All 20 Chapters Covered
h h h
SOLUTION MANUAL
h
, Chapterh 1h Solutions
Radiationh Sources
■ Problemh1.1.h RadiationhEnergyhSpectra:hLinehvs.hContinuous
Lineh(orhdiscretehenergy):ha,hc,hd,he,hf,handhi.hContinuo
ushenergy:hb,hg,handhh.
■ Problemh1.2.h Conversionhelectronhenergieshcompared.
Sincehthehelectronshinhouterhshellshareh boundhlesshtightlyhthanhthosehinhcloserhshells,hconversionhelectronshfromhouterhshellshwillh ha
vehgreaterhemerginghenergies.h Thus,hthehMhshellhelectronhwillhemergehwithhgreaterhenergyhthanhahKhorhLhshellhelectron.
■ Problemh1.3.h Nuclearhdecayhandhpredictedhenergies.
Wehwritehthehconservationhofhenergyhandhmomentumhequationshandhsolvehthemhforhthehenergyhofhthehalphahparticle.h Momentumhis
hgivenhthehsymbolh"p",h andhenergyhish"E".h Forhthehsubscripts,h"al"hstandshforhalpha,hwhileh"b"hdenoteshthehdaughterhnucleus.
pal2 pb2
palhh pbhhh0 E
h al hEb Ealhh Ebh hQ and Qhh5.5hMeV
2hmal 2hmb
Solvinghourhsystemhofhequationshforh Eal,h Eb,h pal,h pb,h wehgeththehsolutionshshownhbelow.h Notehthathwehhavehtwohpossiblehsetshofhsolutio
nsh(thishdoeshnotheffecththehfinalhresult).
mal 5.5hmalh
Ebh h5.5h 1h Ealh
malhhmb malhhmb
3.31662 mal mb 3.31662 mal mb
palhh pbhh
malhhmb malhhmb
Weharehinterestedhinhfindinghthehenergyhofhthehalphahparticlehinhthishproblem,handhsincehwehknowhthehmasshofhthehalphahparticlehan
dhtheh daughterh nucleus,h theh resulth ish easilyh found.h Byhsubstitutinghourh knownhvaluesh ofh malhh4h andh mbhh206h intoh ourh derivedhEa
lequationhwehget:
Ealhh5.395hMeV
Noteh:hWehcanhobtainhsolutionshforhallhthehvariableshbyhsubstitutinghmbhh206handhmalhh4hintohthehderivedhequationshaboveh:
Ealhh5.395hMeV Ebhh0.105hMeV palhhh6.570 amuhMeV pbh hh6.570 amuhMeV
■ Problemh1.4.h Calculationhofh WavelengthhfromhEnergy.
Sincehanhx-rayhmusthessentiallyhbehcreatedhbyhthehde-excitationhofhahsinglehelectron,hthehmaximumhenergyhofhanhx-
rayhemittedhinhahtubehoperatinghathahpotentialhofh195hkVhmusthbeh195hkeV.h Therefore,h wehcanhusehthehequationhE=h,hwhichhishalso
hE=hc/Λ,horhΛ=hc/E.h Plugginghinhourhmaximumhenergyh valuehintohthishequationhgiveshthehminimumhx-rayhwavelength.
hhhc
Λh wherehwehsubstitutehhhh 6.626hh1034hJhhs,h chh299h792h458hmhhshandhEhh195hkeV
E
1
, Chapterh 1h Solutions
1.01869hJ–m
Λh h 0.0636hAngstroms
KeV
■ Problemh 1.5.h h 235hUFissionh EnergyhRelease.
235 117
Usinghtheh reactionh h Uh h h h Snhh118h Sn,h andh massh values,h weh calculateh theh masshdefecth of:
Mh235hUhh h Mh117hSnhhMh118hSnhh M
handhanhexpectedhenerg
yhreleasehofhMc2.
931.5hMeV
hh hh hh h223h MeV
AMU
Thish ish onehofhtheh mosthexothermichreactionsh availableh tohus.h Thish ish onehreasonh why,hofhcourse,h nuclearh powerhfromhuraniumhfis
sionhishsohattractive.
■ Problemh1.6.h SpecifichActivityhofh Tritium.
Here,hwehusehthehtexthequationhSpecifichActivityh=h(ln(2)*Av)/hT12*M),hwherehAvhishAvogadro'shnumber,hT12hishthehhalf-
lifehofhthehisotope,handhMhishthehmolecularhweighthofhthehsample.
ln2hAvogadroh'hshConstant
SpecifichActivityh
T12hM
3hgrams
WehsubstitutehT12hh12.26hyearshandhM= tohgeththehspecifich activityhinhdisintegrations/(gram–year).
mole
1.13492hh1022
SpecifichActivityh
gramh–year
Theh samehresulth expressedh inhtermshofhkCi/ghishshownhbelow
9.73hkCi
SpecifichActivityh
gram
■ Problemh1.7.h Acceleratedhparticlehenergy.
