introduction to quantum physics - knowledge...

40

CHAPTER OUTLINE

40.1 Blackbody Radiation and Planck’s Hypothesis40.2 The Photoelectric Effect40.3 The Compton Effect40.4 Photons and Electromagnetic Waves40.5 The Wave Properties of Particles40.6 The Quantum Particle40.7 The Double-Slit Experiment Revisited40.8 The Uncertainty Principle

Introduction to Quantum Physics

ANSWERS TO QUESTIONS

Q40.1 Planck made two new assumptions: (1) molecular energy isquantized and (2) molecules emit or absorb energy in discreteirreducible packets. These assumptions contradict the classicalidea of energy as continuously divisible. They also imply thatan atom must have a definite structure—it cannot just be asoup of electrons orbiting the nucleus.

Q40.2 The first flaw is that the Rayleigh–Jeans law predicts that theintensity of short wavelength radiation emitted by a blackbodyapproaches infinity as the wavelength decreases. This is knownas the ultraviolet catastrophe. The second flaw is the predictionmuch more power output from a black-body than is shownexperimentally. The intensity of radiation from the blackbodyis given by the area under the red I Tλ ,b g vs. λ curve inFigure 40.5 in the text, not by the area under the blue curve.

Planck’s Law dealt with both of these issues and broughtthe theory into agreement with the experimental data byadding an exponential term to the denominator that depends

on 1λ

. This both keeps the predicted intensity from

approaching infinity as the wavelength decreases and keepsthe area under the curve finite.

Q40.3 Our eyes are not able to detect all frequencies of energy. For example, all objects that are above 0 Kin temperature emit electromagnetic radiation in the infrared region. This describes everything in adark room. We are only able to see objects that emit or reflect electromagnetic radiation in the visibleportion of the spectrum.

Q40.4 Most stars radiate nearly as blackbodies. Vega has a higher surface temperature than Arcturus. Vegaradiates most intensely at shorter wavelengths.

Q40.5 No. The second metal may have a larger work function than the first, in which case the incidentphotons may not have enough energy to eject photoelectrons.

Q40.6 Comparing Equation 40.9 with the slope-intercept form of the equation for a straight line, y mx b= + ,we see that the slope in Figure 40.11 in the text is Planck’s constant h and that the y intercept is −φ ,the negative of the work function. If a different metal were used, the slope would remain the samebut the work function would be different, Thus, data for different metals appear as parallel lines onthe graph.

461

462 Introduction to Quantum Physics

Q40.7 Wave theory predicts that the photoelectric effect should occur at any frequency, provided the lightintensity is high enough. However, as seen in the photoelectric experiments, the light must have asufficiently high frequency for the effect to occur.

Q40.8 The stopping voltage measures the kinetic energy of the most energetic photoelectrons. Each ofthem has gotten its energy from a single photon. According to Planck’s E hf= , the photon energydepends on the frequency of the light. The intensity controls only the number of photons reaching aunit area in a unit time.

Q40.9 Let’s do some quick calculations and see: 1.62 MHz is the highest frequency in the commercial AMband. From the relationship between the energy and the frequency, E hf= , the energy availablefrom such a wave would be 1 07 10 27. × − J , or 6.68 neV. That is 9 orders of magnitude too small toeject electrons from the metal. The only thing this student could gain from this experiment is a heftyfine and a long jail term from the FCC. To get on the order of a few eV from this experiment, shewould have to broadcast at a minimum frequency of 250 Thz, which is in the infrared region.

Q40.10 No. If an electron breaks free from an atom absorbing a photon, we say the atom is ionized.Ionization typically requires energy of several eV. As with the photoelectric effect in a solid metal,the light must have a sufficiently high frequency for a photon energy that is large enough. The gascan absorb energy from longer-wavelength light as it gains more internal energy of random motionof whole molecules.

Q40.11 Ultraviolet light has shorter wavelength and higher photon energy than visible light.

Q40.12 (c) UV light has the highest frequency of the three, and hence each photon delivers more energy to askin cell. This explains why you can become sunburned on a cloudy day: clouds block visible lightand infrared, but not much ultraviolet. You usually do not become sunburned through windowglass, even though you can see the visible light from the Sun coming through the window, becausethe glass absorbs much of the ultraviolet and reemits it as infrared.

Q40.13 The Compton effect describes the scattering of photons from electrons, while the photoelectric effectpredicts the ejection of electrons due to the absorption of photons by a material.

Q40.14 In developing a theory in accord with experimental evidence, Compton assumed that photonsexhibited clear particle-like behavior, and that both energy and momentum are conserved inelectron-photon interactions. Photons had previously been thought of as bits of waves.

Q40.15 The x-ray photon transfers some of its energy to the electron. Thus, its frequency must decrease.

Q40.16 A few photons would only give a few dots of exposure, apparently randomly scattered.

Q40.17 Light has both classical-wave and classical-particle characteristics. In single- and double-slitexperiments light behaves like a wave. In the photoelectric effect light behaves like a particle. Lightmay be characterized as an electromagnetic wave with a particular wavelength or frequency, yet atthe same time light may be characterized as a stream of photons, each carrying a discrete energy, hf.Since light displays both wave and particle characteristics, perhaps it would be fair to call light a“wavicle”. It is customary to call a photon a quantum particle, different from a classical particle.

Chapter 40 463

Q40.18 An electron has both classical-wave and classical-particle characteristics. In single- and double-slitdiffraction and interference experiments, electrons behave like classical waves. An electron has massand charge. It carries kinetic energy and momentum in parcels of definite size, as classical particlesdo. At the same time it has a particular wavelength and frequency. Since an electron displayscharacteristics of both classical waves and classical particles, it is neither a classical wave nor aclassical particle. It is customary to call it a quantum particle, but another invented term, such as“wavicle”, could serve equally well.

Q40.19 The discovery of electron diffraction by Davisson and Germer was a fundamental advance in ourunderstanding of the motion of material particles. Newton’s laws fail to properly describe themotion of an object with small mass. It moves as a wave, not as a classical particle. Proceeding fromthis recognition, the development of quantum mechanics made possible describing the motion ofelectrons in atoms; understanding molecular structure and the behavior of matter at the atomicscale, including electronics, photonics, and engineered materials; accounting for the motion ofnucleons in nuclei; and studying elementary particles.

Q40.20 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

.

Q40.21 Any object of macroscopic size—including a grain of dust—has an undetectably small wavelengthand does not exhibit quantum behavior.

Q40.22 A particle is represented by a wave packet of nonzero width. The width necessarily introducesuncertainty in the position of the particle. The width of the wave packet can be reduced toward zeroonly by adding waves of all possible wavelengths together. Doing this, however, results in loss of allinformation about the momentum and, therefore, the speed of the particle.

Q40.23 The intensity of electron waves in some small region of space determines the probability that anelectron will be found in that region.

