771
Quantum PhysicsCHAPTER OUTLINE 28.1 Blackbody Radiation and
Planck’s Theory 28.2 The Photoelectric Effect 28.3 The Compton Effect 28.4 Photons and
Electromagnetic Waves 28.5 The Wave Properties of
Particles 28.6 The Quantum Particle 28.7 The Double-Slit
Experiment Revisited 28.8 The Uncertainty Principle 28.9 An Interpretation of
Quantum Mechanics 28.10 A Particle in a Box 28.11 The Quantum Particle
Under Boundary Conditions
28.12 The Schrödinger Equation
28.13 Tunneling Through a Potential Energy Barrier
28.14 Context ConnectionThe 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.
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 If we set pm
q V2
2= ∆ , which is the same for both particles, then we see that the electron has the
smaller momentum and therefore the longer wavelength λ =FHGIKJ
hp
.
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 28 773
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 14=σ AeT .
At the higher , P2 24=σ AeT .
Then PP
2
1
2
1
4 44273 38 3
273 371 004 19 1 016 9=
FHGIKJ = +
+FHG
IKJ = =
TT
.. . a f .
The percentage increase in power is 1 016 9 1 0 016 9 1 69%. . . − = =a f , four times larger than the
percentage increase in temperature. P28.2 (a) P = eA Tσ 4, so
TeA
= FHGIKJ = ×
×LNM
OQP × ⋅
L
N
MMMM
O
Q
PPPP= ×
−
Pσ π
1 4 26
2 8
1 4
33 85 10
5 67 105 78 10
.
..
W
1 4 6.96 10 m W m K K
8 2 4e j e j
(b) λmax. .
.= × ⋅ = × ⋅×
= × =− −
−2 898 10 2 898 105 01 10 501
3 37 m K m K
5.78 10 K m nm3T
*P28.3 The peak radiation occurs at approximately 560 nm wavelength. From Wien’s displacement law,
T = × ⋅ = × ⋅×
≈− −
−0 289 8 10 0 289 8 10
560 105 200
2 2
9. .
max
m K m K m
Kλ
.
Clearly, a firefly is not at this temperature, so this is not blackbody radiation .
P28.4 (a) E hf= = × ⋅ ××
FHG
IKJ =
− −−6 626 10 620 10
1 002 5734 12 1
19..
. J s s eV
1.60 10 J eVe je j
(b) E hf= = × ⋅ ××
FHG
IKJ = ×− −
−−6 626 10 3 10 10
1 001 28 1034 9 1 5. .
.. J s s
eV1.60 10 J
eV19e je j
(c) E hf= = × ⋅ ××
FHG
IKJ = ×− −
−−6 626 10 46 0 10
1 001 91 1034 6 1
197. .
.. J s s
eV1.60 10 J
eVe je j
continued on next page
774 Quantum Physics
(d) λ = =××
= × =−cf
3 00 10620 10
4 84 10 4848
127.
. m s Hz
m nm, visible light bluea f
λ
λ
= =××
= × =
= =××
=
−cf
cf
3 00 103 10 10
9 68 10 9 68
3 00 1046 0 10
6 52
8
92
8
6
..
. .
..
.
m s Hz
m cm, radio wave
m s Hz
m, radio wave
P28.5 Each photon has an energy E hf= = × × = ×− −6 626 10 99 7 10 6 61 1034 6 26. . .e je j J .
This implies that there are 150 10
6 61 102 27 10
3
2630×
×= ×−
J s J photon
photons s.
. .
P28.6 Energy of a single 500-nm photon:
E hfhc
γ λ= = =
× ⋅ ×
×= ×
−
−−
6 626 10 3 00 10
500 103 98 10
34 8
919
. ..
J s m s
m J
e je je j
.
The energy entering the eye each second
E t IA t= = = × ×LNM
OQP = ×− − −P∆ ∆ 4 00 10
48 50 10 1 00 2 27 1011 3 2 15. . . . W m m s J2e j e j a fπ
.
The number of photons required to yield this energy
nE
E= = ×
×= ×
−
−γ
2 27 1010
5 71 1015
193.
. J
3.98 J photon photons .
P28.7 We take θ = 0 030 0. radians. Then the pendulum’s total energy is
E mgh mg L L
E
= = −
= − = × −
cos
. . . . .
θa fb ge jb g1 00 9 80 1 00 0 999 5 4 41 10 3 kg m s J2
The frequency of oscillation is fgL
= = =ωπ π2
12
0 498. Hz .
The energy is quantized, E nhf= .
Therefore, nEhf
= = ×× ⋅
= ×
−
− −4 41 10
0 498
1 34 10
3
1
31
.
.
.
J
6.626 10 J s s34e je j
mrg
FIG. P28.7
*P28.8 (a) V r= = = × −43
43
0 02 3 35 103 3 5π π . . m m3a f
m V= = × × =−ρ 7 86 10 3 35 10 0 2633 5. . . kg m m kg3 3
continued on next page
Chapter 28 775
(b) A r= = = × −4 4 0 02 5 03 102 2 3π π . . m m2a f
P = = × × =− −σ AeT 4 8 2 3 45 67 10 5 03 10 0 86 293 1 81. W m . . .K m K W4 2 a f
(c) It emits but does not absorb radiation, so its temperature must drop according to
Q mc T mc T T
dQdt
mcdT
dtdT
dt mc mc
f if
fdQdt
= = − =
= = − =−
⋅ °= − ° = − °
∆ d iP 1 81
0 263 448 0 015 3 0 919
..
. . J s
kg J kg C C s C min
(d) λmax .T = × ⋅−2 898 10 3 m K
λmax.
