mvmax
2
(Equation 6.2), where m is the mass of the electron.
2
SOLUTION Substituting KE max = 1 mvmax into Equation (1), and converting the work
2
function from electron-volts to joules, gives
=
ch
1 mv 2
max
2
+ W0
( 3.00 108
=
1
2
( 9.11 1031
m/s ) ( 6.63 1034 J s )
= 1.9 107 m
2
1.60 1019 J
kg ) ( 1.2 106 m/s ) + ( 2.3 eV )
1 eV
12. REASONING The total energy Q delivered by N photons is NE, where E is the energy
carried by one photon, so that N = Q/E. Equation 29.2 indicates that the photon energy is
E = hf, where h is Plancks constant and f is the frequency. Thus, the number N of photons
can be written as
QQ
N= =
(1)
E hf
Equation 16.1 relates the frequency f of a photon to its wavelength according to f = c/,
where c is the speed of light in a vacuum. Therefore, Equation (1) can be expressed as
Chapter 29 Problems
N=
Q Q
=
hf
hc
238
(2)
According to Equation 12.4, the heat Q required to raise the temperature of a substance by
an amount T is Q = cspecific heatm T, where cspecific heat is the specific heat capacity of the
substance and m is its mass.
SOLUTION Combining Equation 12.4 with Equation (2), the number of photons required
to raise the temperature by an amount T is
N=
Q
hc
or
N=
cspecific heat m T
hc
Applying this result to each type of photon, we obtain
( 6.0 105 m ) 840 J/ ( kg C ) ( 0.50 kg ) ( 2.0 C )
N infrared =
34
( 6.63 10 J s ) ( 3.00 108 m/s )
N blue =
( 4.7 107 m ) 840 J/ ( kg C ) ( 0.50 kg ) ( 2.0 C)
( 6.63 1034 J s ) ( 3.00 108 m/s )
= 2.5 1023
= 2.0 1021
13. SSM WWW REASONING AND SOLUTION
a. According to Equation 24.5b, the electric field can be found from E = S / ( 0 c ) . The
intensity S of the beam is
Energy per
unit time
Nh f
Nh c
S=
=
=
A
A
A
=
( 1.30 1018 photons/s ) ( 6.63 1034 J s ) 3.00 108 m/s
2
514.5 109 m
( 1.00 103 m )
= 1.60 105 W/m 2
where N is the number of photons per second emitted. Then,
E = S / ( 0c ) = 7760 N / C
239 PARTICLES AND WAVES
b. According to Equation 24.3, the average magnetic field is
B = E/c =
2.59 10 5 T
14. REASONING The heat required to melt the ice is given by Q = mL f , where m is the mass
of the ice and L f is the latent heat of fusion for water (see Section 12.8). Since, according
to Equation 29.2, each photon carries an energy of E = hf , the energy content of N photons
is E Total = Nhf . According to Equation 16.1, f = c / , so we have
E Total =
Nhc
If we assume that all of the photon energy is used to melt the ice, then, E Total = Q , so that
Nhc
= mL f
:
E
4
Q
Total
This expression may be solved for N to determine the required number of photons.
SOLUTION
a. We find that
N=
mL f
hc
=
(2.0 kg)(33.5 10 4 J / kg)(620 10 9 m)
= 2.1 10 24 photons
(6.63 10 34 J s)(3.00 10 8 m / s)
b. The number N of molecules in 2.0-kg of water is
N = (2.0 kg)
F 1 mol I F.022 10 molecules I = 6.7 10
6
J
G 10 kg J G 1 mol
K
18
H
KH
23
3
25
molecules
Therefore, on average, the number of water molecules that one photon converts from the ice
phase to the liquid phase is
N 6.7 10 25 molecules
=
= 32 molecules / photon
N
2.1 10 24 photons
15. SSM REASONING The angle through which the X-rays are scattered is related to the
difference between the wavelength of the scattered X-rays and the wavelength of the
incident X-rays by Equation 29.7 as
Chapter 29 Problems
=
240
h(
1 cos )
mc
where h is Plancks constant, m is the mass of the electron, and c is the speed of light in a
vacuum. We can use this relation directly to find the angle, since all the other variables are
known.
SOLUTION Solving Equation 29.7 for the angle , we obtain
cos = 1
mc
( )
h
( 9.111031 kg ) ( 3.00 108 m/s ) ( 0.2703 109 m 0.2685 109 m ) = 0.26
= 1
6.63 1034 J
s
= cos 1 ( 0.26 ) = 75
16. REASONING The momentum of the photon is related to its wavelength and Plancks
constant h. The momentum (nonrelativistic) of the ball depends on its mass m and speed v.
We can set the two momenta equal and solve directly for the speed.
SOLUTION The momentum pphoton of the photon and the momentum pball of the ball are
pphoton =
h
(29.6)
and
pball = mv
(7.2)
Since pphoton = pball, we have
h
=
mv
4
123
4
14 2 4
3
Momentum
of photon
Momentum
of ball
or
v=
h
6.63 1034 J
s
=
= 4.2 1025 m/s
9
3
m 720 10 m 2.2 10 kg
(
)(
)
17. REASONING The frequency f of a photon is related to its energy E by f = E/h
(Equation 29.2), where h is Plancks constant. As discussed in Section 29.4, the energy E is
related to the magnitude p of the photons momentum by E = pc, where c is the speed of
light in a vacuum. By combining these two relations, we see that the frequency can be
expressed in terms of p as f = pc/h.
