Question

The binding energy per nucleon of deuteron $\left({ }_1^2 H \right)$ and helium nucleus ( ${ }_2^4 He$ )is $1.1 MeV$ and $7 MeV$, respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

• A. $13.9 \, MeV$
• B. $26.9 \, MeV$
• C. $23.6 \, MeV$
• D. $19.2 \, MeV$

Answer 1

Answer: (C)

As given,
${ }_1 H ^2+{ }_1 H ^2 \rightarrow{ }_2 He ^4+$ energy
The binding energy per nucleon of a deuteron $H ^2$ ) $=1.1 MeV$
$\therefore$ Total binding energy of one deuteron nucleus $=2 \times 1.1=2.2 MeV$
$\therefore$ The binding energy per nucleon of Helium $He ^4$ ) $=7 MeV$
$\therefore$ Total binding energy $=4 \times 7=28 MeV$
Hence, energy released in the above process $=28-2 \times 2.2$
$ =28-4.4=23.6 MeV $