Question

The mass defect for the nucleus of ${}_{2}H{e}^4$ is $0.0303a.m.u$. The approximate value of binding energy per nucleon in $MeV$ will be $(1a.m.u=931Me\vee )$

Answer 1

Energy released $=$ mass defect $\times c^2$
Binding energy $=0.0303\times 931$
(Since 1 a.m.u$=931$MeV)
Binding energy per nucleon $=\frac{0.0303\times 931}{4}$$=7MeV$