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40
Introduction to Quantum Physics
CHAPTER OUTLINE
40.1
40.2
40.3
40.4
40.5
40.6
40.7
40.8
Blackbody Radiation and
Planck’s Hypothesis
The Photoelectric Effect
The Compton Effect
Photons and
Electromagnetic Waves
The Wave Properties of
Particles
The Quantum Particle
The Double-Slit Experiment
Revisited
The Uncertainty Principle
ANSWERS TO QUESTIONS
Q40.1
Planck made two new assumptions: (1) molecular energy is
quantized and (2) molecules emit or absorb energy in discrete
irreducible packets. These assumptions contradict the classical
idea of energy as continuously divisible. They also imply that
an atom must have a definite structure—it cannot just be a
soup of electrons orbiting the nucleus.
Q40.2
The first flaw is that the Rayleigh–Jeans law predicts that the
intensity of short wavelength radiation emitted by a blackbody
approaches infinity as the wavelength decreases. This is known
as the ultraviolet catastrophe. The second flaw is the prediction
much more power output from a black-body than is shown
experimentally. The intensity of radiation from the blackbody
is given by the area under the red I λ , T vs. λ curve in
Figure 40.5 in the text, not by the area under the blue curve.
Planck’s Law dealt with both of these issues and brought
the theory into agreement with the experimental data by
adding an exponential term to the denominator that depends
1
on . This both keeps the predicted intensity from
b g
λ
approaching infinity as the wavelength decreases and keeps
the area under the curve finite.
Q40.3
Our eyes are not able to detect all frequencies of energy. For example, all objects that are above 0 K
in temperature emit electromagnetic radiation in the infrared region. This describes everything in a
dark room. We are only able to see objects that emit or reflect electromagnetic radiation in the visible
portion of the spectrum.
Q40.4
Most stars radiate nearly as blackbodies. Vega has a higher surface temperature than Arcturus. Vega
radiates most intensely at shorter wavelengths.
Q40.5
No. The second metal may have a larger work function than the first, in which case the incident
photons may not have enough energy to eject photoelectrons.
Q40.6
Comparing Equation 40.9 with the slope-intercept form of the equation for a straight line, y = mx + b ,
we see that the slope in Figure 40.11 in the text is Planck’s constant h and that the y intercept is −φ ,
the negative of the work function. If a different metal were used, the slope would remain the same
but the work function would be different, Thus, data for different metals appear as parallel lines on
the graph.
461
462
Introduction to Quantum Physics
Q40.7
Wave theory predicts that the photoelectric effect should occur at any frequency, provided the light
intensity is high enough. However, as seen in the photoelectric experiments, the light must have a
sufficiently high frequency for the effect to occur.
Q40.8
The stopping voltage measures the kinetic energy of the most energetic photoelectrons. Each of
them has gotten its energy from a single photon. According to Planck’s E = hf , the photon energy
depends on the frequency of the light. The intensity controls only the number of photons reaching a
unit area in a unit time.
Q40.9
Let’s do some quick calculations and see: 1.62 MHz is the highest frequency in the commercial AM
band. From the relationship between the energy and the frequency, E = hf , the energy available
from such a wave would be 1.07 × 10 −27 J , or 6.68 neV. That is 9 orders of magnitude too small to
eject electrons from the metal. The only thing this student could gain from this experiment is a hefty
fine and a long jail term from the FCC. To get on the order of a few eV from this experiment, she
would have to broadcast at a minimum frequency of 250 Thz, which is in the infrared region.
Q40.10
No. If an electron breaks free from an atom absorbing a photon, we say the atom is ionized.
Ionization typically requires energy of several eV. As with the photoelectric effect in a solid metal,
the light must have a sufficiently high frequency for a photon energy that is large enough. The gas
can absorb energy from longer-wavelength light as it gains more internal energy of random motion
of whole molecules.
Q40.11
Ultraviolet light has shorter wavelength and higher photon energy than visible light.
Q40.12
(c) UV light has the highest frequency of the three, and hence each photon delivers more energy to a
skin cell. This explains why you can become sunburned on a cloudy day: clouds block visible light
and infrared, but not much ultraviolet. You usually do not become sunburned through window
glass, even though you can see the visible light from the Sun coming through the window, because
the glass absorbs much of the ultraviolet and reemits it as infrared.
Q40.13
The Compton effect describes the scattering of photons from electrons, while the photoelectric effect
predicts the ejection of electrons due to the absorption of photons by a material.
Q40.14
In developing a theory in accord with experimental evidence, Compton assumed that photons
exhibited clear particle-like behavior, and that both energy and momentum are conserved in
electron-photon interactions. Photons had previously been thought of as bits of waves.
Q40.15
The x-ray photon transfers some of its energy to the electron. Thus, its frequency must decrease.
Q40.16
A few photons would only give a few dots of exposure, apparently randomly scattered.
Q40.17
Light has both classical-wave and classical-particle characteristics. In single- and double-slit
experiments light behaves like a wave. In the photoelectric effect light behaves like a particle. Light
may be characterized as an electromagnetic wave with a particular wavelength or frequency, yet at
the same time light may be characterized as a stream of photons, each carrying a discrete energy, hf.
Since light displays both wave and particle characteristics, perhaps it would be fair to call light a
“wavicle”. It is customary to call a photon a quantum particle, different from a classical particle.
Chapter 40
463
Q40.18
An electron has both classical-wave and classical-particle characteristics. In single- and double-slit
diffraction and interference experiments, electrons behave like classical waves. An electron has mass
and charge. It carries kinetic energy and momentum in parcels of definite size, as classical particles
do. At the same time it has a particular wavelength and frequency. Since an electron displays
characteristics of both classical waves and classical particles, it is neither a classical wave nor a
classical particle. It is customary to call it a quantum particle, but another invented term, such as
“wavicle”, could serve equally well.
Q40.19
The discovery of electron diffraction by Davisson and Germer was a fundamental advance in our
understanding of the motion of material particles. Newton’s laws fail to properly describe the
motion of an object with small mass. It moves as a wave, not as a classical particle. Proceeding from
this recognition, the development of quantum mechanics made possible describing the motion of
electrons in atoms; understanding molecular structure and the behavior of matter at the atomic
scale, including electronics, photonics, and engineered materials; accounting for the motion of
nucleons in nuclei; and studying elementary particles.
Q40.20
If we set
p2
= q∆V , which is the same for both particles, then we see that the electron has the
2m
h
smaller momentum and therefore the longer wavelength λ =
.
p
F
GH
I
JK
Q40.21
Any object of macroscopic size—including a grain of dust—has an undetectably small wavelength
and does not exhibit quantum behavior.
Q40.22
A particle is represented by a wave packet of nonzero width. The width necessarily introduces
uncertainty in the position of the particle. The width of the wave packet can be reduced toward zero
only by adding waves of all possible wavelengths together. Doing this, however, results in loss of all
information about the momentum and, therefore, the speed of the particle.
Q40.23
The intensity of electron waves in some small region of space determines the probability that an
electron will be found in that region.
Q40.24
The wavelength of violet light is on the order of
Q40.25
The spacing between repeating structures on the surface of the feathers or scales is on the order of
1/2 the wavelength of light. An optical microscope would not have the resolution to see such fine
detail, while an electron microscope can. The electrons can have much shorter wavelength.
Q40.26
(a)
1
µm , while the de Broglie wavelength of an
2
electron can be 4 orders of magnitude smaller. Would your height be measured more precisely with
1
an unruled meter stick or with one engraved with divisions down to
mm ?
10
The slot is blacker than any black material or pigment. Any radiation going in through the
hole will be absorbed by the walls or the contents of the box, perhaps after several
reflections. Essentially none of that energy will come out through the hole again. Figure 40.1
in the text shows this effect if you imagine the beam getting weaker at each reflection.
continued on next page
464
Introduction to Quantum Physics
(b)
The open slots between the glowing tubes are brightest. When you look into a slot, you
receive direct radiation emitted by the wall on the far side of a cavity enclosed by the fixture;
and you also receive radiation that was emitted by other sections of the cavity wall and has
bounced around a few or many times before escaping through the slot. In Figure 40.1 in the
text, reverse all of the arrowheads and imagine the beam getting stronger at each reflection.
Then the figure shows the extra efficiency of a cavity radiator. Here is the conclusion of
Kirchhoff’s thermodynamic argument: ... energy radiated. A poor reflector—a good
absorber—avoids rising in temperature by being an efficient emitter. Its emissivity is equal
to its absorptivity: e = a . The slot in the box in part (a) of the question is a black body with
reflectivity zero and absorptivity 1, so it must also be the most efficient possible radiator, to
avoid rising in temperature above its surroundings in thermal equilibrium. Its emissivity in
Stefan’s law is 100% = 1 , higher than perhaps 0.9 for black paper, 0.1 for light-colored paint,
or 0.04 for shiny metal. Only in this way can the material objects underneath these different
surfaces maintain equal temperatures after they come to thermal equilibrium and continue
to exchange energy by electromagnetic radiation. By considering one blackbody facing
another, Kirchhoff proved logically that the material forming the walls of the cavity made
no difference to the radiation. By thinking about inserting color filters between two cavity
radiators, he proved that the spectral distribution of blackbody radiation must be a universal
function of wavelength, the same for all materials and depending only on the temperature.
