ChapterChapter 3939 -- NuclearNuclear PhysicsPhysics AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

© 2007 Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to: •• DefineDefine andand applyapply thethe conceptsconcepts ofof massmass numbernumber,, atomicatomic numbernumber,, andand isotopesisotopes..

•• CalculateCalculate thethe massmass defectdefect andand thethe bindingbinding energyenergy perper nucleonnucleon forfor aa particularparticular isotope.isotope.

•• DefineDefine andand applyapply conceptsconcepts ofof radioactiveradioactive decaydecay andand nuclearnuclear reactionsreactions..

•• StateState thethe variousvarious conservationconservation lawslaws,, andand discussdiscuss theirtheir applicationapplication forfor nuclearnuclear reactions.reactions. CompositionComposition ofof MatterMatter

AllAll ofof mattermatter isis composedcomposed ofof atat leastleast threethree fundamentalfundamental particlesparticles (approximations):(approximations):

Particle Fig. Sym Mass Charge Size

Electron e- 9.11 x 10-31 kg -1.6 x 10-19 C  Proton p 1.673 x 10-27 kg +1.6 x 10-19 C 3 fm Neutron n 1.675 x 10-31 kg 0 3 fm

TheThe massmass ofof thethe protonproton andand neutronneutron areare close,close, butbut theythey areare aboutabout 18401840 timestimes thethe massmass ofof anan electron.electron. TheThe AtomicAtomic NucleusNucleus CompactedCompacted nucleus:nucleus: 44 protonsprotons 55 neutronsneutrons

SinceSince atomatom isis electrielectri-- callycally neutral,neutral, therethere mustmust bebe 44 electrons.electrons.

44 electronselectrons BerylliumBeryllium AtomAtom ModernModern AtomicAtomic TheoryTheory The Bohr atom, which is sometimes shown with electrons as planetary particles, is no longer a valid representation of an atom, but it is used here to simplify our discussion of energy levels.

The uncertain position of an electron is now described as a probability distribution—loosely referred to as an electron cloud. DefinitionsDefinitions AA nucleonnucleon isis aa generalgeneral termterm toto denotedenote aa nuclearnuclear particleparticle -- thatthat is,is, eithereither aa protonproton oror aa neutron.neutron. TheThe atomicatomic numbernumber ZZ ofof anan elementelement isis equalequal toto thethe numbernumber ofof protonsprotons inin thethe nucleusnucleus ofof thatthat element.element. TheThe massmass numbernumber AA ofof anan elementelement isis equalequal toto thethe totaltotal numbernumber ofof nucleonsnucleons (protons(protons ++ neutrons).neutrons).

The mass number A of any element is equal to the sum of the atomic number Z and the number of neutrons N : A = N + Z SymbolSymbol NotationNotation

AAA convenientconvenientconvenient waywayway ofofof describingdescribingdescribing ananan elementelementelement isisis bybyby givinggivinggiving itsitsits massmassmass numbernumbernumber andandand itsitsits atomicatomicatomic number,number,number, alongalongalong withwithwith thethethe chemicalchemicalchemical symbolsymbolsymbol forforfor thatthatthat element.element.element.

A Mass number ZX  Atomic numberSymbol

9 For example, consider beryllium (Be): 4 Be ExampleExample 1:1: DescribeDescribe thethe nucleusnucleus ofof aa lithiumlithium atomatom whichwhich hashas aa massmass numbernumber ofof 77 andand anan atomicatomic numbernumber ofof 3.3. AA == 7;7; ZZ == 3;3; NN == ?? NN == AA –– ZZ == 77 -- 33 neutrons:neutrons: NN == 44 Protons:Protons: ZZ == 33 Electrons:Electrons: SameSame asas ZZ 7 Li 3 LithiumLithium AtomAtom IsotopesIsotopes ofof ElementsElements IsotopesIsotopes areare atomsatoms thatthat havehave thethe samesame numbernumber ofof protonsprotons ((ZZ1= = ZZ2), ), butbut aa differentdifferent numbernumber ofof neutronsneutrons (N).(N). ((AA1  AA2) )

3 4 He 2 He IsotopesIsotopes 2 ofof heliumhelium

HeliumHelium -- 33 HeliumHelium -- 44 NuclidesNuclides BecauseBecause ofof thethe existenceexistence ofof soso manymany isotopes,isotopes, thethe termterm elementelement isis sometimessometimes confusing.confusing. TheThe termterm nuclidenuclide isis better.better.

