All 20 Chapters Covered
gt gt gt
SOLUTION MANUAL
gt
, Chaptergt 1gt Solution
s
Radiationg t Sources
■ Problemgt1.1.g t RadiationgtEnergygtSpectra:gtLinegtvs.gtContinuous
Linegt(orgtdiscretegtenergy):gta,gtc,gtd,gte,gtf,gtandgti.
gtContinuousgtenergy:gtb,gtg,gtandgth.
■ Problemgt1.2.g t Conversiongtelectrongtenergiesgtcompared.
Sincegtthegtelectronsgtingtoutergtshellsgtareg t boundgtlessgttightlygtthangtthosegtingtclosergtshells,gtconversiongtelectronsgtfromgtoutergtshellsgt
willg t havegtgreatergtemerginggtenergies.g t Thus,gtthegtMgtshellgtelectrongtwillgtemergegtwithgtgreatergtenergygtthangtagtKgtorgtLgtshellgtelec
tron.
■ Problemgt1.3.g t Nucleargtdecaygtandgtpredictedgtenergies.
Wegtwritegtthegtconservationgtofgtenergygtandgtmomentumgtequationsgtandgtsolvegtthemgtforgtthegtenergygtofgtthegtalphagtparticle.g t Mome
ntumgtisgtgivengtthegtsymbolgt"p",g t andgtenergygtisgt"E".g t Forgtthegtsubscripts,gt"al"gtstandsgtforgtalpha,gtwhilegt"b"gtdenotesgtthegtdaughte
rgtnucleus.
pal2 pb2
palgtgtpbgt gt0 E
gt al E
gt b Ealgtg t Ebgt gtQ and Qgtgt5.5gtMeV
2gtma 2gtm
b
l
Solvinggtourgtsystemgtofgtequationsgtforgt Eal,g t Eb,g t pal,g t pb,g t wegtgetgtthegtsolutionsgtshowngtbelow.g t Notegtthatgtwegthavegttwogtpossiblegts
etsgtofgtsolutionsgt(thisgtdoesgtnotgteffectgtthegtfinalgtresult).
mal 5.5gtmalg
Ebg t gt5.5g t 1gt Ealgt
gt t malgtgtmb
3.31662 mal mb 3.31662 mal mb
palgtgt pbg t gt
gt gt
Wegtaregtinterestedgtingtfindinggtthegtenergygtofgtthegtalphagtparticlegtingtthisgtproblem,gtandgtsincegtwegtknowgtthegtmassgtofgtthegtalphagtpar
ticlegtandgttheg t daughterg t nucleus,g t theg t resultg t isg t easilygtfound.g t Bygtsubstitutinggtourgtknowngtvaluesg t ofg t malgtgt4g t andg t mbgtgt206g t i
ntogtourgtderivedgtEalequationgtwegtget:
Ealgtgt5.395gtMeV
Notegt:gtWegtcangtobtaingtsolutionsgtforgtallgtthegtvariablesgtbygtsubstitutinggtmbgtgt206gtandgtmalgtgt4gtintogtthegtderivedgtequationsgtabovegt:
Ealgtgt5.395gtMeV Ebgtgt0.105gtMeV palgtgtgt6.570 amugtMeV pbgt gtgt6.570 amugtMeV
■ Problemgt1.4.g t Calculationgtofgt WavelengthgtfromgtEnergy.
Sincegtangtx-raygtmustgtessentiallygtbegtcreatedgtbygtthegtde-excitationgtofgtagtsinglegtelectron,gtthegtmaximumgtenergygtofgtangtx-
raygtemittedgtingtagttubegtoperatinggtatgtagtpotentialgtofgt195gtkVgtmustgtbegt195gtkeV.g t Therefore,g t wegtcangtusegtthegtequationgtE=h,gtwhic
hgtisgtalsogtE=hc/Λ,gtorgtΛ=hc/E.g t Plugginggtingtourgtmaximumgtenergygtvaluegtintogtthisgtequationgtgivesgtthegtminimumgtx-
raygtwavelength.
hgtgtc
Λgt wheregtwegtsubstitutegthgtg t 6.626gtgt1034gtJgtgts,g t cgtgt299gt792gt458gtmgtgtsgtandgtEgtgt195gtkeV
E
1
, Chaptergt 1gt Solution
s
1.01869gtJ–m
gt gt 0.0636gtAngstroms
KeV
■ Problemgt 1.5.g t g t 235gtUFissiongt EnergygtRelease.