SOLUTION
241 PARTICLES AND WAVES
a. Substituting values for p, c, and h, into the relation f = pc/h gives
f=
pc ( 2.3 1029 kg m/s ) ( 3.00 108 m/s )
=
= 1.0 1013 Hz
34
h
6.63 10
J
s
b. An inspection of Figure 24.9 shows that this frequency lies in the infrared region of the
electromagnetic spectrum.
18. REASONING Before the scattering, the electron is at rest and has no momentum. Thus, the
total initial momentum consists only of the photons momentum, which points along the + x
axis. The total initial momentum has no y component. Since the total momentum is
conserved, the total momentum after the scattering must be the same as it was before and,
therefore, has no y component.
The total momentum after the scattering is the sum of the momentum of the scattered
photon and that of the scattered electron, and it only has an x component. But the scattered
photon is moving along the y axis, so its momentum has no x component. Therefore, the
momentum of the electron must have an x component.
The total momentum after the scattering is the sum of the momentum of the scattered
photon and that of the scattered electron, and it has no y component. But the scattered
photon is moving along the y axis, so its momentum points along the y axis. Therefore,
this contribution to the total final momentum must be cancelled by part of the momentum of
the scattered electron, which must have a component along the +y axis.
SOLUTION Since the total momentum is conserved and since the scattered photon has no
momentum in the x direction, the momentum of the scattered electron must have an x
component that equals the momentum of the incident photon. According to Equation 29.6,
the magnitude p of the momentum of the incident photon is p = h/, where h is Plancks
constant and is the wavelength. Therefore, the momentum of the scattered electron has a
component in the +x direction that is
px =
h 6.63 10 34 J s
=
= 7 .37 10 23 kg m / s
9.00 10 12 m
The momentum of the scattered electron has a component along the + y direction. This
component cancels the momentum of the scattered photon that points along the y direction.
To find the momentum of the scattered photon, we first need to determine its wavelength
, which we can do using Equation 29.7:
Chapter 29 Problems
= +
242
h
1 cos 90.0
3
mc 14 2 4
=0
= 9.00 1012 m +
6.63 1034 J s
( 9.11 10
31
)(
8
kg 3.00 10 m/s
)
= 1.14 1011 m
Again using Equation 29.6, we find that the momentum of the scattered electron has a
component in the +y direction that is
py =
h 6.63 10 34 J s
=
= 5.82 10 23 kg m / s
1.14 10 11 m
19. REASONING There are no external forces that act on the system, so the conservation of
linear momentum applies. Since the photon is scattered at = 180 , the collision is
"head-on," and all motion occurs along the horizontal direction, which we take as the x axis.
The incident photon is assumed to be moving along the + x axis. For an initially stationary
electron, the conservation of linear momentum states that:
p
123
Momentum
of incident
photon
=
p + pelectron
14 2 43
14 2 4
3
Momentum
of scattered
photon
Momentum
of recoil
electron
where the momentum of the scattered photon is negative since is moves along the x
direction (the scattering angle is 180). Using the relation p = h/ (Equation 29.6), where h is
Plancks constant and is the wavelength of the photon, we can write the expression for the
momentum of the electron as
pelectron = p + p =
hh
1 1
+
= h +
SOLUTION Substituting numerical values into the equation above, we have
1
1
pelectron = (6.626 10 34 J s)
+
= 4.755 10 24 kg m/s
9
9
0.2750 10 m 0.2825 10 m
243 PARTICLES AND WAVES
20. REASONING The wavelength of the incident X-rays is related to the wavelength of
h
the scattered X-rays by =
( 1 cos ) (Equation 29.7), where h = 6.6261034 Js is
mc
8
Plancks constant, c = 2.99810 m/s is the speed of light in a vacuum, m = 9.1091031 kg
is the mass of an electron, and = 122.0 is the angle at which the X-rays are scattered.
The wavelength of the scattered photon is found from the magnitude p of its
h
momentum via p =
(Equation 29.6).
SOLUTION Solving =
h
( 1 cos ) (Equation 29.7) for , we obtain
mc
=
h
( 1 cos )
mc
(1)
h
h
(Equation 29.6) for yields = , which, on substitution into
p
Equation (1), gives
Solving p =
=
h
h
h
( 1 cos ) = ( 1 cos )
mc
p mc
( 6.626 1034 J s ) ( 1 cos122.0o )
=
1.856 1024 kg m/s ( 9.109 1031 kg ) ( 2.998 108 m/s )
6.626 1034 J
s
= 3.533 1010 m = 0.3533 nm
21. REASONING The change in wavelength that occurs during Compton scattering is given by
Equation 29.7:
=
h
( 1 cos )
mc
or
( ) max =
h
2h
( 1 cos 180) =
mc
mc
( ) max
is the maximum change in the wavelength, and to calculate it we need a value
for the mass m of a nitrogen molecule. This value can be obtained from the mass per mole
M of nitrogen ( N 2 ) and Avogadro's number N A , according to m = M / N A (see
Section 14.1).