Q40.24 The wavelength of violet light is on the order of 12

mµ , while the de Broglie wavelength of an

electron can be 4 orders of magnitude smaller. Would your height be measured more precisely with

an unruled meter stick or with one engraved with divisions down to 1

10 mm?

Q40.25 The spacing between repeating structures on the surface of the feathers or scales is on the order of1/2 the wavelength of light. An optical microscope would not have the resolution to see such finedetail, while an electron microscope can. The electrons can have much shorter wavelength.

Q40.26 (a) The slot is blacker than any black material or pigment. Any radiation going in through thehole will be absorbed by the walls or the contents of the box, perhaps after severalreflections. Essentially none of that energy will come out through the hole again. Figure 40.1in the text shows this effect if you imagine the beam getting weaker at each reflection.

continued on next page


(b) The open slots between the glowing tubes are brightest. When you look into a slot, youreceive 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 hasbounced around a few or many times before escaping through the slot. In Figure 40.1 in thetext, 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 ofKirchhoff’s thermodynamic argument: ... energy radiated. A poor reflector—a goodabsorber—avoids rising in temperature by being an efficient emitter. Its emissivity is equalto its absorptivity: e a= . The slot in the box in part (a) of the question is a black body withreflectivity zero and absorptivity 1, so it must also be the most efficient possible radiator, toavoid rising in temperature above its surroundings in thermal equilibrium. Its emissivity inStefan’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 differentsurfaces maintain equal temperatures after they come to thermal equilibrium and continueto exchange energy by electromagnetic radiation. By considering one blackbody facinganother, Kirchhoff proved logically that the material forming the walls of the cavity madeno difference to the radiation. By thinking about inserting color filters between two cavityradiators, he proved that the spectral distribution of blackbody radiation must be a universalfunction 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 thatphysicists had previously studied separately.

SOLUTIONS TO PROBLEMS

Section 40.1 Blackbody Radiation and Planck’s Hypothesis

P40.1 T =× ⋅×

= ×−

−2 898 10

560 105 18 10

3

93.

. m K m

K

P40.2 (a) λmax.

~.

~=× ⋅ × ⋅− −

−2 898 10 2 898 1010

3 37 m K m K

10 K m4T

ultraviolet

(b) λmax ~.

~2 898 10

103

10× ⋅−− m K

10 K m7 γ − ray

P40.3 Planck’s radiation law gives intensity-per-wavelength. Taking E to be the photon energy and n to bethe number of photons emitted each second, we multiply by area and wavelength range to haveenergy-per-time leaving the hole:

P =−

+ −FH IK= =

+

2 2

2 1

22 1

2

1 25 2 1 2

π λ λ π

λ λ λ λ

hc d

eEn nhf

hc k T

b g b gb g b g B

where E hfhc

= =+

2

1 2λ λ

nE

cd

e

e

ne

hc k T= =

−

+ −FH IK

=× × ×

× −FHG

IKJ

=×

−= ×

+

− −

− × ⋅ × × × ×− − −

P 8

1

8 3 00 10 5 00 10 1 00 10

1 001 10 1

5 90 10

11 30 10

2 22 1

1 24 2

2 8 5 2 9

9 4 2 6 626 10 3 00 10 1 001 10 1.38 10 7 50 10

16

3 8415

1 2

34 8 9 23 3

π λ λ

λ λ

π

λ λ

b gb g

e je j e je j

e j

b g

e je j e je je j

B

m s m m

m

ss

J s m s m J K K

. . .

..

. . .

.

Chapter 40 465

P40.4 (a) P = = × × ⋅ = ×− −eA Tσ 4 4 8 4 41 20 0 10 5 67 10 5 000 7 09 10. . . m W m K K W2 2 4e je jb g

(b) λ λ λmax max max.T = = × ⋅ ⇒ =−5 000 2 898 10 5803 K m K nmb g

(c) We compute: hc

k TB

J s m s

J K K m=

× ⋅ ×

×= ×

−

−−

6 626 10 3 00 10

1 38 10 5 0002 88 10

34 8

236

. .

..

e je je jb g

The power per wavelength interval is P λ λπ

λ λb g b g b g= =

−AI

hc A

hc k T

2

1

2

5 exp B

, and

2 2 6 626 10 3 00 10 20 0 10 7 50 102 34 8 2 4 19π πhc A = × × × = × ⋅− − −. . . .e je j e j J m s4

P 5807 50 10

580 10 2 88 1

1 15 101

7 99 10

19

9 5

13

4.973

10

nm J m s

m m 0.580 m

J m s

W m

4

a fe j b g

=× ⋅

× −=

× ⋅

−

= ×

−

−

.

exp .

.

.

µ µ e

(d)–(i) The other values are computed similarly:

(d)(e)(f)(c)(g)(h)(i)

λλ

πλ

λλhck T

ehc A

B

hc k TB/ ( ), W m

1.00 nm 2882.6 . . .5.00 nm 576.5 . . .400 nm 7.21 1347 . .580 nm 4.97 143.5 . .700 nm 4.12 60.4 . .1.00 mm 0.00288 0.00289 . .10.0 cm . . . .

−

× × ×× × ×

× ×× ×× ××

× × × ×

−

−

−

− − − −

12

7 96 10 7 50 10 9 42 102 40 10 2 40 10 1 00 10

7 32 10 5 44 101 15 10 7 99 104 46 10 7 38 107 50 10 0 260

2 88 10 2 88 10 7 50 10 2 60 10

2

5

1251 26 1226

250 23 227

13 10

13 10

12 10

4

5 5 14 9

P

(j) We approximate the area under the P λb g versus λ curve, between 400 nm and 700 nm, astwo trapezoids:

P =+ × − ×

++ × − ×

−

−

5 44 7 99 10 580 400 10

2

7 99 7 38 10 700 580 10

2

10 9

10 9

. .

. .

a f a f

a f a f

W m m

W m m

P = ×2 13 104. W so the power radiated as visible light is approximately 20 kW .

P40.5 (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 77 10

5 67 105 75 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 04 10 504

3 37 m K m K

5.75 10 K m nm3T


P40.6 E hfhc

= = =× ⋅ ×

×= ×

−

−−

λ

6 626 10 3 00 10

589 3 103 37 10

34 8

919

. .

..

J s m s

m J photon

e je j

nE

= =×

= ×−P 10 0

3 37 102 96 1019

19..

. J s J photon

photons s

P40.7 (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

(d) λ = =×

×= × =−c

f3 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

P40.8 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 .

P40.9 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 are150 10

6 61 102 27 10

3

2630×

×= ×−

J s J photon

photons s.

. .

Chapter 40 467

P40.10 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 jFIG. P40.10

P40.11 The radiation wavelength of ′ =λ 500 nm that is observed by observers on Earth is not the truewavelength, λ, emitted by the star because of the Doppler effect. The true wavelength is related tothe observed wavelength using:

c c v c

v c′=

−

+λ λ1

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 .

