Chapter 27
Quantum Physics
Quick Quizzes
1.
(b). Some energy is transferred to the electron in the scattering process. Therefore, the
scattered photon must have less energy (and hence, lower frequency) than the incident
photon.
2.
(c). Conservation of energy requires the kinetic energy given to the electron be equal to the
difference between the energy of the incident photon and that of the scattered photon.
3.
(c). Two particles with the same de Broglie wavelength will have the same momentum
p = mv . If the electron and proton have the same momentum, they cannot have the same
speed because of the difference in their masses. For the same reason, remembering that
KE = p 2 2m , they cannot have the same kinetic energy. Because the kinetic energy is the
only type of energy an isolated particle can have, and we have argued that the particles
have different energies, the equation f = E h tells us that the particles do not have the
same frequency.
4.
(b). The Compton wavelength, λC = h me c , is a combination of constants and has no
relation to the motion of the electron. The de Broglie wavelength, λ = h me v , is associated
with the motion of the electron through its momentum.
393
394
CHAPTER 27
Answers to Even Numbered Conceptual Questions
2.
A microscope can see details no smaller than the wavelength of the waves it uses to
produce images. Electrons with kinetic energies of several electron volts have wavelengths
of less than a nanometer, which is much smaller than the wavelength of visible light
(having wavelengths ranging from about 400 to 700 nm). Therefore, an electron
microscope can resolve details of much smaller sizes as compared to an optical
microscope.
4.
Measuring the position of a particle implies having photons reflect from it. However,
collisions between photons and the particle will alter the velocity of the particle.
6.
Light has both wave and particle characteristics. In Young’s double-slit experiment, light
behaves as a wave. In the photoelectric effect, it behaves like a particle. Light can be
characterized as an electromagnetic wave with a particular wavelength or frequency, yet
at the same time, light can be characterized as a stream of photons, each carrying a
discrete energy, hf.
8.
(a) particle. Light behaves like a tiny, localized packet of energy, capable of being totally
absorbed by a single electron.
(b) particle. Data from Compton scattering experiments can be fully explained by treating
the scattering like a collision between two particles, conserving both energy and
momentum.
(c) wave. The observed patterns when light passes through a pair of parallel slits have the
same characteristics as the diffraction and interference patterns formed by water
waves passing through closely spaced openings.
10.
Ultraviolet light has a shorter wavelength and higher photon energy than visible light.
12.
Increasing the temperature of the substance increases the average kinetic energy of the
electrons inside the material. This makes it slightly easier for an electron to escape from
the material when it absorbs a photon.
14.
Most stars radiate nearly as blackbodies. Vega has a higher surface temperature than
Arcturus. Vega radiates more intensely at shorter wavelengths.
16.
The red beam. Each photon of red light has less energy (longer wavelength) than a photon
of blue light, so the red beam must contain more photons to carry the same total energy.
Quantum Physics
Answers to Even Numbered Problems
~100 nm, ultraviolet
(b)
~0.1 nm, γ -rays
2.
(a)
4.
red beam, 1.39 × 10 3 photons ; blue beam, 845 photons
6.
9.66 × 10 3 K
8.
5.7 × 10 3 photons sec
(b)
3.3 × 10 −34 J
(b)
0.81 eV
(b)
1.24 × 10 −11 m
4.89 × 10 −4 nm
(b)
268 keV
(a)
0.002 87 nm
(b)
100°
34.
(a)
1.98 × 10 −11 m
(b)
1.98 × 10 −14 m
36.
1.06 × 10 −34 m
38.
(a) ~10 −34 m s
(b) ~10 33 s
(c) No. The minimum transit time that could produce significant diffraction is
~1015 times the age of the Universe.
40.
547 eV
42.
3.5 × 10 −32 m
4.2 × 10 35
10.
(a)
12.
5.4 eV
14.
(a)
16.
8.43 × 10 −12 C
18.
8.7 × 1012 electrons s
20.
(a)
22.
0.124 nm
24.
6.7°
26.
67.5°
28.
1.8 keV, 9.7 × 10 −25 kg ⋅ m s
30.
(a)
32.
only lithium
8.29 × 10 −11 m
(c)
32 keV
395
396
CHAPTER 27
44.
(a)
0.250 m s
(b)
2.25 m
46.
(b)
5.2 MeV
48.
(a)
2.49 × 10 −11 m
(b)
0.285 nm
50.
(a)
2.82 × 10 −37 m
(b)
1.06 × 10 −32 J
52.
191 MeV
54.
0.785 eV
56.
(a)
v ≥ 0.999 9 c
(b)
v ≥ 0.053 c
58.
(a)
7.77 × 10 −12 m
(b)
93.8°
(c)
2.87 × 10 −37 % or more
(c)
35.5
Quantum Physics
397
Problem Solutions
27.1
From Wien’s displacement law,
(a) T =
0.289 8 × 10 −2 m ⋅ K
λmax
0.289 8 × 10 −2 m ⋅ K
970 × 10 −9 m
=
= 2.99 × 10 3 K, or ≈ 3 000 K
(b) T =
0.289 8 × 10 −2 m ⋅ K
λmax
=
0.289 8 × 10 −2 m ⋅ K
145 × 10 −9 m
= 2.00 × 10 4 K, or ≈ 20 000 K
27.2
Using Wien’s displacement law,
(a)
λmax =
0.289 8 × 10 − 2 m ⋅ K
10 4 K
= 2.898 × 10 −7 m ~100 nm
(b) λmax =
27.3
0.289 8 × 10 − 2 m ⋅ K
= 2.898 × 10 −10 m ~10 −1 nm γ -rays
107 K
The wavelength of maximum radiation is given by
λmax
27.4
Ultraviolet
0.289 8 × 10 − 2 m ⋅ K
=
= 5.00 × 10 −7 m= 500 nm
5 800 K
The energy of a photon having wavelength λ is Eγ = hf = hc λ . Thus, the number of
photons delivered by each beam must be:
Red Beam:
( 2 500 eV ) ( 690 × 10−9 m )  1.60 × 10−19 J 
Etotal Etotal λR
3
nR =
=
=