.= × ⋅ = ×−
−2 898 10293
9 89 103
6 m K K
m infrared
(e) E hfhc= = =
× ××
= ×−
−−
λ6 63 10 3 10
9 89 102 01 10
34 8
620.
..
Js m sm
J
(f) The energy output each second is carried by photons according to
P
P
= FHGIKJ
= =×
= ×−
Nt
E
Nt E
∆
∆1 81
2 01 108 98 1020
19..
. J s J photon
photon s
Matter is coupled to radiation, quite strongly, in terms of photon numbers.
Section 28.2 The Photoelectric Effect
P28.9 (a) λφchc= =
× ⋅ ×
×=
−
−
6 626 10 3 00 10
4 20 1 60 10296 nm
34 8
19
. .
. .
J s m s
eV J eV
e je ja fe j
fc
cc
= =××
= ×−λ3 00 10296 10
1 01 108
915.
. m s
m Hz
(b) hc
e VSλφ= + ∆ :
6 626 10 3 00 10
180 104 20 1 60 10 1 60 10
34 8
919 19
. .. . .
× ×
×= × + ×
−
−− −e je j a fe j e j eV J eV ∆VS
Therefore, ∆VS = 2 71. V
776 Quantum Physics
P28.10 K mvmax max . . . .= = 1 × × = × =− −12 2
9 11 10 4 60 10 9 64 10 0 6022 31 5 2 20e je j J eV
(a) φ = − =⋅
− =E Kmax . .1 240
0 602 1 38 eV nm
625 nm nm eV
(b) fhc = =
× ⋅×F
HGIKJ = ×−
−φ 1 38 1 60 103 34 1034
1914. .
. eV
6.626 10 J s J
1 eV Hz
P28.11 (a) e Vhc
S∆ = − → =⋅
− =λ
φ φ 1 2400 376 1 90
nm eV546.1 nm
eV eV. .
(b) e Vhc
VS S∆ ∆= − =⋅
− → =λ
φ 1 2401 90 0 216
nm eV587.5 nm
eV V. .
P28.12 The energy needed is E = = × −1 00 1 60 10 19. . eV J.
The energy absorbed in time interval ∆t is E t IA t= =P ∆ ∆
so ∆tEIA
= = ×
⋅ ×LNM
OQP= × =
−
−
1 60 10
2 82 101 28 10 148
19
15 27.
..
J
500 J s m m s days
2e j e jπ.
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 hfmax = −φ ,
or Kmax
. . .. .=
× ⋅ ×
× ×FHG
IKJ − =
−
− −
6 626 10 3 00 10
200 101 00
104 70 1 51
34 8
9 19
J s m s
m eV
1.60 J eV eV
e je j.
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
Vk Q
re= or Q
rVke
= =× ⋅
× ⋅= ×
−−
5 00 10 1 51
8 99 108 41 10
2
912
. .
..
m N m C
N m C C2 2
e jb g.
Section 28.3 The Compton Effect
P28.14 Ehc= =
× ⋅ ×
×= × =
−
−−
λ
6 626 10 3 00 10
700 102 84 10 1 78
34 8
919
. .. .
J s m s
m J eV
e je j
ph= = × ⋅
×= × ⋅
−
−−
λ6 626 10
109 47 10
34
928.
. J s
700 m kg m s
Chapter 28 777
P28.15 (a) ∆λ θ= −hm ce
1 cosa f : ∆λ = ×× ×
− ° = ×−
−−6 626 10
9 11 10 3 00 101 37 0 4 88 10
34
31 813.
. .cos . .
e je ja f m
(b) Ehc
00
=λ
: 300 10 1 60 106 626 10 3 00 10
3 1934 8
0× × =
× ×−
−
eV J eV m s
e je j e je j.
. .
λ
λ 0124 14 10= × −. m
and ′ = + = × −λ λ λ0124 63 10∆ . m
′ =′=
× ⋅ ×
×= × =
−
−−E
hcλ
6 626 10 3 00 10
4 63 104 30 10 268
34 8
1214
. .
..
J s m s
m J keV
e je j
(c) K E Ee = − ′ = − =0 300 268 5 31 5 keV keV keV. .
P28.16 This is Compton scattering through 180°:
E
hc
hm ce
00
34 8
9 19
12 12
6 626 10 3 00 10
0 110 10 1 60 1011 3
1 2 43 10 1 180 4 86 10
= =× ⋅ ×
× ×=
= − = × − ° = ×
−
− −
− −
λ
λ θ
. .
. ..
cos . cos .
J s m s
m J eV keV
m m
e je je je ja f e ja f∆
FIG. P28.16
′ = + =λ λ λ0 0 115∆ . nm so ′ =′=E
hcλ
10 8. keV.
By conservation of momentum for the photon-electron system, h h
peλ λ0
$ $ $i i i=′− +e j
and p he = +′
FHG
IKJ
1 1
0λ λ
pc
ce = × ⋅×
×
FHGG
IKJJ ×
+×
FHG
IKJ =
−− − −6 626 10
3 00 10
1 60 101
0 110 101
0 115 1022 134
8
19 9 9..
. . ..
J s m s
J eV m m keVe j e j
.
By conservation of system energy, 11 3 10 8. .keV keV= +Ke
so that Ke = 478 eV .
Check: E p c m ce2 2 2 2 4= + or m c K pc m ce e e
2 2 2 2 2+ = +e j b g e j
511 0 478 22 1 511
2 62 10 2 62 10
2 2 2
11 11
keV keV keV keV+ = +
× = ×
. .
