1.

Calculate the energyreleasedin the reaction""_(92)^(235)U +""_(0)^(1)n- to""_(36)^(92) Kr+""_(0)^(1) n+Q Given: Mass of""_(92)""^(235 )U =235.0439amu Mass of""_(56 )^(141)Ba =140.9178 am u Mass of""_(36 )^(92) Kr =91.8854a m uandmass ofneutron=1.008655am u. Expressthe resultin joules .

Answer»

Solution :Given :
Mass of ` ""_(92)""^(235 )U =235.0439amu `
Mass of ` ""_(56 )^(141)Ba =140.9178 am u`
Mass of ` ""_(36 )^(92) Kr =91.8854a m u `
andmass ofneutron`=1.008655am u`. Expressthe resultin joules .
Totalmass ofreactants
`=235.0439+ 1.008655`
`=236.052555a m u`
totalmass ofproducts
`=140.9178+91.8854 +(3 XX 1.008655)`
`=232.8032 +3.025965 `
` =235.829165 am u`
Massdefect,` Deltam=236.0525555 -235 .829165`
`Delta m=0.22339am u `
Energyreleased peratomof `U^(235 )= Delta m xx 931.5 MeV `
`=0.22339 xx 931.5 xx 10^6 xx 1.6 xx 10^(-19)`
`=332.94 xx 10^(-13)`
`=3.3294 xx 10^(-11) J`


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