SOLUTION Using a value of M = 0.0280 kg/mol , we obtain the following result for the
maximum change in the wavelength:
Chapter 29 Problems
( ) max
(
244
)
2 6.63 10 34 J
s
2h
2h
=
=
=
mc
M
0.0280 kg/mol
3.00 108 m/s
N c
23
1
A
6.02 10 mol
(
)
= 9.50 10 17 m
22. REASONING Energy is conserved during the collision. This means that the energy E of
the incident photon must equal the kinetic energy KE of the recoil electron plus the energy
E the scattered photon:
E = KE + E
(1)
The energy E of a photon is related to its frequency f by E = hf (Equation 29.2), where h is
Plancks constant. The frequency, in turn, is related to the wavelength by f = c/
(Equation 16.1), where c is the speed of light in a vacuum. Substituting f = c/ into E = hf
gives
(2)
E = hc/
The wavelength of the scattered photon depends on the wavelength of the incident
photon according to Equation 29.7, so that we have
= +
h
( 1 cos )
mc
Since the photon is scattered straight backward, = 180, and
= +
h
( 1 cos180 ) = + 2h
mc
mc
(3)
SOLUTION The kinetic energy of the recoil electron is given by KE = 1 mv 2
2
(Equation 6.2), where m is its mass and v is its speed. Substituting this expression into
Equation (1), we have E = 1 mv 2 + E . Solving for the speed v of the electron gives
2
v=
2 ( E E)
m
(4)
From Equation (2), we also know that E = hc/ and E = hc / . Substituting this expression
into Equation (4), we find that the speed of the electron can be written as
v=
2h c 1 1
m
245 PARTICLES AND WAVES
Since = +
v=
=
2h
[Equation (3)], the speed of the electron is
mc
2h c 1
1
2h
m
+
mc
2 ( 6.63 1034 J s ) ( 3.00 108 m/s )
9.11 1031 kg
1
1
9
34
2 ( 6.63 10
J s )
0.45000 10 m 0.45000 109 m +
31
8
( 9.11 10 kg ) ( 3.00 10 m/s )
= 3.22 106 m/s
23. REASONING AND SOLUTION
a. We have
= ' (h/mc)(1 cos 163) =
0.1819 nm
b. For the incident photon
E = hf = hc/ =
1.092 10 15 J
c. For the scattered photon
E ' = hf ' = hc/' = 1.064 10 15 J
d. The kinetic energy of the recoil electron is, therefore,
KE = E E ' =
2.8 10 17 J
24. REASONING
a. Consider one square meter of the sails surface. Each of the N photons that strike this
square meter in a one-second interval ( t = 1.0 s) has an initial momentum p that is
h
determined by its wavelength , according to p =
(Equation 29.6), where
34
h = 6.6310 Js is Plancks constant. Each photon is fully reflected, so the final
momentum of a photon is equal to p. The magnitude of the change p in a photons
Chapter 29 Problems
246
momentum, then, is equal to p = |p (p)| = 2p, and the magnitude P of the net
momentum change undergone by all N photons is given by P = N p = 2Np. In order to
cause this momentum change, the sail exerts an impulse of magnitude F t = P
(Equation 7.4) on the photons, where F is the magnitude of the force exerted on the photons
per square meter of the sail. By Newtons Third Law, that force magnitude is equal to the
magnitude of the force exerted by the N photons on one square meter of the sail. From
Equation 7.4, we then, have that
F t = P = 2 Np
(1)
Lastly, the magnitude F = F of the net force necessary for the sail to attain the desired
acceleration of a = 9.8106 m/s2 is given by Newtons Second Law, F = ma (Equation
4.1), where m = 3.0103 kg is the mass of one square meter of the sail. As instructed, we
have ignored all other forces acting on the sail.
b. The intensity S of the laser beam depends on the total energy delivered to the sail by
Total energy
S=
(Equation 24.4), where A is the area of the sail and t is the time
t A
interval. The total energy is equal to the number N of photons that strike the area in one
second times the energy E of a single photon. Therefore, the intensity of the laser beam is
S=
Total energy NE
=
t A
t A
(2)
c
8
(Equation 16.1) is the photons frequency and c = 3.0010 m/s is the speed of light in a
vacuum.
The energy E of a single photon is given by E = hf (Equation 29.2), where f =
SOLUTION
a. Solving Equation (1) for N, we obtain
N=
Substituting p =
F t
2p
h
(Equation 29.6) and F = ma (Equation 4.1) into Equation (3) yields
(3)
247 PARTICLES AND WAVES
N=
F t mat mat
=
=
2p
2h
h
2
( 3.0 103 kg ) ( 9.8 106 m/s2 ) ( 1.00 s ) ( 225 109 m ) = 5.0 1018
=
2 ( 6.63 1034 J s )
hc
c
(Equation 16.1) into E = hf (Equation 29.2) gives E = hf =
.
