The energy released in fusion reaction

${ }_1^2 H+{ }_1^2 H \rightarrow{ }_2^3 He+n$

where Binding Energy of ${ }_1^2 H=2.23 MeV$ and of ${ }_2^3 He=7.73 MeV$, would be

3.27 MeV

The correct answer is Option (2) → 3.27 MeV

${ }_1^2 H+{ }_1^2 H \rightarrow{ }_2^3 He+n$

Binding energy of ${ }_1^2 H=2.23MeV$

Binding energy of ${ }_2^3 He=7.73MeV$

$E_0$, Total Binding energy of products = $2×2.33MeV$

$E_1$, Total Binding energy of reactants = $7.73MeV$

∴ Energy released = $E_1-E_0$

$=7.73MeV-4.46MeV$

$=3.27MeV$