 = 1.39 × 10
8
−34
Eγ , R
hc
1
eV
×
⋅
×
6.63
10
J
s
3.
00
10
m
s
(
)(
)

Blue Beam:
nB =
( 2 500 eV ) ( 420 × 10−9 m )  1.60 × 10−19 J 
Etotal Etotal λB
=
=
= 845
Eγ ,B
hc
( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m s )  1 eV 
398
27.5
CHAPTER 27
E = hf =
hc
λ
=
( 6.63 × 10
which yields
(a)
J ⋅ s ) ( 3.00 × 10 8 m s )  1.00 eV

−19
λ
 1.60 × 10

f
J
1.24 × 10 −6 m ⋅ eV
λ
1.24 × 10 −6 m ⋅ eV
= 2.49 × 10 −5 eV
5.00 × 10 −2 m
(b) E =
1.24 × 10 −6 m ⋅ eV
= 2.49 eV
500 × 10 −9 m
E=
1.24 × 10 −6 m ⋅ eV
= 249 eV
5.00 × 10 −9 m
(c)
27.6
E=
E=
− 34
The wavelength of the peak radiation is
λmax =
c
f max
=
3.00 × 10 8 m s
= 3.00 × 10 −7 m = 300 nm
15
1.00 × 10 Hz
(ultra-violet)
Wien’s displacement law then gives the temperature of the blackbody radiator as
T=
27.7
0.289 8 × 10 −2 m ⋅ K
λmax
=
0.289 8 × 10 −2 m ⋅ K
= 9.66 × 10 3 K
−7
3.00 × 10 m
The energy of a single photon is
Eγ = hf = ( 6.63 × 10 − 34 J ⋅ s ) ( 99.7 × 106 s -1 ) = 6.61 × 10 −26 J
The number of photons emitted in ∆t = 1.00 s is
3
∆E P⋅ ( ∆t ) ( 150 × 10 J s ) ( 1.00 s )
=
=
= 2.27 × 10 30
N=
−26
6.61 × 10 J
Eγ
Eγ
Quantum Physics
27.8
399
The energy entering the eye each second is
2
π
P = I ⋅ A = ( 4.0 × 10 −11 W m 2 )  ( 8.5 × 10 −3 m )  = 2.3 × 10 −15 W
4

The energy of a single photon is
Eγ =
hc
λ
( 6.63 × 10
=
− 34
J ⋅ s ) ( 3.00 × 10 8 m s )
500 × 10
−9
m
= 3.98 × 10 −19 J
so the number of photons entering the eye in ∆t = 1.00 s is
−15
J s ) ( 1.00 s )
∆E P⋅ ( ∆t ) ( 2.3 × 10
=
=
= 5.7 × 10 3
N=
3.98 × 10 −19 J
Eγ
Eγ
27.9
The frequency of the oscillator is f =
1
2π
k
1
=
m 2π
20 N/m
= 0.58 Hz
1.5 kg
−2
1 2 ( 20 N m ) ( 3.0 × 10 m )
= 9.0 × 10 −3 J
and its total energy is E = kA =
2
2
2
(a) From E = nhf , the quantum number is
n=
E
9.0 × 10 −3 J
=
= 2.3 × 10 31
−1
-34
hf ( 6.63 × 10 J ⋅ s )( 0.58 s )
(b) ∆E = ( ∆n ) hf , so the fractional change in energy is
−34
−1
∆E ( 1) ( 6.63 × 10 J ⋅ s )( 0.58 s )
=
= 4.3 × 10 −32
E
9.0 × 10 −3 J
27.10
(a) The total energy of the jungle hero is E =
1
1
2
2
mvmax
= ( 70.0 kg )( 2.0 m s ) = 1.4 × 10 2 J
2
2
According to Planck’s hypotheses, the energy of a harmonic oscillator is quantized
according to En = nhf , where f is the frequency of oscillation. Hence, the quantum
number for this system is
n=
E
1.4 × 10 2 J
=
= 4.2 × 10 35
−34
hf ( 6.63 × 10 J ⋅ s ) ( 0.50 Hz )
400
CHAPTER 27
(b) ∆E = ( ∆n ) hf = ( 1) ( 6.63 × 10 −34 J ⋅ s ) ( 0.50 Hz ) = 3.3 × 10 −34 J
27.11
(a) From the photoelectric effect equation, the work function is
φ=
hc
λ
− KEmax , or
( 6.63 × 10
φ=
J ⋅ s ) ( 3.00 × 108 m s ) 
1 eV

− 19
−9
350 × 10 m
 1.60 × 10
− 34

 − 1.31 eV
J
φ = 2.24 eV
(b) λc =
(c)
27.12
fc =
hc
φ
c
λc
( 6.63 × 10
=
=
− 34
J ⋅ s ) ( 3.00 × 10 8 m s ) 

1 eV

 = 555 nm
− 19
2.24 eV
 1.60 × 10 J 
3.00 × 10 8 m s
= 5.41 × 1014 Hz
555 × 10 −9 m
From Einstein’s photoelectric effect equation,
e ( ∆Vs ) = KEmax = hf − φ
1 eV


Thus, φ = hf − e ( ∆Vs ) = ( 6.63 × 10 −34 J ⋅ s )( 3.0 × 1015 Hz ) 
 − e ( 7.0 V )
−19
 1.6 × 10 eV 
or
27.13
φ = 12.4 eV − 7.0 eV = 5.4 eV
From Einstein’s photoelectric effect equation,
KEmax = hf − φ
Thus, λ =
hc
KEmax + φ
or
=
KEmax =
λ=
λ
−φ
hc
1
2
+φ
me vmax
2
( 6.63 × 10
kg )( 1.00 × 10
−34
or
hc
1
( 9.11 × 10−31
2
λ = 2.34 × 10 -7 m = 234 nm
6
J ⋅ s )( 3.00 × 10 8 m s )
m s ) + ( 2.46 eV ) ( 1.60 × 10 −19 J eV )
2
Quantum Physics
27.14
401
(a) The energy of the incident photons is
Eγ =
hc
λ
( 6.63 × 10
=
J ⋅ s ) ( 3.00 × 10 8 m s ) 
1 eV

