Quantum Physics
CHAPTER OUTLINE
28.1
28.2
28.3
28.4
28.5
28.6
28.7
28.8
28.9
28.10
28.11
28.12
28.13
28.14
Blackbody Radiation and
Planck’s Theory
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
An Interpretation of
Quantum Mechanics
A Particle in a Box
The Quantum Particle
Under Boundary
Conditions
The Schrödinger
Equation
Tunneling Through a
Potential Energy Barrier
Context ConnectionThe
Cosmic Temperature
ANSWERS TO QUESTIONS
Q28.1
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.
Q28.2
(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.
Q28.3
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.
Q28.4
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.
Q28.5
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.
Q28.6
A few photons would only give a few dots of exposure, apparently randomly scattered.
Q28.7
The x-ray photon transfers some of its energy to the electron. Thus, its frequency must decrease.
Q28.8
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.
771
772
Quantum Physics
Q28.9
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.
Q28.10
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.
Q28.11
Any object of macroscopic size—including a grain of dust—has an undetectably small wavelength
and does not exhibit quantum behavior.
Q28.12
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
If we set
F
GH
I
JK
Q28.13
The intensity of electron waves in some small region of space determines the probability that an
electron will be found in that region.
Q28.14
(a)
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 28.1
in the text shows this effect if you imagine the beam getting weaker at each reflection.
(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 28.1 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.
Chapter 28
Q28.15
The motion of the quantum particle does not consist of moving through successive points. The
particle has no definite position. It can sometimes be found on one side of a node and sometimes on
the other side, but never at the node itself. There is no contradiction here, for the quantum particle is
moving as a wave. It is not a classical particle. In particular, the particle does not speed up to infinite
speed to cross the node.
SOLUTIONS TO PROBLEMS
Section 28.1 Blackbody Radiation and Planck’s Theory
*P28.1
This is an example of Stefan’s law.
At the lower temperature, P1 = σ AeT14 .
At the higher , P2 = σ AeT24 .
FG IJ = FG 273 + 38.3 IJ = a1.004 19f = 1.016 9 .
H K H 273 + 37 K
The percentage increase in power is a1.016 9 − 1f = 0.016 9 =
Then
P2
T
= 2
P1
T1
4
4
4
1.69% , four times larger than the
percentage increase in temperature.
P28.2
(a)
P = eAσ T 4 , so
LM
P I
3.85 × 10 W
F
T=G
H eAσ JK = MM 1L4π e6.96 × 10 mj Oe5.67 × 10
PQ
MN NM
14
26
8
(b)
*P28.3
λ max =
2
−8
2
W m ⋅K
4
OP
PP
j PQ
14
= 5.78 × 10 3 K
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
=
= 5.01 × 10 −7 m = 501 nm
3
T
5.78 × 10 K
The peak radiation occurs at approximately 560 nm wavelength. From Wien’s displacement law,
T=
0.289 8 × 10 −2 m ⋅ K
λ max
=
0.289 8 × 10 −2 m ⋅ K
≈ 5 200 K .
560 × 10 −9 m
Clearly, a firefly is not at this temperature, so this is not blackbody radiation .
P28.4
773
e
je
e
je
e
je
(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
continued on next page
jFGH 1.601.00× 10eV
jFGH 1.601.00× 10eV
jFGH 1.601.00× 10eV
−19
−19
−19
IJ = 2.57 eV
JK
IJ = 1.28 × 10
JK
IJ = 1.91 × 10
JK
−5
eV
−7
eV
774
Quantum Physics
(d)
P28.5
λ=
c 3.00 × 10 8 m s
= 4.84 × 10 −7 m = 484 nm, visible light blue
=
f 620 × 10 12 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
a f
Each photon has an energy
This implies that there are
P28.6
e
je
j
E = hf = 6.626 × 10 −34 99.7 × 10 6 = 6.61 × 10 −26 J.
150 × 10 3 J s
= 2.27 × 10 30 photons s .
6.61 × 10 −26 J photon
Energy of a single 500-nm photon:
e6.626 × 10 J ⋅ sje3.00 × 10
= hf =
=
λ
e500 × 10 mj
−34
hc
Eγ
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
−15
J.
The number of photons required to yield this energy
n=
P28.7
2.27 × 10 −15 J
E
=
= 5.71 × 10 3 photons .
Eγ 3.98 × 10 −19 J photon
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
−3
J
g
= 0.498 Hz .
L
The energy is quantized,
E = nhf .
Therefore,
4.41 × 10 −3 J
E
n=
=
hf
6.626 × 10 −34 J ⋅ s 0.498 s −1
e
je
= 1.34 × 10 31
*P28.8
(a)
V=
a
f
4 3 4
π r = π 0.02 m
3
3
3
= 3.35 × 10 −5 m3
m = ρ V = 7.86 × 10 3 kg m3 3.35 × 10 −5 m3 = 0.263 kg
continued on next page
r
mg
FIG. P28.7
j
Chapter 28
(b)
a
f
A = 4π r 2 = 4π 0.02 m
2
= 5.03 × 10 −3 m 2
a
P = σ AeT 4 = 5.67 × 10 −8 W m 2 K 4 5.03 × 10 −3 m 2 0.86 293 K
(c)
f
4
= 1.81 W
It emits but does not absorb radiation, so its temperature must drop according to
d
Q = mc∆T = mc T f − Ti
dT f
dt
(d)
775
=
dQ
dt
mc
=
dT f
dQ
= mc
dt
dt
i
−1.81 J s
−P
=
= −0.015 3 ° C s = −0.919 ° C min
mc 0.263 kg 448 J kg ⋅ C°
λ maxT = 2.898 × 10 −3 m ⋅ K
λ max =
2.898 × 10 −3 m ⋅ K
= 9.89 × 10 −6 m infrared
293 K
hc
6.63 × 10 −34 Js 3 × 10 8 m s
= 2.01 × 10 −20 J
(e)
E = hf =
(f)
The energy output each second is carried by photons according to
λ
=
9.89 × 10 −6 m
P=
FG N IJ E
H ∆t K
1.81 J s
N P
= =
= 8.98 × 10 19 photon s
∆t E 2.01 × 10 −20 J photon
Matter is coupled to radiation, quite strongly, in terms of photon numbers.