Substituting this result into Equation (2), we obtain
b. Substituting f =
S=
NE
Nhc
=
t A t A
(4)
Equation (4) applies to the intensity reaching an area A = 1.0 m2 of the sail in a time
t = 1.0 s. Therefore, in order for the sail to accelerate at the desired rate, the intensity of the
laser must be
S=
Nhc ( 5.0 1018 ) ( 6.63 1034 J s ) ( 3.00 108 m/s )
=
= 4.4 W/m 2
2) (
9
t A
( 1.00 s ) ( 1.00 m 225 10 m )
25. REASONING AND SOLUTION The de Broglie wavelength is given by Equation 29.8 as
= h / p , where p is the magnitude of the momentum of the particle. The magnitude of the
momentum is p = mv , where m is the mass and v is the speed of the particle. Using this
expression in Equation 29.8, we find that = h / ( mv ) , or
v=
6.63 10 34 J s
h
=
= 3.05 10 7 m / s
27
14
m
1.67 10 kg 1.30 10 m
c
h
c
h
The kinetic energy of the proton is
1
1
KE = 2 mv 2 = 2 ( 1.67 10 27 kg)(3.05 10 7 m / s) 2 = 7.77 10 13 J
26. REASONING According to Equation 27.1, the angle that locates the first-order bright
fringes (m = 1) is specified by sin = /d, where is the wavelength and d is the separation
between the slits. The wavelength of the electron is the de Broglie wavelength, which is
given by = h/p (Equation 29.8), where h is Plancks constant and p is the magnitude of the
momentum of the electron.
Chapter 29 Problems
248
SOLUTION Combining Equations 27.1 and 29.8, we find that the angle locating the firstorder bright fringes is specified by
sin =
h
=
d pd
Dividing this result for case A by that for case B, we find
sin A h / ( pA d ) pB
=
=
sin B h / ( pBd ) pA
or
pB =
( 1.2 1022 kg m/s ) sin ( 1.6 104 degrees ) =
pB =
sin ( 4.0 104 degrees )
pA sin A
sin B
4.8 1023 kg m/s
27. REASONING AND SOLUTION The de Broglie wavelength is given by Equation 29.8 as
= h/p, where p is the magnitude of the momentum of the particle. The magnitude of the
momentum is p = mv, where m is the mass and v is the speed of the particle. Using this
expression in Equation 29.8, we find that
=
h
mv
v=
or
6.63 10 34 J s
h
=
= 1.41 10 3 m / s
27
9
m
1.67 10 kg 0.282 10 m
c
h
c
h
28. REASONING AND SOLUTION We know that = h/mv. Solving for the mass yields
m=
6.63 10 34 J s
h
=
=
v
8.4 10 14 m 1.2 10 6 m / s
c
h
c
h
6.6 10 27 kg
29. REASONING AND SOLUTION The average kinetic energy of a helium atom is
23
KE = (3/2)kT = (3/2)(1.38 10
21
J/K)(293 K) = 6.07 10
The speed of the atom corresponding to the average kinetic energy is
v=
6
b g= 2c.07 10 J h= 1.35 10
2 KE
m
The de Broglie wavelength is
21
6.65 10 27 kg
3
m/s
J
249 PARTICLES AND WAVES
=
6.63 10 34 J s
h
=
=
mv
6.65 10 27 kg 1.35 10 3 m / s
c
c
h
h
7 .38 10 11 m
30. REASONING The de Broglie wavelength of a particle is inversely proportional to the
h
magnitude p of its momentum, as we see from =
(Equation 29.8), where
p
h = 6.631034 Js is Plancks constant. The electron is moving at a speed v = 0.88c, which
is close to the speed of light in a vacuum ( c = 3.00108 m/s). Therefore, we will use
mv
p=
v 2 (Equation 28.3) to determine the magnitude of the electrons relativistic
1 2
c
momentum, where m = 9.111031 kg is the electrons mass.
SOLUTION Substituting Equation 28.3 into Equation 9.8, we find that
=
=
h
h
=
p mv
2
1 v
c2
=
h
v2
1 2
mv
c
2
6.63 1034 J s
( 9.11 1031 kg ) ( 0.88) ( 3.00 108 m/s )
0.88 c
12
1
= 1.3 10 m
c
31. SSM REASONING The de Broglie wavelength is related to Plancks constant h and the
magnitude p of the particles momentum. The magnitude of the momentum can be related to
the particles kinetic energy. Thus, using the given wavelength and the fact that the kinetic
energy doubles, we will be able to obtain the new wavelength.
SOLUTION The de Broglie wavelength is
=
h
p
(29.8)
The kinetic energy and the magnitude of the momentum are
1
2
KE = mv 2
(6.2)
p = mv
(7.2)
Chapter 29 Problems
250
where m and v are the mass and speed of the particle. Substituting Equation 7.2 into
Equation 6.2, we can relate the kinetic energy and momentum as follows:
1
2
KE = mv 2 =
m2 v 2 p 2
=
2m
2m
p = 2m ( KE )
or
Substituting this result for p into Equation 29.8 gives
=
h
h
=
p
2m ( KE )
Applying this expression for the final and initial wavelengths f and i, we obtain
f =
h
2m ( KE ) f
h
i =
and
2m ( KE ) i
Dividing the two equations and rearranging reveals that
h
f
=
i
2m ( KE ) f
=
h
2m ( KE ) i
( KE ) i
( KE ) f
f = i
or
( KE ) i
( KE ) f
Using the given value for i and the fact that KE f = 2 ( KEi ) , we find
f = i
( KE ) i
( KE ) f
(
= 2.7 10 10 m
)
(
KE i
2 KEi
)
= 1.9 1010 m
h
34
(Equation 29.8), where p is the magnitude of her momentum and h = 6.6310 Js is
Plancks constant. We will use p = mv (Equation 7.2) to determine the magnitude p of the
womans momentum from her mass m and her speed v at the instant she strikes the water.
