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
Answer & explanation
Correct answer: option 2
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 = $E_1-E_0$
$=7.73MeV-4.46MeV$
$=3.27MeV$