. .
a f a f a f
778 Quantum Physics
P28.17 With K Ee = ′ , K E Ee = − ′0 gives ′ = − ′E E E0
′ =EE0
2 and ′ =
′λ hc
E ′ = = =λ λhc
EhcE0 0
022 2
′ = + −λ λ λ θ0 1C cosa f 2 10 0λ λ λ θ= + −C cosa f
10 001 600 002 43
0− = =cos..
θ λλC
θ = °70 0.
P28.18 (a) K m ve= 12
2 : K = × × = × =− −12
9 11 10 1 40 10 8 93 10 5 5831 6 2 19. . . . kg m s J eVe je j
Ehc
00
1 2401 550= = ⋅ =
λeV nm
0.800 nm eV
E E K′ = −0 and ′ =′= ⋅
−=λ h c
E1 240
1 550 5 580 803
eV nmeV eV
nm.
.
∆λ λ λ= ′− = =0 nm pm0 002 88 2 88. .
(b) ∆λ λ θ= C 1− cosb g: cos..
.θ λλ
= − = − = −1 10 002 880 002 43
0 189∆
C
nm nm
,
so θ = °101
Section 28.4 Photons and Electromagnetic Waves
*P28.19 With photon energy 10 0. eV = hf
f =×
× ⋅= ×
−
−
10 0 1 6 10
6 63 102 41 10
19
3415
. .
..
J
J s Hz
e j.
Any electromagnetic wave with frequency higher than 2 41 1015. × 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
ph
mv6 626 10
1 67 10 1 00 103 97 10
34
27 613.
. ..
J s
kg m s m
e je j
Chapter 28 779
P28.21 (a) Electron: λ = hp
and K m vm v
mpme
e
e e= = =1
2 2 22
2 2 2
so p m Ke= 2
and λ = = × ⋅
× ×
−
− −
hm Ke2
6 626 10
2 9 11 10 3 00 1 60 10
34
31 19
.
. . .
J s
kg Je ja fe j
λ = × =−7 09 10 0 70910. . m nm .
(b) Photon: λ = cf
and E hf= so fEh
=
and λ = =× ⋅ ×
×= × =
−
−−hc
E
6 626 10 3 00 10
3 1 60 104 14 10 414
34 8
197
. .
..
J s m s
J m nm
e je je j
.
P28.22 From the condition for Bragg reflection,
m d dλ θ φ= = FHGIKJ2 2
2sin cos .
But d a= FHGIKJsin
φ2
where a is the lattice spacing.
Thus, with m = 1, λ φ φ φ= FHGIKJFHGIKJ =2
2 2a asin cos sin
FIG. P28.22
λ = =hp
hm Ke2
λ = × ⋅
× × ×= ×
−
− −
−6 626 10
10 54 0 1 60 101 67 10
34
31 19
10.
. ..
J s
2 9.11 kg J m
e je j.
Therefore, the lattice spacing is a = = ×°
= × =−
−λφsin
.sin .
. .1 67 10
50 02 18 10 0 218
1010 m
nm .
P28.23 (a) λ ~10 14− m or less. ph= × ⋅ = ⋅
−
−−
λ~
.6 6 1010
1034
1419 J s
m kg m s or more.
The energy of the electron is E p c m ce= + × + × ×− −2 2 2 4 19 2 8 2 31 2 8 410 3 10 9 10 3 10~ e j e j e j e j
or E ~ ~10 1011 8− J eV or more,
so that K E m ce= − − ×2 8 6 810 0 5 10 10~ . ~ eV eV eVe j or more.
(b) The electric potential energy of the electron-nucleus system would be
U
k q qr
ee
e=× ⋅ −
−−
−1 2
9 19
145
9 10 10
1010~ ~
N m C C
m eV
2 2e je ja f.
With its K Ue+ >> 0 , the electron would immediately escape the nucleus .
780 Quantum Physics
P28.24 (a) The wavelength of the student is λ = =hp
hmv
. If w is the width of the diffracting aperture,
then we need wh
mv≤ = F
HGIKJ10 0 10 0. .λ
so that vh
mw≤ = × ⋅F
HGIKJ = ×
−−10 0 10 0
6 626 1080 0 0 750
1 10 1034
34. ... .
. J s
kg m m sb ga f .
(b) Using ∆tdv
= we get: ∆t ≥×
= ×−0 150
1 36 1033..
m1.10 10 m s
s34 .
(c) No . The minimum time to pass through the door is over 1015 times the age of the
Universe.
P28.25 λ = hp
ph= = × ⋅
×= × ⋅
−
−−
λ6 626 101 00 10
6 63 1034
1123.
..
J sm
kg m s
(a) electrons: Kpme
e= =
×
×=
−
−
2 23 2
312
6 63 10
2 9 11 1015 1
.
..
e je j
J keV
The relativistic answer is more precisely correct:
K p c m c m ce e e= + − =2 2 2 4 2 14 9. keV .
(b) photons: E pcγ = = × × =−6 63 10 3 00 10 12423 8. .e je j keV
Section 28.6 The Quantum Particle
*P28.26 E K mu hf= = =12
2 and λ = hmu
.
v fmu
hh
muu
vphase phase= = = =λ2
2 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 vkp = ω
is not the speed of the particle.
vk
p c m c
mvm v c m c
m vc
cv
ccv
vc
ccv
cvp = =
+=
+= + = + −
FHGIKJ = + − =ω
γγ
γ γ
2 2 2 4 2 2 2 2 2 4
2 2 2
2
2 2
2
2
2
2
2
2
2
1 1 1 1 1h
h
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 28 781
vddk
dd k
dEdp
ddp
m c p c
v m c p c pcp c
p c m c
v cm v
m v m cc
v v c
v v c cc
v v c
v c v v cv
g
g
g
= = = = +
= + + =+
=+
=−
− +=
−
+ − −=
−
ω ω
γγ
h
h2 4 2 2
2 4 2 2 1 2 22 4
2 2 2 4
2 2 2
2 2 2 2 2
2 2 2
2 2 2 2
2 2 2
2 2 2 2 2
12
0 2
1
1
1
1
e j e j
e je j
e je j e j
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:
d msinθ λ= 6 00 100 400200
1 1 20 108 1 10. sin tan.