Section 28.2 The Photoelectric Effect
P28.9
(a)
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
776
P28.10
P28.11
P28.12
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)
e∆VS =
(b)
e∆VS =
e
φ
h
je
j
2
= 9.64 × 10 −20 J = 0.602 eV
1 240 eV ⋅ nm
− 0.602 nm = 1.38 eV
625 nm
F
GH
I
JK
1.38 eV
1.60 × 10 −19 J
= 3.34 × 10 14 Hz
−34
1
eV
6.626 × 10
J⋅s
=
hc
λ
hc
λ
−φ →φ =
−φ =
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
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
so
∆t =
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
s = 148 days .
The gross failure of the classical theory of the photoelectric effect contrasts with the success of
quantum mechanics.
P28.13
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
m
j F 1.00 eV I − 4.70 eV = 1.51 eV .
GH 1.60 × 10 J JK
−19
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
kQ
V= e
r
e
jb
g
5.00 × 10 −2 m 1.51 N ⋅ m C
rV
Q=
=
= 8.41 × 10 −12 C .
9
2
2
ke
8.99 × 10 N ⋅ m C
or
Section 28.3 The Compton Effect
P28.14
E=
p=
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
700 × 10 −9 m
−19
J = 1.78 eV
Chapter 28
P28.15
(a)
∆λ =
(b)
E0 =
777
h
6.626 × 10 −34
1 − cos θ : ∆λ =
1 − cos 37.0° = 4.88 × 10 −13 m
mec
9.11 × 10 −31 3.00 × 10 8
a
hc
λ0
f
e
e
:
je
j
a
f
6.626 × 10 je3.00 × 10
j e
λ
−34
je
8
ms
300 × 10 3 eV 1.60 × 10 −19 J eV =
j
0
λ 0 = 4.14 × 10 −12 m
λ ′ = λ 0 + ∆λ = 4.63 × 10 −12 m
and
e
j
K e = E0 − E ′ = 300 keV − 268.5 keV = 31.5 keV
(c)
P28.16
je
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 4.30 × 10 −14 J = 268 keV
−12
λ′
4.63 × 10
m
E′ =
This is Compton scattering through 180°:
e6.626 × 10 J ⋅ sje3.00 × 10 m sj = 11.3 keV
E =
=
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
−34
hc
0
8
−9
0
−19
−12
−12
m
e
FIG. P28.16
hc
= 10.8 keV .
λ′
λ ′ = λ 0 + ∆λ = 0.115 nm so
E′ =
By conservation of momentum for the photon-electron system,
h $ h $
i=
− i + p e $i
λ0
λ′
and
pe = h
e j
FG 1
Hλ
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
+
1
0.115 × 10
−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
+
em c
E 2 = p 2 c 2 + m e2 c 4 or
e
a511 keV + 0.478 keVf = a22.1 keVf + a511 keVf
2
2.62 × 10 11 = 2.62 × 10 11
2
2
2
+ Ke
j = bpcg + em c j
2
2
e
2 2
778
P28.17
Quantum Physics
With K e = E ′ , K e = E0 − E ′ gives
E′ =
E ′ = E0 − E ′
E0
hc
and λ ′ =
E′
2
λ′ =
a
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 =
P28.18
(a)
K=
1
mev2 :
2
K=
λ′ =
λ 0 0.001 60
=
λ C 0.002 43
1
9.11 × 10 −31 kg 1.40 × 10 6 m s
2
E0 =
E′ = E0 − K and
f
e
hc
λ0
je
=
θ = 70.0°
j
2
= 8.93 × 10 −19 J = 5.58 eV
1 240 eV ⋅ nm
= 1 550 eV
0.800 nm
hc
1 240 eV ⋅ nm
= 0.803 nm
=
E ′ 1 550 eV − 5.58 eV
∆λ = λ ′− λ 0 = 0.002 88 nm = 2.88 pm
(b)
b
g
∆λ = λ C 1 − cos θ : cos θ = 1 −
∆λ
0.002 88 nm
=1−
= − 0.189 ,
λC
0.002 43 nm
θ = 101°
so
Section 28.4 Photons and Electromagnetic Waves
*P28.19
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.
Section 28.5 The Wave Properties of Particles
P28.20
λ=
h
h
6.626 × 10 −34 J ⋅ s
=
=
= 3.97 × 10 −13 m
−27
6
p mv
1.67 × 10
kg 1.00 × 10 m s
e
je
j
Chapter 28
P28.21
(a)
Electron:
λ=
and
λ=
h
p
K=
and
h
2m e K
=
p2
m2v2
1
mev 2 = e
=
2
2m e
2m e
so
6.626 × 10 −34 J ⋅ s
ja fe
e
2 9.11 × 10 −31 kg 3.00 1.60 × 10 −19 J
779
p = 2m e K
j
λ = 7.09 × 10 −10 m = 0.709 nm .