A nuclide is an atom that has a definite mass number A and Z-number. A list of nuclides will include isotopes.

TheThe followingfollowing areare bestbest describeddescribed asas nuclides:nuclides: 3 4 12 13 2 He 2 He 6 C 6 C AtomicAtomic MassMass Unit,Unit, uu OneOne atomicatomic massmass unitunit (1(1 u)u) isis equalequal toto oneone-- twelfthtwelfth ofof thethe massmass ofof thethe mostmost abundantabundant formform ofof thethe carboncarbon atomatom----carboncarbon--1212..

Atomic mass unit: 1 u = 1.6606 x 10-27 kg

Common atomic masses: Proton: 1.007276 u Neutron: 1.008665 u

Electron: 0.00055 u Hydrogen: 1.007825 u ExampeExampe 2:2: TheThe averageaverage atomicatomic massmass ofof BoronBoron--1111 isis 11.00930511.009305 u.u. WhatWhat isis thethe massmass ofof thethe nucleusnucleus ofof oneone boronboron atomatom inin kg?kg? 11 = 11.009305 Electron: 0.00055 u 5 B = 11.009305 Electron: 0.00055 u TheThe massmass ofof thethe nucleusnucleus isis thethe atomicatomic massmass lessless thethe massmass ofof ZZ == 55 electrons:electrons: MassMass == 11.00930511.009305 uu –– 5(0.000555(0.00055 u)u) 11 boronboron nucleusnucleus == 11.0065611.00656 uu

1.6606 x 10-27 kg m 11.00656 u mm == 1.831.83 xx 1010-26-26 kgkg 1 u MassMass andand EnergyEnergy RecallRecall EinsteinEinstein’’ss equivalencyequivalency formulaformula forfor mm andand E:E: Emcc28; 3 x 10 m/s

TheThe energyenergy ofof aa massmass ofof 11 uu cancan bebe found:found: EE == (1(1 u)u)cc2 == (1.66(1.66 xx 1010-27 kg)(3kg)(3 xx 10108 m/s)m/s)2

E = 1.49 x 10-10 J OrOr E = 931.5 MeV

WhenWhen convertingconverting c2  931.5 MeV amuamu toto energy:energy: u ExampleExample 3:3: WhatWhat isis thethe restrest massmass energyenergy ofof aa protonproton (1.007276(1.007276 u)?u)?

EE == mcmc2 == (1.00726(1.00726 u)(931.5u)(931.5 MeV/uMeV/u))

Proton:Proton: EE== 938.3938.3 MeVMeV SimilarSimilar conversionsconversions showshow otherother restrest massmass energies:energies:

Neutron:Neutron: EE== 939.6939.6 MeVMeV

Electron:Electron:Electron: EEE=== 0.5110.5110.511 MeVMeVMeV TheThe MassMass DefectDefect

TheTheThe massmassmass defectdefectdefect isisis thethethe differencedifferencedifference betweenbetweenbetween thethethe restrestrest massmassmass ofofof aaa nucleusnucleusnucleus andandand thethethe sumsumsum ofofof thethethe restrestrest massesmassesmasses ofofof itsitsits constituentconstituentconstituent nucleons.nucleons.nucleons.

TheThe wholewhole isis lessless thanthan thethe sumsum ofof thethe parts!parts! ConsiderConsider thethe carboncarbon--1212 atomatom (12.00000(12.00000 u):u): NuclearNuclear massmass == MassMass ofof atomatom –– ElectronElectron massesmasses == 12.0000012.00000 uu –– 6(0.000556(0.00055 u)u) == 11.99670611.996706 uu TheThe nucleusnucleus ofof thethe carboncarbon--1212 atomatom hashas thisthis mass.mass. (Continued(Continued .. .. .).) MassMass DefectDefect (Continued)(Continued) MassMass ofof carboncarbon--1212 nucleus:nucleus: 11.99670611.996706 Proton:Proton: 1.0072761.007276 uu Neutron:Neutron: 1.0086651.008665 uu TheThe nucleusnucleus containscontains 66 protonsprotons andand 66 neutrons:neutrons:

66 pp == 6(1.0072766(1.007276 u)u) == 6.0436566.043656 uu 66 nn == 6(1.0086656(1.008665 u)u) == 6.0519906.051990 uu TotalTotal massmass ofof parts:parts: == 12.09564612.095646 uu

MassMass defectdefect mmD == 12.09564612.095646 uu –– 11.99670611.996706 uu

m = 0.098940 u mDD = 0.098940 u TheThe BindingBinding EnergyEnergy

TheThe bindingbinding energyenergy EE ofof aa nucleusnucleus isis thethe The binding energy EBB of a nucleus is the energyenergyenergy requiredrequiredrequired tototo separateseparateseparate aaa nucleusnucleusnucleus intointointo itsitsits constituentconstituentconstituent parts.parts.parts.

2 2 EB = mD c where c = 931.5 MeV/u

TheThe bindingbinding energyenergy forfor thethe carboncarbon--1212 exampleexample is:is:

EEB == (0.098940( u)(931.5 MeV/u)

Binding EB for C-12: EB = 92.2 MeV BindingBinding EnergyEnergy perper NucleonNucleon

AnAnAn importantimportantimportant waywayway ofofof comparingcomparingcomparing thethethe nucleinucleinuclei ofofof atomsatomsatoms isisis findingfindingfinding theirtheirtheir bindingbindingbinding energyenergyenergy perperper nucleon:nucleon:nucleon:

Binding energy EB MeV =  per nucleon A nucleon

ForFor ourour CC--1212 exampleexample AA == 1212 and:and:

E 92.2 MeV B 7.68 MeV A 12 nucleon FormulaFormula forfor MassMass DefectDefect TheThe followingfollowing formulaformula isis usefuluseful forfor massmass defect:defect:

MassMass defectdefect mZmNmMDHn   mmD

mmH == 1.0078251.007825 u;u; mmn == 1.0086651.008665 uu ZZ isis atomicatomic number;number; NN isis neutronneutron number;number; MM isis massmass ofof atomatom (including(including electrons).electrons).

ByBy usingusing thethe massmass ofof thethe hydrogenhydrogen atom,atom, youyou avoidavoid thethe necessitynecessity ofof subtractingsubtracting electronelectron masses.masses. 4 ExampleExample 4:4: FindFind thethe massmass defectdefect forfor thethe 2 He nucleusnucleus ofof heliumhelium--4.4. ((MM == 4.0026034.002603 u)u) MassMass defectdefect mZmNmMDHn   mmD

ZmZmH == (2)(1.007825(2)(1.007825 u)u) == 2.0156502.015650 uu

NmNmn == (2)(1.008665(2)(1.008665 u)u) == 2.0173302.017330 uu MM == 4.0026034.002603 uu (From(From nuclidenuclide tables)tables)

mmD == (2.015650(2.015650 uu ++ 2.0173302.017330 u)u) -- 4.0026034.002603 uu

m = 0.030377 u mDD = 0.030377 u ExampleExample 44 (Cont.)(Cont.) FindFind thethe bindingbinding energyenergy perper nucleonnucleon forfor heliumhelium--4.4. ((mmD == 0.0303770.030377 u)u)

E = m c22 where c22 = 931.5 MeV/u EBB = mDD c where c = 931.5 MeV/u

EEB == (0.030377(0.030377 u)(931.5u)(931.5 MeV/uMeV/u)) == 28.328.3 MeVMeV

AA totaltotal ofof 28.328.3 MeVMeV isis requiredrequired ToTo teartear apartapart thethe nucleonsnucleons fromfrom thethe He-4He-4 atom.atom.

