Modern Physics with Modern Computational Methods, Third Edition – Complete Solutions Manual

1
The Wave-Particle Duality - Solutions
p p p p




1. ThepenergypofpphotonspinptermspofpthepwavelengthpofplightpispgivenpbypE
q.p(1.5).pFollowingpExamplep 1.1pandpsubstitutingpλp=p200peVpgives:
hc 1240p eVp ·pnm
Ephotonp= = =p6.2peV
λ 200pnm
2. Thep energyp ofp thep beamp eachp secondp is:
power 100p W
Etotalp = =
=p100pJ
time 1p s
Thepnumberpofpphotonspcomespfromptheptotalpenergypdividedpbypthepenerg
ypofpeachpphotonp(seepProblemp1).pThepphoton’spenergypmustpbepconverted
ptopJoulespusingpthepconstantp1.602p×p10− p J/eVp,pseepExample p 1.5.pThepres
19

ultpis:
N = p Etotalp = 100pJ =p1.01p×p1020
photons
Ephoton 9.93p×p10−19
forp thep numberp ofp photonsp strikingp thep surfacep eachp second.

3. Weparepgivenptheppowerpofptheplaserpinpmilliwatts,pwherep1pmWp=p10−3pW
p.pTheppowerpmaypbepexpressedpas:p1pWp =p1pJ/s.pFollowingpExample p1.1,pt

hepenergypofpapsinglepphotonpis:
hcpE 1240p eVp ·pnm =p1.960peV
photonp =p = 632.8p nm
λp
Wep nowp convertp top SIp unitsp (seep Examplep 1.5):
1.960peVp×p1.602p×p10−19pJ/eVp =p3.14p×p10−19pJ
Followingp thep samep procedurep asp Problemp 2:
1p×p10−3p J/s 15p photons
Ratep ofp emissionp=p =p3.19p×p10
3.14p×p10−19p J/photonp s

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4. ThepmaximumpkineticpenergypofpphotoelectronspispfoundpusingpEq.p(1.6)
pandpthepworkpfunctions,pW,pofpthepmetalsparepgivenpinpTablep1.1.pFollowin

gpProblemp 1,p Ephotonp=phc/λp=p6.20p eVp.p Forp partp (a),p Nap hasp Wp =p2.28p e
Vp:

(KE)maxp=p6.20peVp −p2.28peVp =p3.92peV

Similarly,p forp Alp metalp inp partp (b),p Wp =p4.08p eVp givingp(KE)maxp =p2.12p eV
andpforpAgpmetalpinppartp(c),pWp =p4.73peVp,pgivingp(KE)maxp=p1.47peVp.

5. Thispproblempagainpconcernspthepphotoelectricpeffect.pAspinpProblemp4,pwep
usepEq.p(1.6):
hcp
(KE)maxp = −p
λ
Wp
wherep Wp isp thep workp functionp ofp thep materialp andp thep termp hc/λp describespt
hepenergypofpthepincomingpphotons.pSolvingpforptheplatter:
hc
=p(KE)maxp+pWp =p2.3p eVp +p0.9p eVp =p3.2p eV
λp
Solvingp Eq.p (1.5)p forp thep wavelength:
1240p eVp ·pnm
λp= =p387.5pnm
3.2p eV
6. Appotentialpenergypofp0.72peVpispneededptopstoppthepflowpofpelectrons.pHence,p(
KE)maxpofpthepphotoelectronspcanpbepnopmorepthanp0.72peV.pSolvingpEq.p(1.6)
pforpthepworkpfunction:


hc 1240p eVp ·pnm
Wp =p — (KE)maxp = —p0.72p eVp =p1.98p eV
λ 460pnm
7. Reversingp thep procedurep fromp Problemp 6,p wep startp withp Eq.p (1.6):
hcp
(KE)maxp = −pWp = 1240p eVp ·pnm —p1.98p eVp =p3.19p eV
λ 240pnm
Hence,papstoppingppotentialpofp3.19peVpprohibitspthepelectronspfrompreachingp
thepanode.

