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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 . |
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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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