SinceSince therethere areare fourfour nucleons,nucleons, wewe findfind thatthat E 28.3 MeV B 7.07 MeV A 4 nucleon BindingBinding EnergyEnergy Vs.Vs. MassMass NumberNumber CurveCurve showsshows thatthat 8 EEB increasesincreases withwith A and peaks at A and peaks at 6 AA == 6060.. HeavierHeavier nucleinuclei areare lessless 4 stable.stable. 2 GreenGreen regionregion isis forfor mostmost stablestable atoms.atoms. 50 100 150 200 250 Binding Energy per nucleon Mass number A ForFor heavierheavier nuclei,nuclei, energyenergy isis releasedreleased whenwhen theythey breakbreak upup ((fissionfission).). ForFor lighterlighter nuclei,nuclei, energyenergy isis releasedreleased whenwhen theythey fusefuse togethertogether ((fusionfusion).). StabilityStability CurveCurve

NuclearNuclearNuclear particlesparticlesparticles areareare 140 held together by a

held together by a N held together by a 120 StableStable nuclearnuclearnuclear strongstrongstrong force.force.force. 100 nucleinuclei AA stablestable nucleusnucleus remainsremains 80 forever,forever, butbut asas thethe ratioratio 60 ofof N/ZN/Z getsgets larger,larger, thethe 40 atomsatoms decay.decay. Neutron number ZZ == NN 20 Elements with Z > 82 ElementsElements withwith ZZ >> 8282 20 40 60 80 100 areareare allallall unstable.unstable.unstable. Atomic number Z RadioactivityRadioactivity

AsAs thethe heavierheavier atomsatoms becomebecome   moremore unstable,unstable, particlesparticles andand  photonsphotons areare emittedemitted fromfrom thethe  nucleusnucleus andand itit isis saidsaid toto bebe  radioactiveradioactive.. AllAll elementselements withwith AA >> 8282 areare radioactive.radioactive.

ExamplesExamples are:are: AlphaAlpha particlesparticles   particlesparticles (electrons)(electrons) GammaGamma raysrays   particlesparticles (positrons)(positrons) TheThe AlphaAlpha ParticleParticle

AnAn alphaalpha particleparticle  isis thethe nucleusnucleus ofof aa heliumhelium atomatom consistingconsisting ofof twotwo protonsprotons andand twotwo neutronsneutrons tightlytightly bound.bound.

ChargeCharge == +2+2ee- == 3.23.2 xx 1010-19 CC MassMass == 4.0015064.001506 uu RelativelyRelatively lowlow speedsspeeds (( 0.1c0.1c ))

NotNot veryvery penetratingpenetrating TheThe BetaBeta--minusminus ParticleParticle

AA betabeta--minusminus particleparticle  isis simplysimply anan electronelectron thatthat hashas beenbeen expelledexpelled fromfrom thethe nucleus.nucleus.

- ChargeCharge == ee- == --1.61.6 xx 1010-19 CC - MassMass == 0.000550.00055 uu - HighHigh speedsspeeds (near(near cc))

- VeryVery penetratingpenetrating TheThe PositronPositron

AA betabeta positivepositive particleparticle  isis essentiallyessentially anan electronelectron withwith positivepositive charge.charge. TheThe massmass andand speedsspeeds areare similar.similar.

+ ChargeCharge == ++ee- == 1.61.6 xx 1010-19 CC + MassMass == 0.000550.00055 uu + HighHigh speedsspeeds (near(near cc))

+ VeryVery penetratingpenetrating TheThe GammaGamma PhotonPhoton

AA gammagamma rayray  hashas veryvery highhigh electromagneticelectromagnetic radiationradiation carryingcarrying energyenergy awayaway fromfrom thethe nucleus.nucleus.  ChargeCharge == ZeroZero (0)(0)  MassMass == zerozero (0)(0)   SpeedSpeed == cc (3(3 xx 10108 m/s)m/s)

 MostMost penetratingpenetrating radiationradiation RadioactiveRadioactive DecayDecay AsAs discussed,discussed, whenwhen thethe ratioratio ofof N/ZN/Z getsgets veryvery large,large, thethe nucleusnucleus becomesbecomes unstableunstable andand oftenoften particlesparticles and/orand/or photonsphotons areare emitted.emitted.

4 AlphaAlpha decaydecay 2 resultsresults inin thethe lossloss ofof twotwo protonsprotons andand twotwo neutronsneutrons fromfrom thethe nucleus.nucleus.