(b)
P28.22
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
3 1.60 × 10
J
E = hf
and
e
so
je
e
f=
E
h
j
j
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.
Thus, with m = 1,
λ=
h
=
p
h
2m e K
FG φ IJ cosFG φ IJ = a sin φ
H 2K H 2K
λ = 2 a sin
λ=
P28.23
(a)
6.626 × 10 −34 J ⋅ s
e
2 9.11 × 10
Therefore, the lattice spacing is a =
λ ~ 10 −14 m or less.
FIG. P28.22
−31
λ
sin φ
=
p=
je
kg 54.0 × 1.60 × 10
J
j
= 1.67 × 10 −10 m .
1.67 × 10 −10 m
= 2.18 × 10 −10 = 0.218 nm .
sin 50.0°
h
λ
~
6.6 × 10 −34 J ⋅ s
= 10 −19 kg ⋅ m s or more.
10 −14 m
The energy of the electron is E = p 2 c 2 + m e2 c 4 ~
(b)
−19
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 =
r
10 −14 m
With its K + U e >> 0 ,
the electron would immediately escape the nucleus .
780
P28.24
Quantum Physics
The wavelength of the student is λ =
(a)
FG h IJ
H mv K
F 6.626 × 10 J ⋅ s I =
h
v ≤ 10.0
= 10.0G
mw
H b80.0 kg ga0.750 mf JK
w ≤ 10.0 λ = 10.0
then we need
−34
so that
(b)
Using ∆t =
h
h
=
. If w is the width of the diffracting aperture,
p mv
d
we get:
v
∆t ≥
1.10 × 10 −34 m s .
0.150 m
= 1.36 × 10 33 s .
1.10 × 10 −34 m s
No . The minimum time to pass through the door is over 10 15 times the age of the
(c)
Universe.
P28.25
λ=
h
p
p=
(a)
electrons:
h
λ
=
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
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 .
(b)
photons:
e
je
j
Eγ = pc = 6.63 × 10 −23 3.00 × 10 8 = 124 keV
Section 28.6 The Quantum Particle
*P28.26
E=K =
1
h
mu 2 = hf and λ =
.
2
mu
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.
P28.27
As a bonus, we begin by proving that the phase speed v p =
vp =
ω
k
=
p2c2 + m2c4 h
hγ mv
=
γ 2m 2 v 2 c 2 + m 2 c 4
γ 2m2 v2
= c 1+
ω
k
is not the speed of the particle.
F
GH
c2
c2
v2
=
c
1
+
1
−
v2
c2
γ 2v2
I =c
JK
1+
c2
c2
−
1
=
v
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
Chapter 28
vg =
dω dhω dE d
=
=
=
m2c4 + p2c2
dk
dhk dp dp
vg =
1 2 4
m c + p2c 2
2
vg = c
e
j e0 + 2 pc j =
−1 2
γ 2m 2 v 2
2
2 2
2
=c
2 2
γ m v +m c
p2c4
p2c 2 + m2c4
e
v2 1 − v2 c2
v
2
781
j
e1 − v c j + c
2
2
=c
2
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 28.7 The Double-Slit Experiment Revisited
P28.28
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
P28.29
−8
j FGH
m sin tan −1
LM 0.400 OPIJ = a1fλ = 1.20 × 10
N 200 QK
−10
m
h
λ
m2v2
1
h2
mev 2 = e
=
= e∆V
2
2m e
2 m e λ2
∆V =
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 .
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 +
e
jb
g
FG
H
IJ
K
1
λ , with m = 0 for
2
the first minimum.
Then,
so
(c)
y
= tan θ
L
FG λ IJ = 0.028 4°
H 2d K
y = L tan θ = a10.0 mfbtan 0.028 4°g =
θ = sin −1
4.96 mm .
We cannot say the neutron passed through one slit. We can only say it passed through the slits.
782
*P28.30
Quantum Physics
We find the speed of each electron from energy conservation in the firing process:
0 = K f +U f =
v=
1
mv 2 − eV
2
9.11 × 10
The time of flight is ∆t =
apart is I =
a
2 × 1.6 × 10 −19 C 45 V
2 eV
=
m
−31
kg
f = 3.98 × 10
6
ms
∆x
0.28 m
=
= 7.04 × 10 −8 s . The current when electrons are 28 cm
v 3.98 × 10 6 m s
q e 1.6 × 10 −19 C
=
=
= 2.27 × 10 −12 A .
t ∆t 7.04 × 10 −8 s
Section 28.8 The Uncertainty Principle
P28.31
For the electron,
∆x =
ge
j
h
6.626 × 10 −34 J ⋅ s
= 1.16 mm .
=
4π ∆p 4π 4.56 × 10 −32 kg ⋅ m s
e
j
b
gb
ge
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,
∆x=
P28.32
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
h
= 5.28 × 10 −32 m .
4π ∆ p
h
2
∆v ≥
2π J ⋅ s
h
=
= 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
b
ga
f
ga f
b
1.00 m, we can think of ∆ x growing to 1.00 m + 1.25 m = 2.25 m .
P28.33
∆ y ∆ py
=
x
px
and
d ∆ py ≥
h
.
4π
Eliminate ∆ p y and solve for x.
d
x = 4π p x ∆ y :
h
b g
e
x = 4π 1.00 × 10
−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
The answer, x = 3.79 × 10 28 m , is 190 times greater than the diameter of the observable Universe!