235 117
Usinggtthegtreactiong t gt Ug t g t g t gt Sngtgt118gtSn,g t andgtmassgtvalues,gtwegtcalculategt thegtmassgtdefectgt of:
Mgt235gtUgtgtgt Mgt117gtSngtgtMgt118gtSngtgt Mgtandgtangtexpectedgte
nergygtreleasegtofgtMc2.
931.5gtMeV
gtgt g t gt g t gt gt
AMU
Thisg t isg t onegtofgttheg t mostgtexothermicgtreactionsg t availableg t togtus.g t Thisg t isgt onegtreasong t why,gtofgtcourse,g t nuclearg t powergtfromgt
uraniumgtfissiongtisgtsogtattractive.
■ Problemgt1.6.g t SpecificgtActivitygtofgt Tritium.
Here,gtwegtusegtthegttextgtequationgtSpecificgtActivitygt=gt(ln(2)*Av)/gtT12*M),gtwheregtAvgtisgtAvogadro'sgtnumber,gtT12gtisgtthegthalf-
lifegtofgtthegtisotope,gtandgtMgtisgtthegtmoleculargtweightgtofgtthegtsample.
ln2gtAvogadrogt'gtsgtConstant
SpecificgtActivitygt
T12gtM
3gtgrams
WegtsubstitutegtT12gtgt12.26gtyearsgtandgtM= togtgetgtthegtspecificgt activitygtingtdisintegrations/(gram–year).
mole
1.13492gtgt1022
SpecificgtActivitygt
gramgt–year
Thegt samegtresultgtexpressedgt ingttermsgtofgtkCi/ggtisgtshowngtbelow
9.73gtkCi
SpecificgtActivitygt
gram
■ Problemgt1.7.g t Acceleratedgtparticlegtenergy.
ThegtenergygtofgtagtparticlegtwithgtchargegtqgtfallinggtthroughgtagtpotentialgtVgtisgtqV.g t SincegtV=gt3gtMVgtisgtourgtmaximumgtpotentialgtdi
fference,g t theg t maximumgtenergygtofg t ang t alphag t particleg t hereg t isgt q*(3gt MV),g t whereg t qg t isgt thegt chargeg t ofg t theg t alphag t particleg t (+
2).g t ThegtmaximumgtalphagtparticlegtenergygtexpressedgtingtMeVgtisgtthus:
Energygtgt3gtMegagtVoltsgtgt2gtElectrongtChargesg 6.g t MeV
t gt
2
, Chaptergt 1gt Solution
s
2 1 1
■ Problemgt1.8.g t Photofissiongtofgt deuteriu gt Dgtg t Γg t gt ngt gtpgt+gt Qgt(-2.226gtMeV)
m. 1 0 1
Thegt reactiongtofgtinterestgtisg t g t gtDg t g t gtΓg t g t gtngtg t g t p+gtQgt(-
2 0 1 1
2.226gtMeV).g t Thus,gtthegtΓgtmustgtbringgtangtenergygtofgtatgtleastgt2.226gtMeV
1 0 0 1
ingtordergtforgtthisgtendothermicgtreactiongttogtproceed.g t Interestingly,gtthegtoppositegtreactiongtwillgtbegtexothermic,gtandgtonegtcangtexp
ectgttogtfindgt2.226gtMeVgtgammagtraysgtingtthegtenvironmentgtfromgtstraygtneutronsgtbeinggtabsorbedgtbygthydrogengtnuclei.