P40.12 Planck’s radiation law is I Thc

ehc k Tλ

π

λ λ,b g

e j=

−

2

1

2

5 B.

Using the series expansion e xx xx = + + + +12 3

2 3

! !….

Planck’s law reduces to I Thc

hc k T

hchc k T

ck Tλ

πλ λ

πλ λ

πλ

,b g b g b g=+ + −

≈ =2

1 1

2 22

5

2

5 4B B

B

…

which is the Rayleigh-Jeans law, for very long wavelengths.

Section 40.2 The Photoelectric Effect

P40.13 (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

= =×

×= ×−λ

3 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


P40.14 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

P40.15 (a) λφchc

= Li: λ c =× ⋅ ×

×=

−

−

6 626 10 3 00 10

2 30 1 60 10540

34 8

19

. .

. .

J s m s

eV J eV nm

e je ja fe j

Be: λ c =× ⋅ ×

×=

−

−

6 626 10 3 00 10

3 90 1 60 10318

34 8

19

. .

. .

J s m s

eV J eV nm

e je ja fe j

Hg: λ c =× ⋅ ×

×=

−

−

6 626 10 3 00 10

4 50 1 60 10276

34 8

19

. .

. .

J s m s

eV J eV nm

e je ja fe j

λ λ< c for photo current. Thus, only lithium will exhibit the photoelectric effect.

(b) For lithium,hc

Kλ

φ= + max

6 626 10 3 00 10

400 102 30 1 60 10

1 29 10 0 808

34 8

919

19

. .. .

. .

max

max

× ⋅ ×

×= × +

= × =

−

−−

−

J s m s

m eV

J eV

e je j a fe j K

K

P40.16 From condition (i), hf e VS= +∆ 1 1b g φ and hf e VS= +∆ 2 2b g φ

∆ ∆V VS S1 2 1 48b g b g= + . V .

Then φ φ2 1 1 48− = . eV .

From condition (ii), hf hfc c1 1 2 20 600 0 600= = =φ φ. .

φ φ

φ φ2 2

2 1

0 600 1 48

3 70 2 22

− =

= =

. .

. . .

eV

eV eV

P40.17 (a) e Vhc

S∆ = − → =⋅

− =λ

φ φ1 240

0 376 1 90 nm eV

546.1 nm eV eV. .

(b) e Vhc

VS S∆ ∆= − =⋅

− → =λ

φ1 240

1 90 0 216 nm eV

587.5 nm eV V. .

Chapter 40 469

P40.18 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 ofquantum mechanics.

P40.19 Ultraviolet photons will be absorbed to knock electrons out of the sphere with maximum kineticenergy 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 itspotential approaches 1.51 V, no further electrons will be able to escape, but will fall back onto thesphere. 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.

P40.20 (a) By having the photon source move toward the metal, the incident photons are Dopplershifted to higher frequencies, and hence, higher energy.

(b) If v c= 0 280. , ′ =+−

= × = ×f fv cv c

11

7 00 101 280 720

9 33 1014 14...

.e j Hz .

Therefore, φ = × ⋅ × = × =− −6 626 10 9 33 10 6 18 10 3 8734 14 19. . . . J s Hz J eVe je j .

(c) At v c= 0 900. , f = ×3 05 1015. Hz

and K hfmax . ..

. .= − = × ⋅ ××

FHG

IKJ − =−

−φ 6 626 10 3 05 101 00

103 87 8 7834 15

19 J s Hz eV

1.60 J eV eVe je j .

Section 40.3 The Compton Effect

P40.21 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


P40.22 (a) ∆λ θ= −h

m ce1 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.

. .

λλ 0

124 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. .

P40.23 With K Ee = ′ , K E Ee = − ′0 gives ′ = − ′E E E0

′ =EE0

2 and ′ =

′λ

hcE

′ = = =λ λhc

EhcE0 0

022 2

′ = + −λ λ λ θ0 1C cosa f 2 10 0λ λ λ θ= + −C cosa f1

0 001 600 002 43

0− = =cos..

θλλC

θ = °70 0.

P40.24 This is Compton scattering through 180°:

Ehc

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. P40.24

′ = + =λ λ λ0 0 115∆ . nm so ′ =′=E

hcλ

10 8. keV.

By conservation of momentum for the photon-electron system,h h

peλ λ0i 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

Chapter 40 471

P40.25 (a) Conservation of momentum in the x direction gives: p p peγ γ θ φ= ′ +cos cos

or since θ φ= ,h

ph

eλ λθ

0= +

′FHG

IKJ cos . [1]

Conservation of momentum in the y direction gives: 0 = ′ −p peγ θ θsin sin ,

which (neglecting the trivial solution θ = 0) gives: p ph

e = ′ =′γ λ. [2]

Substituting [2] into [1] gives: h hλ λ

θ0

2=

′cos , or ′ =λ λ θ2 0 cos . [3]

Then the Compton equation is ′ − = −λ λ θ0 1h

m cecosa f

giving 2 10 0λ θ λ θcos cos− = −h

m cea f

or 2 11

10

2cos cosθλ

θ− = −hc

m cea f.

Since Ehc

γ λ=

0, this may be written as: 2 1 12cos cosθ θγ− =

FHGIKJ −

E

m ce

a f

which reduces to: 2 12 2+FHG

IKJ = +

E

m c

E

m ce e

γ γθcos

or cos. .

..θ γ

γ=

+

+=

++

=m c E

m c Ee

e

2

220 511 0 880

0 8800 732

MeV MeV1.02 MeV MeV

so that θ φ= = °43 0. .

(b) Using Equation (3): ′ =′= = =

°= =E

hc hc Eγ

γ

λ λ θ θ0 2 20 880

43 00 602 602

cos cos.cos .

.a f MeV

2 MeV keV .

Then, ′ =′= = × ⋅−p

E

c cγγ 0 602

3 21 10 22..

MeV kg m s .

(c) From Equation (2), p pce = ′ = = × ⋅−

γ0 602

3 21 10 22..

MeV kg m s .

From energy conservation: K E Ee = − ′ = − = =γ γ 0 880 0 602 0 278 278. . . MeV MeV MeV keV .


P40.26 The energy of the incident photon is E p chc

00

= =γ λ.

(a) Conserving momentum in the x direction gives

p p peλ γφ θ= + ′cos cos , or since φ θ= , Ec

p pe0 = + ′γ θd icos . [1]

Conserving momentum in the y direction (with φ = 0) yields

0 = ′ −p peγ θ θsin sin , or p ph

e = ′ =′γ λ. [2]

Substituting Equation [2] into Equation [1] gives

Ec

h h0 =′+

′FHG

IKJλ λ

θcos , or ′ =λ θ2

0

hcE

cos . [3]

By the Compton equation, ′ − = −λ λ θ0 1h

m cecosa f, 2 2

10 0

hcE

hcE

hm ce

cos cosθ θ− = −a f

which reduces to 2 20

20m c E m c Ee e+ = +e jcosθ .