Theh energyhofhah particleh withhchargeh qhfallinghthroughhah potentialhVh ishqV.h Sinceh V=h 3hMVh ishourhmaximumhpotentialhdifference,h theh
maximumhenergyhofh anh alphah particleh hereh ish q*(3h MV),h whereh qh ish theh chargeh ofh theh alphah particleh (+2).h Thehmaximumhalphahparticl
ehenergyhexpressedhinhMeVhishthus:
Energyhh3hMegahVoltshh2hElectronhChargesh h6.h MeV
2
, Chapterh 1h Solutions
■ Problemh1.8.h Photofissionhofh deuterium. 1hDh h Γh
2 1
0hnh
1
1hph+h Qh (-2.226hMeV)
Theh reactionh ofh interesth ish h 2hDh h h 0hΓh h 1hnhh h 1h p+h Qh (-2.226h MeV).h Thus,h theh Γhmusthbringh anh energyhofh ath leasth 2.226h MeV
1 0 0 1
inhorderhforhthishendothermichreactionhtohproceed.h Interestingly,hthehoppositehreactionhwillhbehexothermic,handhonehcanhexpecthtohfi
ndh2.226hMeVhgammahrayshinhthehenvironmenthfromhstrayhneutronshbeinghabsorbedhbyhhydrogenhnuclei.
■ Problemh1.9.h NeutronhenergyhfromhD-Threactionhbyh150hkeVhdeuterons.
Wehwritehdownhthehconservationhofhenergyhandhmomentumhequations,handhsolvehthemhforhthehdesiredhenergieshbyheliminatinghtheh
momenta.h Inhthishsolution,h"a"hrepresentshthehalphahparticle,h"n"hrepresentshthehneutron,handh"d"hrepresentshthehdeuteronh(and,hashb
efore,h"p"hrepresentshmomentum,h"E"hrepresentshenergy,handh"Q"hrepresentshthehQ-valuehofhthehreaction).
pa2 pn2 pdh2
pahh pnhh pd E
h a E
h n Ed
h EahhEnhhEdh hQ
2hma 2hmn 2hmd
Nexth weh wanth toh solvehtheh aboveh equationsh forh theh unknownhenergiesh byheliminatinghtheh momenta.h (Noteh :h Usingh computerhs
oftwarehsuchhashMathematicahishhelpfulhforhpainlesslyhsolvinghthesehequations).
Weh evaluateh theh solutionhbyhplugginghinh theh valueshforh particleh massesh(weh useh approximatehvalueshofh "ma,"h "mn,"andh "md"h inh
AMU,hwhichhishokayhbecausehweharehinterestedhinhobtaininghanhenergyhvaluehaththehend).h WehdefinehallhenergieshinhunitshofhMeV,hn
amelyhtheh Q-value,h andh theh givenhenergyhofh theh deuteronh (bothh energyhvaluesh areh inh MeV).h h Sohweh substituteh mah =h 4,h mnh =h 1,h md
=h2,hQh=h17.6,hEdh =h0.15hintohourhmomentahindependenthequations.h Thishyieldshtwohpossiblehsetshofhsolutionshforhthehenergiesh(inh
MeV).hOnehcorrespondshtohthehneutronhmovinghinhthehforwardhdirection,hwhichhishofhinterest.
Enhh 13.340h MeV Eahh 4.410h MeV
Enhh 14.988h MeV Eahh 2.762h MeV
Nexthwehsolvehforhthehmomentahbyheliminatinghthehenergies.hWhenhwehsubstitutehmah =h4,hmnh =h1,hmdh =h2,hQh=h17.6,hEdh =h0.15hinto
hthesehequationshwehgeththehfollowinghresults.
pd 1 1
pnh h 2 3hpdh2h h352 pah 8hpdh h2 2 3hpdh2h h352
5 5 10
Weh doh knowhtheh initialh momentumhofh theh deuteron,h however,hsinceh weh knowhitsh energy.hWeh canh furtherh evaluateh ourh solutionshfor
pnhandh pahbyhsubstituting:
pdh
Theh particleh momentah(h inhunitshof amuMeV )hforheachh sethofhsolutionshishthus:
pnhh 5.165 pahh 5.940
pnhh 5.475 pahh 4.700
Theh largesthneutronhmomentumhoccurshinhthehforwardh(+)hdirection,hsohthehhighesthneutronhenergyhofh14.98hMeVhcorrespondsh
tohthishdirection.