− 19
−9
400 × 10 m
 1.60 × 10
− 34

 = 3.11 eV
J
For photo-electric emission to occur, it is necessary that Eγ ≥ φ . Thus, of the three
metals given, only lithium will exhibit the photo-electric effect.
hc
(b) For lithium, KEmax =
27.15
λ
− φ = 3.11 eV − 2.30 eV = 0.81 eV
The energy absorbed each second is
2
P= I ⋅ A = ( 500 W m 2 ) π ( 2.82 × 10 −15 m )  = 1.25 × 10 −26 W


The time required to absorb E = 1.00 eV = 1.60 × 10 −19 J is
t=
E
1.60 × 10 −19 J
=
= 1.28 × 107
P 1.25 × 10 −26 J s
1d


s
 = 148 d
4
8.64
10
s
×


This prediction, based on classical theory, is incompatible with observation
27.16
Ultraviolet photons will be absorbed to knock electrons out of the sphere with maximum
hc
− φ , or
kinetic energy KEmax =
λ
KEmax
( 6.63 × 10
=
J ⋅ s ) ( 3.00 × 108 m s )  1.00 eV

− 19
200 × 10 − 9 m
 1.60 × 10
− 34

 − 4.70 eV = 1.52 eV
J
The sphere is left with positive charge and so with positive potential relative to V = 0 at
r = ∞ . As its potential approaches 1.52 V, no further electrons will be able to escape but
will fall back onto the sphere. Its charge is then given by
kQ
V= e
r
or
−2
rV ( 5.00 × 10 m ) ( 1.52 N ⋅ m C )
=
= 8.43 × 10 − 12 C
Q=
8.99 × 10 9 N ⋅ m 2 C 2
ke
402
CHAPTER 27
27.17
The two light frequencies allowed to strike the surface are
f1 =
and
f2 =
c
λ1
=
3.00 × 10 8 m s
= 11.8 × 1014 Hz
254 × 10 −9 m
3.00 × 108 m s
= 6.88 × 1014 Hz
−9
436 × 10 m
KEmax (eV)
The graph you draw should look
somewhat like that given at the
right.
The desired quantities, read
from the axis intercepts of the
graph line, should agree within
their uncertainties with
3.00 eV
0.900 eV
6.88 ´ 10
–f
14
11.8 ´ 1014 f (Hz)
fc
f c = 4.8 × 1014 Hz and φ = 2.0 eV
27.18
The total energy absorbed by an electron is
2
 1.60 × 10 −19 J  1
−31
5
Eγ = φ + KEmax = 3.44 eV 
 + ( 9.11 × 10 kg )( 4.2 × 10 m s )
1 eV

 2
or
Eγ = 6.3 × 10 −19 J
The energy absorbed by a square centimeter of surface in one second is
E = P⋅t = ( I ⋅ A ) ⋅ t = ( 0.055 W m 2 )( 1.00 × 10 −4 m 2 ) ( 1.00 s ) = 5.5 × 10 −6 J
so the number of electrons released per second is
N=
E
5.5 × 10 −6 J
=
= 8.7 × 1012
−19
Eγ 6.3 × 10 J
Quantum Physics
27.19
Assuming the electron produces a single photon as it comes to rest, the energy of that
photon is Eγ = ( KE )i = eV . The accelerating voltage is then
V=
Eγ
e
−34
8
hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s ) 1.24 × 10 −6 V ⋅ m
=
=
λ
eλ
(1.60 × 10-19 C ) λ
=
For λ = 1.0 × 10 −8 m , V =
1.24 × 10 −6 V ⋅ m
= 1.2 × 10 2 V
−8
1.0 × 10 m
and for λ = 1.0 × 10 −13 m , V =
27.20
1.24 × 10 −6 V ⋅ m
= 1.2 × 107 V
1.0 × 10 −13 m
A photon of maximum energy and minimum wavelength is produced when the electron
gives up all its kinetic energy in a single collision.
λmin =
hc
(E )
γ
max
=
−34
8
hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s ) 1.24 × 10 −6 V ⋅ m
=
=
eV
V
(1.60 × 10-19 C )V
(a) If V = 15.0 kV , λmin =
(b) If V = 100 kV , λmin =
27.21
V=
1.24 × 10 −6 V ⋅ m
= 8.29 × 10 −11 m
3
15.0 × 10 V
1.24 × 10 −6 V ⋅ m
= 1.24 × 10 −11 m
100 × 10 3 V
−34
8
KE Eγ hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s )
=
=
=
e
e eλ ( 1.60 × 10 −19 C )( 0.030 0 × 10 −9 m )
= 4.14 × 10 4 V= 41.4 kV
27.22
From Bragg’s law, the wavelength of the reflected x-rays is
λ=
27.23
403
2 d sinθ 2 ( 0.353 nm ) sin 20.5°
=
= 0.124 nm
m
2
Using Bragg’s law, the wavelength is found to be
λ=
2 d sinθ 2 ( 0.296 nm ) sin 7.6°
=
= 0.078 nm
m
1
404
CHAPTER 27
27.24
The first-order constructive interference occurs at the smallest grazing angle. From
Bragg’s law, this angle is


 mλ 
−1 ( 1)( 0.070 nm )
 = sin 
 = 6.7°
2 ( 0.30 )
 2d 


θ = sin −1 
27.25
The interplanar spacing in the crystal is given by Bragg’s law as
d=
27.26
( 1)( 0.140 nm )
mλ
=
= 0.281 nm
2sinθ
2sin14.4°
The scattering angle is given by the Compton shift formula as

θ = cos −1  1 −

λC =
∆λ 
 where the Compton wavelength is
λC 
h
= 0.002 43 nm
me c 2

1.50 × 10 −3 nm 
Thus, θ = cos −1  1 −
 = 67.5°
2.43 × 10 −3 nm 

27.27
Eγ = hf =
p=
h
λ
=
hc
λ
( 6.63 × 10
=
J ⋅ s )( 3.00 × 108 m s ) 
1 eV

−9
−19
700 × 10 m
 1.60 × 10
−34
6.63 × 10 −34 J ⋅ s
= 9.47 × 10 −28 kg ⋅ m s
−9
700 × 10 m