■ Problemgt1.9.g t NeutrongtenergygtfromgtD-Tgtreactiongtbygt150gtkeVgtdeuterons.
Wegtwritegtdowngtthegtconservationgtofgtenergygtandgtmomentumgtequations,gtandgtsolvegtthemgtforgtthegtdesiredgtenergiesgtbygteliminatin
ggtthegtmomenta.g t Ingtthisgtsolution,gt"a"gtrepresentsgtthegtalphagtparticle,gt"n"gtrepresentsgtthegtneutron,gtandgt"d"gtrepresentsgtthegtdeuter
ongt(and,gtasgtbefore,gt"p"gtrepresentsgtmomentum,gt"E"gtrepresentsgtenergy,gtandgt"Q"gtrepresentsgtthegtQ-valuegtofgtthegtreaction).
pa2 pn2 pdgt2
pagtgtpngtgtpd E
gt a E
gt n E
gt d EagtgtEngtgtEdg t gtQ
2gtm 2gtm 2gtm
d
a n
Nextg t weg t wantg t tog t solvegttheg t aboveg t equationsg t forg t theg t unknowngtenergiesg t bygteliminatinggttheg t momenta.g t (Noteg t :g t Usingg t
computergtsoftwaregtsuchgtasgtMathematicagtisgthelpfulgtforgtpainlesslygtsolvinggtthesegtequations).
Weg t evaluateg t theg t solutiongtbygtplugginggting t theg t valuesgtforg t particleg t massesgt(weg t useg t approximategtvaluesgtofg t "ma,"g t "mn,"an
dg t "md"g t ingtAMU,gtwhichgtisgtokaygtbecausegtwegtaregtinterestedgtingtobtaininggtangtenergygtvaluegtatgtthegtend).g t Wegtdefinegtallgtenergie
sgtingtunitsgtofgtMeV,gtnamelygtthegt Q-
value,gtandgt thegtgivengtenergygtofgtthegt deuterongt (bothgtenergygtvaluesgt aregt ingt MeV).g t g t Sogtwegtsubstitutegtmag t =gt 4,gt mng t =gt1,gtmd
=gt2,gtQgt=gt17.6,gtEdg t =gt0.15gtintogtourgtmomentagtindependentgtequations.g t Thisgtyieldsgttwogtpossiblegtsetsgtofgtsolutionsgtforgtthegtener
giesgt(ingtMeV).gtOnegtcorrespondsgttogtthegtneutrongtmovinggtingtthegtforwardgtdirection,gtwhichgtisgtofgtinterest.
Engtg t 13.340gt MeV Eagtg t 4.410gt MeV
Engtg t 14.988gt MeV Eagtg t 2.762gt MeV
Nextgtwegtsolvegtforgtthegtmomentagtbygteliminatinggtthegtenergies.gtWhengtwegtsubstitutegtmag t =gt4,gtmng t =gt1,gtmdg t =gt2,gtQgt=gt17.6,gtEdg t
=gt0.15gtintogtthesegtequationsgtwegtgetgtthegtfollowinggtresults.
pd 1 1
pngt gt 2 3gtpdgt2gt gt352 pagt 8gtpdgt gt2 2 3gtpdgt2gtgt352
5 5 10
Wegtdogtknowgtthegt initialgtmomentumgtofgtthegt deuteron,gt however,gtsincegt wegtknowgtitsgtenergy.gtWegtcangt furthergt evaluategt ourgtsolutions
gtfor
pngt andg t pagtbygtsubstituting:
p dg t gtgt gtgt
Thegt particlegt momentagt(gtingtunitsgtof amuMeV )gtforgteachgt setgtofgtsolutionsgtisgtthus:
pngtg t 5.165 pagtg t 5.940
pngtg t 5.475 pagtg t 4.700
Theg t largestgtneutrongtmomentumgtoccursgtingtthegtforwardgt(+)gtdirection,gtsogtthegthighestgtneutrongtenergygtofgt14.98gtMeVgtcorres
pondsg t togtthisgtdirection.