3
h h h
SOLUTION MANUAL
h
, Chapterh 1h Solutions
Radiationh Sources
■ Problemh1.1.h RadiationhEnergyhSpectra:hLinehvs.hContinuous
Lineh(orhdiscretehenergy):ha,hc,hd,he,hf,handhi.hContinuo
ushenergy:hb,hg,handhh.
■ Problemh1.2.h Conversionhelectronhenergieshcompared.
Sincehthehelectronshinhouterhshellshareh boundhlesshtightlyhthanhthosehinhcloserhshells,hconversionhelectronshfromhouterhshellshwillh ha
vehgreaterhemerginghenergies.h Thus,hthehMhshellhelectronhwillhemergehwithhgreaterhenergyhthanhahKhorhLhshellhelectron.
■ Problemh1.3.h Nuclearhdecayhandhpredictedhenergies.
Wehwritehthehconservationhofhenergyhandhmomentumhequationshandhsolvehthemhforhthehenergyhofhthehalphahparticle.h Momentumhis
hgivenhthehsymbolh"p",h andhenergyhish"E".h Forhthehsubscripts,h"al"hstandshforhalpha,hwhileh"b"hdenoteshthehdaughterhnucleus.
pal2 pb2
palhh pbhhh0 E
h al hEb Ealhh Ebh hQ and Qhh5.5hMeV
2hmal 2hmb
Solvinghourhsystemhofhequationshforh Eal,h Eb,h pal,h pb,h wehgeththehsolutionshshownhbelow.h Notehthathwehhavehtwohpossiblehsetshofhsolutio
nsh(thishdoeshnotheffecththehfinalhresult).
mal 5.5hmalh
Ebh h5.5h 1h Ealh
malhhmb malhhmb
3.31662 mal mb 3.31662 mal mb
palhh pbhh
malhhmb malhhmb
Weharehinterestedhinhfindinghthehenergyhofhthehalphahparticlehinhthishproblem,handhsincehwehknowhthehmasshofhthehalphahparticlehan
dhtheh daughterh nucleus,h theh resulth ish easilyh found.h Byhsubstitutinghourh knownhvaluesh ofh malhh4h andh mbhh206h intoh ourh derivedhEa
lequationhwehget:
Ealhh5.395hMeV
Noteh:hWehcanhobtainhsolutionshforhallhthehvariableshbyhsubstitutinghmbhh206handhmalhh4hintohthehderivedhequationshaboveh:
Ealhh5.395hMeV Ebhh0.105hMeV palhhh6.570 amuhMeV pbh hh6.570 amuhMeV
■ Problemh1.4.h Calculationhofh WavelengthhfromhEnergy.
Sincehanhx-rayhmusthessentiallyhbehcreatedhbyhthehde-excitationhofhahsinglehelectron,hthehmaximumhenergyhofhanhx-
rayhemittedhinhahtubehoperatinghathahpotentialhofh195hkVhmusthbeh195hkeV.h Therefore,h wehcanhusehthehequationhE=h,hwhichhishalso
hE=hc/Λ,horhΛ=hc/E.h Plugginghinhourhmaximumhenergyh valuehintohthishequationhgiveshthehminimumhx-rayhwavelength.
hhhc
Λh wherehwehsubstitutehhhh 6.626hh1034hJhhs,h chh299h792h458hmhhshandhEhh195hkeV
E
1
, Chapterh 1h Solutions
1.01869hJ–m
Λh h 0.0636hAngstroms
KeV
■ Problemh 1.5.h h 235hUFissionh EnergyhRelease.
235 117
Usinghtheh reactionh h Uh h h h Snhh118h Sn,h andh massh values,h weh calculateh theh masshdefecth of:
Mh235hUhh h Mh117hSnhhMh118hSnhh M
handhanhexpectedhenerg
yhreleasehofhMc2.
931.5hMeV
hh hh hh h223h MeV
AMU
Thish ish onehofhtheh mosthexothermichreactionsh availableh tohus.h Thish ish onehreasonh why,hofhcourse,h nuclearh powerhfromhuraniumhfis
sionhishsohattractive.
■ Problemh1.6.h SpecifichActivityhofh Tritium.
Here,hwehusehthehtexthequationhSpecifichActivityh=h(ln(2)*Av)/hT12*M),hwherehAvhishAvogadro'shnumber,hT12hishthehhalf-
lifehofhthehisotope,handhMhishthehmolecularhweighthofhthehsample.
ln2hAvogadroh'hshConstant
SpecifichActivityh
T12hM
3hgrams
WehsubstitutehT12hh12.26hyearshandhM= tohgeththehspecifich activityhinhdisintegrations/(gram–year).
mole
1.13492hh1022
SpecifichActivityh
gramh–year
Theh samehresulth expressedh inhtermshofhkCi/ghishshownhbelow
9.73hkCi
SpecifichActivityh
gram
■ Problemh1.7.h Acceleratedhparticlehenergy.