 = 1.78 eV
J
Quantum Physics
27.28
405
Using the Compton shift formula, the wavelength is found to be
λ = λ0 + ∆λ = λ0 + λC ( 1 − cosθ )
= 0.68 nm + ( 0.002 43 nm ) ( 1 − cos 45° ) = 0.680 7 nm
Therefore,
Eγ =
and
27.29
p=
hc
λ
h
λ
( 6.63 × 10
J ⋅ s )( 3.00 × 108 m s ) 
1 keV

× 10 −16
0.680 7 × 10 −9 m
1.60

=
=
−34

 = 1.8 keV
J
6.63 × 10 −34 J ⋅ s
= 9.7 × 10 −25 kg ⋅ m s
0.680 7 × 10 −9 m
If the scattered photon has energy equal to the kinetic energy of the recoiling electron,
the energy of the incident photon is divided equally between them. Thus,
Eγ =
(E )
γ
0
2
⇒
hc
λ
=
hc
, so λ = 2 λ0 and ∆λ = 2 λ0 − λ0 = 0.001 6 nm
2 λ0
The Compton scattering formula then gives the scattering angle as

θ = cos −1  1 −

27.30
(a)
0.001 6 nm 
∆λ 
−1 
 = 70°
 = cos  1 −
0.002 43 nm 
λC 

∆λ = λC ( 1 − cosθ ) = ( 0.002 43 nm ) ( 1 − cos 37.0° ) = 4.89 × 10 −4 nm
(b) The wavelength of the incident x-rays is
λ0 =
( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m s )  1 nm  = 4.14 × 10−3 nm
hc
=
( Eγ ) ( 300 keV ) (1.60 × 10−16 J keV )  10-9 m 
0
so the scattered wavelength is λ = λ0 + ∆λ = 4.63 × 10 −3 nm
The energy of the scattered photons is then
Eγ =
hc
λ
( 6.63 × 10
( 4.63 × 10
J ⋅ s )( 3.00 × 108 m s ) 

1 keV

 = 268 keV
−16
-9
−3
nm )( 10 m 1 nm )  1.60 × 10 J 
−34
=
406
CHAPTER 27
(c) The kinetic energy of the recoiling electrons is
KE = ( Eγ ) − Eγ = 300 keV − 268 keV = 32 keV
0
27.31
This is Compton scattering with θ = 180° . The Compton
shift is ∆λ = λC ( 1 − cos180° ) = 2 λC = 0.004 86 nm , and the
scattered wavelength is
Incident
Photon
λ = λ0 + ∆λ = ( 0.110 + 0.004 86 ) nm=0.115 nm
Scattered
Photon
The kinetic energy of the recoiling electron is then
 1 1  hc ( ∆λ )
KE = ( Eγ ) − Eγ = hc  −  =
0
λ0 λ
 λ0 λ 
( 6.63 × 10
=
or
−34
J ⋅ s )( 3.00 × 108 m s ) ( 0.004 86 nm )
( 0.110 × 10
−9
m ) ( 0.115 nm )
= 7.65 × 10 −17 J
KE = ( 7.65 × 10 −17 J )( 1 eV 1.60 × 10 -19 J ) = 478 eV
The momentum of the recoiling electron (non-relativistic) is
pe = 2 me ( KE ) = 2 ( 9.11 × 10 −31 kg )( 7.65 × 10 −17 J ) = 1.18 × 10 −23 kg ⋅ m s
27.32
The kinetic energy of the recoiling electron (non-relativistic) is
KE =
2
1
1
me v 2 = ( 9.11 × 10 −31 kg )( 1.40 × 10 6 m s ) = 8.93 × 10 −19 J
2
2
 1 1  hc ( ∆λ ) hc ( ∆λ )
≈
Also, KE = ( Eγ ) − Eγ = hc  −  =
0
λ0 λ
λ02
 λ0 λ 
(a) The Compton shift is then
∆λ =
λ02 ( KE )
hc
( 0.800 × 10
=
( 6.63 × 10
−9
−34
m ) ( 8.93 × 10 −19 J )
2
J ⋅ s )( 3.00 × 10 8 m s )
= 2.87 × 10 −12 m = 0.002 87 nm
Recoiling
Electron
Quantum Physics
407
(b) From the Compton shift formula,

θ = cos −1  1 −

27.33
0.002 87 nm 
∆λ 
−1 
 = 100°
 = cos  1 −
0.002 43 nm 
λC 

(a) The kinetic energy of the recoiling electron is
 1 1  hc ( ∆λ ) hc ( ∆λ ) hc λC ( 1 − cosθ ) 
≈
=
KE = ( Eγ ) − Eγ = hc  −  =
0
λ0 λ
λ02
λ02
 λ0 λ 
=
( 6.63 × 10
−34
J ⋅ s )( 3.00 × 10 8 m s ) ( 0.002 43 nm ) ( 1 − cos 23° )
2
( 0.45 nm ) ( 10−9 m 1 nm )

1 eV
KE = 1.9 × 10 −19 J 
−19
 1.60 × 10

 = 1.2 eV
J
2 ( 1.9 × 10 −19 J )
2 ( KE )
=
= 6.5 × 10 5 m s
(b) v =
-31
me
9.11 × 10 kg
27.34
The de Broglie wavelength of a particle of mass m is λ = h p where the momentum is
given by p = γ mv = mv
1 − ( v c ) . Note that when the particle is not relativistic, then
2
γ ≈ 1 , and this relativistic expression for momentum reverts back to the classical
expression.
(a) For a proton moving at speed v = 2.00 × 10 4 m s , v << c and γ ≈ 1 so
λ=
h
6.63 × 10 −34 J ⋅ s
=
= 1.98 × 10 −11 m
mp v ( 1.67 × 10 -27 kg )( 2.00 × 10 4 m s )
(b) For a proton moving at speed v = 2.00 × 107 m s
λ=
h
h
2
=
1 − (v c)
γ m p v mp v
( 6.63 × 10 J ⋅ s )
=
(1.67 × 10 kg )( 2.00 × 10
−34
-27
2
7
m s)
 2.00 × 107 m s 
−14
1−
 = 1.98 × 10 m
8
3.00
×
10
m
s


