Question

Calculate the energy released in joules and $MeV$ in the following nuclear reaction.
${}_1^{2}H{+}_1^{2}H{\to }_2^{3}He{+}_0^{1}n$
Assume that the masses of ${}_1^{2}H{,}_2^{3}He$ and neutron $\left(n\right)$ respectively are $2.0141,3.0160$ and $1.0087$ in amu.

Answer 1

Energy released $=$ mass defect $\times$ 931.5 Mev

$=(3+2+1-2.014-3.0160-1.0087)\times 931.5$

$=-36.1422$ Mev.