Chapter 28
P28.34
h
2
From the uncertainty principle
∆E∆t ≥
or
∆ mc 2 ∆t =
Therefore,
∆m
h
h
=
=
2
m
4π c ∆t m 4π ∆t ER
e j
h
.
2
a f
a f
∆m
6.626 × 10 −34 J ⋅ s
=
m
4π 8.70 × 10 −17 s 135 MeV
ja
e
P28.35
(a)
FG 1 MeV IJ =
f H 1.60 × 10 J K
−13
b g
A=
where
h
2m
FG h IJ
H 2m∆x K
i
2H
.
g
d i =0
db ∆x g
d ∆x f
To minimize ∆x f , we require
or
1−
i
A
=0
∆xi2
∆xi = A .
so
The minimum width of the impact points is
d i
∆x f
min
FG
H
= ∆xi +
A
∆xi
IJ
K
=2 A =
∆xi = A
L 2 1.054 6 × 10 J ⋅ s O
d∆x i = MM e 5.00 × 10 kg j PP
N
Q
−34
f
−4
min
12
F I
GH JK
2h 2 H
m g
LM 2a2.00 mf OP
MN 9.80 m s PQ
14
.
14
= 5.19 × 10 −16 m
2
Section 28.9 An Interpretation of Quantum Mechanics
P28.36
2.81 × 10 −8 .
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
h
1
2H
∆p x = m∆v x =
. It falls to the floor in a travel time given by H = 0 + gt 2 as t =
, so
2 ∆xi
2
g
the total width of the impact points is
∆x f = ∆xi + ∆v x t = ∆xi +
(b)
783
Probability
P=
z
a
af
ψ x
2
−a
P=
1
π
z π ex + a j FGH πa IJK FGH 1a IJK tan FGH xa IJK
1 Lπ F π I O
1
1 − tan a −1f = M − G − J P =
2
π N 4 H 4KQ
a
=
−a
tan −1
a
2
−1
2
dx =
−1
a
−a
A
2H
= ∆xi +
∆xi
g
784
P28.37
Quantum Physics
(a)
ψ x = Ae e
af
i 5.00 ×10 10 x
j = A cose5 × 10 10 xj + Ai sine5 × 10 10 xj = A cosa kxf + Ai sinakxf goes through
a full cycle when x changes by λ and when kx changes by 2π . Then kλ = 2π where
2π
2π m
k = 5.00 × 10 10 m −1 =
. Then λ =
= 1.26 × 10 −10 m .
λ
5.00 × 10 10
e
h
6.626 × 10 −34 J ⋅ s
= 5.27 × 10 −24 kg ⋅ m s
1.26 × 10 −10 m
(b)
p=
(c)
m e = 9.11 × 10 −31 kg
λ
=
j
e
j
j
5.27 × 10 −24 kg ⋅ m s
m2v2 p2
K= e
=
=
2m e
2m
2 × 9.11 × 10 −31 kg
e
2
= 1.52 × 10 −17 J =
1.52 × 10 −17 J
= 95.5 eV
1.60 × 10 −19 J eV
Section 28.10 A Particle in a Box
P28.38
For an electron wave to “fit” into an infinitely deep potential well, an
integral number of half-wavelengths must equal the width of the well.
nλ
= 1.00 × 10 −9 m
2
(a)
(b)
e
so
λ=
j
2.00 × 10 −9 h
=
n
p
Since
h 2 λ2
p2
h2
n2
K=
=
=
2m e
2m e
2m e 2 × 10 −9
j
For
K ≈ 6 eV
n=4
With
n = 4,
K = 6.03 eV
e
2
e
j
= 0.377n 2 eV
FIG. P28.38
Chapter 28
P28.39
(a)
We can draw a diagram that parallels our treatment of standing
mechanical waves. In each state, we measure the distance d
from one node to another (N to N), and base our solution upon
that:
Since
d N to N =
p=
(b)
h
λ
=
λ
2
and λ =
h
p
h
.
2d
LM e6.626 × 10
MM 8e9.11 × 10
N
−34
j OP .
kg j P
PQ
J⋅s
Next,
p2
h2
1
K=
=
=
2m e 8m e d 2 d 2
Evaluating,
K=
In state 1,
d = 1.00 × 10 −10 m
K 1 = 37.7 eV .
In state 2,
d = 5.00 × 10 −11 m
K 2 = 151 eV .
In state 3,
d = 3.33 × 10 −11 m
K 3 = 339 eV .
In state 4,
d = 2.50 × 10 −11 m
K 4 = 603 eV .
6.02 × 10 −38 J ⋅ m 2
d2
K=
−31
2
3.77 × 10 −19 eV ⋅ m 2
.
d2
FIG. P28.39
When the electron falls from state 2 to state 1, it puts out energy
E = 151 eV − 37.7 eV = 113 eV = hf =
hc
λ
into emitting a photon of wavelength
λ=
e
je
j
6.626 × 10 −34 J ⋅ s 3.00 × 10 8 m s
hc
=
= 11.0 nm .
E
113 eV 1.60 × 10 −19 J eV
a
fe
j
The wavelengths of the other spectral lines we find similarly:
Transition
E eV
λ nm
a f
a f
4→3
264
4.71
4→ 2
452
2.75
4→1
565
2.20
3→2
188
6.60
3→1
302
4.12
2→1
113
11.0
785
786
P28.40
Quantum Physics
The confined proton can be described in the same way as a
standing wave on a string. At level 1, the node-to-node
distance of the standing wave is 1.00 × 10 −14 m , so the
h
wavelength is twice this distance: = 2.00 × 10 −14 m.
p
The proton’s kinetic energy is
e
j
2
6.626 × 10 −34 J ⋅ s
p2
h2
1
2
=
=
K = mv =
2
2m 2mλ2 2 1.67 × 10 −27 kg 2.00 × 10 −14 m
e
=
je
j
2
FIG. P28.40
−13
J
3.29 × 10
= 2.05 MeV
1.60 × 10 −19 J eV
In the first excited state, level 2, the node-to-node distance is half as long as in state 1. The
momentum is two times larger and the energy is four times larger: K = 8.22 MeV .