.× LNMOQP
FHG
IKJ = = ×− − − m me j a fλ
λ = hm ve
so m vh
e =λ
and K m vm v
mh
me Ve
e
e e
= = = =12 2 2
22 2 2
2λ∆
∆Vh
eme
=2
22 λ ∆V =
× ⋅
× × ×=
−
− − −
6 626 10
2 1 60 10 9 11 10 1 20 10105
34 2
19 31 10 2
.
. . .
J s
C kg m V
e je je je j
.
P28.29 (a) λ = = × ⋅×
= ×−
−−h
mv6 626 10
1 67 10 0 4009 92 10
34
277.
. ..
J s
kg m s m
e jb g
(b) For destructive interference in a multiple-slit experiment, d msinθ λ= +FHGIKJ
12
, with m = 0 for
the first minimum.
Then, θ λ= FHGIKJ = °−sin .1
20 028 4
d
so yL= tanθ y L= = ° =tan . tan . .θ 10 0 0 028 4 4 96 m mma fb g .
(c) We cannot say the neutron passed through one slit. We can only say it passed through the slits.
782 Quantum Physics
*P28.30 We find the speed of each electron from energy conservation in the firing process:
012
2 2 1 6 109 11 10
3 98 10
2
19
316
= + = −
= =× ×
×= ×
−
−
K U mv eV
veVm
f f
..
. C 45 V
kg m s
a f
The time of flight is ∆ ∆t
xv
= =×
= × −0 287 04 10 8..
m3.98 10 m s
s6 . The current when electrons are 28 cm
apart is Iqt
et
= = = ××
= ×−
−−
∆1 6 10
2 27 1019
812.
. C
7.04 10 s A .
Section 28.8 The Uncertainty Principle P28.31 For the electron, ∆ ∆p m ve= = × × = × ⋅− − −9 11 10 500 1 00 10 4 56 1031 4 32. . . kg m s kg m se jb ge j
∆∆
xh
p= = × ⋅
× ⋅=
−
−46 626 10
1 1634
32π π.
. J s
4 4.56 10 kg m s mm
e j.
For the bullet, ∆ ∆p m v= = × = × ⋅− −0 020 0 500 1 00 10 1 00 104 3. . . kg m s kg m sb gb ge j
∆∆
xh
p= = × −
45 28 10 32
π. m .
P28.32 (a) ∆ ∆ ∆ ∆p x m v x= ≥ h
2 so ∆
∆v
hm x
≥ = ⋅ =4
24 2 00 1 00
0 250π
ππ
J s kg m
m s. .
.b ga f .
(b) The duck might move by 0 25 5 1 25. . m s s mb ga f = . With original position uncertainty of
1.00 m, we can think of ∆ x growing to 1 00 1 25 2 25. . . m m m+ = .
P28.33 ∆ ∆yx
p
py
x= and d p
hy∆ ≥
4π.
Eliminate ∆ py and solve for x.
x p ydhx= 4π ∆b g : x = × ×
×
× ⋅− −
−
−4 1 00 10 100 1 00 10
2 00 10
6 626 103 2
3
34π . .
.
. kg m s m
m
J se jb ge j e j
e j
The answer, x = ×3 79 1028. m , is 190 times greater than the diameter of the observable Universe!
Chapter 28 783
P28.34 From the uncertainty principle ∆ ∆E t ≥ h
2
or ∆ ∆mc t2
2e j = h.
Therefore, ∆
∆ ∆m
mh
c t mht ER
= =4 42π πa f a f
∆mm
= × ⋅× ×
FHG
IKJ = ×
−
− −−6 626 10
135
12 81 10
34
17 138.
. J s
4 8.70 10 s MeV
MeV1.60 10 Jπe ja f .
P28.35 (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
∆ ∆∆
p m vxx x
i= = h
2. It falls to the floor in a travel time given by H gt= +0
12
2 as tHg
= 2, so
the total width of the impact points is
∆ ∆ ∆ ∆∆
∆∆
x x v t xm x
Hg
xAxf i x i
ii
i= + = +
FHG
IKJ = +b g h
22
where Am
Hg
= h
22
.
To minimize ∆x f , we require d x
d xf
i
∆
∆d ib g = 0 or 1 02− =A
xi∆
so ∆x Ai = .
The minimum width of the impact points is
∆ ∆∆
∆
x xAx
Am
Hgf i
i x Ai
d imin
= +FHG
IKJ = =
FHGIKJ=
22 2
1 4h
.
(b) ∆x fd i e j a fmin
.
.
.
..=
× ⋅
×
L
NMM
O
QPPLNMM
OQPP
= ×−
−−
2 1 054 6 10
5 00 10
2 2 00
9 805 19 10
34
4
1 2 1 416
J s
kg
m
m s m2
Section 28.9 An Interpretation of Quantum Mechanics
P28.36 Probability P xa
x adx
aa
xaa
a
a
a
a
a
= =+
= FHGIKJFHGIKJFHGIKJ
− −
−
−z zψ
π πa f
e j2
2 211
tan
P = − − = − −FHGIKJ
LNM
OQP =
− −11 1
14 4
12
1 1
π ππ π
tan tan a f
784 Quantum Physics
P28.37 (a) ψ x Ae A x Ai x A kx Ai kxi xa f e j e j a f a fe j= = × + × = +×5 00 10 10 10
10
5 10 5 10.
cos sin cos sin goes through
a full cycle when x changes by λ and when kx changes by 2π . Then kλ π= 2 where
k = × =−5 00 10210 1. mπλ
. Then λ π=×
= × −2
5 00 101 26 10
1010m
m.
