Solution Manual for Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by John Morrison, ISBN: 9780128177907, All 15 Chapters Covered, Verified Latest Edition

SOLUTION MANUAL Modern Physics with Modern
Computational Methods: for Scientists and
Engineers 3rd Edition by Morrison Chapters 1- 15

,Table of contents
n n



1.nThenWave-ParticlenDuality

2.nThenSchrödingernWavenEquation

3.nOperatorsnandnWaves

4.nThenHydrogennAtom

5.nMany-ElectronnAtoms

6.nThenEmergencenofnMasersnandnLasers

7.nDiatomicnMolecules

8.nStatisticalnPhysics

9.nElectronicnStructurenofnSolids

10.nChargenCarriersninnSemiconductors

11.nSemiconductornLasers

12.nThenSpecialnTheorynofnRelativity

13.nThenRelativisticnWavenEquationsnandnGeneralnRelativity

14.nParticlenPhysics

15.nNuclearnPhysics

,1

Then Wave-Particlen Dualityn -n Solutions




1. Thenenergynofnphotonsninntermsnofnthenwavelengthnofnlightni
sngivennbynEq.n(1.5).nFollowingnExamplen 1.1nandnsubstitutingn
λn=n200neVngives:
hc 1240n eVn ·nnm
= =n6.2neV
Ephotonn= λ 200nnm
2. Then energyn ofn then beamn eachn secondn is:
power 100n W
= =n100nJ
Etotaln= time 1n s
Thennumbernofnphotonsncomesnfromnthentotalnenergyndividednb
ynthenenergynofneachnphotonn(seenProblemn1).nThenphoton’snene
rgynmustnbenconvertedntonJoulesnusingnthenconstantn1.602n×n1
0−19nJ/eVn,nseenExamplen1.5.nThenresultnis:
nEtotaln
N = = 100nJ =n1.01n×n1020
photons E
pho
ton 9.93n×n10−19
forn then numbern ofn photonsn strikingn then surfacen eachn second.
3.Wenarengivennthenpowernofnthenlaserninnmilliwatts,nwheren1nm
Wn=n10−3nWn.nThenpowernmaynbenexpressednas:n1nWn=n1nJ/s.n
FollowingnExamplen1.1,nthenenergynofnansinglenphotonnis:
1240n eVn ·nnm
hcn =n1.960neV
Ephotonn = 632.8n nm
n =
λn
Wen nownconvertn tonSIn unitsn (seen Examplen 1.5):
1.960neVn×n1.602n×n10−19nJ/eVn =n3.14n×n10−19nJ
FollowingnthensamenprocedurenasnProblemn2:
1n×n10−3nJ/s 15n photons
Ratenofn emissionn=n = n3.19n×n10
3.14n×n10−19n J/photonn s

, 2

4. Thenmaximumnkineticnenergynofnphotoelectronsnisnfoundnus
ingnEq.n(1.6)nandnthenworknfunctions,nW,nofnthenmetalsnarengiv
enninnTablen1.1.nFollowingnProblemn 1,n Ephotonn=nhc/λn=n6.20n eV
n.n Forn partn (a),n Nan hasn Wn =n2.28n eV n:


(KE)maxn=n6.20neVn−n2.28neVn =n3.92neV
Similarly,nfornAlnmetalninnpartn(b),nWn =n4.08neVn givingn(KE)maxn=n2.12ne
V
andnfornAgnmetalninnpartn(c),nWn=n4.73neVn,ngivingn(KE)maxn=n1.47neVn.

5.Thisnproblemnagainnconcernsnthenphotoelectricneffect.nAsninnPr
oblemn4,nwenusenEq.n(1.6):
hcn−n
(KE)maxn =
Wnλ
wheren Wn isn then workn functionn ofn then materialn andn then termn hc/λ
n describesnthenenergynofnthenincomingnphotons.nSolvingnfornthenlatt

er:
hc
=n(KE)maxn+nWn =n2.3n eVn +n0.9neVn =n3.2n eV
λn
SolvingnEq.n(1.5)nfornthenwavelength:
1240n eVn ·nnm
λn= =n387.5nnm
3.2n e
V
6. Anpotentialnenergynofn0.72neVnisnneededntonstopnthenflownofnelectro
ns.nHence,n(KE)maxnofnthenphotoelectronsncannbennonmorenthann0.
72neV.nSolvingnEq.n(1.6)nfornthenworknfunction:
hc 1240n eVn ·nn —n0.72n eVn =n1.98n eV
W n =n —
λ m
(KE)maxn
=
460nnm
7. Reversingn then proceduren fromn Problemn 6,n wen startn withn Eq.n (1.6):
hcn 1240n eVn ·nn
(KE)maxn = −nWn —n1.98n eVn =n3.19n eV
= m
λ
240nnm
Hence,nanstoppingnpotentialnofn3.19neVnprohibitsnthenelectronsnfro
mnreachingnthenanode.

8. Justn atn threshold,n then kineticn energyn ofn then electronn isn z
ero.n Settingn(KE)maxn=n0n inn Eq.n (1.6),
hc
Wn= = 1240n eVn ·nn =n3.44n eV
λ0 m

360nnm