AA44 ZZX 22Yenergy

XX isis parentparent atomatom andand YY isis daughterdaughter atomatom TheThe energyenergy isis carriedcarried awayaway primarilyprimarily byby thethe K.E.K.E. ofof thethe alphaalpha particle.particle. ExampleExample 5:5: WriteWrite thethe reactionreaction thatthat occursoccurs whenwhen radiumradium--226226 decaysdecays byby alphaalpha emission.emission.

AA44 ZZX 22Yenergy

226 226 4 4 88RaYenergy 88 2 2 FromFrom tables,tables, wewe findfind ZZ andand AA forfor nuclides.nuclides. TheThe daughterdaughter atom:atom: ZZ == 86,86, AA == 222222 226 222 4 88RaRnenergy 86 2

Radium-226Radium-226 decaysdecays intointo radon-222.radon-222. BetaBeta--minusminus DecayDecay

BetaBeta--minusminus  decaydecay resultsresults whenwhen aa neutronneutron decaysdecays intointo aa protonproton andand anan electron.electron. Thus,Thus, thethe ZZ--numbernumber increasesincreases byby one.one.

AA0 ZZX 11Y energy

XX isis parentparent atomatom andand YY isis daughterdaughter atomatom

TheThe energyenergy isis carriedcarried awayaway primarilyprimarily - byby thethe K.E.K.E. ofof thethe electron.electron. BetaBeta--plusplus DecayDecay

BetaBeta--plusplus  decaydecay resultsresults whenwhen aa protonproton decaysdecays intointo aa neutronneutron andand aa positron.positron. Thus,Thus, thethe ZZ--numbernumber decreasesdecreases byby one.one.

AA0 ZZX 11Y energy

XX isis parentparent atomatom andand YY isis daughterdaughter atomatom

TheThe energyenergy isis carriedcarried awayaway primarilyprimarily + byby thethe K.E.K.E. ofof thethe positron.positron. RadioactiveRadioactive MaterialsMaterials TheThe raterate ofof decaydecay forfor radioactiveradioactive substancessubstances isis expressedexpressed inin termsterms ofof thethe activityactivity RR,, givengiven by:by:

N N = Number of Activity R  N = Number of t undecayedundecayed nucleinuclei

OneOneOne becquerelbecquerelbecquerel (((Bq)BqBq)) isisis ananan activityactivityactivity equalequalequal tototo oneoneone disintegrationdisintegrationdisintegration perperper secondsecondsecond (1(1(1 sss--11).).).

OneOneOne curiecuriecurie (((Ci)CiCi)) isisis thethethe activityactivityactivity ofofof aaa radioactiveradioactiveradioactive materialmaterialmaterial thatthatthat decaysdecaysdecays atatat thethethe rateraterate ofofof 3.73.73.7 xxx 1010101010 BqBqBq ororor 3.73.73.7 xxx 1010101010 disintegrationsdisintegrationsdisintegrations perperper second.second.second. TheThe HalfHalf--LifeLife TheThe halfhalf--lifelife TT ofof 1/2 No anan isotopeisotope isis thethe timetime inin whichwhich oneone-- halfhalf ofof itsits unstableunstable N nucleinuclei willwill decay.decay. 0 2 n 1 N0 NN 0  2 4

WhereWhere nn isis numbernumber Number Undecayed Nuclei 1 2 3 4 ofof halfhalf--liveslives Number of Half-lives HalfHalf--LifeLife (Cont.)(Cont.) TheThe samesame reasoningreasoning willwill applyapply toto activityactivity RR oror toto amountamount ofof materialmaterial.. InIn general,general, thethe followingfollowing threethree equationsequations cancan bebe appliedapplied toto radioactivity:radioactivity: NucleiNuclei RemainingRemaining ActivityActivity RR

n n 1 1 NN 0  RR 0  2 2

MassMass RemainingRemaining NumberNumber ofof HalfHalf--lives:lives: n t 1 n  mm 0  T1 2 2 ExampleExample 6:6: AA samplesample ofof iodineiodine--131131 hashas anan initialinitial activityactivity ofof 55 mCimCi.. TheThe halfhalf--lifelife ofof II--131131 isis 88 daysdays.. WhatWhat isis thethe activityactivity ofof thethe samplesample 3232 daysdays later?later? FirstFirst wewe determinedetermine thethe numbernumber ofof halfhalf--lives:lives: t 32 d n  nn == 44 halfhalf--liveslives T1/2 8 d