Thus, φ θ= =++

FHG

IKJ

−cos 12

02

02m c Em c E

e

e

.

(b) From Equation [3], ′ = =++

FHG

IKJλ θ

2 220 0

20

20

hcE

hcE

m c Em c E

e

e

cos .

Therefore, ′ =′=

+ +=

++

FHG

IKJE

hc hc

hc E m c E m c E

E m c Em c Ee e

e

eγ λ 2 2 2

2

02

02

0

02

02

0b ge j e j,

and ′ =′=

++

FHG

IKJp

E

cE

cm c E

m c Ee

eγ

γ 02

02

022

.

(c) From conservation of energy, K E E EE m c E

m c Eee

e

= − ′ = −++

FHG

IKJ0 0

02

02

022

γ

or KE m c E m c E

m c EE

m c Ee

e e

e e

=+ − −

+

FHG

IKJ = +

02

02

02

0

02

202

2 2 2

2e j.

Finally, from Equation (2), p pE

cm c E

m c Eee

e

= ′ =++

FHG

IKJγ

02

02

022

.

Chapter 40 473

*P40.27 The electron’s kinetic energy is

K mv= = × × = ×− −12

12

9 11 10 2 16 102 31 2 18. . kg 2.18 10 m s J6e j .

This is the energy lost by the photon, hf hf0 − ′

hc hcλ λ0

182 16 10−′= × −. J . We also have

′ − = − =×

× ×− °

′ = + ×

−

−

−

λ λ θ

λ λ

0

34

31

013

16 63 10

1 17 4

1 11 10

hm ce

cos.

cos .

.

a fe j

a f Js s

9.11 10 kg 3 10 m

m

8

(a) Combining the equations by substitution,

1 10 111

2 16 101 09 10

0 1110 111

1 09 10

0 111 1 09 10 1 21 10

1 09 10 1 21 10 1 11 10 0

1 21 10 10 4 1 09 10 1 11 10

2

0 0

18

347

0 0

02

0

7

702 6

0

702 6 13

0

6 6 2 7 13

λ λ

λ λλ λ

λ λ

λ λ

λ

−+

=×

× ×= ×

+ −+

= ×

= × + ×

× + × − × =

=− × ± × − × − ×

−

−

−

− −

− − −

..

.

..

.

. . .

. . .

. . .

pm J s

6.63 10 Js 3 10 mm

pm pm

m

pm m

m m

m 1.21 m m

8

02

2

e j

b ge j

e j e je j1 09 107. ×e j

only the positive answer is physical: λ 0101 01 10= × −. m .

(b) Then ′ = × + × = ×− − −λ 1 01 10 1 11 10 1 01 1010 13 10. . . m m m.Conservation of momentum in the transverse direction:

0

6 63 1017 4

9 11 10

1 2 18 10 3 10

1 96 10 1 99 10 81 1

34

10

31

6 8 2

24 24

=′

−

× ⋅×

°=× ×

− × ×

× = × = °

−

−

−

− −

hm veλ

θ γ φ

φ

φ φ

sin sin

.sin .

. sin

.

. . sin .

J s1.01 10 m

kg 2.18 10 m s6e je j


P40.28 (a) Thanks to Compton we have four equations in the unknowns φ, v, and ′λ :

hc hcm c m ce eλ λ

γ0

2 2=′+ − (energy conservation) [1]

h hm veλ λ

φ γ φ0

2=′

+cos cos (momentum in x direction) [2]

0 2=′

−h

m veλφ γ φsin sin (momentum in y direction) [3]

′ − = −λ λ φ0 1 2h

m cecosb g (Compton equation). [4]

Using sin sin cos2 2φ φ φ= in Equation [3] gives γλ

φm vh

e =′

2cos .

Substituting this into Equation [2] and using cos cos2 2 12φ φ= − yields

h h h hλ λ

φλ

φλ

φ0

2 2 22 12

4 1=′

− +′

=′

−cos cos cose j e j,

or ′ = −λ λ φ λ4 02

0cos . [5]

Substituting the last result into the Compton equation gives

4 2 1 2 1 2 102

02

22λ φ λ φ φcos cos cos− = − − = −

hm c

hcm ce e

e j e j.

With the substitution λ 00

=hcE

, this reduces to

cos22

02

0212

φ =++

=++

m c Em c E

xx

e

e

where xE

m ce

≡ 02 .

For x = =0 700

1 37.

. MeV

0.511 MeV, this gives φ =

++

= °−cos .1 12

33 0xx

.

FIG. P40.28(a)

(b) From Equation [5], ′ = − =++FHGIKJ −

LNM

OQP =

++FHGIKJλ λ φ λ λ0

20 04 1 4

12

12 32

cose j xx

xx

.

Then, Equation [1] becomes

hc hc xx

m c m ce eλ λγ

0 0

2 222 3

=++FHGIKJ + − or

Em c

Em c

xxe e

02

02

22 3

1−++FHGIKJ + = γ .

Thus, γ = + −++FHGIKJ1

22 3

x xxx

, and with x = 1 37. we get γ = 1 614. .

Therefore, vc= − = − =−1 1 0 384 0 7852γ . . or v c= 0 785. .

Chapter 40 475

P40.29 ′ − = −λ λ θh

m ce1 cosa f

′′ − ′ = − −

′′ − = − − + −

λ λ π θ

λ λ π θ θ

hm ch

m ch

m ch

m ch

m c

e

e e e e

1 cos

cos cos

a f

a f

Now cos cosπ θ θ− = −a f , so ′′ − = =λ λ 2 0 004 86h

m ce. nm . FIG. P40.29

P40.30 Maximum energy loss appears as maximum increase in wavelength, which occurs for scattering

angle 180°. Then ∆λ = − ° FHGIKJ =1 180

2cosa f h

mch

mc where m is the mass of the target particle. The

fractional energy loss is

E EE

hc hchc

h mch mc

0

0

0

0

0

0 0

22

− ′=

− ′=

′ −′

=+

=+

λ λλ

λ λλ

λλ λ λ∆∆

.

Further, λ 00

=hcE

, so E E

Eh mc

hc E h mcE

mc E0

0 0

02

0

22

22

− ′=

+=

+.

(a) For scattering from a free electron, mc2 0 511= . MeV , so

E EE

0

0

2 0 5110 511 2 0 511

0 667− ′

=+

=.

. ..

MeV MeV MeVa fa f .

(b) For scattering from a free proton, mc2 938= MeV, and

E EE

0

0

2 0 511938 2 0 511

0 001 09− ′

=+

=.

