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SOLUTION MANUAL Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by Morrison Chapters 1- 15 $18.09
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1. Themenergymofmphotonsminmtermsmofmthemwavelengthmofmlight
mismgivenmbymEq.m(1.5).mFollowingmExamplem 1.1mandmsubstituti
ngmλm=m200meVmgives:
hc 1240m eVm ·mnm
= =m6.2meV
Ephotonm= λ 200mnm
2. Them energym ofm them beamm eachm secondm is:
power 100m W
= =m100mJ
Etotalm= time 1m s
Themnumbermofmphotonsmcomesmfrommthemtotalmenergymdivided
mbymthemenergy mofmeachmphotonm(see mProblemm1).mThemphoton’s
2m×m10−19mJ/eVm,mseemExamplem1.5.mThemresultmis:
mEtotal m
N = = 100mJ =m1.01m×m1020
photons E
pho
ton 9.93m×m10−19
form them numberm ofm photonsm strikingm them surfacem eachm second.
3. Wemaremgivenmthempowermofmthemlaserminmmilliwatts,mwherem1m
mWm=m10−3mWm.mThempowermmaymbemexpressedmas:m1mWm=m1mJ
/s.mFollowingmExamplem1.1,mthemenergymofmamsinglemphotonmi
s:
1240m eVm ·mnm
hcm =m1.960meV
Ephotonm = 632.8m nm
=
λm
m
Wem nowm convertm tom SIm unitsm (seem Examplem 1.5):
1.960meVm×m1.602m×m10−19mJ/eVm =m3.14m×m10−19mJ
Followingm them samem procedurem asm Problemm 2:
1m×m10−3mJ/s 15m photons
Ratemofm emissionm=m = m3.19m× m10
3.14m×m10−19m J/photonm s
, 2
4. Themmaximummkineticmenergymofmphotoelectronsmismfoundmu
singmEq.m(1.6)mandmthemworkmfunctions,mW,mofmthemmetalsmarem
givenminmTablem1.1.mFollowingmProblemm 1,m Ephotonm=mhc/λm=m6.
20m eVm.m Form partm (a),m Nam hasm Wm =m2.28m eVm:
(KE)maxm=m6.20meVm−m2.28meVm =m3.92meV
Similarly,m form Alm metalm inm partm (b),m Wm =m4.08m eVm givingm(KE)maxm=m2.12
m eV
5.Thismproblemmagainmconcernsmthemphotoelectricmeffect.mAsminmP
roblemm4,mwemusemEq.m(1.6):
hcm−m
(KE)maxm =
Wmλ
wherem Wm ism them workm functionm ofm them materialm andm them termm hc
/λm describesmthemenergymofmthemincomingmphotons.mSolvingmformthem
latter:
hc
=m(KE)maxm+mWm =m2.3m eVm +m0.9m eVm =m3.2m eV
λm
Solvingm Eq.m (1.5)m form them wavelength:
1240m eVm ·mnm
λm= =m387.5mnm
3.2m e
V
6. Ampotentialmenergymofm0.72meVmismneededmtomstopmthemflowmofmelect
rons.mHence,m(KE)maxmofmthemphotoelectronsmcanmbemnommoremtha
nm0.72meV.mSolvingmEq.m(1.6)mformthemworkmfunction:
hc 1240m eVm ·m —m0.72m eVm =m1.98m eV
W m =m —
λ nm
(KE)maxm
=
460mnm
7. Reversingm them procedurem fromm Problemm 6,m wem startm withm Eq.m (1.6):
hcm 1240m eVm ·m
(KE)maxm = −mWm —m1.98m eVm =m3.19m eV
= nm
λ
240mnm
Hence,mamstoppingmpotentialmofm3.19meVmprohibitsmthemelectronsmfr
ommreachingmthemanode.
8. Justm atm threshold,m them kineticm energym ofm them electronm ism
zero.m Settingm(KE)maxm=m0m inm Eq.m (1.6),
hc
Wm= = 1240m eVm ·m =m3.44m eV
λ0 nm
360mnm
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