Question

Binding energy per nucleon of ${}_1{H}^2$ and ${}_{2}H{e}^4$ are $1.1MeV$ and $7.0MeV$, respectively. Energy released in the process ${}_1{H}^2{+}_1{H}^2{⟶}_{2}H{e}^4$ is

• A. $20.8MeV$
• B. $16.6MeV$
• C. $25.2MeV$
• D. $23.6MeV$

Answer 1

The correct option is D $23.6MeV$

${}_{2}H{e}^4{\to }_1{H}^2{+}_1{H}^2$
No. of nucleon $=$ No of protons $+$ No of neutrons
Mass of ${}_{2}H{e}^4=4\times 7$
Mass of ${}_1{H}^2=2\times 1.1$
Energy released $=4\times 7-2(2\times 1.1)=23.6MeV$