Energy released when two deuterons \(({{}_{1}H}^{2})\) fuse to form a helium nucleus \(({{}_{2}He}^{4})\) is : (Given : …

Binding energy per nucleon of deuteron \( \left(^2_1\text{H}\right) = 1.1\, \text{MeV} \)
Binding energy per nucleon of helium-4 \( \left(^4_2\text{He}\right) = 7.0\, \text{MeV} \)

Two deuterons \( \Rightarrow \) total nucleons = 2 × 2 = 4

\[
\text{BE}_{\text{reactants}} = 2 \times (1.1 \times 2) = 4.4\, \text{MeV}
\]

\[
\text{BE}_{\text{product}} = 4 \times 7.0 = 28.0\, \text{MeV}
\]

\[
Q = \text{BE}_{\text{product}} - \text{BE}_{\text{reactants}} = 28.0 - 4.4 = \boxed{23.6\, \text{MeV}}
\]