Question

The amount of energy released in the fusion of ${}_1{H}^2$ to form a ${}_{2}H{e}^4$ nucleus will be :
[Binding energy per nucleon of ${}_1{H}^2=1.1MeV$ and ${}_{2}H{e}^4=7MeV$]

• A. $8.1MeV$
• B. $5.9MeV$
• C. $23.6MeV$
• D. $2MeV$

Answer 1

The correct option is C $23.6MeV$

Nuclear reaction: $2{}_1{H}^2\to {}_{2}H{e}^4$

Energy released $=$ Binding energy of products $-$ Binding energy of reactants.

Total Binding energy $=$ nucleons x binding energy per nucleon

$4\times 7-2\times 2\times 1.1=23.6MeV$