408
CHAPTER 27
27.35
(a) From λ = h p = h mv , the speed is
v=
(b) λ =
27.36
h
6.63 × 10 −34 J ⋅ s
=
= 1.46 × 10 3 m s = 1.46 km s
me λ ( 9.11 × 10 −31 kg )( 5.00 × 10 −7 m )
h
6.63 × 10 −34 J ⋅ s
=
= 7.28 × 10 −11 m
me v ( 9.11 × 10 −31 kg )( 1.00 × 107 m s )
After falling freely with acceleration ay = − g = −9.80 m s 2 for 50.0 m, starting from rest,
the speed of the ball will be
v = v02 + 2 ay ( ∆y ) = 0 + 2 ( −9.80 m s ) ( −50.0 m ) = 31.3 m s
2
so the de Broglie wavelength is
λ=
27.37
h
h
6.63 × 10 −34 J ⋅ s
=
=
= 1.06 × 10 −34 m
p mv ( 0.200 kg )( 31.3 m s )
(a) The momentum of the electron would be
p=
h
λ
≥
6.63 × 10 −34 J ⋅ s
~ 7 × 10 −20 kg ⋅ m s
10 -14 m
If the electron is nonrelativistic, then its speed would be
v=
p 7 × 10 −20 kg ⋅ m s
~
~ 8 × 1010 m s >> c
−31
me
9.11 × 10 kg
which is impossible. Thus, a relativistic calculation is required.
With a rest energy of ER = 0.511 MeV ≈ 8 × 10 -14 J , its kinetic energy is
KE = E − ER = p 2 c 2 + ER2 − ER
Thus, KE ~
( 7 × 10
−20
kg ⋅ m s ) ( 3 × 10 8 m s ) + ( 8 × 10 −14 J ) − 8 × 10 −14 J
 1 MeV
or KE ~ 2 × 10 −11 J 
-13
 1.60 × 10
2
2

2
 → ~10 MeV or more
J
2
Quantum Physics
409
(b) The negative electrical potential energy of the electron (that is, binding energy)
would be
9
2
2
−19
k e q1 q2 ( 9 × 10 N ⋅ m C )( 10 C ) ( − e )
~
= − 10 5 eV= − 10 -1 MeV
PE =
10 −14 m
r
With its kinetic energy much greater than the magnitude of its negative potential
energy, the electron would immediately escape from the nucleus.
27.38
(a) The de Broglie wavelength is λ = h p = h mv , so v = h mλ
With λmin = w 10 = 0.075 m ~ 10 −1 m and m = 80 kg ~ 10 2 kg , we find that the
maximum speed allowing significant diffraction is
vmax =
h
mλmin
~
6 × 10 −34 J ⋅ s
= 6 × 10 −35 m s or ~ 10 −34 m s
2
−1
10
kg
10
m
(
)(
)
(b) With d = 0.15 m ~ 10 −1 m , the minimum time to pass through the door and have
significant diffraction occur is
tmin =
(c)
d
vmax
~
10 −1 m
or ~ 10 33 s
−34
10 m s
No. The minimum transit time that could produce significant diffraction is
~1015 times the age of the Universe.
27.39
For relativistic particles, p =
E2 − ER2
c
and λ =
h
hc
=
p
E2 − ER2
For 3.00 MeV electrons, E = KE + ER = 3.00 MeV + 0.511 MeV=3.51 MeV
so
λ=
( 6.63 × 10
J ⋅ s )( 3.00 × 10 8 m s )  1 MeV
−13
2
2 
( 3.51 MeV ) − ( 0.511 MeV )  1.60 × 10
−34

−13
 = 3.58 × 10 m
J
410
CHAPTER 27
27.40
From Chapter 24, the minima (or dark fringes) in a single slit diffraction pattern occur
where sin θ = mλ a for m = ±1, ± 2, ± 3,… Here, λ is the wavelength of the wave passing
through the slit of width a. When the fringes are observed on a screen at distance L from
the slit, the distance from the central maximum to the minima of order m is given by
ym = L tan θ m ≈ L sin θ m = mλ ( L a ) . The spacing between successive minima is then
L
∆y = ym+1 − ym = λ  
a
Hence, if ∆y = 2.10 cm when L = 20.0 cm and a = 0.500 nm , the de Broglie wavelength of
the electrons passing through the slit must be
 0.500 × 10 −9 m 
a
−11
λ = ∆y   = ( 2.10 × 10 −2 m ) 
 = 5.25 × 10 m
−2
L
 20.0 × 10
m 
The momentum of one of these electrons is then
p=
h
λ
=
6.63 × 10 −34 J ⋅ s
= 1.263 × 10 −23 kg ⋅ m s
-11
5.25 × 10 m
and, assuming the electron is non-relativistic, its kinetic energy is
(1.263 × 10−23 kg ⋅ m s )  1 eV
p2
KE =
=

-19
2me
2 ( 9.11 × 10 −31 kg )
 1.60 × 10
2

 = 547 eV
J
Note that if the particle had been relativistic, its kinetic energy would have been
computed from
KE = E − ER = p 2 c 2 + ER2 − ER
Quantum Physics
27.41
(a) The required electron momentum is
p=
h
λ
=
6.63 × 10 −34 J ⋅ s  1 keV

1.0 × 10 -11 m  1.60 × 10 -16

−7 keV ⋅ s
 = 4.1 × 10
J
m
and the total energy is
E = p 2 c 2 + ER2
2
keV ⋅ s 
2

8
=  4.1 × 10 −7
 ( 3.00 × 10 m s ) + ( 511 keV ) = 526 keV
m


2
The kinetic energy is then,
KE = E − ER = 526 keV − 511 keV = 15 keV
(b) Eγ =
hc
λ
( 6.63 × 10
=
27.42
J ⋅ s )( 3.00 × 10 8 m s )  1 keV