Blackbody radiation is a fundamental connection between the matter and the energy that
physicists had previously studied separately.
SOLUTIONS TO PROBLEMS
Section 40.1
Blackbody Radiation and Planck’s Hypothesis
2.898 × 10 −3 m ⋅ K
= 5.18 × 10 3 K
560 × 10 −9 m
P40.1
T=
P40.2
(a)
λ max =
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
~
~ 10 −7 m
T
10 4 K
(b)
λ max ~
2.898 × 10 −3 m ⋅ K
~ 10 −10 m
10 7 K
P40.3
ultraviolet
γ − ray
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=
n=
=
n=
bλ
b
gb g
2
2π hc 2 λ 2 − λ 1 π d 2
1
g FH e
+λ2 2
5
2 hc
bλ
b
1 +λ 2
gk T
B
IK
−1
g
= En = nhf
E = hf =
where
2 hc
λ1 + λ 2
8π 2 cd 2 λ 2 − λ 1
P
=
4
2 hc b λ 1 + λ 2 g k BT
E
λ1 + λ 2 e
−1
g FH
b
IK
e
je
j e1.00 × 10
8π 2 3.00 × 10 8 m s 5.00 × 10 −5 m
e1 001 × 10
−9
5.90 × 10 16 s
e
j
e 3.84 − 1
j FGH e
m
4
e
2 6.626 ×10
−34
= 1.30 × 10 15 s
je
8
J⋅s 3.00 ×10 m s
2
j e1 001×10
−9
je
−9
m 1.38 × 10
j
m
−23
je
J K 7 .50 × 10 3 K
j
IJ
K
−1
Chapter 40
P40.4
e
je
b5 000 K g = 2.898 × 10
jb
(a)
P = eAσT 4 = 1 20.0 × 10 −4 m 2 5.67 × 10 −8 W m 2 ⋅ K 4 5 000 K
(b)
λ maxT = λ max
(c)
We compute:
e
−3
je
e
e
je
2π hc 2 A = 2π 6.626 × 10 −34 3.00 × 10 8
f
= 7.09 × 10 4 W
j
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 2.88 × 10 −6 m
−23
kBT
1.38 × 10
J K 5 000 K
2π hc 2 A
5
a
4
m ⋅ K ⇒ λ max = 580 nm
g
jb
The power per wavelength interval is P bλ g = AI bλ g =
λ
P 580 nm =
g
465
j e20.0 × 10 j = 7.50 × 10
2
−4
7.50 × 10 −19 J ⋅ m 4 s
e580 × 10
−9
j
m
5
b
b
g
exp 2.88 µm 0.580 µm − 1
g
exp hc λk B T − 1
=
−19
, and
J ⋅ m4 s
1.15 × 10 13 J m ⋅ s
e 4.973 − 1
= 7.99 × 10 10 W m
(d)–(i) The other values are computed similarly:
(d)
(e)
(f)
(c)
(g)
(h)
(i)
(j)
1.00 nm
7.96 × 10 1251
5.00 nm
576.5
2.40 × 10 250
400 nm
7.21
1347
580 nm
4.97
143.5
700 nm
4.12
60.4
10.0 cm
2.88 × 10
0.00289
−5
2.88 × 10 −5
bg
a5.44 + 7.99f × 10
10
a7.99 + 7.38f × 10
2
10
a
f
a
f
W m 580 − 400 × 10 −9 m
W m 700 − 580 × 10 −9 m
2
4
P = 2.13 × 10 W
so the power radiated as visible light is approximately 20 kW .
P = eAσT 4 , so
LM
P I
3.77 × 10 W
F
T=G
H eAσ JK = MM 1L4π e6.96 × 10 mj Oe5.67 × 10
PQ
MN MN
14
26
8
(b)
P ( λ ), W m
λ5
7.50 × 10 26 9.42 × 10 −1226
2.40 × 10 23 1.00 × 10 −227
7.32 × 10 13 5.44 × 10 10
1.15 × 10 13
7.99 × 10 10
12
4.46 × 10
7.38 × 10 10
7.50 × 10 −4 0.260
7.50 × 10 −14 2.60 × 10 −9
We approximate the area under the P λ versus λ curve, between 400 nm and 700 nm, as
two trapezoids:
+
(a)
e hc / λkBT − 1
1.00 mm 0.00288
P=
P40.5
2π hc 2 A
hc
λk B T
2882.6
λ
λ max =
2
−8
2
W m ⋅K
4
OP
P
j PPQ
14
= 5.75 × 10 3 K
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
466
P40.6
Introduction to Quantum Physics
E = hf =
n=
P40.7
P40.8
hc
λ
e6.626 × 10
=
−34
je
J ⋅ s 3.00 × 10 8 m s
589.3 × 10
−9
m
j = 3.37 × 10
−19
J photon
10.0 J s
P
=
= 2.96 × 10 19 photons s
E 3.37 × 10 −19 J photon
e
je
e
je
e
je
jFGH 1.601.00× 10eV
jFGH 1.601.00× 10eV
jFGH 1.601.00× 10eV
IJ = 2.57 eV
JK
I = 1.28 × 10
J
JK
I = 1.91 × 10
J
JK
(a)
E = hf = 6.626 × 10 −34 J ⋅ s 620 × 10 12 s −1
(b)
E = hf = 6.626 × 10 −34 J ⋅ s 3.10 × 10 9 s −1
(c)
E = hf = 6.626 × 10 −34 J ⋅ s 46.0 × 10 6 s −1
(d)
λ=
c 3.00 × 10 8 m s
=
= 4.84 × 10 −7 m = 484 nm, visible light blue
12
f
620 × 10 Hz
λ=
c 3.00 × 10 8 m s
=
= 9.68 × 10 −2 m = 9.68 cm, radio wave
f
3.10 × 10 9 Hz
λ=
c 3.00 × 10 8 m s
=
= 6.52 m, radio wave
f
46.0 × 10 6 Hz
−19
−19
−19
−5
eV
−7
eV
a f
Energy of a single 500-nm photon:
Eγ = hf =
hc
λ
e6.626 × 10 J ⋅ sje3.00 × 10
e500 × 10 mj
−34
=
8
ms
−9
j = 3.98 × 10
−19
J.
The energy entering the eye each second
e
E = P∆t = IA∆t = 4.00 × 10 −11 W m 2
jLMN π4 e8.50 × 10
−3
j OPQa1.00 sf = 2.27 × 10
m
2
The number of photons required to yield this energy
n=
P40.9
2.27 × 10 −15 J
E
=
= 5.71 × 10 3 photons .
Eγ 3.98 × 10 −19 J photon
Each photon has an energy
This implies that there are
e
je
j
E = hf = 6.626 × 10 −34 99.7 × 10 6 = 6.61 × 10 −26 J .
150 × 10 3 J s
6.61 × 10
−26
J photon
= 2.27 × 10 30 photons s .
−15
J.
Chapter 40
P40.10
467
We take θ = 0.030 0 radians. Then the pendulum’s total energy is
a
f
E = b1.00 kg ge9.80 m s jb1.00 − 0.999 5g = 4.41 × 10
E = mgh = mg L − L cos θ
2
The frequency of oscillation is f =
ω
1
=
2π 2π
The energy is quantized,
E = nhf .
Therefore,
n=
−3
J
g
= 0.498 Hz .
L
4.41 × 10 −3 J
E
=
hf
6.626 × 10 −34 J ⋅ s 0.498 s −1
e
je
FIG. P40.10
j
= 1.34 × 10 31
P40.11
The radiation wavelength of λ ′ = 500 nm that is observed by observers on Earth is not the true
wavelength, λ, emitted by the star because of the Doppler effect. The true wavelength is related to
the observed wavelength using:
c
λ′
=
c
λ
b g:
1 + bv cg
1− v c
λ = λ′
P40.12
2.898 × 10 −3 m ⋅ K
λ max
T=
:
b g
Planck’s radiation law is
I λ, T =
Using the series expansion
ex = 1 + x +
Planck’s law reduces to
I λ, T =
b g
a
a
f
f
1 − 0.280
= 375 nm .
1 + 0.280
λ maxT = 2.898 × 10 −3 m ⋅ K :
The temperature of the star is given by
T=
b g = a500 nmf
1 + bv c g
1− v c
e
2.898 × 10 −3 m ⋅ K
= 7.73 × 10 3 K .
−9
375 × 10
2π hc 2
λ5 e hc λk BT − 1
j
.
x 2 x3
+
+ ….
2! 3!
2π hc 2
5
λ
b1 + hc λk T + …g − 1
B
≈
2π hc 2
5
b
λ hc λk B T
g
=
2π ck B T
λ4
which is the Rayleigh-Jeans law, for very long wavelengths.