The proton has mass, has charge, moves slowly compared to light in a standing wave state, and
stays inside the nucleus. When it falls from level 2 to level 1, its energy change is
2.05 MeV − 8.22 MeV = −6.16 MeV .
Therefore, we know that a photon (a traveling wave with no mass and no charge) is emitted at the
speed of light, and that it has an energy of +6.16 MeV .
e
je
j
Its frequency is
f=
6.16 × 10 6 eV 1.60 × 10 −19 J eV
E
= 1.49 × 10 21 Hz .
=
h
6.626 × 10 −34 J ⋅ s
And its wavelength is
λ=
c 3.00 × 10 8 m s
=
= 2.02 × 10 −13 m .
f 1.49 × 10 21 s −1
This is a gamma ray, according to the electromagnetic spectrum chart in Chapter 24.
*P28.41
(a)
The energies of the confined electron are En =
jump from state 1 to state 4 is
h2
n 2 . Its energy gain in the quantum
8m e L2
h2
4 2 − 1 2 and this is the photon energy:
8m e L2
e
j
FG
H
h 2 15
hc
15 hλ
= hf = . Then 8m e cL2 = 15 hλ and L =
2
λ
8m e c
8m e L
(b)
IJ
K
12
Let λ ′ represent the wavelength of the photon emitted:
e
j
2
2
5
hc λ ′ h 15 8m e L
Then
=
= and λ ′ = 1.25 λ .
2
2
λ hc 8m e L 12 h
4
.
hc
h2
h2
12 h 2
42 −
22 =
.
=
2
2
λ ′ 8m e L
8m e L
8m e L2
Chapter 28
Section 28.11 The Quantum Particle Under Boundary Conditions
Section 28.12 The Schrödinger Equation
P28.42
af
ψ x = A cos kx + B sin kx
∂ψ
= − kA sin kx + kB cos kx
∂x
∂ 2ψ
= − k 2 A cos kx − k 2 B sin kx
∂x 2
−
a
a
f
2m
2mE
E − U ψ = − 2 A cos kx + B sin kx
h
h
f
Therefore the Schrödinger equation is satisfied if
FG
H
IJ a
K
∂ 2ψ
2m
= − 2 E − U ψ or
∂x 2
h
f
f FGH
a
− k 2 A cos kx + B sin kx = −
This is true as an identity (functional equality) for all x if E =
P28.43
∂ψ
= ikψ
∂x
ψ = Aeib kx −ω t g
We have
k2 =
f
h2k2
.
2m
and
∂ 2ψ
= − k 2ψ .
∂x 2
∂ 2ψ
2m
= − k 2ψ = − 2 E − U ψ .
2
∂x
h
a
Schrödinger’s equation:
Since
IJ a
K
2mE
A cos kx + B sin kx .
h2
a 2π f = b 2π p g
2
λ2
2
=
h2
p2
h2
and
E −U =
f
p2
.
2m
Thus this equation balances.
P28.44
(a)
af
ψ x = Aeikx
With
d
d 2ψ
= − Ak 2 eikx .
Aeikx = Aikxeikx and
dx
dx 2
(b)
P28.45
(a)
p2
m2v2
1
h 2 d 2ψ
h2 k 2
h 2 4π 2
ikx
=
+
=
ψ
=
ψ
=
ψ = mv 2ψ = Kψ .
Ae
2
2 2
2m dx
2m
2m
2m
2
4π λ 2m
Then
−
With
ψ x =
Then
−
af
z
L
x = x
0
x =
continued on next page
FG
H
IJ
K
nπ x
dψ
d 2ψ
2
sin
= A sin kx ,
= Ak cos kx and
= − Ak 2 sin kx
L
L
dx
dx 2
p2
h 2 d 2ψ
h2
h 2 4π 2
2
=
+
sin
=
ψ
=
ψ = Kψ .
Ak
kx
2m dx 2
2m
2m
4π 2 λ2 2m
FG
H
IJ
K
z FGH
IJ
K
LM 4π x sin 4π x + cos 4π x OP
L Q
NL L
L
2π x
4π x
2
2
1 1
sin 2
dx =
x − cos
dx
L
L
L
L0 2 2
1 x2
L 2
L
−
0
1 L2
L 16π 2
L
=
0
L
2
787
788
Quantum Physics
(b)
Probability =
FG
H
z
0.510 L
Probability= 0.020 −
(c)
Probability
(d)
P28.46
IJ
K
LM
N
4π x
2π x
2
1
1 L
sin 2
sin
dx =
x−
4
L
L
L
L
π
L
0. 490 L
a
0. 490 L
f
1
sin 2.04π − sin 1.96π = 5.26 × 10 −5
4π
LM x − 1 sin 4π x OP
N L 4π L Q
0. 260 L
= 3.99 × 10 −2
0. 240 L
In the n = 2 graph in Figure 28.23 (b), it is more probable to find the particle either near
L
3L
x = or x =
than at the center, where the probability density is zero.
4
4
L
Nevertheless, the symmetry of the distribution means that the average position is .
