A deuterium reaction that occurs in an experimental fusion reactor is in two stage:
(a) Two deuterium `(._1^2D)` nuclei fuse together to form a tritium nucleus, with a proton as a by product written as `D(D,p)T`.
(b) A tritium nucleus fuses with another deuterium nucleus to form a helium `._2^4He` nucleus with neutron as a by - product, written as T (D,n) `._2^4He`.
Compute (a) the energy released in each of the two stages, (b) the energy released in the combined reaction per deutrium. (c ) What percentage of the mass energy of the initial deuterium is released. Given,
`{:(._1^2D=2.014102 am u),(._1^3 T=3.016049), (._2^4 He =4.002603 am u),(._1^1 H =1.007825 am u),(._0^1 n =1.00665 am u):}`.

Explanation

(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`