3
gt gt gt
SOLUTION MANUAL
gt
, Chaptergt 1gt Solution
s
Radiationg t Sources
■ Problemgt1.1.g t RadiationgtEnergygtSpectra:gtLinegtvs.gtContinuous
Linegt(orgtdiscretegtenergy):gta,gtc,gtd,gte,gtf,gtandgti.
gtContinuousgtenergy:gtb,gtg,gtandgth.
■ Problemgt1.2.g t Conversiongtelectrongtenergiesgtcompared.
Sincegtthegtelectronsgtingtoutergtshellsgtareg t boundgtlessgttightlygtthangtthosegtingtclosergtshells,gtconversiongtelectronsgtfromgtoutergtshellsgt
willg t havegtgreatergtemerginggtenergies.g t Thus,gtthegtMgtshellgtelectrongtwillgtemergegtwithgtgreatergtenergygtthangtagtKgtorgtLgtshellgtelec
tron.
■ Problemgt1.3.g t Nucleargtdecaygtandgtpredictedgtenergies.
Wegtwritegtthegtconservationgtofgtenergygtandgtmomentumgtequationsgtandgtsolvegtthemgtforgtthegtenergygtofgtthegtalphagtparticle.g t Mome
ntumgtisgtgivengtthegtsymbolgt"p",g t andgtenergygtisgt"E".g t Forgtthegtsubscripts,gt"al"gtstandsgtforgtalpha,gtwhilegt"b"gtdenotesgtthegtdaughte
rgtnucleus.
pal2 pb2
palgtgtpbgt gt0 E
gt al E
gt b Ealgtg t Ebgt gtQ and Qgtgt5.5gtMeV
2gtma 2gtm
b
l
Solvinggtourgtsystemgtofgtequationsgtforgt Eal,g t Eb,g t pal,g t pb,g t wegtgetgtthegtsolutionsgtshowngtbelow.g t Notegtthatgtwegthavegttwogtpossiblegts
etsgtofgtsolutionsgt(thisgtdoesgtnotgteffectgtthegtfinalgtresult).
mal 5.5gtmalg
Ebg t gt5.5g t 1gt Ealgt
gt t malgtgtmb
3.31662 mal mb 3.31662 mal mb
palgtgt pbg t gt
gt gt
Wegtaregtinterestedgtingtfindinggtthegtenergygtofgtthegtalphagtparticlegtingtthisgtproblem,gtandgtsincegtwegtknowgtthegtmassgtofgtthegtalphagtpar
ticlegtandgttheg t daughterg t nucleus,g t theg t resultg t isg t easilygtfound.g t Bygtsubstitutinggtourgtknowngtvaluesg t ofg t malgtgt4g t andg t mbgtgt206g t i
ntogtourgtderivedgtEalequationgtwegtget:
Ealgtgt5.395gtMeV
Notegt:gtWegtcangtobtaingtsolutionsgtforgtallgtthegtvariablesgtbygtsubstitutinggtmbgtgt206gtandgtmalgtgt4gtintogtthegtderivedgtequationsgtabovegt:
Ealgtgt5.395gtMeV Ebgtgt0.105gtMeV palgtgtgt6.570 amugtMeV pbgt gtgt6.570 amugtMeV
■ Problemgt1.4.g t Calculationgtofgt WavelengthgtfromgtEnergy.
Sincegtangtx-raygtmustgtessentiallygtbegtcreatedgtbygtthegtde-excitationgtofgtagtsinglegtelectron,gtthegtmaximumgtenergygtofgtangtx-
raygtemittedgtingtagttubegtoperatinggtatgtagtpotentialgtofgt195gtkVgtmustgtbegt195gtkeV.g t Therefore,g t wegtcangtusegtthegtequationgtE=h,gtwhic
hgtisgtalsogtE=hc/Λ,gtorgtΛ=hc/E.g t Plugginggtingtourgtmaximumgtenergygtvaluegtintogtthisgtequationgtgivesgtthegtminimumgtx-
raygtwavelength.
hgtgtc
Λgt wheregtwegtsubstitutegthgtg t 6.626gtgt1034gtJgtgts,g t cgtgt299gt792gt458gtmgtgtsgtandgtEgtgt195gtkeV
E
1
, Chaptergt 1gt Solution
s
1.01869gtJ–m
gt gt 0.0636gtAngstroms
KeV
■ Problemgt 1.5.g t g t 235gtUFissiongt EnergygtRelease.