..

MeV MeV MeVa fa f .

Section 40.4 Photons and Electromagnetic Waves

*P40.31 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.


*P40.32 The photon energy is Ehc

= =× ⋅ ×

×= ×

−

−−

λ

6 63 10

633 103 14 10

34

919

..

J s 3 10 m s

m J

8e j. The power carried by the

beam is 2 10 3 14 10 0 62818 19× × =− photons s J photon We je j. . . Its intensity is the average Poynting

vector I Sr

= = =×

= ×−

av2 W 4

m W m

Pπ π

2 3 250 628

1 75 102 61 10

.

..

a fe j

.

(a) S E BE B

av rms rms= °=1

901

2 20 0µ µsin max max . Also E B cmax max= . So S

Ecav = max

2

02µ.

E cS

B

max

max

.

.

..

= = × × ×

= ×

=×

×= ×

−

−

2 2 4 10 3 10 2 61 10

1 40 10

1 40 103 10

4 68 10

01 2 7 8 5 1 2

4

4

85

µ πav2 Tm A m s W m

N C

N C m s

T

b g e je je je j

(b) Each photon carries momentum Ec

. The beam transports momentum at the rate Pc

. It

imparts momentum to a perfectly reflecting surface at the rate2 2 0 628

3 104 19 108

9Pc

= =×

= × −force W

m s N

..

a f.

(c) The block of ice absorbs energy mL t= P∆ melting

mt

L= =

×

×= × −P∆ 0 628

3 33 101 02 105

2.

..

W 1.5 3 600 s

J kg kg

b g.

Section 40.5 The Wave Properties of Particles

P40.33 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

P40.34 (a)pm

219

250 0 1 60 10= × −. .a fe j J

php

= × ⋅

= =

−3 81 10

0 174

24.

.

kg m s

nmλ

(b)pm

23 19

250 0 10 1 60 10= × × −. .e je j J

php

= × ⋅

= = ×

−

−

1 20 10

5 49 10

22

12

.

.

kg m s

mλ

The relativistic answer is slightly more precise:

λ = =+ −L

NMOQP

= × −hp

hc

mc K m c2 2 2 41 2

125 37 10

e j. m .

Chapter 40 477

P40.35 (a) Electron: λ =hp

and K m vm v

mpme

e

e e= = =

12 2 2

22 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 jλ = × =−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

.

P40.36 (a) The wavelength of a non-relativistic particle of mass m is given by λ = =hp

hmK2

where the

kinetic energy K is in joules. If the neutron kinetic energy Kn is given in electron volts, its

kinetic energy in joules is K Kn= × −1 60 10 19. J eVe j and the equation for the wavelength

becomes

λ = =× ⋅

× ×=

×−

− −

−hmK K K

n n26 626 10

2 1 67 10 1 60 10

2 87 1034

27 19

11.

. .

. J s

kg J eV m

e je jwhere Kn is expressed in electron volts.

(b) If Kn = =1 00 1 000. keV eV, then λ =×

= × =−

−2 87 101 000

9 07 10 90711

13.. m m fm .

P40.37 (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

Uk q q

r

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 .


P40.38 From the condition for Bragg reflection,

m d dλ θφ

= = FHGIKJ2 2

2sin cos .

But d a= FHGIKJsin

φ2

where a is the lattice spacing.

Thus, with m = 1, λφ φ

φ= FHGIKJFHGIKJ =2

2 2a asin cos sin

FIG. P40.38

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 = =×

°= × =

−−λ

φsin.

sin .. .

1 67 1050 0

2 18 10 0 21810

10 m nm .

P40.39 (a) E p c m c2 2 2 2 4= +

with E hf= , ph

=λ

and mch

=λC

so h fh c h c2 2

2 2

2

2 2

= +λ λC

2 andfcFHGIKJ = +

2

21 1λ λC

2 (Eq. 1).

(b) For a photonfc=

1λ

.

The third term 1λC

in Equation 1 for electrons and other massive particles shows that

they will always have a different frequency from photons of the same wavelength .

*P40.40 For the massive particle, K mc= −γ 1 2b g and λγm

hp

hmv

= = . For the photon (which we represent as γ),

E K= and λγγ = = = =−

cf

chE

chK

chmc1 2b g . Then the ratio is

λλ

γγ

γγ

γ

m

ch mvmc h

vc

=−

=−1 12b g .

(a)λλγ

m=

− −FH IK −LNM

OQP=

1 0 9

1 0 9 1 1 0 9 11 60

2 2

.

. ..

a f

(b)λλγ

m=

− −FH

IK −

LNM

OQP= ×

1 0 001

1 0 001 1 1 0 001 12 00 10

2 2

3.

. ..

a fa f a f

(c) As vc→ 1, γ →∞ and γ − 1 becomes nearly equal to γ. Then

λλ

γγ

γ

m→ =1 1 .

(d) As vc→ 0 , 1 1 1

12

112

2

2

1 2 2

2

2

2−FHGIKJ − ≈ − −FHG

IKJ − =

−vc

vc

vc

and λλγ

m

v c

v c

cv

→ = → ∞11 2

22 2b ge j

.

Chapter 40 479

P40.41 λ =hp

ph

= =× ⋅×

= × ⋅−

−−

λ6 626 101 00 10

6 63 1034

1123.

..

J s m

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

P40.42 (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≤ =

× ⋅FHG

IKJ = ×

−−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.

Section 40.6 The Quantum Particle

*P40.43 E K mu hf= = =12

2 and λ =h

mu.

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.

*P40.44 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 = + − =

ωγ

γ

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 1

In fact, the phase speed is larger than the speed of light. A point of constant phase in the wavefunction carries no mass, no energy, and no information.

Now for the group speed:



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

2 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 40.7 The Double-Slit Experiment Revisited

P40.45 (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, θλ

= 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.

P40.46 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 =h

m ve so m v

he =

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

.

*P40.47 We find the speed of each electron from energy conservation in the firing process:

012

2 2 1 6 10

9 11 103 98 10

2

19

316

= + = −

= =× ×

×= ×

−

−

K U mv

veVm

f f eV

C 45 V

kg m s

.

..

a f

The time of flight is ∆∆

tx

v= =

×= × −0 28

7 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 .

Chapter 40 481

Section 40.8 The Uncertainty Principle

P40.48 (a) ∆ ∆ ∆ ∆p x m v x= ≥2

so ∆∆

vhm 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+ = .

P40.49 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∆

∆x

hp

= =× ⋅

× ⋅=

−

−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∆

∆x

hp

= = × −

45 28 10 32

π. m .

P40.50∆ ∆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 jThe answer, x = ×3 79 1028. m , is 190 times greater than the diameter of the Universe!