n 4 11  RR0 5 mCi  R = 0.313 mCi 22  R = 0.313 mCi

ThereThere wouldwould alsoalso bebe 1/161/16 remainingremaining ofof thethe massmass andand 1/161/16 ofof thethe numbernumber ofof nuclei.nuclei. NuclearNuclear ReactionsReactions ItIt isis possiblepossible toto alteralter thethe structurestructure ofof aa nucleusnucleus byby bombardingbombarding itit withwith smallsmall particles.particles. SuchSuch eventsevents areare calledcalled nuclearnuclear reactions:reactions: General Reaction: x + X  Y + y

ForFor example,example, ifif anan alphaalpha particleparticle bombardsbombards aa nitrogennitrogen--1414 nucleusnucleus itit producesproduces aa hydrogenhydrogen atomatom andand oxygenoxygen--17:17:

4141 17 27 NHO 1 8 ConservationConservation LawsLaws

ForFor anyany nuclearnuclear reaction,reaction, therethere areare threethree conservationconservation lawslaws whichwhich mustmust bebe obeyed:obeyed:

ConservationConservation ofof Charge:Charge: TheThe totaltotal chargecharge ofof aa systemsystem cancan neitherneither bebe increasedincreased nornor decreased.decreased. ConservationConservation ofof Nucleons:Nucleons: TheThe totaltotal numbernumber ofof nucleonsnucleons inin aa reactionreaction mustmust bebe unchanged.unchanged. ConservationConservation ofof MassMass Energy:Energy: TheThe totaltotal massmass-- energyenergy ofof aa systemsystem mustmust notnot changechange inin aa nuclearnuclear reaction.reaction. ExampleExample 7:7: UseUse conservationconservation criteriacriteria toto determinedetermine thethe unknownunknown elementelement inin thethe followingfollowing nuclearnuclear reaction:reaction: 17 4 A 13HLiHeXenergy 2 Z ChargeCharge beforebefore == +1+1 ++ 33 == +4+4 ChargeCharge afterafter == +2+2 ++ ZZ == +4+4 ZZ == 44 –– 22 == 22 (Helium(Helium hashas ZZ == 22))

NucleonsNucleons beforebefore == 11 ++ 77 == 88 NucleonsNucleons afterafter == 44 ++ AA == 88 (Thus,(Thus, AA == 44))

17 4 4 13HLiHeHeenergy 2  2  ConservationConservation ofof MassMass--EnergyEnergy ThereThere isis alwaysalways massmass--energyenergy associatedassociated withwith anyany nuclearnuclear reaction.reaction. TheThe energyenergy releasedreleased oror absorbedabsorbed isis calledcalled thethe QQ--valuevalue andand cancan bebe foundfound ifif thethe atomicatomic massesmasses areare knownknown beforebefore andand after.after.

17 4 4 13HLiHeHeQ  2 2

17 4 4 Q13 H Li  2 He  2 He QQ isis thethe energyenergy releasedreleased inin thethe reaction.reaction. IfIf QQ isis positivepositive,, itit isis exothermicexothermic.. IfIf QQ isis negativenegative,, itit isis endothermicendothermic.. ExampleExample 8:8: CalculateCalculate thethe energyenergy releasedreleased inin thethe bombardmentbombardment ofof lithiumlithium--77 withwith hydrogenhydrogen--1.1.

17 4 4 13HLiHeHeQ  2 2

17 4 4 Q13 H Li  2 He  2 He 1 4 1 H 1.007825 u 2 He  4.002603 u 7 Li  7.016003 u 4 3 2 He  4.002603 u

SubstitutionSubstitution ofof thesethese massesmasses gives:gives:

QQ == 0.0186220.018622 u(931.5u(931.5 MeV/uMeV/u)) QQ=17.3=17.3 MeVMeV TheThe positivepositive QQ meansmeans thethe reactionreaction isis exothermic.exothermic. SummarySummary FundamentalFundamental atomicatomic andand nuclearnuclear particlesparticles