Theh energyhofhah particleh withhchargeh qhfallinghthroughhah potentialhVh ishqV.h Sinceh V=h 3hMVh ishourhmaximumhpotentialhdifference,h theh
maximumhenergyhofh anh alphah particleh hereh ish q*(3h MV),h whereh qh ish theh chargeh ofh theh alphah particleh (+2).h Thehmaximumhalphahparticl
ehenergyhexpressedhinhMeVhishthus:
Energyhh3hMegahVoltshh2hElectronhChargesh h6.h MeV
2
, Chapterh 1h Solutions
■ Problemh1.8.h Photofissionhofh deuterium. 1hDh h Γh
2 1
0hnh
1
1hph+h Qh (-2.226hMeV)
Theh reactionh ofh interesth ish h 2hDh h h 0hΓh h 1hnhh h 1h p+h Qh (-2.226h MeV).h Thus,h theh Γhmusthbringh anh energyhofh ath leasth 2.226h MeV
1 0 0 1
inhorderhforhthishendothermichreactionhtohproceed.h Interestingly,hthehoppositehreactionhwillhbehexothermic,handhonehcanhexpecthtohfi
ndh2.226hMeVhgammahrayshinhthehenvironmenthfromhstrayhneutronshbeinghabsorbedhbyhhydrogenhnuclei.
■ Problemh1.9.h NeutronhenergyhfromhD-Threactionhbyh150hkeVhdeuterons.
Wehwritehdownhthehconservationhofhenergyhandhmomentumhequations,handhsolvehthemhforhthehdesiredhenergieshbyheliminatinghtheh
momenta.h Inhthishsolution,h"a"hrepresentshthehalphahparticle,h"n"hrepresentshthehneutron,handh"d"hrepresentshthehdeuteronh(and,hashb
efore,h"p"hrepresentshmomentum,h"E"hrepresentshenergy,handh"Q"hrepresentshthehQ-valuehofhthehreaction).
pa2 pn2 pdh2
pahh pnhh pd E
h a E
h n Ed
h EahhEnhhEdh hQ
2hma 2hmn 2hmd
Nexth weh wanth toh solvehtheh aboveh equationsh forh theh unknownhenergiesh byheliminatinghtheh momenta.h (Noteh :h Usingh computerhs
oftwarehsuchhashMathematicahishhelpfulhforhpainlesslyhsolvinghthesehequations).
Weh evaluateh theh solutionhbyhplugginghinh theh valueshforh particleh massesh(weh useh approximatehvalueshofh "ma,"h "mn,"andh "md"h inh
AMU,hwhichhishokayhbecausehweharehinterestedhinhobtaininghanhenergyhvaluehaththehend).h WehdefinehallhenergieshinhunitshofhMeV,hn
amelyhtheh Q-value,h andh theh givenhenergyhofh theh deuteronh (bothh energyhvaluesh areh inh MeV).h h Sohweh substituteh mah =h 4,h mnh =h 1,h md
=h2,hQh=h17.6,hEdh =h0.15hintohourhmomentahindependenthequations.h Thishyieldshtwohpossiblehsetshofhsolutionshforhthehenergiesh(inh
MeV).hOnehcorrespondshtohthehneutronhmovinghinhthehforwardhdirection,hwhichhishofhinterest.
Enhh 13.340h MeV Eahh 4.410h MeV
Enhh 14.988h MeV Eahh 2.762h MeV
Nexthwehsolvehforhthehmomentahbyheliminatinghthehenergies.hWhenhwehsubstitutehmah =h4,hmnh =h1,hmdh =h2,hQh=h17.6,hEdh =h0.15hinto
hthesehequationshwehgeththehfollowinghresults.
pd 1 1
pnh h 2 3hpdh2h h352 pah 8hpdh h2 2 3hpdh2h h352
5 5 10
Weh doh knowhtheh initialh momentumhofh theh deuteron,h however,hsinceh weh knowhitsh energy.hWeh canh furtherh evaluateh ourh solutionshfor
pnhandh pahbyhsubstituting:
pdh
Theh particleh momentah(h inhunitshof amuMeV )hforheachh sethofhsolutionshishthus:
pnhh 5.165 pahh 5.940
pnhh 5.475 pahh 4.700
Theh largesthneutronhmomentumhoccurshinhthehforwardh(+)hdirection,hsohthehhighesthneutronhenergyhofh14.98hMeVhcorrespondsh
tohthishdirection.
3
