The binding energy per nucleon is 1.1 MeV for ${ }_1^2 \mathrm{H}$ and 7.0 MeV for ${ }_2^4 He$. The energy released in the given fusion reaction will be

${ }_1^2 H+{ }_1^2 H \rightarrow{ }_2^4 He$

23.6 MeV

The correct answer is Option (3) → 23.6 MeV

${ }_1^2 H+{ }_1^2 H \rightarrow{ }_2^4 He$

${ }_1^2 H$, Binding energy per nucleon = $1.1 MeV$

${ }_2^4 He$, Binding energy per nucleon = $7.0 MeV$

Total binding energy before reaction = $2×2×1.1MeV=4.4MeV$

Total binding energy after reaction = $4×7.0MeV=28.0MeV$

Energy released = $28.0 - 4.4 = 23.6MeV$