32. REASONING
The de Broglie wavelength of the woman is found from p =
Once the woman jumps from the cliff, she is in free fall with an initial speed of v0 = 0 m/s,
and an acceleration a = 9.8 m/s2. Since we have taken upward to be the positive direction,
her displacement during the fall is H = 9.5 m. Her final speed v, then, is given by
251 PARTICLES AND WAVES
2
2
v = v0 + 2aH
(2.9)
h
h
(Equation 29.8) for yields = . Substituting p = mv
p
(Equation 7.2) into this result, we obtain
SOLUTION
Solving p =
=
h
h
=
p mv
(1)
Substituting v0 = 0 m/s into Equation (2.9) and taking the square root of both sides, we find
that
v 2 = ( 0 m/s ) + 2aH
2
or
v 2 = 2aH
or
v = 2aH
(2)
Substituting Equation (2) into Equation (1), we find the de Broglie wavelength of the
woman at the instant she strikes the water to be:
h
h
6.63 1034 J s
=
=
=
= 1.2 1036 m
2
mv m 2aH ( 41 kg ) 2 ( 9.8 m/s ) ( 9.5 m )
33. REASONING When the electron is at rest, it has electric potential energy, but no kinetic
energy. The electric potential energy EPE is given by EPE = eV (Equation 19.3), where e is
the magnitude of the charge on the electron and V is the potential difference. When the
electron reaches its maximum speed, it has no potential energy, but its kinetic energy is
1 mv 2
. The conservation of energy states that the final total energy of the electron equals the
2
initial total energy:
1 mv 2 =
eV
12
12 2 4
4 3 Initial 3
total
Final total
energy
energy
Solving this equation for the potential difference gives V = mv 2 / ( 2e ) .
The speed of the electron can be expressed in terms of the magnitude p of its momentum by
v = p/m (Equation 7.2). The magnitude of the electrons momentum is related to its
de Broglie wavelength by p = h/ (Equation 29.8), where h is Plancks constant. Thus, the
speed can be written as v = h/(m). Substituting this expression for v into V = mv 2 / ( 2e )
gives V = h 2 / ( 2me 2 ) .
Chapter 29 Problems
252
SOLUTION The potential difference that accelerates the electron is
( 6.63 1034 J s )
h2
V=
=
= 1.86 104 V
2me 2 2 ( 9.11 1031 kg ) ( 1.60 1019 C ) ( 0.900 1011 m ) 2
2
34. REASONING The linear momentum p of a particle is given by p = mv (Equation 7.2),
where m and v are its mass and velocity. Since particle A is initially at rest, its momentum is
zero. The initial momentum of particle B is p0B = mBv0B. This is also the total initial linear
momentum of the two-particle system.
After the collision the combined mass of the two particles is mA+ mB, and the common
velocity is vf .Thus, the total linear momentum of the system after the collision is
pf = (mA+ mB)vf . From Section 7.2, we know that the total linear momentum of an isolated
system is conserved. An isolated system is one in which the vector sum of the external
forces acting on the system is zero. Since there are no external forces acting on the particles,
the two-particle system is an isolated system. Thus, the total linear momentum of the system
after the collision equals the total linear momentum before the collision.
The de Broglie wavelength is inversely related to the magnitude p of a particles
momentum by = h/p (Equation 29.8), where h is Plancks constant.
SOLUTION The de Broglie wavelength f of the object that moves off after the collision is
given by f = h/pf (Equation 29.8). Since momentum is mass times velocity, the magnitude
of the momentum that the object has after the collision is pf = (mA + mB)vf, where vf is the
common speed of the two particles. We can evaluate this momentum by using the law of
conservation of momentum, which indicates that the total momentum after the collision is
the same as it is before the collision. Before the collision only particle B is moving, so that
the magnitude of the total momentum at that time has a value of mBv0B, where v0B is the
initial speed of particle B. Assuming the particles travel along the + x axis, we write the
conservation of linear momentum as follows:
+ ( mA + mB ) vf = +mBv0B
14243
1 4 4 2 4 43
Total momentum
after collision
Total momentum
before collision
Using this result, we find that the desired de Broglie wavelength is
f =
h
h
h
=
=
pf ( mA + mB ) vf mBv0B
253 PARTICLES AND WAVES
But the term on the far right is just the given de Broglie wavelength of the incident
particle B. Therefore, we conclude that f = 2.0 1034 m .
35. SSM REASONING The de Broglie wavelength of the electron is related to the
magnitude p of its momentum by = h/p (Equation 29.8), where h is Plancks constant. If
the speed of the electron is much less than the speed of light, the magnitude of the electrons
momentum is given by p = mv (Equation 7.2). Thus, the de Broglie wavelength can be
written as = h/(mv).