P40.51 With ∆x = × −2 10 15 m m, the uncertainty principle requires ∆∆

pxx ≥ = × ⋅−

22 6 10 20. kg m s .

The average momentum of the particle bound in a stationary nucleus is zero. The uncertainty inmomentum measures the root-mean-square momentum, so we take prms ≈ × ⋅−3 10 20 kg m s . For anelectron, the non-relativistic approximation p m ve= would predict v ≈ ×3 1010 m s , while v cannotbe greater than c.

Thus, a better solution would be E m c pc m ce e= +LNM

OQP ≈ =2 2 2

1 2256e j b g MeV γ

γ ≈ =−

1101

1 2 2v cso v c≈ 0 999 96. .

For a proton, vpm

= gives v = ×1 8 107. m s, less than one-tenth the speed of light.

*P40.52 (a) K mvmv

mpm

= = =12 2 2

22 2a f

(b) To find the minimum kinetic energy, think of the minimum momentum uncertainty, andmaximum position uncertainty of 10 15− = m ∆x . We model the proton as moving along a

straight line with ∆ ∆p x =2

, ∆∆

px

=2

. The average momentum is zero. The average squared

momentum is equal to the squared uncertainty:

Kpm

p

m x m x m= = = = =

× ⋅

×= ×

=

−

− −

−2 2 2

2

2

2 2

34 2

2 15 2 27

13

2 2 4 2 32

6 63 10

32 10 1 67 108 33 10

5 21

∆

∆ ∆

b ga f a f

e je jπ π

.

..

.

J s

m kg J

MeV


P40.53 (a) At the top of the ladder, the woman holds a pellet inside a small region ∆xi . Thus, theuncertainty principle requires her to release it with typical horizontal momentum

∆ ∆∆

p m vxx x

i= =

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

22

where Am

Hg

=2

2.

To minimize ∆x f , we required x

d xf

i

∆

∆

d ib g = 0 or 1 02− =

Axi∆

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 4

.

(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

Additional Problems

P40.54 ∆Vhe

feS = FHG

IKJ −

φ

From two points on the graph 0 4 1 1014= FHGIKJ × −

he e

. Hze j φ

and 3 3 12 1014. V Hz= FHGIKJ × −

he ee j φ

.

Combining these two expressions we find:

(a) φ = 1 7. eV

(b)he= × ⋅−4 2 10 15. V s

FIG. P40.54

(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

Chapter 40 483

P40.55 We want an Einstein plot of Kmax versus f

λ , nm , Hz , eV588505445

5.105.946.74

0.670.981.35

f K1014max

399 7.52 1.63

(a) slope = ±0 402

8%. eV

10 Hz14

(b) e V hfS∆ = −φ

h =× ⋅F

HGIKJ = × ⋅ ±

−−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. P40.55

P40.56 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

= − = −hc

Khc e B R

memax

2 2 2

2.

P40.57 ∆λ θ= − =× ⋅

× ×= ×

−

−−h

m cp1

6 626 10

1 67 10 3 00 100 234 3 09 10

34

27 816cos

.

. .. .a f e j

e je ja f J s

kg m s m

λ

λ λ λ

00

34 8

1315

015

6 626 10 3 00 10

200 1 60 106 20 10

6 51 10

= =× ⋅ ×

×= ×

′ = + = ×

−

−−

−

hcE

. .

..

.

J s m s

MeV J MeV m

m

e je ja fe j

∆

(a) Ehc

γ λ=

′= 191 MeV

(b) Kp = 9 20. MeV


P40.58 Isolate the terms involving φ in Equations 40.13 and 40.14. Square and add to eliminate φ.

h m ve2

02 2

0

2 2 21 1 2λ λ

θλ λ

γ+′−

′

LNM

OQP=

cos

Solve for vc

b

b c

2

2 2=

+e j: b

hme

= +′−

′

LNM

OQP

2

202 2

0

1 1 2λ λ

θλ λcos

.

Substitute into Eq. 40.12: 11 1

10

2

1 2 2

2+FHGIKJ −

′LNM

OQP= = −

+FHG

IKJ =

+−hm c

bb c

c bce λ λ

γ .

Square each side: chc

mhm

chme e e

2

0

2

20

22

2

202 2

0

2 1 1 1 1 1 1 2+ −

′LNM

OQP+ −

′LNM

OQP

= +FHGIKJ +

′−

′

LNM

OQPλ λ λ λ λ λ

θλ λcos

.

From this we get Eq. 40.11: ′ − =FHGIKJ −λ λ θ0 1

hm ce

cos .

P40.59 Show that if all of the energy of a photon is transmitted to an electron, momentum will not beconserved.

Energy:hc hc

K m ce eλ λγ

0

2 1=′+ = −b g if hc

′=

λ0 (1)

Momentum:h h

m v m ve eλ λγ γ

0=

′+ = if ′ = ∞λ (2)

From (1), γλ

= +hm ce0

1 (3)

v cm c

h m ce

e= −

+FHG

IKJ1 0

0

2λλ

(4)

Substitute (3) and (4) into (2) and show the inconsistency:

h hm c

m cm c

h m cm c h h h m c

h m c

h h m che

ee

e

e e

e

e

λ λλλ

λλ

λ

λ λλ

0 0

0

0

20

0

0

02

0

01 12 2

= +FHG

IKJ −

+FHG

IKJ =

+ +

+=

+b gb g

.

This is impossible, so all of the energy of a photon cannot be transmitted to an electron.

P40.60 Begin with momentum expressions: ph

=λ

, and p mv mcvc

= = FHGIKJγ γ .

Equating these expressions, γλ

λλ

vc

hmc

FHGIKJ =FHGIKJ =

1 C .

Thus,v c

v c

b gb g

2

2

2

1 −= FHGIKJ

λλC

orvc

vc

FHGIKJ = FHG

IKJ − FHG

IKJFHGIKJ

2 2 2 2λλ

λλ

C C

vc

2

2

2

2 21

1

1=

+=

+

λ λ

λ λ λ λC

C C

b gb g b g

giving vc

=+1

2λ λCb g.

Chapter 40 485

P40.61 (a) Starting with Planck’s law, I Thc

ehc k Tλ

π

λ λ,b g =

−

2

1

2

5 B

the total power radiated per unit area I T dhc

ed

hc k Tλ λ

π

λλ

λ,b g

0

2

50

2

1

∞ ∞z z=−B

.

Change variables by letting xhck T

=λ B

and dxhcd

k T= −

λλB

2 .

Note that as λ varies from 0→∞ , x varies from ∞→ 0 .

Then I T dk T

h cx

edx

k Th cx

λ λπ π π

,b ge j0

4

3 2

30 4

3 2

42

1

215

∞

∞z z= −

−=

FHGIKJ

B4

B4

.

Therefore, I T dk

h cT Tλ λ

πσ,b g

0

5

3 24 42

15

∞

z =FHG

IKJ =B

4

.

