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

SOLUTION MANUAL Modern Physics with Modern
s s s s s




s Computational Methods: for Scientists and
s s s s




s Engineers 3rd Edition by Morrison Chapters 1- 15
s s s s s s s

,Table of contents
s s



1. sThe sWave-Particle sDuality

2. sThe sSchrödinger sWave sEquation

3. sOperators sand sWaves

4. sThe sHydrogen sAtom

5. sMany-Electron sAtoms

6. sThe sEmergence sof sMasers sand sLasers

7. sDiatomic sMolecules

8. sStatistical sPhysics

9. sElectronic sStructure sof sSolids

10. sCharge sCarriers sin sSemiconductors

11. sSemiconductor sLasers

12. sThe sSpecial sTheory sof sRelativity

13. sThe sRelativistic sWave sEquations sand sGeneral sRelativity

14. sParticle sPhysics

15. sNuclear sPhysics

,1

The s Wave-Particle s Duality s - s Solutions




1. The senergy sof sphotons sin sterms sof sthe swavelength sof slight
sis sgiven sby sEq. s(1.5). sFollowing sExample s 1.1 sand
ssubstituting sλ s= s200 seV sgives:

hc 1240 s eV s · snm
= = s6.2 seV
Ephoton s= λ 200 snm
2. The s energy s of s the s beam s each s second s is:
power 100 s W
= = s100 sJ
Etotal s= time 1 ss
The snumber sof sphotons scomes sfrom sthe stotal senergy sdivided
sby sthe senergy sof seach sphoton s(see sProblem s1). sThe sphoton’s

senergy smust sbe sconverted sto sJoules susing sthe sconstant s1.602

s× s10−19 sJ/eV s, ssee sExample s1.5. sThe sresult sis:


N =sEtotal s = 100 sJ = s1.01 s× s1020
photons E
pho
ton 9.93 s× s10−19
for s the s number s of s photons s striking s the s surface s each s second.
3. We sare sgiven sthe spower sof sthe slaser sin smilliwatts, swhere s1
smW s= s10−3 sW s. sThe spower smay sbe sexpressed sas: s1 sW s= s1
sJ/s. sFollowing sExample s1.1, sthe senergy sof sa ssingle sphoton

sis:

1240 s eV s · snm
hc = s1.960 seV
sEphoton s = 632.8 s nm
s =
λs
We s now s convert s to s SI s units s (see s Example s 1.5):
1.960 seV s× s1.602 s× s10−19 sJ/eV s = s3.14 s× s10−19 sJ
Following sthe ssame sprocedure sas sProblem s2:
1 s× s10−3 sJ/s 15 s photons
Rate sof s emission s= s = s3.19 s× s10
3.14 s× s10−19 s J/photon s s

, 2

4. The smaximum skinetic senergy sof sphotoelectrons sis sfound
susing sEq. s(1.6) sand sthe swork sfunctions, sW, sof sthe smetals sare

sgiven sin sTable s1.1. sFollowing sProblem s 1, s E photon s= shc/λ s=

s6.20 s eV s. s For s part s (a), s Na s has s W s = s2.28 s eV s:


(KE)max s= s6.20 seV s− s2.28 seV s = s3.92 seV
Similarly, sfor sAl smetal sin spart s(b), sW s = s4.08 seV s giving s(KE)max s=
s2.12 s eV

and sfor sAg smetal sin spart s(c), sW s = s4.73 seV s, sgiving s(KE)max s= s1.47 seV s.

This sproblem sagain sconcerns sthe sphotoelectric seffect. sAs sin
5.
Problem s4, swe suse sEq. s(1.6):
s


hc s−
(KE)max s =
λ
sW s

where s W s is s the s work s function s of s the s material s and s the s term
s hc/λ s describes sthe senergy sof sthe sincoming sphotons. sSolving sfor

sthe slatter:

hc
= s(KE)max s+ sW s = s2.3 s eV s + s0.9 seV s = s3.2 seV
λs
Solving sEq. s (1.5) sfor s the swavelength:
1240 s eV s · snm
λ s= = s387.5 snm
3.2
s eV

6. A spotential senergy sof s0.72 seV sis sneeded sto sstop sthe sflow sof
selectrons. sHence, s(KE)max sof sthe sphotoelectrons scan sbe sno smore

sthan s0.72 seV. sSolving sEq. s(1.6) sfor sthe swork sfunction:

hc 1240 s eV s · — s0.72 s eV s = s1.98 s eV
W s= s — (KE)max
λ snm
s =


460 snm
7. Reversing s the s procedure s from s Problem s 6, s we s start s with s Eq. s (1.6):
hc s
(KE)max s = − sW 1240 s eV s · — s1.98 s eV s = s3.19 s eV
snm
s=
λ
240 snm
Hence, sa sstopping spotential sof s3.19 seV sprohibits sthe selectrons
sfrom sreaching sthe sanode.



8. Just s at s threshold, s the s kinetic s energy s of s the s electron
s is s zero. s Setting s(KE)max s= s0 s in s Eq. s (1.6),
hc
W s= = 1240 s eV s · = s3.44 s eV
λ0 snm



360 snm