When the electron is at rest, it has electric potential energy, but no kinetic energy. The
electric potential energy EPE is given by EPE = eV (Equation 19.3), where e is the
magnitude of the charge on the electron and V is the potential difference. When the electron
reaches its maximum speed, it has no potential energy, but its kinetic energy is 1 mv 2 . The
2
conservation of energy states that the final total energy of the electron equals the initial total
energy:
1 mv 2 =
eV
12
12 2 4
4 3 Initial 3
total
Final total
energy
energy
Solving this equation for the final speed gives v = 2eV / m . Substituting this expression
for v into = h/(mv) gives = h / 2meV .
SOLUTION After accelerating through the potential difference, the electron has a
de Broglie wavelength of
=
h
2meV
=
6.63 1034 J s
2 ( 9.11 1031 kg ) ( 1.60 1019 C ) ( 418 V )
= 6.01 1011 m
36. REASONING AND SOLUTION The energy of the photon is E = hf = hc/photon , while the
kinetic energy of the particle is KE = (1/2)mv2 = h2/(2m 2). Equating the two energies and
rearranging the result gives photon / = (2mc/h). Now the speed of the particle is v = 0.050c,
so = h/(0.050 mc), and
photon / = 2/0.050 = 4.0 10 1
37. REASONING AND SOLUTION According to the uncertainty principle, the minimum
uncertainty in the momentum can be determined from p y y = h / ( 4 ) . Since p y = mv y , it
Chapter 29 Problems
254
follows that p y = mv y . Thus, the minimum uncertainty in the velocity of the oxygen
molecule is given by
h
6.63 10 34 J s
v y =
=
= 8.3 10 6 m/s
26
3
4 m y 4 5.3 10 kg 0.12 10 m
(
)(
)
38. REASONING When particles pass through the slit the great majority fall on the screen
between the first dark fringes on either side of the central bright fringe. The first dark fringes
are located by the angles (below the midpoint) and + (above the midpoint), so this is
the minimum range of angles over which the particles spread out. As Figure 29.15 shows,
p y
1
the angle is found from = tan
p (Equation 1.4), where py is the uncertainty in
x
the y component py of a particles momentum, and px is the x component of a particles
momentum. The given de Broglie wavelength = 0.200 mm of the particles is the
wavelength they possess before passing through the slit, when their momentum has only the
x component px. Therefore, the x component of each particles momentum is given by
px =
h
(Equation 29.8), where h = 6.631034 Js is Plancks constant.
We are given that the uncertainty in the position of each particle along the y direction is
equal to one-half the width of the slit: y = 1 W . The minimum uncertainty in the y
2
component py of a particles momentum, then, is found from the Heisenberg uncertainty
principle:
h
p y ( y ) =
(29.10)
4
(
)
SOLUTION Solving Equation 29.10 for py and substituting y = 1 W , we obtain
2
p y =
Substituting Equation (1) and px =
we find that
h
h
h
=
=
1W
4 ( y ) 4
2 W
2
(
)
p y
h
1
(Equation 29.8) into = tan
p
x
(1)
(Equation 1.6),
255 PARTICLES AND WAVES
p y
= tan 1
p
x
h
1 2 W
= tan
h
= tan 1
2 W
633 109 m
= tan 1
3
2 ( 0.200 10 m )
= 0.0289o
Therefore, the particles spread out over the range 0.0289 to +0.0289 .
39. SSM WWW REASONING The uncertainty in the electrons position is y = 3.0
15
10
m. The minimum uncertainty py in the y component of the electrons momentum is
given by the Heisenberg uncertainty principle as p y = h / ( 4 y ) (Equation 29.10).
15
SOLUTION Setting y = 3.0 10
m in the relation p y = h / ( 4 y ) gives
h
6.63 1034 J s
p y =
=
= 1.8 1020 kg m/s
15
4 y 4 ( 3.0 10
m)
40. REASONING Suppose the object is moving along the + y axis. The uncertainty in the
objects position is y = 2.5 m. The minimum uncertainty py in the objects momentum is
specified by the Heisenberg uncertainty principle (Equation 29.10) in the form ( py)
( y) = h/(4). Since momentum is mass m times velocity v, the uncertainty in the velocity
v is related to the uncertainty in the momentum by v = ( py)/m.
SOLUTION
a. Using the uncertainty principle, we find the minimum uncertainty in the momentum as
follows:
h
p y ( y ) =
4
(
)
py =
h
6.63 1034 J
s
=
= 2.11035 kg m/s
4 y
4 ( 2.5 m )
b. For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocity that is
given by
p y 2.1 1035 kg m/s
v y =
=
= 4.7 1034 m/s
m
0.045 kg
Chapter 29 Problems
256
c. For an electron this uncertainty in momentum corresponds to an uncertainty in velocity that is
given by
p y 2.11035 kg m/s
v y =
=
= 2.3 105 m/s
m
9.111031 kg
41. SOLUTION The minimum uncertainty y in the position of the particle is related to the
minimum uncertainty p y in the momentum via the Heisenberg uncertainty principle. To
cast this relationship into a form that gives us the desired percentage for the minimum
uncertainty in the speed, we note that the minimum uncertainty in the position is specified as
the de Broglie wavelength . We can then express the de Broglie wavelength in terms of
Plancks constant h and the magnitude py of the particles momentum. The magnitude of the
momentum is related to the mass m and the speed vy of the particle.
