Question

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

• A. $2.8MeV$
• B. $16.6MeV$
• C. $25.2MeV$
• D. $23.6MeV$

Answer 1

The correct option is D $23.6MeV$

Binding Energy of ${}_1{H}^2=2\left(1.1\right)=2.2MeV$
For 2 atoms of ${}_1{H}^2$ Binding Energy is $4.4MeV$
Total binding energy of ${}_{2}H{e}^4=28MeV$

Energy released would be
$\Delta E$ = total binding energy of ${}_{2}H{e}^4-2$ (total binding energy of ${}_1{H}^2)$
$=4\times 7.0-4.4=23.6MeV$