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The nuclear fusion reaction between deut...

The nuclear fusion reaction between deuterium and tritium takes place

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(i) The reaction is `_(1)D^(2)+_(1)D^(2) rarr _(1)T^(3)+_(1)p^(1)+E_(1)`,`Delta m=[2(2.014102)-3.016049-1.007825]=0.00433a.m.u`,`therefore` Energy released ,`E_(1) =0.00433xx931 MeV`,=4.031 MeV. `_(1)T^(3)+_(1)D^(2) rarr _(2)He^(4)+_(0)n^(1)+E_(2)`,`Delta m=[(3.016049+2.014102)-(4.002603+1.008655)]=0.0188a.m.u`,`therefore` energy released `E_(2)=0.01888xx931 MeV=17.580MeV`,Total energy released in both processes=17.580+4.031=21.611MeV (ii)Energy released per `_(1)D^(2)=(21.611)/(3)=7.206 MeV`
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