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