2
2
ψ dx = 1
z
L
or
z
L
A 2 sin 2
0
A 2 sin 2
0
all space
FG nπ x IJ dx = A FG L IJ = 1
H 2K
H LK
2
or
z
L4
The desired probability is
P=
2
ψ dx =
0
sin 2 θ =
where
P28.48
0.510 L
Normalization requires
z
P28.47
OP
Q
F x 1 sin 4π x IJ
Thus, P = G −
H L 4π L K
F xI
(a)
ψ a x f = AG 1 − J
H LK
L4
=
0
A=
2
.
L
z
FG 2π x IJ dx
H LK
L4
2
L
FG nπ x IJ dx = 1
H LK
0
sin 2
1 − cos 2θ
.
2
FG 1 − 0 − 0 + 0IJ =
H4
K
2
2
0.250 .
d 2ψ
2A
=− 2
dx 2
L
2 Ax
dψ
=− 2
dx
L
Schrödinger’s equation
d 2ψ
2m
= − 2 E −U ψ
2
dx
h
becomes
2 2
2
2
2A
2m
2m − h x A 1 − x L
x2
− 2 = − 2 EA 1 − 2 + 2
L
L
h
h
mL2 L2 − x 2
a
f
F
GH
I
JK
e
je
e
1
mE mEx 2 x 2
=− 2 + 2 2 − 4 .
2
L
h
h L
L
1 mE
=
L2 h 2
mE
1
−
=0
h 2 L2 L4
−
This will be true for all x if both
and
both these conditions are satisfied for a particle of energy
continued on next page
E=
h2
L2 m
.
j
j
Chapter 28
789
F xI
F 2x + x I dx
For normalization,
1 = z A G 1 − J dx = A z G 1 −
H LK
H L L JK
L 2x + x OP = A LL − 2 L + L + L − 2 L + L O = A FG 16L IJ
15
1 = A Mx −
A=
.
M
P
H
K
3
5
3
5
15
16
L
N
Q
N 3L 5L Q
F1 − 2 x + x I dx = 15 LMx − 2 x + x OP = 30 LM L − 2L + L OP
15
P = z ψ dx =
z
GH L L JK 16L N 3L 5L Q 16L N 3 81 1 215 Q
16L
L
(b)
2
2
3
5
2
2
2
−L
2
2
L
−L
2
2
4
4
L
2
4
2
−L
L3
(c)
L3
2
−L 3
P=
2
2
−L 3
4
3
4
2
5
5
L3
−L 3
47
= 0.580
81
Section 28.13 Tunneling Through a Potential Energy Barrier
P28.49
T = e −2CL where C =
2CL =
P28.50
e
a
2m U − E
f
h
je
2 2 9.11 × 10 −31 8.00 × 10 −19
1.055 × 10
j
−34
e2.00 × 10 j = 4.58
−10
(a)
T = e −4.58 = 0.010 3 , a 1% chance of transmission.
(b)
R = 1 − T = 0.990 , a 99% chance of reflection.
C=
ja
e
fe
j
2 9.11 × 10 −31 5.00 − 4.50 1.60 × 10 −19 kg ⋅ m s
1.055 × 10
−34
e
J⋅s
je
FIG. P28.49
= 3.62 × 10 9 m −1
a
j
T = e −2CL = exp −2 3.62 × 10 9 m −1 950 × 10 −12 m = exp −6.88
f
T = 1.03 × 10 −3
*P28.51
FIG. P28.50
The original tunneling probability is T = e −2CL where
c2maU − Efh
C=
h
12
=
a
e
f
2π 2 × 9.11 × 10 −31 kg 20 − 12 1.6 × 10 −19 J
6.626 × 10 −34 J ⋅ s
j
12
= 1.448 1 × 10 10 m −1 .
1 240 eV ⋅ nm
= 2.27 eV , to make the electron’s new kinetic energy
λ
546 nm
12 + 2.27 = 14.27 eV and its decay coefficient inside the barrier
The photon energy is hf =
C′ =
hc
=
a
e
f
2π 2 × 9.11 × 10 −31 kg 20 − 14.27 1.6 × 10 −19 J
6.626 × 10
−34
J⋅s
j
12
= 1.225 5 × 10 10 m −1 .
Now the factor of increase in transmission probability is
−9
e −2C ′L
= e 2 LaC −C ′ f = e 2 ×10
−2CL
e
m× 0. 223 × 10 10 m −1
= e 4.45 = 85.9 .
790
Quantum Physics
Section 28.14 Context Connection
The Cosmic Temperature
P28.52
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
The temperature of the star is given by
T=
P28.53
2.898 × 10 −3 m ⋅ K
(a)
(b)
P28.54
b g = a500 nmf
1 + bv cg
1− v c
λ = λ′
λ max
:
a
a
f
f
1 − 0.280
= 375 nm .
1 + 0.280
λ maxT = 2.898 × 10 −3 m ⋅ K :
T=
2.898 × 10 −3 m ⋅ K
= 7.73 × 10 3 K .
375 × 10 −9
Wien’s law:
λ maxT = 2.898 × 10 −3 m ⋅ K .
Thus,
λ max =
2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K
=
= 1.06 × 10 −3 m = 1.06 mm
T
2.73 K
.
This is a microwave .