Section 40.2
P40.13
(a)
The Photoelectric Effect
λc
e6.626 × 10 J ⋅ sje3.00 × 10 m sj =
=
=
φ
a4.20 eVfe1.60 × 10 J eVj
(b)
λ
8
−19
fc =
hc
−34
hc
c
λc
=
3.00 × 10 8 m s
296 × 10 −9 m
= 1.01 × 10 15 Hz
e6.626 × 10 je3.00 × 10 j = a4.20 eVfe1.60 × 10
−34
= φ + e∆VS :
Therefore,
296 nm
180 × 10 −9
∆VS = 2.71 V
8
−19
j e
j
J eV + 1.60 × 10 −19 ∆VS
468
P40.14
P40.15
Introduction to Quantum Physics
K max =
1
1
2
mv max
= 9.11 × 10 −31 4.60 × 10 5
2
2
(a)
φ = E − K max =
(b)
fc =
(a)
λc =
e
je
j
2
= 9.64 × 10 −20 J = 0.602 eV
1 240 eV ⋅ nm
− 0.602 nm = 1.38 eV
625 nm
I
JK
F
GH
φ
1.38 eV
1.60 × 10 −19 J
=
= 3.34 × 10 14 Hz
−34
1
eV
h 6.626 × 10
J⋅s
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 540 nm
a2.30 eVfe1.60 × 10 J eVj
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 318 nm
=
a3.90 eVfe1.60 × 10 J eVj
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 276 nm
=
a4.50 eVfe1.60 × 10 J eVj
−34
hc
φ
Li:
λc =
Be:
λc
Hg:
λc
8
−19
−34
8
−19
−34
8
−19
λ < λ c for photo current. Thus, only lithium will exhibit the photoelectric effect.
(b)
For lithium,
hc
λ
= φ + K max
e6.626 × 10
−34
je
J ⋅ s 3.00 × 10 8 m s
−9
400 × 10 m
= 1.29 × 10 −19 J = 0.808 eV
K max
P40.16
From condition (i),
and
b g
g = b∆V g + 1.48 V .
hf = e ∆VS1 + φ 1
b ∆V
S1
j = a2.30 eVfe1.60 × 10 j + K
−19
b
g
hf = e ∆VS 2 + φ 2
S2
Then
φ 2 − φ 1 = 1.48 eV .
From condition (ii),
hf c 1 = φ 1 = 0.600 hf c 2 = 0.600φ 2
φ 2 − 0.600φ 2 = 1.48 eV
φ 2 = 3.70 eV
P40.17
(a)
e∆VS =
(b)
e∆VS =
hc
λ
hc
λ
−φ →φ =
−φ =
φ 1 = 2.22 eV .
1 240 nm ⋅ eV
− 0.376 eV = 1.90 eV
546.1 nm
1 240 nm ⋅ eV
− 1.90 eV → ∆VS = 0. 216 V
587.5 nm
max
Chapter 40
P40.18
The energy needed is
E = 1.00 eV = 1.60 × 10 −19 J .
The energy absorbed in time interval ∆t is
E = P∆t = IA∆t
∆t =
so
E
1.60 × 10 −19 J
=
IA
500 J s ⋅ m 2 π 2.82 × 10 −15 m
= 1.28 × 10
O
j QP
jLNM e
e
7
2
469
s = 148 days .
The gross failure of the classical theory of the photoelectric effect contrasts with the success of
quantum mechanics.
P40.19
Ultraviolet photons will be absorbed to knock electrons out of the sphere with maximum kinetic
energy K max = hf − φ ,
or
K max
e6.626 × 10
=
−34
je
J ⋅ s 3.00 × 10 8 m s
200 × 10
−9
j F 1.00 eV I − 4.70 eV = 1.51 eV .
GH 1.60 × 10 J JK
−19
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=
P40.20
k eQ
r
Q=
or
jb
e
g
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
(a)
By having the photon source move toward the metal, the incident photons are Doppler
shifted to higher frequencies, and hence, higher energy.
(b)
If v = 0.280 c ,
f′= f
Therefore,
φ = 6.626 × 10 −34 J ⋅ s 9.33 × 10 14 Hz = 6.18 × 10 −19 J = 3.87 eV .
At v = 0.900 c ,
f = 3.05 × 10 15 Hz
(c)
1+v c
= 7.00 × 10 14
1−v c
e
e
e
j
1.28
= 9.33 × 10 14 Hz .
0.720
je
j
je
and K max = hf − φ = 6.626 × 10 −34 J ⋅ s 3.05 × 10 15 Hz
Section 40.3
P40.21
E=
p=
jFGH 1.601.00× 10eV J IJK − 3.87 eV =
The Compton Effect
hc
λ
h
λ
=
=
e6.626 × 10
−34
je
J ⋅ s 3.00 × 10 8 m s
700 × 10
−9
m
j = 2.84 × 10
6.626 × 10 −34 J ⋅ s
= 9.47 × 10 −28 kg ⋅ m s
−9
700 × 10 m
−19
J = 1.78 eV
−19
8.78 eV .
470
P40.22
Introduction to Quantum Physics
(a)
∆λ =
(b)
E0 =
h
6.626 × 10 −34
1 − cos 37.0° = 4.88 × 10 −13 m
1 − cos θ : ∆λ =
mec
9.11 × 10 −31 3.00 × 10 8
a
hc
λ0
f
e
e
:
f
6.626 × 10 je3.00 × 10
j e
λ
−34
je
8
ms
j
0
λ ′ = λ 0 + ∆λ = 4.63 × 10 −12 m
e
je
j
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 4.30 × 10 −14 J = 268 keV
λ′
4.63 × 10 −12 m
K e = E0 − E ′ = 300 keV − 268.5 keV = 31.5 keV
(c)
With K e = E ′ , K e = E0 − E ′ gives
E0
hc
and λ ′ =
E′
2
a
E ′ = E0 − E ′
λ′ =
λ ′ = λ 0 + λ C 1 − cos θ
f
hc
hc
=2
= 2λ 0
E0 2
E0
a
2 λ 0 = λ 0 + λ C 1 − cos θ
1 − cos θ =
P40.24
a
300 × 10 3 eV 1.60 × 10 −19 J eV =
E′ =
E′ =
j
λ 0 = 4.14 × 10 −12 m
and
P40.23
je
f
λ 0 0.001 60
=
λ C 0.002 43
θ = 70.0°
This is Compton scattering through 180°:
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 11.3 keV
λ
e0.110 × 10 mje1.60 × 10 J eVj
h
∆λ =
a1 − cosθ f = e2.43 × 10 mja1 − cos 180°f = 4.86 × 10
m c
hc
E0 =
−34
=
8
−9
0
−19
−12
−12
m
FIG. P40.24
e
hc
= 10.8 keV .
λ′
h
h
By conservation of momentum for the photon-electron system,
i=
−i + pe i
λ0
λ′
λ ′ = λ 0 + ∆λ = 0.115 nm so
E′ =
e j
FG 1
Hλ
pe = h
and
e
p e = 6.626 × 10
−34
F e3.00 × 10
J ⋅ sjG
GH 1.60 × 10
8
j IJ FG 1
J eV JK H 0.110 × 10
ms c
−19
−9
m
+
−9
1
λ′
IJ
K
IJ =
mK
22.1 keV
.
c
11.3 keV = 10.8 keV + K e
By conservation of system energy,
K e = 478 eV .
so that
Check:
0
1
0.115 × 10
+
em c
E 2 = p 2 c 2 + m e2 c 4 or
a511 keV + 0.478 keVf = a22.1 keVf + a511 keVf
2
2.62 × 10 11 = 2.62 × 10 11
2
e
2
2
+ Ke
j = bpcg + em c j
2
2
e
2 2
Chapter 40
P40.25
(a)
Conservation of momentum in the x direction gives:
pγ = pγ′ cos θ + p e cos φ
or since θ = φ ,
h
h
= pe +
cos θ .
λ0
λ′
Conservation of momentum in the y direction gives:
0 = pγ′ sin θ − p e sin θ ,
which (neglecting the trivial solution θ = 0 ) gives:
p e = pγ′ =
IJ
K
FG
H
h
2h
=
cos θ , or
λ0 λ′
Substituting [2] into [1] gives:
[2]
λ ′ = 2 λ 0 cos θ .
[3]
a
h
1 − cos θ
mec
f
giving
2 λ 0 cos θ − λ 0 =
or
2 cos θ − 1 =
hc 1
1 − cos θ .
λ 0 mec 2
2 cos θ − 1 =
FE
GH m c
hc
λ0
, this may be written as:
or cos θ =
m e c 2 + Eγ
2
2m e c + Eγ
f
a
f
2
I a1 − cos θ f
JK
F 2 + E I cos θ = 1 + E
GH m c JK
m c
γ
=
γ
2
e
2
0.511 MeV + 0.880 MeV
= 0.732 so that θ = φ = 43.0° .
1.02 MeV + 0.880 MeV
Using Equation (3): Eγ′ =
Then,
a
h
1 − cos θ
mec
γ
e
e
(c)
h
.
λ′
λ′ − λ0 =
which reduces to:
(b)
[1]
Then the Compton equation is
Since Eγ =
471
pγ′ =
Eγ
hc
hc
0.880 MeV
=
=
=
= 0.602 MeV = 602 keV .
λ ′ λ 0 2 cos θ
2 cos 43.0°
2 cos θ
a
Eγ′
c
=
From Equation (2), p e = pγ′ =
f
0.602 MeV
= 3.21 × 10 −22 kg ⋅ m s .
c
0.602 MeV
= 3.21 × 10 −22 kg ⋅ m s .
c
From energy conservation: K e = Eγ − Eγ′ = 0.880 MeV − 0.602 MeV = 0.278 MeV = 278 keV .
472
P40.26
Introduction to Quantum Physics
The energy of the incident photon is E0 = pγ c =
(a)
hc
.