.e j
.
(b) ph= = × ⋅
×= × ⋅
−
−−
λ6 626 101 26 10
5 27 1034
1024.
..
J s m
kg m s
(c) me = × −9 11 10 31. kg
Km v
mpm
e
e= = =
× ⋅
× ×= × = ×
×=
−
−−
−
−
2 2 2 24 2
3117
17
192 2
5 27 10
2 9 11 101 52 10
1 52 101 60 10
95 5.
..
..
. kg m s
kg J
J J eV
eVe je j
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.
nλ2
1 00 10 9= × −. m so λ = × =−2 00 10 9.
nhp
(a) Since Kpm
h
mhm
nn
e e e= = =
×=
−
2 2 2 2 2
9 22
2 2 2 2 100 377
λe je j
e j. eV
For K ≈ 6 eV n = 4
(b) With n = 4, K = 6 03. eV
FIG. P28.38
Chapter 28 785
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 dN to N = λ2
and λ = hp
ph h
d= =λ 2
.
Next, Kpm
hm d de e
= = =× ⋅
×
L
NMMM
O
QPPP
−
−
2 2
2 2
34 2
312 81 6 626 10
8 9 11 10
.
.
J s
kg
e je j
.
Evaluating, Kd
= × ⋅−6 02 10 38
2. J m2
Kd
= × ⋅−3 77 10 19
2. eV m2
.
In state 1, d = × −1 00 10 10. m K1 37 7= . eV .
In state 2, d = × −5 00 10 11. m K2 151= eV.
In state 3, d = × −3 33 10 11. m K3 339= eV .
In state 4, d = × −2 50 10 11. m K4 603= eV .
FIG. P28.39
(b) When the electron falls from state 2 to state 1, it puts out energy
E hfhc= − = = =151 37 7 113 eV eV eV.λ
into emitting a photon of wavelength
λ = =× ⋅ ×
×=
−
−hcE
6 626 10 10
113 1 60 1011 0
34 8
19
.
..
J s 3.00 m s
eV J eV nm
e je ja fe j
.
The wavelengths of the other spectral lines we find similarly:
Transition 4 3→ 4 2→ 4 1→ 3 2→ 3 1→ 2 1→ E eVa f 264 452 565 188 302 113 λ nma f 4.71 2.75 2.20 6.60 4.12 11.0
786 Quantum Physics
P28.40 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
wavelength is twice this distance: hp= × −2 00 10 14. m.
The proton’s kinetic energy is
K mvpm
hm
= = = =× ⋅
× ×
= ××
=
−
− −
−
−
12 2 2
6 626 10
2 1 67 10 2 00 10
3 29 102 05
22 2
2
34 2
27 14 2
13
19
λ
.
. .
..
J s
kg m
J1.60 10 J eV
MeV
e je je j
FIG. P28.40
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 8 22 6 16. . . MeV MeV 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 .
Its frequency is fEh
= =× ×
× ⋅= ×
−
−
6 16 10 1 60 10
6 626 101 49 10
6 19
3421
. .
..
eV J eV
J s Hz
e je j.
And its wavelength is λ = =××
= ×−−c
f3 00 101 49 10
2 02 108
21 113.
..
m s s
m .
This is a gamma ray, according to the electromagnetic spectrum chart in Chapter 24.
*P28.41 (a) The energies of the confined electron are Eh
m Lnn
e
=2
22
8. Its energy gain in the quantum
jump from state 1 to state 4 is h
m Le
2
22 2
84 1−e j and this is the photon energy:
hm L
hfhc
e
2
215
8= =
λ. Then 8 152m cL he = λ and L
hm ce
=FHGIKJ
158
1 2λ
.
(b) Let ′λ represent the wavelength of the photon emitted: hc h
m Lh
m Lh
m Le e e′= − =
λ
2
22
2
22
2
284
82
128
.
Then hc
hc
h m L
m L he
eλλ′ = =
2 2
2 2
15 8
8 1254
e j and ′ =λ λ1 25. .
Chapter 28 787
Section 28.11 The Quantum Particle Under Boundary Conditions Section 28.12 The Schrödinger Equation
P28.42 ψ x A kx B kxa f = +cos sin ∂∂
= − +ψx
kA kx kB kxsin cos
∂∂
= − −2
22 2ψ
xk A kx k B kxcos sin − − = − +2 2
2m
E UmE
A kx B kxh ha f a fψ cos sin
Therefore the Schrödinger equation is satisfied if
∂∂
= −FHGIKJ −
2
2 22ψ ψ
xm
E Uha f or − + = −FHG
IKJ +k A kx B kx
mEA kx B kx2
22
cos sin cos sina f a fh
.
This is true as an identity (functional equality) for all x if Ekm
= h2 2
2.
P28.43 We have ψ ψ ψω= ∂∂
=−Aex
iki kx tb g and ∂∂
= −2
22ψ ψ
xk .
Schrödinger’s equation: ∂∂
= − = − −2
22
22ψ ψ ψ
xk
mE U
ha f .
Since kp
hp2
2
2
2
2
2
2
2 2= = =
πλ
πa f b gh
and E Upm
− =2
2.
Thus this equation balances.