We suppose that the fireball of the Big Bang is a black body.
ja
e
I = eσ T 4 = (1) 5.67 × 10 −8 W m 2 ⋅ K 4 2.73 K
f
4
= 3.15 × 10 −6 W m 2
As a bonus, we can find the current power of direct radiation from the Big Bang in the section of the
universe observable to us. If it is fifteen billion years old, the fireball is a perfect sphere of radius
fifteen billion light years, centered at the point halfway between your eyes:
e
2
P = IA = I ( 4π r ) = 3.15 × 10
−6
W m
2
ja4π fe15 × 10
9
F 3 × 10 m s I 3.156 × 10
ly j G
H 1 ly yr JK e
2
8
2
7
s yr
j
2
P = 7.98 × 10 47 W .
Additional Problems
a ∆V f = a ∆V f
2
*P28.55
The condition on electric power delivered to the filament is P = I ∆V =
so r =
F P ρl I
GH π a∆V f JK
2
12
R
2
ρl
A
a ∆V f π r
2
=
ρl
. Here P = 75 W, ρ = 7.13 × 10 −7 Ω ⋅ m , and ∆V = 120 V . As the filament radiates in
steady state, it must emit all of this power through its lateral surface area P = σ eAT 4 = σ e 2π rlT 4 .
We combine the conditions by substitution:
continued on next page
2
Chapter 28
F P ρl I
P = σ e 2π G
H π a∆V f JK
12
lT 4
2
a∆V fP
12
= σ e 2π 1 2 ρ 1 2 l 3 2 T 4
F a∆V fP I = FG
120 V a75 W f m K
l=G
J
G
H σ e2π ρ T K H 5.67 × 10 W 0.450a2fπ e7.13 × 10 Ω ⋅ mj a2 900 K f
= e0.192 m j = 0.333 m = l
F P ρ l I = F 75 W 7.13 × 10 Ω ⋅ m 0.333 mI = r = 1.98 × 10 m .
and r = G
JK
π a120 V f
H π a∆V f JK GH
F hI φ
∆V = G J f −
H eK e
φ
F hI
From two points on the graph
0 = G J e 4.1 × 10 Hzj −
H eK
e
φ
F hI
and
3.3 V = G J e12 × 10 Hzj − .
H eK
e
23
12
12
12
12
4
−8
12
2
4
12
−7
32 23
12
12
−7
2
P28.56
−5
2
S
14
14
Combining these two expressions we find:
(a)
φ = 1.7 eV
(b)
h
= 4.2 × 10 −15 V ⋅ s
e
(c)
At the cutoff wavelength
e
je
791
hc
λc
=φ =
λ c = 4.2 × 10 −15 V ⋅ s 1.6 × 10 −19 C
FIG. P28.56
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
4
I
JJ
K
23
792
P28.57
Quantum Physics
We want an Einstein plot of K max versus f =
λ , nm
588
505
445
399
f , 10 14 Hz K max , eV
0.67
5.10
0.98
5.94
1.35
6.74
7.52
1.63
0.402 eV
± 8%
10 14 Hz
(a)
slope =
(b)
e∆VS = hf − φ
a
h = 0.402
(c)
c
λ
fFGH 1.60 ×1010
−19
J⋅s
14
I=
JK
f (THz)
6. 4 × 10 −34 J ⋅ s ± 8%
FIG. P28.57
K max = 0
at f ≈ 344 × 10 12 Hz
φ = hf = 2.32 × 10 −19 J = 1.4 eV
P28.58
From the path the electrons follow in the magnetic field, the maximum kinetic energy is seen to be:
K max =
P28.59
e2B2 R 2
.
2m e
From the photoelectric equation,
K max = hf − φ =
Thus, the work function is
φ=
(a)
mgyi =
hc
λ
hc
λ
− K max =
−φ .
hc
λ
−
e2B2 R2
.
2m e
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
h
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%
2
E
75.0 kg 9.80 m s 50.0 m
b
ge
ja
f
Chapter 28
P28.60
P28.61
(a)
λ = 2L = 2.00 × 10 −10 m
(b)
p=
(c)
E=
(a)
h
λ
=
6.626 × 10 −34 J ⋅ s
= 3.31 × 10 −24 kg ⋅ m s
2.00 × 10 −10 m
p2
= 0.172 eV
2m
(b)
See the figure.
See the figure.
FIG. P28.61(a)
(c)
FIG. P28.61(b)
ψ is continuous and ψ → 0 as x → ±∞ . The function can be normalized. It describes a
particle bound near x = 0 , by a very deep, very narrow well of potential energy.
(d)
Since ψ is symmetric,
z
∞
z
∞
2
0
−∞
or
2
ψ dx = 2 ψ dx = 1
F 2 A I ee
GH −2α JK
z
∞
2
2 A 2 e −2α x dx =
0
−∞
j
− e0 = 1 .
This gives A = α .
(e)
P28.62
(a)
Pb −1 2α g→b1 2α g = 2
e aj
2
z
1 2α
e −2α x dx =
x=0
FG 2α IJ ee
H −2α K
− 2α 2α
j e
j
− 1 = 1 − e −1 = 0.632
Use Schrödinger’s equation
∂ 2ψ
2m
= − 2 E −U ψ
∂x 2
h
a
f
with solutions
continued on next page
ψ 1 = Ae ik1 x + Be −ik1 x
[region I]
ψ 2 = Ce ik 2 x
[region II].
FIG. P28.62(a)
793
794
Quantum Physics
2mE
h
Where
k1 =
and
k2 =
a
2m E − U
h
f.
Then, matching functions and derivatives at x = 0
bψ g = bψ g
F dψ IJ = FG dψ IJ
and G
H dx K H dx K
1 0
2 0
1
2
0
A+B=C
gives
k1 A − B = k 2C .
a
0
f
Then
B=
1 − k 2 k1
A
1 + k 2 k1
and
C=
2
A.