λ0
Conserving momentum in the x direction gives
p λ = p e cos φ + pγ′ cos θ , or since φ = θ ,
E0
= p e + pγ′ cos θ .
c
d
i
[1]
Conserving momentum in the y direction (with φ = 0 ) yields
0 = pγ′ sin θ − p e sin θ , or p e = pγ′ =
h
.
λ′
[2]
Substituting Equation [2] into Equation [1] gives
FG
H
IJ
K
E0
2 hc
h
h
=
+
cos θ .
cos θ , or λ ′ =
λ′ λ′
E0
c
By the Compton equation, λ ′ − λ 0 =
[3]
a
f
a
2 hc
2 hc
h
=
1 − cos θ
cos θ −
E0
E0
mec
h
1 − cos θ ,
mec
e 2m c
which reduces to
e
2
j
+ E0 cos θ = m e c 2 + E0 .
φ = θ = cos −1
Thus,
F m c +E I
GH 2m c + E JK
e
e
(b)
λ′ =
2 hc
2 hc m e c 2 + E0
cos θ =
.
E0
E0 2 m e c 2 + E0
Therefore,
Eγ′ =
hc
hc
=
2
λ ′ 2 hc E0 m e c + E0
pγ′ =
b
Eγ′
c
=
ge
F
GH
E0 2 m e c 2 + E0
2 c m e c 2 + E0
From conservation of energy, K e = E0 − Eγ′ = E0 −
or
Ke =
I
JK
j e 2m c
e
2
+ E0
j
=
F
GH
.
F
GH
E0 2 m e c 2 + E0
2 m e c 2 + E0
F
GH
I
JK
F
GH
2
0
.
0
E 0 2 m e c 2 + E0
2 m e c 2 + E0
I
JK
E 0 2 m e c 2 + 2 E0 − 2 m e c 2 − E 0
E02
=
2
2
m e c + E0
2 m e c 2 + E0
Finally, from Equation (2), p e = pγ′ =
2
I
JK
From Equation [3],
and
(c)
F
GH
f
E0 2 m e c 2 + E 0
2 c m e c 2 + E0
I
JK
.
e
j
.
I
JK
,
Chapter 40
*P40.27
The electron’s kinetic energy is
K=
1
1
mv 2 = 9.11 × 10 −31 kg 2.18 × 10 6 m s
2
2
e
j
2
= 2.16 × 10 −18 J .
This is the energy lost by the photon, hf 0 − hf ′
hc hc
−
= 2.16 × 10 −18 J. We also have
λ0 λ′
λ′ − λ0 =
h
6.63 × 10 −34 Js s
1 − cos θ =
1 − cos 17.4°
me c
9.11 × 10 −31 kg 3 × 10 8 m
a
f
e
j
a
f
λ ′ = λ 0 + 1.11 × 10 −13 m
(a)
Combining the equations by substitution,
1
λ0
−
1
2.16 × 10 −18 J s
=
= 1.09 × 10 7 m
λ 0 + 0.111 pm 6.63 × 10 −34 Js 3 × 10 8 m
e
j
λ 0 + 0.111 pm − λ 0
= 1.09 × 10 7 m
λ20 + λ 0 0.111 pm
b
e
g
j
0.111 pm = 1.09 × 10 7 m λ20 + 1.21 × 10 −6 λ 0
1.09 × 10 7 λ20 + 1.21 × 10 −6 mλ 0 − 1.11 × 10 −13 m 2 = 0
λ0 =
−1. 21 × 10 −6 m ±
e1.21 × 10
j − 4e1.09 × 10 je−1.11 × 10
2e1.09 × 10 j
−6
m
2
7
7
only the positive answer is physical: λ 0 = 1.01 × 10 −10 m .
(b)
Then λ ′ = 1.01 × 10 −10 m + 1.11 × 10 −13 m = 1.01 × 10 −10 m.
Conservation of momentum in the transverse direction:
0=
h
sin θ − γm e v sin φ
λ′
e
j
9.11 × 10 −31 kg 2.18 × 10 6 m s sin φ
6.63 × 10 −34 J ⋅ s
°
=
sin
17
.
4
2
1.01 × 10 −10 m
1 − 2.18 × 10 6 3 × 10 8
e
1.96 × 10 −24 = 1.99 × 10 −24 sin φ
φ = 81.1°
j
−13
m2
j
473
474
P40.28
Introduction to Quantum Physics
(a)
Thanks to Compton we have four equations in the unknowns φ, v, and λ ′ :
hc hc
=
+ γ me c 2 − me c 2
λ0 λ′
h
λ0
0=
=
(energy conservation)
h
cos 2φ + γ m e v cos φ
λ′
h
sin 2φ − γ m e v sin φ
λ′
λ′ − λ0 =
b
h
1 − cos 2φ
mec
g
[1]
(momentum in
x direction)
[2]
(momentum in
y direction)
[3]
(Compton equation).
[4]
2h
cos φ .
λ′
Using sin 2φ = 2 sin φ cos φ in Equation [3] gives γ m e v =
Substituting this into Equation [2] and using cos 2φ = 2 cos 2 φ − 1 yields
h
h
h
2h
=
2 cos 2 φ − 1 +
cos 2 φ =
4 cos 2 φ − 1 ,
λ0 λ′
λ′
λ′
e
or
j
e
j
λ ′ = 4λ 0 cos 2 φ − λ 0 .
[5]
Substituting the last result into the Compton equation gives
4λ 0 cos 2 φ − 2 λ 0 =
h
hc
1 − 2 cos 2 φ − 1 = 2
1 − cos 2 φ .
mec
mec 2
e
With the substitution λ 0 =
cos 2 φ =
For x =
m e c 2 + E0
2
2 m e c + E0
j
e
j
hc
, this reduces to
E0
=
E
1+x
where x ≡ 0 2 .
2+x
me c
1+ x
= 33.0° .
2+x
0.700 MeV
= 1.37 , this gives φ = cos −1
0.511 MeV
FIG. P40.28(a)
(b)
e
j
LM FG 1 + x IJ − 1OP = λ FG 2 + 3x IJ .
N H 2 + xK Q H 2 + x K
From Equation [5], λ ′ = λ 0 4 cos 2 φ − 1 = λ 0 4
0
Then, Equation [1] becomes
FG 2 + x IJ + γ m c − m c or E − E FG 2 + x IJ + 1 = γ .
λ
λ H 2 + 3x K
m c
m c H 2 + 3x K
F 2 + x IJ , and with x = 1.37 we get γ = 1.614 .
Thus, γ = 1 + x − xG
H 2 + 3x K
hc
=
0
Therefore,
hc
0
e
2
e
2
0
e
2
0
e
2
v
= 1 − γ −2 = 1 − 0.384 = 0.785 or v = 0.785 c .
c
Chapter 40
P40.29
λ′ − λ =
a
h
1 − cos θ
me c
475
f
a
f
h
1 − cos π − θ
mec
h
h
h
h
λ ′′ − λ =
−
−
cos π − θ +
cos θ
me c me c
mec mec
λ ′′ − λ ′ =
a
a
f
f
Now cos π − θ = − cos θ , so λ ′′ − λ = 2
P40.30
h
= 0.004 86 nm .
mec
FIG. P40.29
Maximum energy loss appears as maximum increase in wavelength, which occurs for scattering
2h
h
=
angle 180°. Then ∆λ = 1 − cos 180°
where m is the mass of the target particle. The
mc
mc
fractional energy loss is
fFGH IJK
a
2 h mc
E0 − E ′ hc λ 0 − hc λ ′ λ ′ − λ 0
∆λ
=
=
.
=
=
λ′
λ 0 + ∆λ λ 0 + 2 h mc
E0
hc λ 0
Further, λ 0 =
(a)
2 h mc
E − E′
2 E0
hc
=
=
, so 0
.
2
E0
E0
hc E0 + 2 h mc mc + 2E0
For scattering from a free electron, mc 2 = 0.511 MeV , so
a
f
2 0.511 MeV
E0 − E ′
=
= 0.667 .
E0
0.511 MeV + 2 0.511 MeV
(b)
a
f
For scattering from a free proton, mc 2 = 938 MeV , and
a
f
2 0.511 MeV
E0 − E ′
=
= 0.001 09 .
E0
938 MeV + 2 0.511 MeV
Section 40.4
*P40.31
a
f
Photons and Electromagnetic Waves
With photon energy 10.0 eV = hf
f=
e
10.0 1.6 × 10 −19 J
6.63 × 10
−34
J⋅s
j = 2.41 × 10
15
Hz .
Any electromagnetic wave with frequency higher than 2.41 × 10 15 Hz counts as ionizing radiation.
This includes far ultraviolet light, x-rays, and gamma rays.
476
*P40.32
Introduction to Quantum Physics
hc
e
Sav =
=
−9
je
beam is 2 × 10 18
af
j
0.628 W 4
P
=
π r 2 π 1.75 × 10 −3 m
vector I = Sav =
j = 3.14 × 10
−19
J . The power carried by the
λ
633 × 10 m
photons s 3.14 × 10 −19 J photon = 0.628 W . Its intensity is the average Poynting
The photon energy is E =
(a)
e
6.63 × 10 −34 J ⋅ s 3 × 10 8 m s
e
1
µ0
Erms Brms sin 90° =
b
Emax = 2 µ 0 cSav
g
12
j
2
= 2.61 × 10 5 W m 2 .
