Explanation

(c) As given
$\ _{1}H^{2}\ + \ _{1}H^{2} \rightarrow \ _{2}He^{4}\ + energy$
The binding energy per nucleon of a deuteron$(\ _{1}H^{2})$
$= 1.1MeV$
$\therefore$ Total binding energy of one deuteron nucleus
$= \ 2\ \times \ 1.1 = 2.2MeV$
$\therefore$ The binding energy per nucleon of Helium$(\ _{2}He^{4})$
=7MeV
$\therefore$ Total binding energy
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ 4\ \times \ 7\ = 28MeV$
Hence, energy released in the above process
$\ \ \ \ \ \ \ \ \ \ \ = 28 - 2\ \times 2.2$
= $28\ –\ 4.4\ = \ 23.6\ MeV$