Question

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

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

Answer 1

Answer: (C)

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