1 Emax Bmax
µ0
2
2
ee
. Also Emax = Bmax c . So Sav =
je
je
2
Emax
.
2µ 0 c
= 2 4π × 10 −7 Tm A 3 × 10 8 m s 2.61 × 10 5 W m 2
jj
12
= 1. 40 × 10 4 N C
Bmax =
(b)
1.40 × 10 4 N C
8
3 × 10 m s
P
E
. The beam transports momentum at the rate . It
c
c
imparts momentum to a perfectly reflecting surface at the rate
2 0.628 W
2P
= force =
= 4.19 × 10 −9 N .
c
3 × 10 8 m s
Each photon carries momentum
a
(c)
Section 40.5
P40.33
λ=
P40.34
(a)
f
The block of ice absorbs energy mL = P∆t melting
P∆t 0.628 W 1.5 × 3 600 s
m=
=
= 1.02 × 10 −2 kg .
L
3.33 × 10 5 J kg
b
g
The Wave Properties of Particles
h
h
6.626 × 10 −34 J ⋅ s
=
=
= 3.97 × 10 −13 m
p mv 1.67 × 10 −27 kg 1.00 × 10 6 m s
e
je
p2
= 50.0 1.60 × 10 −19 J
2m
p = 3.81 × 10 −24 kg ⋅ m s
a fe
λ=
(b)
= 4.68 × 10 −5 T
j
j
h
= 0.174 nm
p
p2
= 50.0 × 10 3 1.60 × 10 −19 J
2m
p = 1. 20 × 10 −22 kg ⋅ m s
e
je
j
h
= 5.49 × 10 −12 m
p
The relativistic answer is slightly more precise:
λ=
λ=
h
=
p
LMemc
N
hc
2
+K
j
2
− m2c4
OP
Q
12
= 5.37 × 10 −12 m .
477
Chapter 40
P40.35
(a)
Electron:
λ=
and
λ=
h
p
K=
and
h
2m e K
=
p2
m2v2
1
=
mev 2 = e
2
2m e
2m e
p = 2m e K
so
6.626 × 10 −34 J ⋅ s
ja fe
e
2 9.11 × 10 −31 kg 3.00 1.60 × 10 −19 J
j
λ = 7.09 × 10 −10 m = 0.709 nm .
(b)
P40.36
(a)
Photon:
λ=
c
f
and
λ=
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 4.14 × 10 −7 m = 414 nm .
−19
E
J
3 1.60 × 10
E = hf
and
e
je
e
f=
so
E
h
j
j
h
h
=
where the
p
2mK
kinetic energy K is in joules. If the neutron kinetic energy K n is given in electron volts, its
The wavelength of a non-relativistic particle of mass m is given by λ =
e
j
kinetic energy in joules is K = 1.60 × 10 −19 J eV K n and the equation for the wavelength
becomes
λ=
h
2mK
=
6.626 × 10 −34 J ⋅ s
e
je
j
2 1.67 × 10 −27 kg 1.60 × 10 −19 J eV K n
=
2.87 × 10 −11
Kn
m
where K n is expressed in electron volts.
P40.37
(b)
If K n = 1.00 keV = 1 000 eV , then
(a)
λ ~ 10 −14 m or less.
p=
λ=
h
λ
~
2.87 × 10 −11
6.6 × 10 −34 J ⋅ s
= 10 −19 kg ⋅ m s or more.
−14
m
10
The energy of the electron is E = p 2 c 2 + m e2 c 4 ~
(b)
m = 9.07 × 10 −13 m = 907 fm .
1 000
e10 j e3 × 10 j + e9 × 10 j e3 × 10 j
−19 2
8 2
−31 2
or
E ~ 10 −11 J ~ 10 8 eV or more,
so that
K = E − m e c 2 ~ 10 8 eV − 0.5 × 10 6 eV ~ 10 8 eV or more.
e
8 4
j
The electric potential energy of the electron-nucleus system would be
e
je
ja f
9 × 10 9 N ⋅ m 2 C 2 10 −19 C − e
k e q1 q 2
~
~ −10 5 eV .
Ue =
−14
r
10
m
With its K + U e >> 0 ,
the electron would immediately escape the nucleus .
478
P40.38
Introduction to Quantum Physics
From the condition for Bragg reflection,
mλ = 2d sin θ = 2d cos
FG φ IJ
H 2K
FG φ IJ .
H 2K
d = a sin
But
where a is the lattice spacing.
λ=
h
=
p
FG φ IJ cosFG φ IJ = a sin φ
H 2K H 2K
λ = 2 a sin
Thus, with m = 1,
h
6.626 × 10 −34 J ⋅ s
λ=
2m e K
e
2 9.11 × 10
Therefore, the lattice spacing is a =
P40.39
(a)
je
−31
kg 54.0 × 1.60 × 10
−19
J
j
= 1.67 × 10 −10 m .
λ
1.67 × 10 −10 m
=
= 2.18 × 10 −10 = 0.218 nm .
sin φ
sin 50.0°
E2 = p 2 c 2 + m 2c 4
with E = hf ,
(b)
FIG. P40.38
p=
h
λ
so
h2 f 2 =
For a photon
f 1
= .
c λ
The third term
h2 c 2
λ
2
+
h2 c 2
λ2C
and
mc =
and
FG f IJ
H cK
h
λC
2
=
1
1
+
λ2 λ2C
(Eq. 1).
1
in Equation 1 for electrons and other massive particles shows that
λC
they will always have a different frequency from photons of the same wavelength .
*P40.40
h
h
=
. For the photon (which we represent as γ),
p γ mv
λγ
chγ mv
γ v
c ch ch
ch
=
=
=
=
. Then the ratio is
.
E = K and λ γ = =
2
2
λm
f E K
γ − 1 mc
γ − 1 mc h γ − 1 c
b g
For the massive particle, K = γ − 1 mc 2 and λ m =
(a)
(b)
λγ
λm
λγ
λm
=
=
b g
1a0.9f
1 − 0.9 2
LMFH1
N
b g
IK OP =
Q
1 − 0.9 2 − 1
a
1 0.001
L
1 − a0.001f MF 1
NH
2
1.60
f
O=
1 − a0.001f I − 1P
K Q
2
2.00 × 10 3
(c)
As
λγ
γ
v
→ 1, γ → ∞ and γ − 1 becomes nearly equal to γ. Then
→ 1= 1 .
λm
γ
c
(d)
As
v
v2
→ 0, 1− 2
c
c
F
GH
I
JK
−1 2
FG 1 IJ v
H 2K c
−1≈1− −
2
2
−1 =
λγ
vc
2c
1 v2
→1
=
→ ∞ .
and
2
2
2
λm
2 c
v
12 v c
b ge
j
Chapter 40
P40.41
λ=
h
p
p=
(a)
h
λ
=
479
6.626 × 10 −34 J ⋅ s
= 6.63 × 10 −23 kg ⋅ m s
1.00 × 10 −11 m
e
j
2
6.63 × 10 −23
p2
=
Ke =
J = 15.1 keV
2m e 2 9.11 × 10 −31
electrons:
e
j
The relativistic answer is more precisely correct:
K e = p 2 c 2 + m e2 c 4 − m e c 2 = 14.9 keV .
P40.42
e
je
photons:
(a)
The wavelength of the student is λ =
then we need
Using ∆t =
h
h
=
. If w is the width of the diffracting aperture,
p mv
FG h IJ
H mv K
F 6.626 × 10 J ⋅ s I =
h
= 10.0G
v ≤ 10.0
mw
H b80.0 kg ga0.750 mf JK
w ≤ 10.0 λ = 10.0
−34
so that
(b)
j
Eγ = pc = 6.63 × 10 −23 3.00 × 10 8 = 124 keV
(b)
d
we get:
v
∆t ≥
1.10 × 10 −34 m s .
0.150 m
= 1.36 × 10 33 s .
−34
1.10 × 10
ms
No . The minimum time to pass through the door is over 10 15 times the age of the
(c)
Universe.
Section 40.6
*P40.43
The Quantum Particle
E=K =
h
1
.
mu 2 = hf and λ =
mu
2
v phase = fλ =
mu 2 h
u
=
= v phase .
2 h mu
2
This is different from the speed u at which the particle transports mass, energy, and momentum.
*P40.44
As a bonus, we begin by proving that the phase speed v p =
vp =
ω
is not the speed of the particle.
k
F
GH
I
JK
p2c2 + m2c4
γ 2m 2 v2 c 2 + m 2 c 4
ω
c2
c2
v2
c2
c2
=
=
= c 1+ 2 2 = c 1+ 2 1− 2 = c 1+ 2 −1 =
k
v
γ mv
γ v
v
c
v
γ 2m2 v2
In fact, the phase speed is larger than the speed of light. A point of constant phase in the wave
function carries no mass, no energy, and no information.
Now for the group speed:
continued on next page
480
Introduction to Quantum Physics
vg =
dω d ω dE d
=
=
=
m2c4 + p2c2
dk
d k dp dp
vg =
1 2 4
m c + p2c2
2
e
p2c4
j e0 + 2 pc j =
−1 2
2
p2c2 + m2c4
e
j
v2 1 − v2 c2
γ 2m 2 v 2
=c 2
=c
vg = c 2 2 2
γ m v + m2c2
v 1 − v2 c2 + c2
e
j
e
v2 1 − v2 c2
ev
2
2
+c −v
2
j
j e1 − v c j
2
2
=v
It is this speed at which mass, energy, and momentum are transported.