(b) From part (a), σπ π

= =×

× ⋅ ×

−

−

215

2 1 38 10

15 6 626 10 3 00 10

5

3 2

5 23 4

34 3 8 2k

h cB4 J K

J s m s

.

. .

e je j e j

σ = × ⋅−5 67 10 8. W m K2 4 .

P40.62 Planck’s law states I Thc

ehc e

hc k Thc k Tλ

π

λπ λ

λλ,b g =

−= −− −2

12 1

2

52 5 1

B

B .

To find the wavelength at which this distribution has a maximum, compute

dId

hc e e ehck T

dId

hc

e

hck T

e

e

hc k T hc k T hc k T

hc k T

hc k T

hc k T

λπ λ λ

λ

λπ

λ λ

λ λ λ

λ

λ

λ

= − − − − −FHG

IKJ

RS|T|

UV|W|=

=−

− +−

RS|T|

UV|W|=

− − − −2 5 1 1 0

2

15

10

2 6 1 5 2

2

2

6

B B B

B

B

B

B

B

Letting xhck T

=λ B

, the condition for a maximum becomes xe

e

x

x −=

15 .

We zero in on the solution to this transcendental equation by iterations as shown in the table below.The solution is found to be

x xe ex x − 1e j x xe ex x − 1e j4.000 00 4.074 629 4 4.964 50 4.999 403 04.500 00 4.550 552 1 4.965 50 5.000 374 95.000 00 5.033 918 3 4.965 00 4.999 889 04.900 00 4.936 762 0 4.965 25 5.000 132 04.950 00 4.985 313 0 4.965 13 5.000 015 34.975 00 5.009 609 0 4.965 07 4.999 957 04.963 00 4.997 945 2 4.965 10 4.999 986 24.969 00 5.003 776 7 4.965 115 5.000 000 84.966 00 5.000 860 9

xhc

k T= =λmax

.B

4 965 115 and λmax .T

hck

=4 965 115 B

.



Thus, λmax

. .

. ..T =

× ⋅ ×

×= × ⋅

−

−−

6 626 075 10 2 997 925 10

4 965 115 1 380 658 102 897 755 10

34 8

233

J s m s

J K m K

e je je j

.

This result is very close to Wien’s experimental value of λmax .T = × ⋅−2 898 10 3 m K for this constant.

P40.63 (a) Planck’s radiation law predicts maximum intensity at a wavelength λmax we find from

dId

dd

hc e

hc e ehck T

hc e

hc k T

hc k T hc k T hc k T

λ λπ λ

π λλ

π λ

λ

λ λ λ

= = −RST

UVW= − −

−FHG

IKJ + − −

−−

−−

−−

0 2 1

0 2 1 1 2 5 1

2 51

2 52

22 6

1

B

B B B

B

b g

b g b g b ga f a f

or−

−+

−=

hce

k T e e

hc k T

hc k T hc k T

λ

λ λλ λ

B

B B

B

b g

b g b g72 6

1

5

10

which reduces to 5 1λ λ λk T

hce ehc k T hc k TB B BF

HGIKJ − =b g b g .

Define xhck T

=λ B

. Then we require 5 5e xex x− = .

Numerical solution of this transcendental equation gives x = 4 965. to four digits. So

λmax .=

hck T4 965 B

, in agreement with Wien’s law.

The intensity radiated over all wavelengths is I T d A Bhc d

e hc k Tλ λ

π λ

λ λ,b g b g

0

2

50

2

1

∞ ∞

z z= + =−B

.

Again, define xhck T

=λ B

so λ =hc

xk TB and d

hcx k T

dxλ = − 2B

.

Then, A Bhc x k T hcdx

h c x k T e

k Th c

x dx

exx

x+ =

−

−=

−=∞

∞

z z2

1

2

1

2 5 5 5

5 5 2

0 4

3 2

3

0

π πB

B

B4

e j e j.

The integral is tabulated as π 4

15, so (in agreement with Stefan’s law) A B

k Th c

+ =215

5 4

3 2π B

4

.

The intensity radiated over wavelengths shorter than λmax is

I T d Ahc d

e hc k Tλ λ

π λ

λ

λ

λ

λ

,max max

b g b g0

2

50

2

1z z= =

−B

.

With xhck T

=λ B

, this similarly becomes Ak T

h cx dxe

Bx=−

∞z21

3

4.965

π 4 4

3 2 .

So the fraction of power or of intensity radiated at wavelengths shorter than λmax is

AA B

k T h c x dx e

k T h cx dxe

x

x+=

− −LNMM

OQPP= −

−

zz

2 15 1

2 151

151

4 3 2 4 3

0

4.965

5 4 3 2 4

3

0

4.965π π

π π

B4

B4

e j e j.


Chapter 40 487

(b) Here are some sample values of the integrand, along with a sketch of the curve:

x x ex3 1

3

2

10 000 0 000 100 9 51 100 200 3 61 101 00 0 5822 00 1 253 00 1 424 00 1 194 90 0 883

4 965 0 860

−

××

−

−

−

e j. .. .. .. .. .. .. .. .. .

FIG. P40.63(b)

Approximating the integral by trapezoids gives A

A B+≈ − =1

154 870 0 250 14π. .a f .

P40.64 p mv mE= = = × ×− −2 2 1 67 10 0 040 0 1 60 1027 19. . . kg eV J eVe jb ge jλ = = × =−h

mv1 43 10 0 14310. . m nm

This is of the same order of magnitude as the spacing between atoms in a crystal so diffractionshould appear.

P40.65 λC =h

m ce and λ =

hp

:λλC = =

h m ch p

pm c

e

e;

E c p m ce2 2 2 2 2= + e j : p

Ec

m ce= −2

22b g

λλC = − = −

LNM

OQP=FHGIKJ −

1 11

2

22

2

2

22

2

2

m cEc

m cm c

Ec

m cE

m cee

ee

eb g b g b g

P40.66 (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 ≥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


P40.67 From the uncertainty principle ∆ ∆E t ≥2

or ∆ ∆mc t2

2e j = .

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 Jπ e ja f .

P40.68 ∆λ θ λ λ= − = ′ −h

m ce1 0cosa f

′ =′=

+= + −LNM

OQP

′ = + −LNM

OQP

′ = + −LNM

OQP

= + −LNM

OQP

−

−

− −

Ehc hc

hch

m c

Ehc hc

m c

Ehc hc

m cE

Em c

e

e

e e

λ λ λλ θ

λ λθ

λ λθ θ

00

1

02

0

1

02

0

1

00

2

1

1

1 1

1 1 1 1

∆cos

cos

cos cos

a f

a f

a f a f

P40.69 (a) The light is unpolarized. It contains both horizontal and vertical field oscillations.

(b) The interference pattern appears, but with diminished overall intensity.