P28.44 (a) With ψ x Aeikxa f =
ddx
Ae Aikxeikx ikx= and ddx
Ak eikx2
22ψ = − .
Then − = + = = = = =h h2 2
2
2 2 2
2
2
2
2 2 22
2 2 44
2 2 212m
ddx
km
Aeh
mpm
m vm
mv Kikxψπ
πλ
ψ ψ ψ ψ ψ .
(b) With ψ πx
Ln x
LA kxa f = F
HGIKJ =
2sin sin ,
ddx
Ak kxψ = cos and
ddx
Ak kx2
22ψ = − sin
Then − = + = = =h h2 2
2
22
2 2
2 2
2
2 24
4 2 2mddx m
Ak kxh
mpm
Kψ π
π λψ ψ ψsin .
P28.45 (a) x xL
xL
dxL
xx
Ldx
L L
= FHGIKJ = −FHG
IKJz z2 2 2 1
212
42
0 0
sin cosπ π
xL
xL
L xL
xL
xL
LL L
= − +LNM
OQP =1
21
164 4 4
2
2
0
2
20π
π π πsin cos
continued on next page
788 Quantum Physics
(b) Probability= FHGIKJ = −LNM
OQPz 2 2 1 1
442
0 490
0 510
0 490
0 510
Lx
Ldx
Lx
LL x
LL
L
L
L
sin sin.
.
.
.ππ
π
Probability= − − = × −0 0201
42 04 1 96 5 26 10 5. sin . sin . .
ππ πa f
(c) ProbabilityxL
xL L
L
−LNMOQP = × −1
44
3 99 100 240
0 2602
ππ
sin ..
.
(d) In the n = 2 graph in Figure 28.23 (b), it is more probable to find the particle either near
xL=4
or xL= 34
than at the center, where the probability density is zero.
Nevertheless, the symmetry of the distribution means that the average position is L2
.
P28.46 Normalization requires
ψ 2 1dxall spacez = or A
n xL
dxL
2 2
0
1sinπFHGIKJ =z
An x
Ldx A
LL2 2
0
2
21sin
πFHGIKJ = FHG
IKJ =z or A
L= 2
.
P28.47 The desired probability is P dxL
xL
dxL L
= = FHGIKJz zψ
π2
0
42
0
42 2sin
where sincos2 1 22
θ θ= −.
Thus, PxL
xL
L
= −FHGIKJ = − − +FHG
IKJ =
14
4 14
0 0 0 0 2500
4
ππ
sin . .
P28.48 (a) ψ x AxL
a f = −FHGIKJ1
2
2 ddx
AxL
ψ = − 22
ddx
AL
2
2 22ψ = −
Schrödinger’s equation ddx
mE U
2
2 22ψ ψ= − −ha f
becomes − = − −FHGIKJ +
− −
−2 2
12 1
2 2
2
2 2
2 2 2 2
2 2 2
AL
mEA
xL
m x A x L
mL L xh h
he j e je j
− = − + −12 2
2
2 2
2
4LmE mEx
LxLh h
.
This will be true for all x if both 12 2L
mE=h
and mE
L Lh2 2 41
0− =
both these conditions are satisfied for a particle of energy EL m
= h2
2 .
continued on next page
Chapter 28 789
(b) For normalization, 1 1 122
2
2
22
2
2
4
4= −FHGIKJ = − +
FHG
IKJ− −
z zAxL
dx Ax
LxL
dxL
L
L
L
123 5
23 5
23 5
1615
1516
23
2
5
42 2= − +
LNM
OQP
= − + + − +LNM
OQP =FHGIKJ =
−
A xxL
xL
A L LL
L LL
AL
AL
L
L
.
(c) P dxL
xL
xL
dxL
xxL
xL L
L L L
L
L
L
L
L
L
= = − +FHG
IKJ = − +
LNM
OQP
= − +LNM
OQP− − −
z zψ 2
3
3 2
2
4
43
3 3
2
5
53
315
161
2 1516
23 5
3016 3
281 1 215
P = =4781
0 580.
Section 28.13 Tunneling Through a Potential Energy Barrier
P28.49 T e CL= −2 where Cm U E
=−2 a f
h
22 2 9 11 10 8 00 10
1 055 102 00 10 4 58
31 19
3410CL =
× ×
×× =
− −
−−
. .
.. .
e je je j
(a) T e= =−4.58 0 010 3. , a 1% chance of transmission.
(b) R T= − =1 0 990. , a 99% chance of reflection.
FIG. P28.49
P28.50 C =× − × ⋅
× ⋅= ×
− −
−−
2 9 11 10 5 00 4 50 1 60 10
1 055 103 62 10
31 19
349 1
. . . .
..
e ja fe j kg m s
J s m
T e
T
CL= = − × × = −
= ×
− − −
−
2 9 1 12
3
2 3 62 10 950 10 6 88
1 03 10
exp . exp .
.
m me je j a f
FIG. P28.50
*P28.51 The original tunneling probability is T e CL= −2 where
Cm U E
=−
=× × − ×
× ⋅= ×
− −
−−2 2 2 9 11 10 12 1 6 10
6 626 101 448 1 10
1 2 31 19 1 2
3410 1a fc h a fe j
h
π . .
..
kg 20 J
J s m .
The photon energy is hfhc= =
⋅=
λ1 240
2 27eV nm
546 nm eV. , to make the electron’s new kinetic energy
12 2 27 14 27+ =. . eV and its decay coefficient inside the barrier
′ =× × − ×
× ⋅= ×
− −
−−C
2 2 9 11 10 20 14 27 1 6 10
6 626 101 225 5 10
31 19 1 2
3410 1
π . . .