8. Justp atp threshold,p thep kineticp energyp ofp thep electronp isp zero.p Settin
gp(KE)maxp=p0p inp Eq.p (1.6),
hc 1240p eVp ·pnm
Wp= = =p3.44p eV
λ0 360pnm
9. Apfrequencypofp1200p THzpispequalp top 1200p×p1012p Hz.p Usingp Eq.p (1.10),

Ephotonp=phfp =p4.136p×p10−15p eVp ·psp×p1.2p×p1015p Hzp =p4.96p eV

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Next,pusingpthepworkpfunctionpforpsodiump(Na)pmetalpandpEq.p(1.6),p(KE)

maxp=pEphotonp−pWp =p4.96p evp−p2.28p eVp =p2.68p eV

10. WepstartpfrompEq.p(1.8)pforpthepcasepofpmp=p2:
pp p
1p p1p p1p

λ =pR 22 −pn2

Nowp invertp thep equationp andp plugp inp forp thep Rydbergp constant,p R:
p p −1
λp=p 1 1 — p1p
1.0971p×p105p cm− 4 n2
1
Subtractp thep fractionsp byp gettingp ap commonp denominator:
pp p −1
λp=p 1pcm n2p−p4
1.0971p×p10 4n2
5
Invertp thep termp inp thep parenthesisp andp factorp outp thep commonp factorp ofp 4

λp=p
1.0971p×p10 n2p −p
5
p
4
p
4pcm n2

Doingp thep division,p wep getp Eq.p (1.7)p forp thep Balmerp formula:
p
n2p
λp=p(3645.6p×p10−8 cm)
n2p−p
4
11. Followingp Examplep 1.2,
pp p p
13.6pe 13.6p eV
∆Ep=p−p — −p =p2.86p eV
V
52 22

Usingp Eq.p (1.12):
hc 1240p eVp ·pnm
λp= = =p434p nm
∆E 2.86p eV
12. Sincep thep initialp statep hasp mp=p2,p wep canp usep Eq.p (1.7)p withp np=p4:
p
42
λp=p(364.56pnm) =p486.1pnm
42p −p
4
Topgetpthepenergypofpthepphoton,pusepEq.p(1.5):

hcpE 1240p eVp ·pnm =p2.551peV
photonp =p = 486.1p nm
λp

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13. Asp inp Problemp 12,p thep initialp statep hasp mp=p2,p sop wep usep Eq.p (1.7)p with
np =p3:
p
32
λp=p(364.56pnm) =p656.2pnm
32p −p
4
14. FrompFigurep1.6,pthepionizationpenergypofpaphydrogenpatompinpthepnp=p
2pstatepisp−3.4peVp.pSopitptakespapphotonpofp3.4peVp topjustpionizepthispatom.p
Topgetpthepwavelengthpofplight,pjustpinvertpEq.p(1.5):
hc 1240p eVp ·pnm
λp= = =p364.7pnm
Ephoton 3.40p eV
15. Startingp withp Eq.p (1.5)p withp ap wavelengthp ofp 200p nm:

hcpE 1240p eVp ·pnm =p6.20p eV
photonp =p
= 200pnm
λp
FrompFigurep1.6,paphydrogenpatompinpthepnp=p2pstatephaspthepelectronpboun
dpwithppotentialpenergyp(PE) p =p−3.4peVp.pFollowingpExamplep1.3,

(KE)p=p6.20peVp −p3.40peVp =p2.80peV

16. Startingp withp Eq.p (1.5)p withp ap wavelengthp ofp 45p nm:

hcpE 1240p eVp ·pnm =p27.6p eV
photonp =p = 45pnm
λp
Forp ap hydrogenp atomp inp thep groundp state,p thep electronp isp boundp withp potenti
alpenergyp(PE) p =p−13.6peVp.pFollowingpExamplep1.3,

(KE)p=p27.6peVp −p13.6peVp =p14.0peV

Topfindp thepelectron’spvelocity,pconvertp top SIpunitsp (seepExamplep 1.5):

1.6p×p10−19J
(KE)p=p14.0p eVp =p2.24p×p10−18pJ
· 1p eV

FrompAppendixp A,panpelectronphaspmasspmp=p9.11p×p10−31p kg.pUsingpthe
2
pwell pknownpformulapKEp =p(1/2)mv ,pandpsolvingpforpv:

rp s m
−18p J
4.48p×p10−31
vp=p
2(KE)p
= =p2.22p×p106p
m 9.11p×p10 kg s

17. FrompFigurep1.6,pwepseepthatpthepfirstptransitionpforpthepLymanpseriespi
spbetweenpnp=p2ptopnp=p1,pandpsimilarlypwepcanpgetptheptransitionspforpthep
BalmerpandpPaschenpseries.pUsingpEq.p(1.8):
p p
1p p 1p p p1p

λ =pR m2 −pn2