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%