2

 = 1.2 × 10 keV
−11
-16
1.0 × 10 m
 1.60 × 10 J 
−34
px = mvx , and ∆px = m ( ∆vx ) assuming m is without uncertainty.
Since ∆vx = 1.0 × 10 −3 v = 3.0 × 10 −2 m s , we have
∆px = ( 50.0 × 10 −3 kg )( 3.0 × 10 −2 m s ) = 1.5 × 10 −3 kg ⋅ m s
and
∆x ≥
h
6.63 × 10 −34 J ⋅ s
=
= 3.5 × 10 −32 m
−3
4π ( ∆px ) 4π ( 1.5 × 10 kg ⋅ m s )
411
412
CHAPTER 27
27.43
From the uncertainty principle, the minimum uncertainty in the momentum of the
electron is
∆px =
h
6.63 × 10 −34 J ⋅ s
=
= 5.3 × 10 −25 kg ⋅ m s
4π ( ∆x ) 4π ( 0.10 × 10 −9 m )
so the uncertainty in the speed of the electron is
∆vx =
∆px 5.3 × 10 −25 kg ⋅ m s
=
= 5.8 × 10 5 m s or ~ 106 m s
m
9.11 × 10 −31 kg
Thus, if the speed is on the order of the uncertainty in the speed, then v ~ 10 6 m s
27.44
(a) With uncertainty ∆x in position, the minimum uncertainty in the speed is
∆vx =
∆px
2π J ⋅ s
h
≥
=
= 0.250 m s
4π m ( ∆x ) 4π ( 2.00 kg ) ( 1.00 m )
m
(b) If we knew Fuzzy’s initial position and velocity exactly, his final position after an
elapsed time t would be given by x f = xi + vx t . Since we have uncertainty in both
the initial position and the speed, the uncertainty in the final position is
∆x f = ∆xi + ( ∆vx ) t = 1.00 m + ( 0.250 m s ) ( 5.00 s ) = 2.25 m
27.45
With ∆x = 5.00 × 10 −7 m s , the minimum uncertainty in the speed is
∆vx =
∆px
h
6.63 × 10 −34 J ⋅ s
≥
=
= 116 m s
me
4π me ( ∆x ) 4π ( 9.11 × 10 -31 kg )( 5.00 × 10 -7 m )
p2
( mv )
1
mv 2 =
=
2
2m
2m
2
27.46
(a) For a non-relativistic particle, KE =
Quantum Physics
413
(b) From the uncertainty principle,
∆px ≥
h
6.63 × 10 −34 J ⋅ s
=
= 5.3 × 10 −20 kg ⋅ m s
4π ( ∆x ) 4π ( 1.0 × 10 -15 m )
Since the momentum must be at least as large as its own uncertainty, the minimum
kinetic energy is
( 5.3 × 10−20 kg ⋅ m s )  1 MeV  = 5.2 MeV
p2
= min =


−13
2m
2 ( 1.67 × 10 −27 kg )  1.60 × 10 J 
2
KEmin
27.47
The peak radiation occurs at approximately 560 nm wavelength. From Wien’s
displacement law,
T=
0.289 8 × 10 −2 m ⋅ K
λmax
=
0.289 8 × 10 −2 m ⋅ K
≈ 5 200 K
560 × 10 −9 m
Clearly, a firefly is not at this temperature,
so
27.48
this is not blackbody radiation
(a) Minimum wavelength photons are produced when an electron gives up all its
kinetic energy in a single collision. Then, Eγ = 50 000 eV and
λmin =
−34
8
hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s )
=
= 2.49 × 10 −11 m
4
−19
Eγ ( 5.00 × 10 eV )( 1.60 × 10
J eV )
(b) From Bragg’s law, the interplanar spacing is
( 1) ( 2.49 × 10 m )
mλ
d=
=
= 2.85 × 10 −10 m = 0.285 nm
2sin θ
2sin ( 2.50° )
−11
27.49
The x-ray wavelength is λ = hc Eγ , so Bragg’s law yields
 m hc
 mλ 
−1
 = sin 
 2d 
 2 d Eγ
θ = sin −1 
or





2 ( 6.63 × 10 −34 J ⋅ s )( 3.00 × 10 8 m s )
 = 18.2°
θ = sin 
−9
−16
J keV ) 
 2 ( 0.352 × 10 m ) ( 11.3 keV ) ( 1.60 × 10
−1
414
CHAPTER 27
27.50
(a) From v 2 = v02 + 2 ay ( ∆y ) , Johnny’s speed just before impact is
v = 2 g ∆y = 2 ( 9.80 m s ) ( 50.0 m ) = 31.3 m s
and his de Broglie wavelength is
λ=
6.63 × 10 −34 J ⋅ s
h
=
= 2.82 × 10 −37 m
mv ( 75.0 kg )( 31.3 m s )
(b) The energy uncertainty is
∆E ≥
h
6.63 × 10 −34 J ⋅ s
=
= 1.06 × 10 −32 J
4π ( ∆t ) 4π ( 5.00 × 10 −3 s )
(c) % error =
∆E
( 100% )
mg ∆y
(1.06 × 10 J ) (100% ) ≥
≥
( 75.0 kg ) ( 9.80 m s ) ( 50.0 m )
−32
2
27.51
2.87 × 10 −35 %
The magnetic force supplies the centripetal acceleration for the electrons, so m
or p = mv = qrB
The maximum kinetic energy is then KEmax
or
KEmax
(1.60 × 10
=
−19
p 2 q2 r 2 B2
=
=
2m
2m
J ) ( 0.200 m ) ( 2.00 × 10 −5 T )
2
2
2 ( 9.11 × 10
−31
2
kg )
= 2.25 × 10 −19 J
The work function of the surface is given by φ = Eγ − KEmax = hc λ − KEmax
or
φ=
( 6.63 × 10
−34
J ⋅ s )( 3.00 × 108 m s )
450 × 10 −9 m

