Question

The binding energies per nucleon for a deuteron and an $\alpha $ -particle are $1.1MeV$ and $7.1MeV$ respectively. The energy released in the fusion reaction ${ }_{1} H ^{2}+{ }_{1} H ^{2} \rightarrow{ }_{2} He ^{4}$ is $Q$ . Then find the value of $\frac{Q}{3}$ in $MeV$ .

Answer 1

The formula of $Q$ value in terms of binding energy,
$Q=B.E.=\Delta mc^{2}$ , where $\Delta m$ is known as the mass defect in reaction,
given values of binding energy per nucleon,
$\left(\frac{B . E .}{A}\right)_{H^{2}}=1.1MeV\&\left(\frac{B . E .}{A}\right)_{\left(He\right)^{4}}=7.1MeV$
it means, $Q=\left(7 . 1 \times 4 - 1 . 1 \times 4\right)MeV$
So, $\frac{Q}{3}=8MeV$