Nuclear Chemistry - Mizzou University of Missouri核化学-美国密苏里堪萨斯大学

1NuclearChemistryChapter19Copyright

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Permissionrequiredforreproductionordisplay.2XAZMassNumber

AtomicNumberElementSymbolAtomicnumber(Z)=numberofprotonsinnucleusMassnumber(A)=numberofprotons+numberofneutrons=atomicnumber(Z)+numberofneutronsAZ1p11H1orproton1n0neutron0e-10b-1orelectron0e+10b+1orpositron4He24a2oraparticle11100-10+142Review3BalancingNuclearEquationsConservemassnumber(A).Thesumofprotonsplusneutronsintheproductsmustequalthesumofprotonsplusneutronsinthereactants.1n0U23592+Cs13855Rb96371n0++2235+1=138+96+2x1Conserveatomicnumber(Z)ornuclearcharge.Thesumofnuclearchargesintheproductsmustequalthesumofnuclearchargesinthereactants.1n0U23592+Cs13855Rb96371n0++292+0=55+37+2x04519.1Balancethefollowingnuclearequations(thatis,identifytheproductX):(a)

(b)619.1Solution(a)

Themassnumberandatomicnumberare212and84,respectively,ontheleft-handsideand208and82,respectively,ontheright-handside.Thus,Xmusthaveamassnumberof4andanatomicnumberof2,whichmeansthatitisanαparticle.ThebalancedequationisStrategy

Inbalancingnuclearequations,notethatthesumofatomicnumbersandthatofmassnumbersmustmatchonbothsidesoftheequation.7(b)Inthiscase,themassnumberisthesameonbothsidesoftheequation,buttheatomicnumberoftheproductis1morethanthatofthereactant.Thus,Xmusthaveamassnumberof0andanatomicnumberof-1,whichmeansthatitisaβparticle.TheonlywaythischangecancomeaboutistohaveaneutronintheCsnucleustransformedintoaprotonandanelectron;thatis,(notethatthisprocessdoesnotalterthemassnumber).Thus,thebalancedequationis19.18Check

Notethattheequationin(a)and(b)arebalancedfornuclearparticlesbutnotforelectricalcharges.Tobalancethecharges,wewouldneedtoaddtwoelectronsontheright-handsideof(a)andexpressbariumasacation(Ba+)in(b).19.19NuclearStabilityCertainnumbersofneutronsandprotonsareextrastablenorp=2,8,20,50,82and126Likeextrastablenumbersofelectronsinnoblegases(e-=2,10,18,36,54and86)NucleiwithevennumbersofbothprotonsandneutronsaremorestablethanthosewithoddnumbersofneutronsandprotonsAllisotopesoftheelementswithatomicnumbershigherthan83areradioactiveAllisotopesofTcandPmareradioactive1011n/ptoolargebetadecayXn/ptoosmallpositrondecayorelectroncaptureY12NuclearStabilityandRadioactiveDecayBetadecay14C14N+0b67-140K40Ca+0b1920-11n1p+0b01-1Decrease#ofneutronsby1Increase#ofprotonsby1Positrondecay11C11B+0b

65+138K38Ar+0b

1918+11p1n+0b10+1Increase#ofneutronsby1Decrease#ofprotonsby113ElectroncapturedecayIncreasenumberofneutronsby1Decreasenumberofprotonsby1NuclearStabilityandRadioactiveDecay37Ar+0e

