Question

The binding energies per nucleon for deutron ${(}_1{H}^2)$ and helium ${(}_{2}H{e}^4)$ are $1.1$ MeV and $7.0$ MeV respectively. Calculate the energy released, when two deutrons fuse to form a helium nucleus ${(}_{2}H{e}^4)$.

Answer 1

${}_1{H}^2+{}_1{H}^2\to {}_{2}H{e}^4$

So, Energy released = Energy of ${}_{2}H{e}^4$ $-2$(Energy of ${}_1{H}^2$)

$=7.0-2\times 1.1=4.8MeV$