Particle Fig. Sym Mass Charge Size

Electron e 9.11 x 10-31 kg -1.6 x 10-19 C  Proton p 1.673 x 10-27 kg +1.6 x 10-19 C 3 fm Neutron n 1.675 x 10-31 kg 0 3 fm

The mass number A of any element is equal to the sum of the protons (atomic number Z) and the number of neutrons N : A = N + Z SummarySummary Definitions:Definitions: AA nucleonnucleon isis aa generalgeneral termterm toto denotedenote aa nuclearnuclear particleparticle -- thatthat is,is, eithereither aa protonproton oror aa neutron.neutron. TheThe massmass numbernumber AA ofof anan elementelement isis equalequal toto thethe totaltotal numbernumber ofof nucleonsnucleons (protons(protons ++ neutrons).neutrons). IsotopesIsotopes areare atomsatoms thatthat havehave thethe samesame numbernumber ofof protonsprotons ((ZZ1= = ZZ2), ), butbut aa differentdifferent numbernumber ofof neutronsneutrons (N).(N). ((AA1  AA2) )

AA nuclidenuclide isis anan atomatom thatthat hashas aa definitedefinite massmass numbernumber AA andand ZZ--number.number. AA listlist ofof nuclidesnuclides willwill includeinclude isotopes.isotopes. SummarySummary (Cont.)(Cont.)

SymbolicSymbolic notationnotation AX  Mass number Symbol forfor atomsatoms Z Atomic number MassMass defectdefect mZmNmMDHn   mmD   BindingBinding E = m c2 where c2 = 931.5 MeV/u energyenergy B D

Binding Energy EB MeV =  per nucleon A nucleon SummarySummary (Decay(Decay Particles)Particles) AnAn alphaalpha particleparticle  isis thethe nucleusnucleus ofof aa heliumhelium atomatom consistingconsisting ofof twotwo protonsprotons andand twotwo tightlytightly boundbound neutrons.neutrons. AA betabeta--minusminus particleparticle  isis simplysimply anan electronelectron thatthat hashas beenbeen expelledexpelled fromfrom thethe nucleus.nucleus. AA betabeta positivepositive particleparticle  isis essentiallyessentially anan electronelectron withwith positivepositive charge.charge. TheThe massmass andand speedsspeeds areare similar.similar. AA gammagamma rayray  hashas veryvery highhigh electromagneticelectromagnetic radiationradiation carryingcarrying energyenergy awayaway fromfrom thethe nucleus.nucleus. SummarySummary (Cont.)(Cont.) AlphaAlpha Decay:Decay:

AA44 ZZX 22Yenergy

BetaBeta--minusminus Decay:Decay:

AA0 ZZX 11Y energy BetaBeta--plusplus Decay:Decay:

AA0 ZZX 11Y energy SummarySummary (Radioactivity)(Radioactivity)

TheThe halfhalf--lifelife TT ofof anan isotopeisotope isis thethe timetime inin The half-life T1/21/2 of an isotope is the time in whichwhichwhich oneoneone-half--halfhalf ofofof itsitsits unstableunstableunstable nucleinucleinuclei willwillwill decay.decay.decay.

NucleiNuclei RemainingRemaining ActivityActivity RR

n n 1 1 NN 0  RR 0  2 2

MassMass RemainingRemaining NumberNumber ofof HalfHalf--lives:lives: n t 1 n  mm 0  T1 2 2 SummarySummary (Cont.)(Cont.)

Nuclear Reaction: x + X  Y + y + Q

ConservationConservation ofof Charge:Charge: TheThe totaltotal chargecharge ofof aa systemsystem cancan neitherneither bebe increasedincreased nornor decreased.decreased. ConservationConservation ofof Nucleons:Nucleons: TheThe totaltotal numbernumber ofof nucleonsnucleons inin aa reactionreaction mustmust bebe unchanged.unchanged. ConservationConservation ofof MassMass Energy:Energy: TheThe totaltotal massmass-- energyenergy ofof aa systemsystem mustmust notnot changechange inin aa nuclearnuclear reaction.reaction. (Q(Q--valuevalue == energyenergy released)released) CONCLUSION:CONCLUSION: ChapterChapter 3939 NuclearNuclear PhysicsPhysics