1 eV
= 2.17 × 10 −19 J 
−19
 1.60 × 10
− 2.25 × 10 −19 J

 = 1.36 eV
J
v2
= qvB ,
r
Quantum Physics
27.52
415
The incident wavelength is λ0 = hc Eγ , and the Compton shift is
∆λ =
h
hc
hc
1 − cosθ ) =
( 1 − cosθ ) =
( 1 − cosθ )
2 (
mp c
mp c
( ER ) proton
 1 1 − cosθ 
 , or
The scattered wavelength is λ = λ0 + ∆λ = hc  +
 Eγ ( ER ) proton 
( 6.63 × 10
λ=
J ⋅ s ) ( 3.00 × 10 8 m s ) 
1 − cos ( 40.0° ) 
1
−15
+

 = 6.53 × 10 m
−13
1.60 × 10
J MeV
939 MeV 
 200 MeV
−34
The energy of the scattered photon is then
Eγ′ =
27.53
hc
λ
( 6.63 × 10
=
J ⋅ s )( 3.00 × 108 m s )  1 MeV

−13
6.53 × 10 −15 m
 1.60 × 10
−34
From the photoelectric effect equation, KEmax = Eγ − φ =
For λ = λ0 , KEmax = 1.00 eV
For λ =
λ0
2
, KEmax = 4.00 eV
so
1.00 eV =
giving
4.00 eV =
hc
λ
hc
λ0

 = 191 MeV
J
−φ
−φ
2 hc
λ0
−φ
(1)
(2)
Multiplying equation (1) by a factor of 2 and subtracting
the result from equation (2) gives the work function as φ = 2.00 eV
27.54
From the photoelectric effect equation, KEmax = Eγ − φ =
hc
λ
−φ
hc
−φ
670 nm
For λ = 670 nm, KEmax = E1
so
E1 =
For λ = 520 nm, KEmax = 1.50 E1
giving
1.50 E1 =
hc
−φ
520 nm
(1)
(2)
416
CHAPTER 27
Multiplying equation (1) by a factor of 1.50 and subtracting the result from equation (2)
gives the work function as
2.00  1 nm 
 3.00
−
  −9 
 670 nm 520 nm   10 m 
φ = hc 


1 eV
 = 0.785 eV
−19
 1.60 × 10 J 
φ = 1.26 × 10 −19 J 
27.55
(a) If KE =
( 6.63 × 10
=
hc
λ
−34
J ⋅ s )( 3.00 × 108 m s )
1.00 × 10 −8 m
= 1.99 × 10 −17 J = 124 eV
the electron is non-relativistic and
2 ( 1.99 × 10 −17 J )
2 ( KE )
=
v=
9.11 × 10 −31 kg
me

m 
c

=  6.61 × 10 6
 = 0.022 0 c

8
s   3.00 × 10 m s 

(b) When
KE =
hc
λ
( 6.63 × 10
=
J ⋅ s )( 3.00 × 108 m s )  1 MeV 

 = 12.4 MeV
−13
1.00 × 10 −13 m
 1.60 × 10 J 
−34
the electron is highly relativistic and KE = ( γ − 1) ER
or γ = 1 +
12.4 MeV
KE
= 1+
= 25.3
0.511 MeV
ER
Then, v = c 1 − 1 γ 2 = c 1 − 1 ( 25.3 ) = 0.999 2 c
2
27.56
∆p ≥
h
6.63 × 10 −34 J ⋅ s  1 MeV
=

4π ( ∆x ) 4π ( 2.0 × 10 −15 m )  1.60 × 10 −13

−7 MeV ⋅ s
 = 1.65 × 10
J
m
Since the momentum must be at least as large as its own uncertainty,
p ≥ 1.65 × 10 −7
MeV ⋅ s
m
Quantum Physics
417
(a) If the confined particle is an electron, ER = 0.511 MeV and
2
MeV ⋅ s 
2
2
8
( pc ) + ER2 ≥  1.65 × 10−7
 ( 3.00 × 10 m s ) + ( 0.511 MeV )
m


2
E=
or E ≥ 49.5 MeV . Then, γ =
E
49.5 MeV
≥
= 96.8 and
ER 0.511 MeV
v = c 1 − 1 γ 2 ≥ c 1 − 1 ( 96.8 ) = 0.999 9c (highly relativistic)
2
(b) If the confined particle is a proton, ER = 939 MeV and
E=
( pc )
2
2
MeV ⋅ s 
2

8
+ ER2 ≥  1.65 × 10 −7
 ( 3.00 × 10 m s ) + ( 939 MeV )
m 

or E ≥ 940 MeV . Then, γ =
2
E 940 MeV
≥
= 1.001 and
ER 939 MeV
v = c 1 − 1 γ 2 ≥ c 1 − 1 ( 1.001) = 0.053c (non-relativistic)
2
27.57
(a) From conservation of energy, E + ER = φ +
( pc )
2
+ ER2 , where E is the photon energy
and ER = me c 2 .
The de Broglie wavelength of the electron is λ =
h
, giving
p
pc =
hc
λ
If λ is also the wavelength of the incident photon, then
E = hc λ and pc = E
The energy conservation equation then becomes
E + ER − φ = E2 + ER2
Squaring both sides and simplifying yields 2 ER E − 2φ ( E + ER ) + φ 2 = 0 , which
reduces to
E=
2
φ ( 2 ER − φ ) φ ( me c − φ 2 )
=
2 ( ER − φ )
me c 2 − φ
418
CHAPTER 27
(b) From the photoelectric effect equation, KEmax = E − φ . Using the result from above,
and the fact that me c 2 = 0.511 MeV , gives
KEmax =
φ ( me c 2 − φ 2 )
me c 2 − φ
−φ =
φ2
2 ( me c 2 − φ )
=
( 6.35 eV )
2
2 ( 5.11 × 10 5 eV − 6.35 eV )
KEmax = 3.95 × 10 −5 eV = 6.31 × 10 −24 J
or
Therefore,
2 ( KEmax )
v≤
27.58
me
=
2 ( 6.31 × 10 −24 J )
9.11 × 10
−31
kg
= 3.72 × 10 3
m
= 3.72 km s
s
(a) From conservation of energy, ( Eγ ) + ER = Eγ + ( ER + KE ) or
0
(E )
γ
0
= Eγ + KE = 120 keV + 40.0 keV=160 keV
Therefore, the wavelength of the incident photon is
λ0 =
( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m s ) = 7.77 × 10−12 m
hc
=
( Eγ ) (160 keV ) (1.60 × 10−16 J keV )
0
(b) The wavelength of the scattered photon is
−34
8
hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s )
=
= 1.04 × 10 −11 m
λ=
Eγ
( 120 keV ) ( 1.60 × 10 −16 J keV )
so the Compton shift is ∆λ = λ − λ0 = 2.59 × 10 −12 m
The Compton shift formula then gives the photon scattering angle as