Section 40.7
P40.45
The Double-Slit Experiment Revisited
h
6.626 × 10 −34 J ⋅ s
=
= 9.92 × 10 −7 m
mv 1.67 × 10 −27 kg 0.400 m s
(a)
λ=
(b)
For destructive interference in a multiple-slit experiment, d sin θ = m +
jb
e
FG
H
the first minimum.
FG λ IJ = 0.028 4°
H 2d K
y = L tan θ = a10.0 mfbtan 0.028 4°g =
so
P40.46
y
= tan θ
L
4.96 mm .
We cannot say the neutron passed through one slit. We can only say it passed through the slits.
Consider the first bright band away from the center:
e6.00 × 10
d sin θ = mλ
λ=
h
so
me v
me v =
K=
and
h2
∆V =
2 em e λ2
*P40.47
IJ
K
1
λ , with m = 0 for
2
θ = sin −1
Then,
(c)
g
−8
j FGH
m sin tan −1
h
LM 0.400 OPIJ = a1fλ = 1.20 × 10
N 200 QK
m
λ
m2v2
1
h2
=
= e∆V
mev 2 = e
2
2m e
2 m e λ2
∆V =
−10
e
2 1.60 × 10 −19
e6.626 × 10
C je9.11 × 10
−34
−31
j
kg je1.20 × 10
J⋅s
2
−10
j
m
2
= 105 V .
We find the speed of each electron from energy conservation in the firing process:
1
0 = K f + U f = mv 2 − eV
2
v=
2 eV
=
m
9.11 × 10
The time of flight is ∆t =
apart is I =
a
2 × 1.6 × 10 −19 C 45 V
−31
kg
f = 3.98 × 10
6
ms
∆x
0.28 m
=
= 7.04 × 10 −8 s . The current when electrons are 28 cm
6
v
3.98 × 10 m s
q
e 1.6 × 10 −19 C
=
=
= 2.27 × 10 −12 A .
t ∆t 7.04 × 10 −8 s
Chapter 40
Section 40.8
P40.48
481
The Uncertainty Principle
∆v ≥
h
2π J ⋅ s
=
= 0. 250 m s .
4π m∆x 4π 2.00 kg 1.00 m
(a)
∆p∆x = m∆v∆x ≥
(b)
The duck might move by 0. 25 m s 5 s = 1.25 m . With original position uncertainty of
so
2
b
f
ga
ga f
b
1.00 m, we can think of ∆x growing to 1.00 m + 1.25 m = 2.25 m .
P40.49
J⋅s
6.626 × 10
h
=
= 1.16 mm .
−
32
4π ∆p 4π 4.56 × 10
kg ⋅ m s
b
e
gb
j
ge
j
h
= 5.28 × 10 −32 m .
4π ∆p
∆x =
∆y ∆p y
h
=
and
d∆p y ≥
.
x
px
4π
Eliminate ∆p y and solve for x.
b g
d
x = 4π p x ∆y :
h
e
x = 4π 1.00 × 10
The answer, x = 3.79 × 10
28
−3
jb
ge
kg 100 m s 1.00 × 10
−2
e2.00 × 10
mj
e6.626 × 10
j
−3
m
−34
J⋅s
j
m , is 190 times greater than the diameter of the Universe!
With ∆x = 2 × 10 −15 m m, the uncertainty principle requires ∆p x ≥
= 2.6 × 10 −20 kg ⋅ m s .
2 ∆x
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 p rms ≈ 3 × 10 −20 kg ⋅ m s . For an
electron, the non-relativistic approximation p = m e v would predict v ≈ 3 × 10 10 m s , while v cannot
be greater than c.
Thus, a better solution would be
LMe
N
E = me c 2
γ ≈ 110 =
For a proton, v =
*P40.52
j
∆p = m∆v = 0.020 0 kg 500 m s 1.00 × 10 −4 = 1.00 × 10 −3 kg ⋅ m s
For the bullet,
P40.51
ge
−34
∆x =
P40.50
jb
e
∆p = m e ∆v = 9.11 × 10 −31 kg 500 m s 1.00 × 10 −4 = 4.56 × 10 −32 kg ⋅ m s
For the electron,
j + bpcg OPQ
2
2
12
1
≈ 56 MeV = γ m e c 2
so
1 − v2 c2
v ≈ 0.999 96 c .
p
gives v = 1.8 × 10 7 m s, less than one-tenth the speed of light.
m
a f
2
mv
p2
1
mv 2 =
=
2
2m
2m
(a)
K=
(b)
To find the minimum kinetic energy, think of the minimum momentum uncertainty, and
maximum position uncertainty of 10 −15 m = ∆x . We model the proton as moving along a
. The average momentum is zero. The average squared
, ∆p =
2 ∆x
2
momentum is equal to the squared uncertainty:
straight line with ∆p∆x =
b g
e
j
2
2
2
2
6.63 × 10 −34 J ⋅ s
∆p
p2
= 8.33 × 10 −13 J
K=
=
=
=
=
2
2
2
2
−
−
2
15
27
2m
2m
∆
∆
4 x 2m 32π
x m 32π 10
m 1.67 × 10
kg
= 5.21 MeV
a f
a f
e
j
482
P40.53
Introduction to Quantum Physics
(a)
At the top of the ladder, the woman holds a pellet inside a small region ∆xi . Thus, the
uncertainty principle requires her to release it with typical horizontal momentum
2H
1
, so
∆p x = m∆v x =
. It falls to the floor in a travel time given by H = 0 + gt 2 as t =
2
g
2 ∆x i
the total width of the impact points is
b g
∆x f = ∆xi + ∆v x t = ∆xi +
A=
where
2m
A
2H
= ∆x i +
g
∆x i
i
2H
.
g
d i =0
d b ∆x g
d ∆x f
To minimize ∆x f , we require
FG
IJ
m
x
2
∆
H
K
1−
or
i
A
=0
∆xi2
∆xi = A .
so
The minimum width of the impact points is
d∆x i = FGH ∆x + ∆Ax IJK
f
min
i
i
=2 A =
∆xi = A
LM 2e1.054 6 × 10 J ⋅ sj OP L 2a2.00 mf O
∆x
=
d i M 5.00 × 10 kg P MM 9.80 m s PP
Q
Q N
N
12
−34
(b)
F I
GH JK
2 2H
m g
f
−4
min
2
14
.
14
= 5.19 × 10 −16 m
Additional Problems
P40.54
∆VS =
FG h IJ f − φ
H eK e
From two points on the graph
FG h IJ e4.1 × 10 Hzj − φ
H eK
e
φ
F hI
3.3 V = G J e12 × 10 Hzj − .
H eK
e
14
0=
14
and
Combining these two expressions we find:
FIG. P40.54
(a)
φ = 1.7 eV
(b)
h
= 4. 2 × 10 −15 V ⋅ s
e
(c)
At the cutoff wavelength
e
je
hc
λc
=φ =
λ c = 4.2 × 10 −15 V ⋅ s 1.6 × 10 −19 C
FG h IJ ec
H eK λ
c
3 × 10 m s
j a1.7 eVe fe1.6 × 10 j J eVj =
8
−19
730 nm
Chapter 40
P40.55
483
We want an Einstein plot of K max versus f
λ , nm
588
505
445
399
f , 10 14 Hz K max , eV
5.10
0.67
5.94
0.98
6.74
1.35
7.52
1.63
0.402 eV
± 8%
10 14 Hz
(a)
slope =
(b)
e∆VS = hf − φ
a
h = 0.402
(c)
fFGH 1.60 ×1010
−19
J⋅s
14
I=
JK
6. 4 × 10 −34 J ⋅ s ± 8%
f (THz)
K max = 0
FIG. P40.55
at f ≈ 344 × 10 12 Hz
φ = hf = 2.32 × 10 −19 J = 1.4 eV
P40.56
From the path the electrons follow in the magnetic field, the maximum kinetic energy is seen to be:
K max =
P40.57
e2B2 R 2
.
2m e
From the photoelectric equation,
K max = hf − φ =
Thus, the work function is
φ=
∆λ =
e
hc
λ
− K max =
j
hc
λ
−φ .
hc
λ
−
e2B2 R2
.
2m e
6.626 × 10 −34 J ⋅ s
h
1 − cos θ =
0.234 = 3.09 × 10 −16 m
−27
8
mp c
kg 3.00 × 10 m s
1.67 × 10
a
e
f
e
je
je
j
j
a
f
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
λ0 =
=
= 6.20 × 10 −15 m
E0
200 MeV 1.60 × 10 −13 J MeV
a
fe
λ ′ = λ 0 + ∆λ = 6.51 × 10
−15
m
hc
= 191 MeV
λ′
(a)
Eγ =
(b)
K p = 9.20 MeV
j
484
P40.58
Introduction to Quantum Physics
Isolate the terms involving φ in Equations 40.13 and 40.14. Square and add to eliminate φ.
h2
Solve for
v2
b
=
:
2
c
b + c2
e
j
LM 1
Nλ
OP
Q
1
2 cos θ
−
= γ 2 m e2 v 2
2
λ 0λ ′
λ′
+
2
0
2
OP
Q
F h IJ LM 1 − 1 OP = γ = FG 1 − b IJ = c + b .