..
kg J
J s m
a fe j.
Now the factor of increase in transmission probability is
ee
e e eC L
CLL C C
− ′
−− ′ × × ×= = = =
− −2
22 2 10 0 223 10 4.459 10 1
85 9a f m m. . .
790 Quantum Physics
Section 28.14 Context ConnectionThe 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 c v c
v c′=
−+λ λ
1
1b gb g : λ λ= ′
−+
=−+
=1
1500
1 0 2801 0 280
375v c
v cb gb g a f a f
a f nm nm..
.
The temperature of the star is given by λmax .T = × ⋅−2 898 10 3 m K :
T = × ⋅−2 898 10 3.
max
m Kλ
: T = × ⋅×
= ×−
−2 898 10
375 107 73 10
3
93.
. m K
K .
P28.53 (a) Wien’s law: λmax .T = × ⋅−2 898 10 3 m K .
Thus, λmax. .
. .= × ⋅ = × ⋅ = × =− −
−2 898 10 2 898 101 06 10 1 06
3 33 m K m K
2.73 K m mm
T .
(b) This is a microwave .
P28.54 We suppose that the fireball of the Big Bang is a black body.
I e T= = × ⋅ = ×− −σ 4 8 4 61 5 67 10 2 73 3 15 10( ) . . . W m K K W m2 4 2e ja f
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:
P = = = × ××FHG
IKJ ×−IA I r( ) . .4 3 15 10 4 15 10
3 103 156 102 6 9 2 8 2
7 2π π W m ly
m s1 ly yr
s yr2e ja fe j e j
P = ×7 98 1047. W. Additional Problems
*P28.55 The condition on electric power delivered to the filament is P = = = =I VVR
V A V r∆
∆ ∆ ∆a f a f a f2 2 2 2
ρπ
ρl l
so rV
=FHG
IKJ
P ρπ
l
∆a f21 2
. 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 σ πeAT e r T4 42 l . We combine the conditions by substitution:
continued on next page
Chapter 28 791
PP
P
P
=FHG
IKJ
=
=FHG
IKJ =
× × ⋅
F
HGG
I
KJJ
= = =
− −
σ π ρπ
σ π ρ
σ π ρ π
eV
T
V e T
Ve T
2
2
2120 75
5 67 10 0 450 2 7 13 10 2 900
0 192 0 333
2
1 2
4
1 2 1 2 1 2 3 2 4
1 2
1 2 1 2 4
2 3 1 2
8 1 2 7 1 2 4
2 3
3 2 2 3
ll
l
l
l
∆
∆
∆
Ω
a fa f
a f a fa f e j a f
e j
V W m K
W m K
m m
2 4
. . .
. .
and rV
r=FHG
IKJ
= × ⋅FHG
IKJ
= = ×−
−P ρπ π
l
∆Ω
a f a f2
1 27
2
1 2
575 7 13 10 0 333
120 V1 98 10
W m m m
. .. .
P28.56 ∆Vhe
feS = FHG
IKJ − φ
From two points on the graph 0 4 1 1014= FHGIKJ × −h
e e. Hze j φ
and 3 3 12 1014. V Hz= FHGIKJ × −h
e ee j φ.
Combining these two expressions we find: (a) φ = 1 7. eV
(b) he= × ⋅−4 2 10 15. V s
FIG. P28.56
(c) At the cutoff wavelength hc h
eec
c cλφ
λ= = FHG
IKJ
λ c = × ⋅ ××
×=− −
−4 2 10 1 6 10
3 10
1 7 1 6 1073015 19
8
19. .
. . V s C
m s
eV J eV nme je j e j
a fe j
792 Quantum Physics
P28.57 We want an Einstein plot of Kmax versus fc=λ
λ , nm , Hz , eV588505445
5.105.946.74
0.670.981.35
f K1014max
399 7.52 1.63
(a) slope = ±0 4028%
. eV10 Hz14
(b) e V hfS∆ = −φ
h = × ⋅FHG
IKJ = × ⋅ ±
−−0 402
1 60 106 4 10 8%
1934.
..a f J s
10 J s14
(c) Kmax = 0
at f ≈ ×344 1012 Hz
φ = = × =−hf 2 32 10 1 419. . J eV
f (THz)
FIG. P28.57
P28.58 From the path the electrons follow in the magnetic field, the maximum kinetic energy is seen to be:
Ke B R
memax =
2 2 2
2.
From the photoelectric equation, K hfhc
max = − = −φλ
φ .
Thus, the work function is φλ λ
= − = −hcK
hc e B Rme
max
2 2 2
2.
P28.59 (a) mgy mvi f= 12
2
v gy
hmv
f i= = =
= = × ⋅ = ×−
−
2 2 9 80 50 0 31 3
6 626 1031 3
2 82 1034
37
. . .
..
.
m s m m s
J s75.0 kg m s
m not observable
2e ja f
b gb g a fλ
(b) ∆ ∆E t ≥ h
2
so ∆E ≥ × ⋅×
= ×−
−−6 626 10
1 06 1034
332.
. J s
4 5.00 10 s J
π e j
(c) ∆EE
= × = ×−
−1 06 10
9 80 50 02 87 10
3235.
. .. %
J
75.0 kg m s m2b ge ja f
Chapter 28 793
P28.60 (a) λ = = × −2 2 00 10 10L . m
(b) ph= = × ⋅
×= × ⋅
−
−−
λ6 626 10
3 31 1034
1024.
. J s
2.00 10 m kg m s
(c) Epm
= =2
20 172. eV
P28.61 (a) See the figure.
FIG. P28.61(a)
(b) See the figure.