(c) The results are the same in each case.

(d) The interference pattern appears and disappears as the polarizer turns, with alternatelyincreasing and decreasing contrast between the bright and dark fringes. The intensity on thescreen is precisely zero at the center of a dark fringe four times in each revolution, when thefilter axis has turned by 45°, 135°, 225°, and 315° from the vertical.

(e) Looking at the overall light energy arriving at the screen, we see a low-contrast interferencepattern. After we sort out the individual photon runs into those for trial 1, those for trial 2,and those for trial 3, we have the original results replicated: The runs for trials 1 and 2 formthe two blue graphs in Figure 40.24 in the text, and the runs for trial 3 build up the redgraph.

Chapter 40 489

P40.70 Let ′u represent the final speed of the electron and let

′ = −′F

HGIKJ−

γ 12

2

1 2uc

. We must eliminate β and ′u from the

three conservation equations:

hcm c

hcm ce eλ

γλ

γ0

2 2+ =′+ ′ [1]

hm u

hm ue eλ

γλ

θ γ β0+ −

′= ′ ′cos cos [2]

hm ue′

= ′ ′λ

θ γ βsin sin [3]

Square Equations [2] and [3] and add:

FIG. P40.70

hm u

h h m u h h m um u

h hm u

h m u h m u h m uu c

ee e

e

ee e e

2

02

2 2 22

20

2

0

2 2 2

2

02

2

22 2 2

0

2

0

2 2

2 2

2 2 2

2 2 21

λγ

λγλ

θλ λ

γ θλ

γ

λ λγ

γλ

γ θλ

θλ λ

+ +′+ −

′−

′= ′ ′

+′+ + −

′−

′=

′− ′

cos cos

cos cos

Call the left-hand side b. Then bbuc

m ue−′

= ′2

22 2 and ′ =

+=

+u

bm b c

c bm c be e

22 2

2

2 2 .

Now square Equation [1] and substitute to eliminate ′γ :

hm c

h h m c h h m c m cu c

m c bee e e

e

2

02

2 2 22

20

2

0

2 2

2 22 22 2 2

1λγ

λγλ λ λ

γλ

+ +′+ −

′−

′=

− ′= + .

So we have h h

m ch m c h m c h

ee e

2

02

2

22 2 2

0

2

0

2 2 2λ λ

γγλ

γλ λ λ

+′+ + −

′−

′

= + +′+ + −

′−

′m c

h hm u

h m u h m u he e

e e22

02

2

22 2 2

0

2

0

2 2 2λ λ

γγλ

γ θλ

θλ λ

cos cos

Multiply through by λ λ0

2 2′

m ce

λ λ γλ γ λ γ

λ λλ λ γ λ γ γλ θ θ

λ λ γγ γλ γλ θ

θ

02 0

2

2 2 00

2 2

2 20

2

2

2 2

02

2 2

20

2

2 2

2 2 2 2 2 2

12

12

12

1

′ +′− − = ′ +

′+

′− −

′ − −FHG

IKJ +

′−FHGIKJ = −FHG

IKJ + −

hm c

hm c

hm c

um c

h um c

h um c

hm c

uc

hm c

uc

hm c

uc

hm c

e e e e e e e

e e e

cos cos

coscosa f

The first term is zero. Then ′ =−

−

FHG

IKJ + −

FHG

IKJ −

−

λ λθ γ

θ0

111

11

1u c

u chm c u ce

coscos

a f a f.

Since γ − = − FHGIKJ = −FHG

IKJ +FHGIKJ

12

1 1 1uc

uc

uc

this result may be written as ′ =−

−

FHG

IKJ +

+−

−λ λθ

θ01

111

1u c

u ch

m cu cu ce

coscos

a f a f .

It shows a specific combination of what looks like a Doppler shift and a Compton shift. This problemis about the same as the first problem in Albert Messiah’s graduate text on quantum mechanics.


ANSWERS TO EVEN PROBLEMS

P40.2 (a) ~10 7− m ultraviolet ; P40.30 (a) 0 667. ; (b) 0 001 09.(b) ~10 10− m gamma ray

P40.32 (a) 14.0 kV/m, 46 8. Tµ ; (b) 4.19 nN;(c) 10.2 gP40.4 (a) 70 9. kW; (b) 580 nm;

(c) 7 99 1010. × W m;P40.34 (a) 0.174 nm; (b) 5.37 pm or 5.49 pm

ignoring relativistic correction(d) 9 42 10 1226. W m× − ;(e) 1 00 10 227. W m× − ; (f) 5 44 1010. W m× ;(g) 7 38 1010. W m× ; (h) 0 260. W m; P40.36 (a) see the solution; (b) 907 fm(i) 2 60 10 9. W m× − ; (j) 20 kW

P40.38 0 218. nm

P40.6 2 96 1019. × photons sP40.40 (a) 1.60; (b) 2 00 103. × ; (c) 1; (d) ∞

P40.8 5 71 103. × photonsP40.42 (a) 1 10 10 34. × − m s ; (b) 1 36 1033. × s ; (c) no

P40.10 1 34 1031. × P40.44 see the solution

P40.12 see the solution P40.46 105 V

P40.14 (a) 1 38. eV ; (b) 334 THz P40.48 (a) 0 250. m s ; (b) 2 25. m

P40.16 Metal one: 2 22. eV , Metal two: 3 70. eVP40.50 3 79 1028. × m, much larger than the

diameter of the observable UniverseP40.18 148 d, the classical theory is a gross failure

P40.52 (a) see the solution; (b) 5.21 MeVP40.20 (a) The incident photons are Dopplershifted to higher frequencies, and hence,higher energy; (b) 3 87. eV ; (c) 8 78. eV P40.54 (a) 1 7. eV ; (b) 4 2 10 15. × ⋅− V s ; (c) 730 nm

P40.56hc e B R

meλ−

2 2 2

2P40.22 (a) 488 fm ; (b) 268 keV ; (c) 31 5. keV

P40.24 pce =

22 1. keV; Ke = 478 eV

P40.58 see the solution

P40.26 (a) cos−++

FHG

IKJ

12

02

02m c Em c E

e

e

;P40.60 see the solution


(b) ′ =++

FHG

IKJE

E m c Em c E

e

eγ

02

02

022

,

′ =++

FHG

IKJp

Ec

m c Em c E

e

eγ

02

02

022

;

P40.64 0 143. nm, comparable to the distancebetween atoms in a crystal, so diffractioncan be observed

(c) KE

m c Ee

e

=+

02

202e j

,

pE

cm c E

m c Eee

e

=++

FHG

IKJ

02

02

022

P40.66 (a) 2 82 10 37. × − m; (b) 1 06 10 32. × − J ;(c) 2 87 10 35. %× −


P40.70 see the solutionP40.28 (a) 33 0. ° ; (b) 0 785. c

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