Question

A deuterium reaction that occurs in a hypothetical experimental fusion reactor occurs in two stages
(i) two deuterium $\left({}_1{D}^2\right)$ nuclei fuse together to form a tritium nucleus, with a proton as a by-product: $D(D,p)T$.
(ii) a tritium nucleus fuses with another deuterium nucleus to form a $\left({}_2{He}^4\right)$ nucleus with neutron as a by-product $T(D,n{)}_2{He}^4$.
Compute (a) the energy released in each of two the stages and the energy released in the combined reaction.
(b) the energy released per deuterium nucleus.
Given $\left({}_1{D}^2\right)=2.014102$ a.m.u
${}_1{T}^3=3.016049a.m.u$
${}_2{He}^4=4.002603a.m.u$
${}_1{H}^1=1.007825a.m.u$
${}_0{n}^1=1.008665a.m.u$

Answer 1

(a) (i) The reaction is
${}_1{D}^2{+}_1{D}^2{\to }_1{T}^3{+}_1{P}^1{E}_1$
$\Delta m=[2\left(2.014102\right)-3.016049-1.007825]=0.00433$
$\therefore$ Energy released, $E_1=0.00433\times 931$ MeV
$=4.031$ MeV
(ii) ${}_1{T}^3{+}_1{D}^2{\to }_2{He}^4{+}_0{n}^1+E_2$
$\Delta m=[(3.016049+2.014102)-(4.002603+1.008665)]=0.0188a.m.u$
$\therefore$ Energy released $E_2=0.01888\times 931$ Me$V$ $=17.580$ Me$V$
Total energy released in both processes
$=17.580+4.031=21.611$ Me$V$
(b) Energy released per ${}_1{D}^2=\frac{21.611}{3}=7.206$ Me$V$