SOLUTION The percentage minimum uncertainty in the speed is
Percentage =
v y
vy
100%
(1)
According to the Heisenberg uncertainty principle, the minimum uncertainty p y in the
momentum and the minimum uncertainty y in the position of the particle are related
according to
h
p y ( y ) =
(29.10)
4
(
)
We know that y is equal to the de Broglie wavelength = h / p y (Equation 29.8), where
the magnitude of the momentum is p y = mv y (Equation 7.2). Thus, we have
y = =
h
h
=
p y mv y
Substituting this result for y into Equation 29.10, we obtain
h
( p y ) ( y ) = ( p y ) mv
y
h
=
4
(2)
The last step in our transformation of the uncertainty principle is to realize that
p y = mv y = mv y , since the mass is constant. Substituting this expression for p y
(
)
into Equation (2) shows that
257 PARTICLES AND WAVES
(
h
py
mv y
)
= mv y
(
h
mv y
)
h
=
4
v y
or
vy
=
1
4
Using this result in Equation (1), we find that
Percentage =
v y
vy
100% =
1
100% = 8.0%
4
42. REASONING The mass m of the particle is related to its rest energy E0 by E0 = mc 2
(Equation 28.5). Therefore, if there is a minimum uncertainty E0 in measuring the rest
energy of the particle, there will be a corresponding uncertainty m in measuring its mass:
E0 = ( m ) c 2
m =
or
E0
(1)
c2
The minimum uncertainty E0 in the particles rest energy is related to the length of time t
the particle exists in a state by the Heisenberg uncertainty principle:
( E0 ) ( t ) = 4h
(Equation 29.11), where h = 6.631034 Js is Plancks constant.
SOLUTION Solving Equation 29.11 for the minimum uncertainty E0, we obtain
E0 =
h
4 ( t )
(2)
Substituting Equation (2) into Equation (1), we find that
m =
E0
c
2
h
6.63 1034 J
s
=
=
2
4 ( t ) c
4 7.4 1020 s 3.00 108 m/s
(
)(
)
2
= 7.9 1033 kg
Chapter 29 Problems
258
43. REASONING In order for the person to diffract to the same extent as the sound wave, the
de Broglie wavelength of the person must be equal to the wavelength of the sound wave.
SOLUTION
a. Since the wavelengths are equal, we have that
sound = person
sound =
h
mperson v person
Solving for v person , and using the relation sound = v sound / f sound (Equation 16.1), we have
v person =
=
h
mperson ( v sound / f sound )
=
h f sound
mperson v sound
( 6.63 10 34 J s)(128 Hz )
=
( 55.0 kg)( 343 m / s)
4 .50 10 36 m / s
b. At the speed calculated in part (a), the time required for the person to move a distance of
one meter is
t=
x
1.0 m
=
v
4.50 10 36 m / s
F1.0 h I F1 day I F 1 year I =
G s J G h J G days J
H KH KH K
3600
24.0
365.25
7.05 10 27 years
Factors to convert
seconds to years
44. REASONING The energy of a photon is related to its frequency and Plancks constant.
The frequency, in turn, is related to the speed and wavelength of the light. Thus, we can
relate the energy to the wavelength. The given relationship between the wavelengths will
then allow us to determine the unknown energy.
SOLUTION The energy E of a photon with frequency f is
E = hf
(29.2)
where h is Plancks constant. The frequency is related to the speed c and wavelength of
the light according to
c
f=
(16.1)
Substituting this expression for f into Equation 29.2 gives
259 PARTICLES AND WAVES
E = hf = h
c
Applying this result to both sources, we have
EB = h
c
B
EA = h
and
c
A
Dividing the two expressions gives
c
EB
B A
=
=
EA h c
B
A
h
Using the given value for EA and the fact that B = 3A in this result shows that
EB = EA
A
= 2.1 1018 J A = 7.0 10 19 J
3
B
A
(
)
45. REASONING The de Broglie wavelength is related to Plancks constant h and the
magnitude p of the particles momentum. The magnitude of the momentum is related to the
mass m and the speed v at which the bacterium is moving. Since the mass and the speed are
given, we can calculate the wavelength directly.