1 + k 2 k1
Incident wave Ae
(b)
gives
ikx
reflects Be
− ikx
, with probability
b
b
1 − k 2 k1
B2
R= 2 =
A
1 + k 2 k1
With
E = 7.00 eV
and
U = 5.00 eV
k2
E −U
=
=
k1
E
P28.63
x2 =
z
a1 − 0.535f
a1 + 0.535f
The probability of transmission is
T = 1 − R = 0.908 .
2
−∞
For a one-dimensional box of width L, ψ n =
Thus, x 2 =
z
FG
H
IJ
K
FG
H
IJ
K
nπ x
2
sin
.
L
L
L
nπ x
2 2
L2
L2
x sin 2
− 2 2
dx =
3 2n π
L0
L
(from integral tables).
2
=
bk
bk
2
g
+k g
1
− k2
1
2
2
R=
x 2 ψ dx
2
2.00
= 0.535 .
7.00
The reflection probability is
∞
g
g
= 0.092 0 .
2
2
.
Chapter 28
*P28.64
(a)
The requirement that
bpcg + emc j
2 2
2
E=
K n = En − mc 2 =
(b)
795
nλ
h nh
= L so p = =
is still valid.
2
λ 2L
FG nhc IJ + emc j
H 2L K
2
⇒ En =
FG nhc IJ + emc j
H 2L K
2
2 2
2 2
− mc 2
Taking L = 1.00 × 10 −12 m, m = 9.11 × 10 −31 kg , and n = 1, we find K 1 = 4.69 × 10 −14 J .
e
j
2
6.626 × 10 −34 J ⋅ s
h2
=
Nonrelativistic, E1 =
8mL2 8 9.11 × 10 −31 kg 1.00 × 10 −12 m
e
je
j
2
= 6.02 × 10 −14 J .
Comparing this to K 1 , we see that this value is too large by 28.6% .
P28.65
For a particle with wave function
af
ψ x =
2 −x a
e
a
and
0
(a)
ψ x
(b)
Prob x < 0 =
for x > 0
for x < 0 .
af
2
a
= 0, x < 0
f
z
0
af
ψ x
2
Normalization
2 −2 x a
e
, x>0
a
FIG. P28.65
za f
0
dx =
−∞
(c)
af
ψ2 x =
and
0 dx = 0
−∞
z af z
z z FGH IJK
∞
ψ x
2
dx =
−∞
0
0
z
∞
2
0
−∞
0dx +
∞
0
−∞
a
2
ψ dx + ψ dx = 1
2 −2 x a
e
dx = 0 − e −2 x a
a
f
z
a
2
Prob 0 < x < a = ψ dx =
0
z FGH IJK
a
0
∞
0
e
j
= − e −∞ − 1 = 1
2 −2 x a
e
dx = − e −2 x a
a
a
0
= 1 − e −2 = 0.865
This wave function would require the potential energy to be +∞ for x < 0 and −∞ for x = 0 .
This potential energy function cannot be realistic in detail.
796
Quantum Physics
ANSWERS TO EVEN PROBLEMS
P28.2
(a) 5.78 × 10 3 K ; (b) 501 nm
P28.38
(a) n = 4; (b) 6.03 eV
P28.4
(a) 2.57 eV; (b) 12.8 µeV ; (c) 191 neV;
(d) 484 nm visible, 9.68 cm and 6.52 m
radio waves
P28.40
6.16 MeV, 202 fm, a gamma ray
P28.42
see the solution, E =
P28.44
see the solution
P28.46
see the solution
P28.48
(a) E =
P28.6
5.71 × 10 3 photons
P28.8
(a) 0.263 kg; (b) 1.81 W;
(c) −0.015 3 ° C s = −0.919 ° C min ;
(d) 9.89 µm ; (e) 2.01 × 10 −20 J ;
(f) 8.98 × 10 19 photon s
P28.10
P28.12
(a) 1.38 eV; (b) 334 THz
z
L
(b) requiring
148 days, absurdly large
−28
h2
;
mL2
A=
FG 15 IJ
H 16L K
A2 1 −
−L
12
F
GH
h2k 2
2m
x2
L2
I
JK
2
dx = 1 gives
;
P28.14
1.78 eV, 9.47 × 10
P28.16
22.1 keV
, 478 eV
c
P28.50
1.03 × 10 −3
P28.18
(a) 2.88 pm; (b) 101°
P28.52
7.73 × 10 3 K
P28.20
397 fm
P28.54
3.15 µ W m 2
P28.22
0.218 nm
P28.56
(a) 1.7 eV; (b) 4.2 fV⋅s; (c) 730 nm
P28.24
(a) 1.10 × 10 −34 m s ; (b) 1.36 × 10 33 s ;
(c) no, the time is over 10 15 times the age
of the universe
kg ⋅ m s
u
2
P28.26
v phase =
P28.28
105 V
P28.30
2.27 pA
P28.32
(a) 0.250 m s ; (b) 2.25 m
P28.34
2.81 × 10 −8
P28.36
1
2
(c)
P28.58
hc
λ
47
= 0.580
81
−
e2B2 R 2
2m e
P28.60
(a) 2.00 × 10 −10 m;
(b) 3.31 × 10 −24 kg ⋅ m s ;
(c) 0.172 eV
P28.62
(a) see the solution;
(b) R = 0.092 0 , T = 0.908
P28.64
(a)
FG nhc IJ
H 2L K
2
+ m 2 c 4 − mc 2 ;
(b) 46.9 fJ, 28.6%