235 117
Usinggtthegtreactiong t gt Ug t g t g t gt Sngtgt118gtSn,g t andgtmassgtvalues,gtwegtcalculategt thegtmassgtdefectgt of:
Mgt235gtUgtgtgt Mgt117gtSngtgtMgt118gtSngtgt Mgtandgtangtexpectedgte
nergygtreleasegtofgtMc2.
931.5gtMeV
gtgt g t gt g t gt gt
AMU
Thisg t isg t onegtofgttheg t mostgtexothermicgtreactionsg t availableg t togtus.g t Thisg t isgt onegtreasong t why,gtofgtcourse,g t nuclearg t powergtfromgt
uraniumgtfissiongtisgtsogtattractive.
■ Problemgt1.6.g t SpecificgtActivitygtofgt Tritium.
Here,gtwegtusegtthegttextgtequationgtSpecificgtActivitygt=gt(ln(2)*Av)/gtT12*M),gtwheregtAvgtisgtAvogadro'sgtnumber,gtT12gtisgtthegthalf-
lifegtofgtthegtisotope,gtandgtMgtisgtthegtmoleculargtweightgtofgtthegtsample.
ln2gtAvogadrogt'gtsgtConstant
SpecificgtActivitygt
T12gtM
3gtgrams
WegtsubstitutegtT12gtgt12.26gtyearsgtandgtM= togtgetgtthegtspecificgt activitygtingtdisintegrations/(gram–year).
mole
1.13492gtgt1022
SpecificgtActivitygt
gramgt–year
Thegt samegtresultgtexpressedgt ingttermsgtofgtkCi/ggtisgtshowngtbelow
9.73gtkCi
SpecificgtActivitygt
gram
■ Problemgt1.7.g t Acceleratedgtparticlegtenergy.
ThegtenergygtofgtagtparticlegtwithgtchargegtqgtfallinggtthroughgtagtpotentialgtVgtisgtqV.g t SincegtV=gt3gtMVgtisgtourgtmaximumgtpotentialgtdi
fference,g t theg t maximumgtenergygtofg t ang t alphag t particleg t hereg t isgt q*(3gt MV),g t whereg t qg t isgt thegt chargeg t ofg t theg t alphag t particleg t (+
2).g t ThegtmaximumgtalphagtparticlegtenergygtexpressedgtingtMeVgtisgtthus:
Energygtgt3gtMegagtVoltsgtgt2gtElectrongtChargesg 6.g t MeV
t gt
2
, Chaptergt 1gt Solution
s
2 1 1
■ Problemgt1.8.g t Photofissiongtofgt deuteriu gt Dgtg t Γg t gt ngt gtpgt+gt Qgt(-2.226gtMeV)
m. 1 0 1
Thegt reactiongtofgtinterestgtisg t g t gtDg t g t gtΓg t g t gtngtg t g t p+gtQgt(-
2 0 1 1
2.226gtMeV).g t Thus,gtthegtΓgtmustgtbringgtangtenergygtofgtatgtleastgt2.226gtMeV
1 0 0 1
ingtordergtforgtthisgtendothermicgtreactiongttogtproceed.g t Interestingly,gtthegtoppositegtreactiongtwillgtbegtexothermic,gtandgtonegtcangtexp
ectgttogtfindgt2.226gtMeVgtgammagtraysgtingtthegtenvironmentgtfromgtstraygtneutronsgtbeinggtabsorbedgtbygthydrogengtnuclei.
