Question

The binding energies per nucleon of deuteron $\left({}_1{H}^2\right)$ and helium atom $\left({}_{2}H{e}^4\right)$ are $1.1\text{MeV}$ and $7\text{MeV}$. If two deuteron atoms react to form a single helium atom, then the energy released is

• A. $13.9\text{MeV}$
• B. $26.9\text{MeV}$
• C. $23.6\text{MeV}$
• D. $19.2\text{MeV}$

Answer 1

The correct option is C $23.6\text{MeV}$

The nuclear reaction is,
${}_1^{2}H{+}_1^{2}H{\to }_2^{4}He+Q$

Total binding energy of helium nucleus $=4\times 7=28MeV$

Total binding energy of each deuteron $=2\times 1.1=2.2MeV$

Hence, energy released when two deuteron fuse to form helium, $=28-2\times 2.2=28-4.4=23.6MeV$