Question

The average binding energies of deuterium and $\alpha$ -particle are 1.12 MeV and 7.1MeV respectively. The energy that liberates in the reaction ${}_1{H}^2$+${}_1{H}^2$ $\to$ ${}_{2}H{e}^4$ is $\underset{–}{}$ MeV.

• A. $23.92$
• B. $25$
• C. $25.6$
• D. $15.8$

Answer 1

The correct option is A $23.92$

Average binding energy of deuterium ${}_1{H}^2$ per nucleon $E_d=1.12$ MeV

Average binding energy of $\alpha$-particle ${}_2{H}^4$ per nucleon $E_{\alpha }=7.1$ MeV

Energy liberated in the given reaction $Q=4\left(E_{\alpha }\right)-2\left(E_d\right)-2\left(E_d\right)$

$\therefore$ $Q=4\left(7.1\right)-2\left(1.12\right)-2\left(1.12\right)=23.92$ MeV