1+G
H m cKNλ λ′Q H b + c K
c
F h I LM 1
2 hc L 1
1O h L 1
1O
− P = c +G
− P+
c +
M
M
λ′ Q m Nλ
λ′Q
m Nλ
H m JK N λ
F h IJ 1 − cosθ .
λ′ − λ = G
H m cK
b=
LM
N
1
1
2 cos θ
h
+ 2 −
.
2
2
λ 0λ ′
me λ 0 λ ′
−1 2
Substitute into Eq. 40.12:
e
0
e
From this we get Eq. 40.11:
P40.59
0
2
2
e
0
2
2
2
2
Square each side:
2
2
0
2
2
e
2
0
+
e
Show that if all of the energy of a photon is transmitted to an electron, momentum will not be
conserved.
hc hc
hc
=0
=
+ K e = m e c 2 γ − 1 if
Energy:
λ0 λ′
λ′
h
h
=
+ γ m e v = γ m e v if λ ′ = ∞
Momentum:
λ0 λ′
h
γ =
+1
From (1),
λ 0mec
b g
v = c 1−
FG λ m c IJ
H h + λ m cK
Substitute (3) and (4) into (2) and show the inconsistency:
h
λ0
F h IJ m c
= G1 +
H λ m cK
0
e
e
F λ m c IJ
1−G
H h+ λ m cK
e
0
0
2
=
e
b
b
g
λ 0 m e c + h h h + 2λ 0 m e c
h
=
2
λ0
λ0
h + λ 0 me c
g
h + 2λ 0 m e c
.
h
This is impossible, so all of the energy of a photon cannot be transmitted to an electron.
P40.60
Begin with momentum expressions:
p=
Equating these expressions,
γ
h
λ
, and p = γ mv = γ mc
C
2
2
C
2
2
or
C
2
C
2
2
2
2
C
v=
2
C
2
giving
FG v IJ .
H cK
FG v IJ = FG h IJ 1 = λ .
H c K H mc K λ λ
bv cg = FG λ IJ
HλK
1 − b v cg
FG v IJ = FG λ IJ − FG λ IJ FG v IJ
H cK H λ K H λ K H cK
bλ λ g = 1
v
=
c
1 + bλ λ g
bλ λ g + 1
2
Thus,
(2)
(3)
(4)
e
0
(1)
2
e
0
OP
Q
1
2 cos θ
−
.
λ 0λ ′
λ ′2
C
c
b
1 + λ λC
g
2
.
Chapter 40
P40.61
(a)
b g
I λ, T =
Starting with Planck’s law,
2π hc 2
λ5 e hc λk BT − 1
zb g
∞
the total power radiated per unit area
I λ , T dλ =
0
z
∞
0
2π hc 2
λ5 e hc λk BT − 1
hc
λk B T
hcdλ
and
dx = −
.
k B Tλ2
Note that as λ varies from 0 → ∞ , x varies from ∞ → 0 .
∞
2π k B4 T 4
Then
I λ , T dλ = −
h3 c 2
0
dλ .
x=
Change variables by letting
zb g
zb
∞
Therefore,
g
I λ , T dλ =
0
(b)
485
σ=
From part (a),
2π 5 k B4
15 h 3 c 2
=
ze
0
∞
F
I
GH 15h c JK T
2π 5 k B4
3 2
x3
x
e −1
4
j
dx =
F I
GH JK
2πk B4 T 4 π 4
.
15
h3 c 2
= σ T4 .
e
2π 5 1.38 × 10 −23 J K
e
15 6.626 × 10 −34 J ⋅ s
j
4
j e3.00 × 10
3
8
ms
j
2
σ = 5.67 × 10 −8 W m 2 ⋅ K 4 .
P40.62
b g
Planck’s law states I λ , T =
2π hc 2
λ5 e hc λk BT − 1
= 2π hc 2 λ−5 e hc λk BT − 1
−1
.
To find the wavelength at which this distribution has a maximum, compute
R|
S|
T
dI
= 2π hc 2 −5 λ−6 e hc λk BT − 1
dλ
2π hc 2
dI
=
dλ λ6 e hc λk BT − 1
−1
R| hc
S|−5 + λk T
T
B
− λ−5 e hc λk BT − 1
−2 hc λk T
B
e
U|
V=0
−1 |
W
F − hc I U|V = 0
GH λ k T JK |W
2
B
e hc λk BT
e hc λk BT
xe x
hc
= 5.
, the condition for a maximum becomes x
λk B T
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
Letting x =
e
x
xe x e x − 1
4.000 00
4.500 00
5.000 00
4.900 00
4.950 00
4.975 00
4.963 00
4.969 00
4.966 00
4.074 629 4
4.550 552 1
5.033 918 3
4.936 762 0
4.985 313 0
5.009 609 0
4.997 945 2
5.003 776 7
5.000 860 9
x=
hc
= 4.965 115
λ max k B T
continued on next page
j
and
e
x
xe x e x − 1
4.964 50
4.965 50
4.965 00
4.965 25
4.965 13
4.965 07
4.965 10
4.965 115
4.999 403 0
5.000 374 9
4.999 889 0
5.000 132 0
5.000 015 3
4.999 957 0
4.999 986 2
5.000 000 8
λ maxT =
hc
.
4.965 115 k B
j
486
Introduction to Quantum Physics
Thus, λ max
e6.626 075 × 10
T=
−34
je
J ⋅ s 2.997 925 × 10 8 m s
e
4.965 115 1.380 658 × 10
−23
JK
j
j=
2.897 755 × 10 −3 m ⋅ K .
This result is very close to Wien’s experimental value of λ maxT = 2.898 × 10 −3 m ⋅ K for this constant.
P40.63
(a)
Planck’s radiation law predicts maximum intensity at a wavelength λ max we find from
RS
T
0 = 2π hc λ a −1f e b
−1
dI
d
hc λk B T g
2π hc 2 λ−5 e b
=0=
−1
dλ
dλ
2 −5
hc λkB T
g − 1 −2 eb hc λk T g F
7
λ kB
which reduces to
Define x =
5
hc
.
λk B T
GH
B
− hce b
or
UV
W
hc λkB T
I
JK
g
2
hc λk T g
−1
T eb
+
B
FG λk T IJ eb
H hc K
B
hc λkB T
a f
− hc
hc λk B T g
+ 2π hc 2 −5 λ−6 e b
−1
λ2 k B T
g −1
5
hc λk T g
−1
eb
6
λ
−1
=0
B
= eb
hc λk BT
g.
Then we require 5 e x − 5 = xe x .
Numerical solution of this transcendental equation gives x = 4.965 to four digits. So
hc
λ max =
, in agreement with Wien’s law.
4.965 k B T
zb g
∞
0
0
z
0
2π hc 2 dλ
λ5 e b hc λk BT g − 1
.
hc
hc
hc
so λ =
and dλ = − 2
dx .
xk B T
λk B T
x kBT
Again, define x =
Then, A + B =
z
∞
The intensity radiated over all wavelengths is I λ , T dλ = A + B =
−2π hc 2 x 5 k B5 T 5 hcdx
x =∞
e
j
h5 c 5 x 2 kBT e x − 1
The integral is tabulated as
=
2π k B4 T 4
h3 c 2
ze
∞
0
x 3 dx
ex − 1
j
.
2π 5 k B4 T 4
π4
.
, so (in agreement with Stefan’s law) A + B =
15
15 h 3 c 2
The intensity radiated over wavelengths shorter than λ max is
zb g
z
λ max
λ max
0
0
I λ , T dλ = A =
With x =
2π hc 2 dλ
λ5 e b hc λkBT g − 1
.
2π k B4 T 4
hc
, this similarly becomes A =
λk B T
h3 c 2
z
∞
x 3 dx
.
x
4.965 e − 1
So the fraction of power or of intensity radiated at wavelengths shorter than λ max is
A
=
A+B
e 2π k T
continued on next page
4
B
4
L
jMMN
h 3 c 2 π 4 15 −
z
4.965
e
O
j PPQ
x 3 dx e x − 1
0
2π 5 k B4 T 4 15 h 3 c 2
= 1−
15
π4
z
4.965
0
x 3 dx
.
ex − 1
Chapter 40
(b)
Here are some sample values of the integrand, along with a sketch of the curve:
e
j
x3 ex − 1
x
0.000
0.00
−1
0.100
9.51 × 10 −3
0.200
3.61 × 10 −2
1.00
0.582
2.00
1.25
3.00
1.42
4.00
1.19
FIG. P40.63(b)
4.90
0.883
4.965
0.860
Approximating the integral by trapezoids gives
P40.64
jb
e
a
ge
p = mv = 2mE = 2 1.67 × 10 −27 kg 0.040 0 eV 1.60 × 10 −19 J eV
λ=
f
A
15
≈ 1 − 4 4.870 = 0. 250 1 .
A+B
π
j
h
= 1.43 × 10 −10 m = 0.143 nm
mv
This is of the same order of magnitude as the spacing between atoms in a crystal so diffraction
should appear.
