Question

The binding energy of deuteron is $2.2 \, \text{MeV}$ and that of $_{2}^{4} \text{He}$ is $28 \, \text{MeV} \text{.}$ If two deuterons are fused to form one $_{2}^{4} \text{He}$ , then the energy released is

• A. $30.2MeV$
• B. $25.8MeV$
• C. $23.6MeV$
• D. $19.2MeV$

Answer 1

Answer: (C)

Corresponding nuclear reaction is given by,
$_{1}^{2} \text{H} + _{1}^{2} \text{H} \rightarrow _{2}^{4} \text{He} + \text{energy}$ , this reaction is also called nuclear fusion of deuterium to helium atom.
Energy released $= B . E .$ of $_{2}^{4} H e - 2 \left(\right. B . E .$ of $_{1}^{2} H \left.$
$= 28 - 2 \left(\right. 2.2 \left.\right) = 28 - 4.4 = 23.6 \, \text{MeV} \text{.}$