37Cl1817-155Fe+0e

55Mn2625-11p+0e

1n10-1AlphadecayDecreasenumberofneutronsby2Decreasenumberofprotonsby2212Po4He+208Pb84282Spontaneousfission252Cf2125In+21n9849014Nuclearbindingenergyistheenergyrequiredtobreakupanucleusintoitscomponentprotonsandneutrons.Nuclearbindingenergy+19F91p+101n9109x(pmass)+10x(nmass)=19.15708amuDE=(Dm)c2Dm=18.9984amu–19.15708amuDm=-0.1587amuDE=-2.37x10-11JDE=-0.1587amux(3.00x108m/s)2=-1.43x1016amum2/s2Usingconversionfactors:1kg=6.022x1026amu1J=kgm2/s215=2.37x10-11J19nucleons=1.25x10-12J/nucleonbindingenergypernucleon=bindingenergynumberofnucleonsDE=(-2.37x10-11J)x(6.022x1023/mol)DE=-1.43x1013J/molDE=-1.43x1010kJ/molNuclearbindingenergy=1.43x1010kJ/mol16Nuclearbindingenergypernucleonvsmassnumbernuclearstabilitynuclearbindingenergynucleon1719.2Theatomicmassofis126.9004amu.Calculatethenuclearbindingenergyofthisnucleusandthecorrespondingnuclearbindingenergypernucleon.1819.2Strategy

Tocalculatethenuclearbindingenergy,wefirstdeterminethedifferencebetweenthemassofthenucleusandthemassofalltheprotonsandneutrons,whichgivesusthemassdefect.Next,weapplyEquation(19.2)[ΔE=(Δm)c2].SolutionThereare53protonsand74neutronsintheiodinenucleus.Themassof53atomis53x1.007825amu=53.41473amuandthemassof74neutronsis74x1.008665amu=74.64121amu1919.2Therefore,thepredictedmassforis53.41473+74.64121=128.05594amu,andthemassdefectisΔm=126.9004amu-128.05594amu=-1.1555amuTheenergyreleasedisΔE=(Δm)c2=(-1.1555amu)(3.00x108m/s)2=-1.04x1017amu·m2/s22019.2Let’sconverttoamorefamiliarenergyunitofjoules.Recallthat1J=1kg·m2/s2.Therefore,weneedtoconvertamutokg:Thus,thenuclearbindingenergyis1.73x10-10J.Thenuclearbindingenergypernucleonisobtainedasfollows:21KineticsofRadioactiveDecayNdaughterrate=lNN=thenumberofatomsattimetN0=thenumberofatomsattimet=0listhedecayconstantNtlnN0-lt==t½l0.6932223RadiocarbonDating14N+1n14C+1H716014C14N+0b+n67-1t½=5730yearsUranium-238Dating238U206Pb+84a+60b92-1822t½=4.51x109years24NuclearTransmutation14N+4a

17O+1p728127Al+4a

30P+1n13215014N+1p11C+4a716225Writethebalancedequationforthenuclearreactionwheredrepresentsthedeuteriumnucleus(thatis,).19.326Strategy

Towritethebalancednuclearequation,rememberthatthefirstisotopeisthereactantandthesecondisotopeistheproduct.Thefirstsymbolinparentheses(d)isthebombardingparticleandthesecondsymbolinparentheses(α)istheparticleemittedasaresultofnucleartransmutation.19.32719.3Solution

Theabbreviationtellsusthatwheniron-56isbombardedwithadeuteriumnucleus,itproducesthemanganese-54nucleusplusanαparticle.Thus,theequationforthisreactionisCheckMakesurethatthesumofmassnumbersandthesumofatomicnumbersarethesameonbothsidesoftheequation.28NuclearTransmutation29NuclearFission235U+1n90Sr+143Xe+31n+Energy92543800Energy=[mass235U+massn–(mass90Sr+mass143Xe+3xmassn)]xc2Energy=3.3x10-11Jper235U=2.0x1013Jpermole235UCombustionof1tonofcoal=5x107J30NuclearFission235U+1n90Sr+143Xe+31n+Energy92543800Representativefissionreaction3132NuclearFissionNuclearchainreactionisaself-sustainingsequenceofnuclearfissionreactions.Theminimummassoffissionablematerialrequiredtogenerateaself-sustainingnuclearchainreactionisthecriticalmass.33SchematicofanAtomicBomb34SchematicDiagramofaNuclearReactorU3O8refueling35ChemistryInAction:Nature’sOwnFissionReactorNaturalUranium0.7202%U-23599.2798%U-238MeasuredatOklo0.7171%U-23536NuclearFusion2H+2H3H+1H