SOLUTION The de Broglie wavelength is
h
p
=
(29.8)
The magnitude of the momentum is p = mv (Equation 7.2), which we can substitute into
Equation 29.8 to show that the de Broglie wavelength of the bacterium is
=
h
h
6.63 1034 J
s
=
=
= 11018 m
15
p mv
2 10
kg ( 0.33 m/s )
(
)
Chapter 29 Problems
260
46. REASONING AND SOLUTION
a. We know E = hc/ for a photon. The energy of the photon is
F 10 J I = 8.0 10
1.60
E = 5.0 eV G
H 1 eV J
K
19
19
J
The wavelength is
c
h
c
6.63 10 34 J s 3.00 10 8 m / s
hc
=
=
E
8.0 10 19 J
h=
2 .5 10 7 m
b. The speed of the 5.0-eV electron is
v=
2E
=
m
c
2 8.0 10 19 J
9.11 10
31
h= 1.3 10
kg
6
m/s
The de Broglie wavelength is
=
6.63 10 34 J s
h
=
= 5.6 10 10 m
31
6
mv
9.11 10 kg 1.3 10 m / s
c
c
h
h
47. REASONING AND SOLUTION In the first case, the energy of the incident photon is given
by Equation 29.3 as
hf = KE max + W0 = 0.68 eV + 2.75 eV = 3.43 eV
In the second case, a rearrangement of Equation 29.3 yields
KE max = hf W0 = 3.43 eV 2.17 eV =
1.26 eV
48. REASONING The speed v of a particle is related to the magnitude p of its momentum by
v = p/m (Equation 7.2). The magnitude of the momentum is related to the particles
de Broglie wavelength by p = h/ (Equation 29.8), where h is Plancks constant. Thus,
the speed of a particle can be expressed as v = h/(m) . We will use this relation to find the
speed of the proton.
SOLUTION The speeds of the proton and electron are
vproton =
h
mproton proton
and
velectron =
h
melectron electron
261 PARTICLES AND WAVES
Dividing the first equation by the second equation, and noting that electron= proton, we
obtain
vproton
m
m
= electron electron = electron
velectron
mproton proton
mproton
Using values for melectron and mproton taken from the inside of the front cover, we find that
the speed of the proton is
m
vproton = velectron electron
mproton
9.111031 kg
(
6
3
= 2.45 10 m/s
= 4.50 10 m/s )
27
kg
1.67 10
49. REASONING The width of the central bright fringe in the diffraction patterns will be
identical when the electrons have the same de Broglie wavelength as the wavelength of the
photons in the red light. The de Broglie wavelength of one electron in the beam is given by
Equation 29.8, electron = h / p , where p = mv .
SOLUTION Following the reasoning described above, we find
red light = electron
red light =
h
melectron v electron
Solving for the speed of the electron, we have
v electron =
h
6.63 10 34 J s
=
= 1.10 10 3 m / s
melectron red light (9.11 10 31 kg)(661 10 9 m)
50. REASONING We will first calculate the potential energy of the system at each of the two
separations, and then find the energy difference for the two configurations. Since the
electric potential energy lost by the system is carried off by a photon that is emitted during
the process, the energy difference must be equal to the energy of the photon. The
wavelength of the photon can then by found using Equation 29.2 with Equation 16.1:
E = hc / .
Chapter 29 Problems
262
SOLUTION The initial potential energy of the system is (see Equations 19.3 and 19.6)
FI
kq
GJ
r
HK
L 10
(8.99
= ( 1.6 10 C) M
N
EPE 1 = eV1 = e
1
19
9
O
P
Q
N m 2 / C 2 )( 8.30 10 6 C)
= 2 .84 10 14 J
0.420 m
The final potential energy is
EPE 2 = eV2 = ( 1.6 10 19 C)
L 10
(8.99
M
N
O
P
Q
N m 2 / C 2 )( 8.30 10 6 C)
= 7 .56 10 15 J
1.58 m
9
The energy difference, and therefore the energy of the emitted photon, is
E = EPE 1 EPE 2 = 2 .84 10 14 J 7 .56 10 15 J = 2 .08 10 14 J
The wavelength of this photon is
=
hc (6.63 10 34 J s)(3.00 10 8 m / s)
=
= 9.56 10 12 m
E
2.08 10 14 J
51. REASONING Since the net external force acting on the system (the photon and the
electron) is zero, the conservation of linear momentum applies. In addition, there are no
nonconservative forces, so the conservation of total energy applies as well. Since the
photon scatters at an angle of = 180.0 in Figure 29.10, the collision is "head-on." Thus,
the motion takes place entirely along the horizontal direction, which we will take as the x
axis, with the right as being the positive direction.
The conservation of linear momentum gives rise to Equation 29.7, which relates the
difference between the scattered and incident X-ray photon wavelengths to the
scattering angle of the electron as
=
b
g
b
g
h
h
2h
1 cos =
1 cos 180.0 =
mc
mc
mc
(1)
The conservation of total energy is written as
hc
4
Energy
of incident
photon
+
0=
{
Initial
kinetic
energy
of electron
hc
4
Energy
of scattered
photon
+
1
mv 2
2
Final
kinetic
energy
of electron
(2)
263 PARTICLES AND WAVES
Equations (1) and (2) will permit us to find the wavelength of the incident X-ray photon.
SOLUTION Solving Equation (1) for and substituting the result into Equation (2) gives
hc
=
hc
+
2h
+
mc
1
2
mv 2
Algebraically rearranging this result, we obtain a quadratic equation for :
2 +
2
Fh I
GJ
H K
mc
4 .85 10 12 m
2h 2
c h=0
m 1 mv 2
2
9 . 70 10 20 m 2
34
31
8
where we have used h = 6.63 10 J s, m = 9.11 10 kg, c = 3.00 10 m/s, and
v = 4.67 106 m/s. Solving this quadratic equation for , we obtain
= 3.09 10 10 m