Question

Binding energy per nucleon of ${}_1{H}^2$ and ${}_{2}H{e}^4$ are $1.1$ MeV and $7.0$ MeV respectively. Energy released in the process ${}_1{H}^2{+}_1{H}^2⟶{}_{2}H{e}^4$ is -

• A. 20.8 MeV
• B. 16.6 MeV
• C. 25.2 MeV
• D. 23.6 MeV

Answer 1

The correct option is D 23.6 MeV

Energy released
$\Delta E$ = Total binding energy of ${}_{2}H{e}^4$ - 2(total binding energy of ${}_1{H}^2)$
$=4\times 7.0-2\left(1.1\right)\left(2\right)$
= 23.6 MeV