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

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

,Table of contents
6 6




1.6The6Wave-Particle6Duality

2.6The6Schrödinger6Wave6Equation

3.6Operators6and6Waves

4.6The6Hydrogen6Atom

5.6Many-Electron6Atoms

6.6The6Emergence6of6Masers6and6Lasers

7.6Diatomic6Molecules

8.6Statistical6Physics

9.6Electronic6Structure6of6Solids

10.6Charge6Carriers6in6Semiconductors

11.6Semiconductor6Lasers

12.6The6Special6Theory6of6Relativity

13.6The6Relativistic6Wave6Equations6and6General6Relativity

14.6Particle6Physics

15.6Nuclear6Physics

,1

The6 Wave-Particle6 Duality6 -6 Solutions




1. The6energy6of6photons6in6terms6of6the6wavelength6of6light6is6
given6by6Eq.6(1.5).6Following6Example6 1.16and6substituting6λ6
=62006eV6gives:
hc 12406 eV6 ·6nm
= =66.26eV
Ephoton6= λ 2006nm
2. The6 energy6 of6 the6 beam6 each6 second6 is:
power 1006 W
= =61006J
Etotal6= time 16 s
The6number6of6photons6comes6from6the6total6energy6divided6by
6the6energy6of6each6photon6(see6Problem61).6The6photon’s6energ

y6must6be6converted6to6Joules6using6the6constant61.6026×610−196
J/eV6,6see6Example61.5.6The6result6is:
N =6Etotal6 = 1006J =61.016×61020
photons E
pho
ton 9.936×610−19
for6 the6 number6 of6 photons6 striking6 the6 surface6 each6 second.
3.We6are6given6the6power6of6the6laser6in6milliwatts,6where616mW
=610−36W6.6The6power6may6be6expressed6as:616W6=616J/s.6Follo
6

wing6Example61.1,6the6energy6of6a6single6photon6is:
12406 eV6 ·6nm
hc6 =61.9606eV
Ephoton6 = 632.86 nm
=
λ6
6


We6 now6 convert6 to6 SI6 units6 (see6 Example6 1.5):
1.9606eV6×61.6026×610−196J/eV6 =63.146×610−196J
Following6the6 same6procedure6 as6 Problem62:
16×610−36J/s 156 photons
Rate6of6 emission6=6 = 63.196×610
3.146×610−196 J/photon6 s

, 2

4. The6maximum6kinetic6energy6of6photoelectrons6is6found6usi
ng6Eq.6(1.6)6and6the6work6functions,6W,6of6the6metals6are6given6
in6Table61.1.6Following6Problem6 1,6 Ephoton6=6hc/λ6=66.206 eV6.6 Fo
r6 part6 (a),6 Na6 has6 W6 =62.286 eV6:
(KE)max6=66.206eV6−62.286eV6 =63.926eV
Similarly,6for6Al6metal6in6part6(b),6W6 =64.086eV6 giving6(KE)max6=62.126eV
and6for6Ag6metal6in6part6(c),6W6=64.736eV6,6giving6(KE)max6=61.476eV6.

5.This6problem6again6concerns6the6photoelectric6effect.6As6in6Pro
blem64,6we6use6Eq.6(1.6):
hc6−6
(KE)max6=
W6λ
where6 W6 is6 the6 work6 function6 of6 the6 material6 and6 the6 term6 hc/λ6 d
escribes6the6energy6of6the6incoming6photons.6Solving6for6the6latter:
hc
=6(KE)max6+6W6 =62.36 eV6 +60.96 eV6 =63.26 eV
λ6
Solving6Eq.6(1.5)6for6the6wavelength:
12406 eV6 ·6nm
λ6= =6387.56nm
3.26 e
V
6. A6potential6energy6of60.726eV6is6needed6to6stop6the6flow6of6electron
s.6Hence,6(KE)max6of6the6photoelectrons6can6be6no6more6than60.726
eV.6Solving6Eq.6(1.6)6for6the6work6function:
hc 12406 eV6 ·6n —60.726 eV6 =61.986 eV
W6 =6
λ m
(KE)max6
=
4606nm
7. Reversing6 the6 procedure6 from6 Problem6 6,6 we6 start6 with6 Eq.6 (1.6):
hc6 12406 eV6 ·6n
(KE)max6= −6W6 —61.986 eV6 =63.196 eV
= m
λ
2406nm
Hence,6a6stopping6potential6of63.196eV6prohibits6the6electrons6fro
m6reaching6the6anode.

8. Just6 at6 threshold,6 the6 kinetic6 energy6 of6 the6 electron6 is6 zer
o.6 Setting6(KE)max6=606 in6 Eq.6 (1.6),
hc
W6= = 12406 eV6 ·6n =63.446 eV
λ0 m

3606nm
9. A6frequency6of612006THz6is6equal6to612006×610126Hz.6Using6Eq.6(1.10),