umber(Z)=numberofprotonsinnucleusMassnumber(A)=numberofprotons+numberofneutrons=umber(Z)+numberofneutronsAZ1p11H1orproton1n0neutron0e-10b-1orelectron0e+10b+1orpositron4He24a2oraparticle11100-10+142Review2BalancingNuclearEquationsConservemassnumber(A).+Cs13855Rb96371n0++2235+1=138+96+umber(Z)+Cs13855Rb96371n0++292+0=55+37+(thatis,identifytheproductX):(a)(b)(a)umberare212and84,respectively,ontheleft-handsideand208and82,respectively,ontheright-,umberof2,,(b)Inthiscase,themassnumberisthesameonbothsidesoftheequation,,umberof-1,;thatis,(notethatthisprocessdoesnotalterthemassnumber).Thus,(a)and(b),wewouldneedtoaddtwoelectronsontheright-handsideof(a)andexpressbariumasacation(Ba+)in(b).=2,8,20,50,82and126Likeextrastablenumbersofelectronsinnoblegases(e-=2,10,18,36,54and86)umbershigherthan83areradioactiveAllisotopesofTcandPmareradioactive910n/ptoolargebetadecayXn/ptoosmallpositrondecayorelectroncaptureY
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