Question

The binding energy per nucleon in deuterium and helium nuclei are $1.1MeV$ and $7.0MeV$, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is :

• A. $23.6MeV$
• B. $2.2MeV$
• C. $28.0MeV$
• D. $30.2MeV$

Answer 1

Answer: (A)

${}_1{H}^2{+}_1{H}^2{\to }_{2}H{e}^4+\Delta E$
The binding energy per nucleon of a deuteron $=1.1MeV$
$\therefore$ Total binding energy = $2\times 1.1$ $=2.2MeV$
The binding energy per nucleon of a helium nuclei $=7MeV$
$\therefore$ Total binding energy = $4\times 7=28MeV$
$\therefore$ Hence, energy released
$\Delta E=(28-2\times 2.2)$ $=23.6MeV$