P40.65
λC =
p
λ C h me c
=
=
;
λ
hp
mec
h
h
and λ = :
mec
p
e
E2 = c 2 p 2 + me c 2
j
2
p=
:
λC
1
E2
=
− mec
λ
me c c 2
b g
P40.66
(a)
mgyi =
2
E2
− mec
c2
b g
2
LM E − bm cg OP = F E I
bm c g N c
Q GH m c JK
2
1
=
2
e
e
2
2
e
2
2
−1
1
mv 2f
2
ja
e
f
v f = 2 gyi = 2 9.80 m s 2 50.0 m = 31.3 m s
λ=
(b)
h
6.626 × 10 −34 J ⋅ s
=
= 2.82 × 10 −37 m
mv
75.0 kg 31.3 m s
b
∆E∆t ≥
so ∆E ≥
(c)
gb
g
anot observablef
2
6.626 × 10 −34 J ⋅ s
e
4π 5.00 × 10
−3
s
j
= 1.06 × 10 −32 J
∆E
1.06 × 10 −32 J
=
= 2.87 × 10 −35%
E
75.0 kg 9.80 m s 2 50.0 m
b
ge
ja
f
487
488
P40.67
Introduction to Quantum Physics
From the uncertainty principle
∆E∆t ≥
or
∆ mc 2 ∆t =
Therefore,
∆m
h
h
=
=
2
m
4π c ∆t m 4π ∆t ER
2
e j
2
.
a f
a f
∆m
6.626 × 10 −34 J ⋅ s
=
m
4π 8.70 × 10 −17 s 135 MeV
ja
e
P40.68
∆λ =
a
−13
2.81 × 10 −8 .
f
h
1 − cos θ = λ ′ − λ 0
mec
LM
N
a
hc
hc
h
=
= hc λ 0 +
E′ =
1 − cos θ
mec
λ ′ λ 0 + ∆λ
LM1 + hc a1 − cos θ fOP
E′ =
λ N m c λ
Q
O
hc L
hc
E′ =
1+
1 − cos θ fP
a
M
λ N m c λ
Q
hc
0
0
P40.69
FG 1 MeV IJ =
f H 1.60 × 10 J K
e
e
2
2
−1
0
0
−1
fOP
Q
−1
L E a1 − cosθ fOP
= E M1 +
N mc
Q
0
0
e
−1
2
(a)
The light is unpolarized. It contains both horizontal and vertical field oscillations.
(b)
The interference pattern appears, but with diminished overall intensity.
(c)
The results are the same in each case.
(d)
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.
(e)
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 40.24 in the text, and the runs for trial 3 build up the red
graph.
Chapter 40
P40.70
489
Let u ′ represent the final speed of the electron and let
I
JK
F
GH
−1 2
u′ 2
. We must eliminate β and u ′ from the
c2
three conservation equations:
γ ′ = 1−
hc
hc
+ γ mec 2 =
+ γ ′m e c 2
λ0
λ′
h
h
+ γm eu − cos θ = γ ′m eu ′ cos β
λ0
λ′
h
sin θ = γ ′m eu ′ sin β
λ′
Square Equations [2] and [3] and add:
h2
λ20
h
2
λ20
+ γ 2 m e2 u 2 +
+
h
2
λ ′2
h2
λ ′2
+
[2]
FIG. P40.70
[3]
2 hγ m e u
−
2 h 2 cos θ 2 hγ m eu cos θ
−
= γ ′ 2 m e2 u ′ 2
λ 0λ ′
λ′
2 hγ m e u
−
2 hγ m eu cos θ 2 h 2 cos θ
m e2 u ′ 2
−
=
λ′
λ 0λ ′
1 − u′ 2 c 2
λ0
+ γ 2 m e2 u 2 +
[1]
λ0
Call the left-hand side b. Then b −
bu ′ 2
b
c 2b
2 2
2
m
u
u
=
′
′
=
=
and
.
e
c2
m e2 + b c 2 m e2 c 2 + b
Now square Equation [1] and substitute to eliminate γ ′ :
h2
λ20
+ γ 2 m e2 c 2 +
So we have
h2
λ20
h2
2
λ′
+
2 hγ m e c
+
λ0
h2
h2
λ20
Multiply through by
λ 0 λ ′γ 2 +
F
GH
2h2
λ 0λ ′
+ γ 2 m e2 c 2 +
λ ′2
= me c 2 +
−
+
h2
−
2 hγ m e c
m e2 c 2
=
= m e2 c 2 + b .
λ′
1 − u′ 2 c 2
2 hγ m e c
λ0
+ γ 2 m e2 u 2 +
λ ′2
−
2 hγ m e c 2 h 2
−
λ′
λ 0λ ′
2 hγ m e u
λ0
−
2 hγ m e u cos θ 2 h 2 cos θ
−
λ′
λ 0λ ′
λ 0λ ′
m e2 c 2
λ λ ′γ 2 u 2 2 hλ ′uγ 2 hγλ 0u cos θ 2 h 2 cos θ
2 hλ ′γ 2 hλ 0γ
2h2
−
− 2 2 = λ 0λ ′ + 0
+
−
−
mec
me c
me c
mec 2
me c 2
mec 2
m e2 c 2
λ 0λ ′ γ 2 − 1 −
I
JK
FG
H
IJ
K
FG
H
IJ
K
2 hγλ 0
γ 2u 2
u
u cos θ
2 hγλ ′
2h2
+
=
+ 2 2 1 − cos θ
1−
1−
2
mec
c
me c
c
c
me c
The first term is zero. Then
Since
this result may be written as
a
f
F 1 − au cosθ f c I + hγ F 1 I a1 − cosθ f .
λ′ = λ G
H 1 − u c JK m c GH 1 − u c JK
F uI F uI F uI
γ = 1 − G J = G1 − J G1 + J
H c K H c KH c K
F 1 − au cos θ f c I + h 1 + u c a1 − cosθ f .
λ′ = λ G
H 1 − u c JK m c 1 − u c
−1
0
e
2
−1
0
e
It shows a specific combination of what looks like a Doppler shift and a Compton shift. This problem
is about the same as the first problem in Albert Messiah’s graduate text on quantum mechanics.
490
Introduction to Quantum Physics
ANSWERS TO EVEN PROBLEMS
P40.2
(a) ~ 10 −7 m ultraviolet ;
(b) ~ 10 −10 m gamma ray
P40.4
(a) 70.9 kW; (b) 580 nm;
(c) 7.99 × 10 10 W m;
(d) 9.42 × 10 −1226 W m ;
P40.30
(a) 0.667 ; (b) 0.001 09
P40.32
(a) 14.0 kV/m, 46.8 µT ; (b) 4.19 nN;
(c) 10.2 g
P40.34
(a) 0.174 nm; (b) 5.37 pm or 5.49 pm
ignoring relativistic correction
P40.36
(a) see the solution; (b) 907 fm
P40.38
0.218 nm
P40.40
(a) 1.60; (b) 2.00 × 10 3 ; (c) 1; (d) ∞
P40.42
(a) 1.10 × 10 −34 m s ; (b) 1.36 × 10 33 s ; (c) no
(e) 1.00 × 10 −227 W m ; (f) 5.44 × 10 10 W m ;
(g) 7.38 × 10 10 W m ; (h) 0.260 W m;
(i) 2.60 × 10
19
−9
W m ; (j) 20 kW
P40.6
2.96 × 10
P40.8
5.71 × 10 3 photons
P40.10
1.34 × 10 31
P40.44
see the solution
P40.12
see the solution
P40.46
105 V
P40.14
(a) 1.38 eV ; (b) 334 THz
P40.48
(a) 0.250 m s ; (b) 2.25 m
P40.16
Metal one: 2.22 eV , Metal two: 3.70 eV
P40.50
P40.18
148 d, the classical theory is a gross failure
3.79 × 10 28 m , much larger than the
diameter of the observable Universe
P40.20
(a) The incident photons are Doppler
shifted to higher frequencies, and hence,
higher energy; (b) 3.87 eV ; (c) 8.78 eV
P40.52
(a) see the solution; (b) 5.21 MeV
P40.54
(a) 1.7 eV ; (b) 4.2 × 10 −15 V ⋅ s ; (c) 730 nm
P40.22
(a) 488 fm ; (b) 268 keV ; (c) 31.5 keV
P40.56
P40.24
22.1 keV
; K e = 478 eV
pe =
c
P40.58
see the solution
P40.26
(a) cos −1
P40.60
see the solution
P40.62
see the solution
P40.64
0.143 nm, comparable to the distance
between atoms in a crystal, so diffraction
can be observed
P40.66
(a) 2.82 × 10 −37 m; (b) 1.06 × 10 −32 J ;
(c) 2.87 × 10 −35%
P40.68
see the solution
P40.70
see the solution
photons s
F m c +E I;
GH 2m c + E JK
E F 2m c + E I
(b) E ′ =
,
2 GH m c + E JK
E F 2m c + E I
p′ =
;
2 c GH m c + E JK
2
e
e
0
e
0
γ
0
2
e
γ
e
0
e
(c) K e =
pe
P40.28
2
0
2
0
2
2
0
j
E F 2m c + E I
=
2 c GH m c + E JK
2 m e c 2 + E0
e
0
e
2
2
λ
−
e2B2 R 2
2m e
0
E02
e
hc
0
0
(a) 33.0° ; (b) 0.785 c
,