θ = cos −1  1 −

2.59 × 10 −12 m 
∆λ 
−1 
cos
1
=
−

 = 93.8°

2.43 × 10 −12 m 
λC 

Quantum Physics
pγ =
(c) The momentum of the scattered photon is
Eγ
c
= 120
419
keV
c
The rest energy of an electron is ER = 0.511 MeV = 511 keV , so the total energy of
the recoiling electron is
E = ER + KE = 511 keV + 40.0 keV = 551 keV
The momentum of the electron is then
E2 − ER2
pe =
c
=
2
2
( 551 keV ) − ( 511 keV )
c
=206
keV
c
Taking the direction of the incident photon to be the x-axis, conservation of
momentum in the y direction requires that pγ sin θ = pe sin φ , where φ is the recoil
angle of the electron. Thus,
 120 keV c ) sin 93.8° 
 pγ sin θ 
−1 (
 = 35.5°
 = sin 
206 keV c
 pe 


φ = sin −1 
27.59
(a) The woman tries to release the pellets from rest directly above the spot on the floor.
However, there is some uncertainty ∆xi in their horizontal position when released.
Also, the uncertainty principle ( ∆x ) ( ∆px ) ≥ h 2 states that the minimum uncertainty
that can exist in their horizontal velocity at the instant of release is given by
∆vx =
∆px
h
=
2m ( ∆xi )
m
Thus, the uncertainty in the horizontal position of the pellet when it reaches the
floor after falling for a time t = 2H g is
∆x f = ∆xi + ( ∆vx ) t = ∆xi +
A
∆xi
where
A≡
h 2H
2m g
To determine the minimum value ∆x f can take on, rewrite this equation in the form
∆x f
A
=
∆xi
A
+
x
∆
A
i
or
y=x+
1
x
where
y≡
∆x f
A
and x ≡
∆xi
A
CHAPTER 27
y = x + 1¤x
12
10
8
y
420
6
4
2
0
0
1
2
3
4
5
6
x
The graph of y vs x for x > 0 given above shows that y has a minimum value of 2 at
x = 1 . Thus,
 ∆x f 

 = 2 or
 A min
( ∆x )
f
min
=2 A
We then have that
( ∆x )
f
or
( ∆x )
f
min
min
 h 2H 
= 2

 2m g 
12
12
 2h   2 H 
=   

m  g 
14
(b) If H = 2.00 m and m = 0.500 g , then
1
 2 ( 1.055 × 10 −34 J ⋅ s )  2  2 ( 2.00 m )  4
 
∆x ≥ 
= 5.19 × 10 −16 m
2 
−3
 0.500 × 10 kg   9.80 m s 
1
Quantum Physics
27.60
The Compton wavelength is λC =
421
h
, where m is the mass of the particle imagined to be
mc
scattering the photon.
The de Broglie wavelength of the particle is
λ=
1
h
h
=
, where γ =
2
p γ mv
1 − (v c)
Therefore, h = ( mc ) λC = ( γ mv ) λ which gives γ ( v c ) = λC λ
(v c)
2
1 − (v c)
2
Squaring this result yields
λ 
=  C  , which simplifies to
 λ 
2
( λC λ ) =
1
c
v
or v =
  =
2
2
2
 c  ( λC λ ) + 1 1 + ( λ λC )
1 + ( λ λC )
2
27.61
2
(a) The density of iron is ρ Fe = 7.86 × 10 3 kg m 3 , so the mass of the sphere must be
3
4
4

m = ρ FeV = ρ Fe  π r 3  = ( 7.86 × 10 3 kg m 3 ) π ( 2.00 × 10 −2 m ) = 0.263 kg
3
3

(b) From Stefan’s law (see Chapter 11), the rate at which the sphere radiates energy is
P = σ AeT 4 where σ = 5.669 6 × 10 −8 W m 2 ⋅ K 4 , A = 4π r 2 is the surface area,
e = 0.860 is the emissivity, and T = 20 + 273 = 293 K is the absolute temperature.
Thus,
P = ( 5.669 6 × 10 −8 W m 2 ⋅ K 4 ) 4π ( 2.00 × 10 −2 m ) ( 0.860 )( 293 K ) = 1.81 W
2
4
(c) If the sphere radiates a quantity of energy ∆Q , its temperature will decrease by
∆T = ∆Q mcFe , where cFe = 448 J kg ⋅ °C is the specific heat of iron. The rate that the
temperature of the isolated sphere would change is then
∆Q ∆ t
1.81 J s
P
∆T
=−
=−
=−
= − 0.015 3 °C s
mcFe
mcFe
∆t
( 0.263 kg )( 448 J kg ⋅ °C )
or
∆T
 60 s 
= ( − 0.015 3 °C s ) 
 = − 0.919 °C min
∆t
 1 min 
422
CHAPTER 27
(d) From Wien’s displacement law, the wavelength of peak radiation from an object
with an absolute temperature of 293 K is
λmax =
0.289 8 × 10 −2 m ⋅ K 0.289 8 × 10 −2 m ⋅ K
=
= 9.89 × 10 −6 m = 9.89 µ m
293 K
T
(e) The energy of a photon having this wavelength is
Eγ = hf =
(f)
hc
λmax
=
( 6.63 × 10
−34
J ⋅ s )( 3.00 × 108 m s )
9.89 × 10 −6 m
= 2.01 × 10 −20 J
If all of the radiated energy was at this wavelength, the number of photons the
sphere would radiate each second would be
N=
1.81 J s
P
=
= 8.98 × 1019 photons s
Eγ 2.01 × 10 −20 J photon