FIG. P28.61(b)
(c) 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,
ψ ψ2 2
0
2 1dx dx−∞
∞ ∞
z z= =
or 22
212 2
0
20A e dx
Ae ex−
∞−∞z =
−FHGIKJ − =α
α e j .
This gives A = α .
(e) P a e dx e ex
x− →
−
=
− −= =−FHGIKJ − = − =z1 2
2 2
0
1 22 2 12
22
1 1 0 632α αα
αα αα
αb g b g e j e j e j1 2 .
P28.62 (a) Use Schrödinger’s equation
∂∂
= − −2
2 22ψ ψ
xm
E Uha f
with solutions
ψ 11 1= + −Ae Beik x ik x [region I]
ψ 22= Ceik x [region II].
FIG. P28.62(a)
continued on next page
794 Quantum Physics
Where kmE
12=h
and km E U
22
=−a f
h.
Then, matching functions and derivatives at x = 0
ψ ψ1 0 2 0b g b g= gives A B C+ =
and ddx
ddx
ψ ψ1
0
2
0
FHGIKJ = FHG
IKJ gives k A B k C1 2− =a f .
Then Bk kk k
A=−+
11
2 1
2 1
and Ck k
A=+
21 2 1
.
Incident wave Aeikx reflects Be ikx− , with probability RBA
k k
k k
k k
k k= =
−
+=
−
+
2
22 1
2
2 12
1 22
1 22
1
1
b gb g
b gb g
.
(b) With E = 7 00. eV
and U = 5 00. eV
kk
E UE
2
1
2 007 00
0 535= − = =..
. .
The reflection probability is R =−
+=
1 0 535
1 0 5350 092 0
2
2
.
..
a fa f .
The probability of transmission is T R= − =1 0 908. .
P28.63 x x dx2 2 2=−∞
∞
z ψ
For a one-dimensional box of width L, ψ πn L
n xL
= FHGIKJ
2sin .
Thus, xL
xn x
Ldx
L Ln
L2 2 2
0
2 2
2 22
3 2= F
HGIKJ = −z sin
ππ
(from integral tables).
Chapter 28 795
*P28.64 (a) The requirement that n
Lλ2
= so ph nh
L= =λ 2
is still valid.
E pc mc Enhc
Lmc
K E mcnhc
Lmc mc
n
n n
= + ⇒ = FHGIKJ +
= − = FHGIKJ + −
b g e j e j
e j
2 2 2 22 2
22
2 2 2
2
2
(b) Taking L = × −1 00 10 12. m, m = × −9 11 10 31. kg , and n = 1, we find K1144 69 10= × −. J .
Nonrelativistic, EhmL1
2
2
34 2
31 12 214
8
6 626 10
8 9 11 10 1 00 106 02 10= =
× ⋅
× ×= ×
−
− −
−.
. ..
J s
kg m J
e je je j
.
Comparing this to K1, we see that this value is too large by 28 6%. .
P28.65 For a particle with wave function
ψ xa
e x aa f = −2 for x > 0
and 0 for x < 0 .
(a) ψ xa f 2 0= , x < 0 and ψ 2 22x
ae x aa f = − , x > 0
(b) Prob x x dx dx< = = =−∞ −∞z z0 0 0
20 0
a f a f a fψ
FIG. P28.65
(c) Normalization ψ ψ ψx dx dx dxa f 2 20
2
0
1−∞
∞
−∞
∞
z z z= + =
02
0 1 1
02
1 0 865
02
0
20
2
0
2
0
20
2
dxa
e dx e e
x a dxa
e dx e e
x a x a
ax a
ax a a
−∞
−∞
− ∞ −∞
− − −
z zz z
+ FHGIKJ = − = − − =
< < = = FHGIKJ = − = − =
e j
a fProb ψ .
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 78 103. × K ; (b) 501 nm P28.4 (a) 2.57 eV; (b) 12 8. eVµ ; (c) 191 neV; (d) 484 nm visible, 9.68 cm and 6.52 m
radio waves P28.6 5 71 103. × photons P28.8 (a) 0.263 kg; (b) 1.81 W; (c) − ° = − °0 015 3 0 919. . C s C min ; (d) 9.89 mµ ; (e) 2 01 10 20. × − J; (f) 8 98 1019. × photon s P28.10 (a) 1.38 eV; (b) 334 THz P28.12 148 days, absurdly large P28.14 1.78 eV, 9 47 10 28. × ⋅− kg m s
P28.16 22 1. keV
c, 478 eV
P28.18 (a) 2.88 pm; (b) 101° P28.20 397 fm P28.22 0.218 nm P28.24 (a) 1 10 10 34. × − m s ; (b) 1 36 1033. × s ; (c) no, the time is over 1015 times the age
of the universe
P28.26 vu
phase =2
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 12
P28.38 (a) n = 4; (b) 6.03 eV P28.40 6.16 MeV, 202 fm, a gamma ray
P28.42 see the solution, Ekm
= h2 2
2
P28.44 see the solution P28.46 see the solution
P28.48 (a) EmL
= h2
2 ;
(b) requiring AxL
dxL
L2
2
2
2
1 1−FHGIKJ =
−z gives
AL
= FHGIKJ
1516
1 2
;
(c) 4781
0 580= .
P28.50 1 03 10 3. × − P28.52 7 73 103. × K P28.54 3 15. W m2µ P28.56 (a) 1.7 eV; (b) 4.2 fV⋅s; (c) 730 nm
P28.58 hc e B R
meλ−
2 2 2
2
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) nhc
Lm c mc
2
22 4 2F
HGIKJ + − ;
(b) 46.9 fJ, 28.6%