■ Problemgt1.9.g t NeutrongtenergygtfromgtD-Tgtreactiongtbygt150gtkeVgtdeuterons.
Wegtwritegtdowngtthegtconservationgtofgtenergygtandgtmomentumgtequations,gtandgtsolvegtthemgtforgtthegtdesiredgtenergiesgtbygteliminatin
ggtthegtmomenta.g t Ingtthisgtsolution,gt"a"gtrepresentsgtthegtalphagtparticle,gt"n"gtrepresentsgtthegtneutron,gtandgt"d"gtrepresentsgtthegtdeuter
ongt(and,gtasgtbefore,gt"p"gtrepresentsgtmomentum,gt"E"gtrepresentsgtenergy,gtandgt"Q"gtrepresentsgtthegtQ-valuegtofgtthegtreaction).
pa2 pn2 pdgt2
pagtgtpngtgtpd E
gt a E
gt n E
gt d EagtgtEngtgtEdg t gtQ
2gtm 2gtm 2gtm
d
a n
Nextg t weg t wantg t tog t solvegttheg t aboveg t equationsg t forg t theg t unknowngtenergiesg t bygteliminatinggttheg t momenta.g t (Noteg t :g t Usingg t
computergtsoftwaregtsuchgtasgtMathematicagtisgthelpfulgtforgtpainlesslygtsolvinggtthesegtequations).
Weg t evaluateg t theg t solutiongtbygtplugginggting t theg t valuesgtforg t particleg t massesgt(weg t useg t approximategtvaluesgtofg t "ma,"g t "mn,"an
dg t "md"g t ingtAMU,gtwhichgtisgtokaygtbecausegtwegtaregtinterestedgtingtobtaininggtangtenergygtvaluegtatgtthegtend).g t Wegtdefinegtallgtenergie
sgtingtunitsgtofgtMeV,gtnamelygtthegt Q-
value,gtandgt thegtgivengtenergygtofgtthegt deuterongt (bothgtenergygtvaluesgt aregt ingt MeV).g t g t Sogtwegtsubstitutegtmag t =gt 4,gt mng t =gt1,gtmd
=gt2,gtQgt=gt17.6,gtEdg t =gt0.15gtintogtourgtmomentagtindependentgtequations.g t Thisgtyieldsgttwogtpossiblegtsetsgtofgtsolutionsgtforgtthegtener
giesgt(ingtMeV).gtOnegtcorrespondsgttogtthegtneutrongtmovinggtingtthegtforwardgtdirection,gtwhichgtisgtofgtinterest.
Engtg t 13.340gt MeV Eagtg t 4.410gt MeV
Engtg t 14.988gt MeV Eagtg t 2.762gt MeV
Nextgtwegtsolvegtforgtthegtmomentagtbygteliminatinggtthegtenergies.gtWhengtwegtsubstitutegtmag t =gt4,gtmng t =gt1,gtmdg t =gt2,gtQgt=gt17.6,gtEdg t
=gt0.15gtintogtthesegtequationsgtwegtgetgtthegtfollowinggtresults.
pd 1 1
pngt gt 2 3gtpdgt2gt gt352 pagt 8gtpdgt gt2 2 3gtpdgt2gtgt352
5 5 10
Wegtdogtknowgtthegt initialgtmomentumgtofgtthegt deuteron,gt however,gtsincegt wegtknowgtitsgtenergy.gtWegtcangt furthergt evaluategt ourgtsolutions
gtfor
pngt andg t pagtbygtsubstituting:
p dg t gtgt gtgt
Thegt particlegt momentagt(gtingtunitsgtof amuMeV )gtforgteachgt setgtofgtsolutionsgtisgtthus:
pngtg t 5.165 pagtg t 5.940
pngtg t 5.475 pagtg t 4.700
Theg t largestgtneutrongtmomentumgtoccursgtingtthegtforwardgt(+)gtdirection,gtsogtthegthighestgtneutrongtenergygtofgt14.98gtMeVgtcorres
pondsg t togtthisgtdirection.
3
