In the reaction ${ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 n$, if the…
In the reaction ${ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 n$, if the binding energies of ${ }_1^2 \mathrm{H},{ }_1^3 \mathrm{H}$ and ${ }_2^4 \mathrm{He}$ are respectively $a, b$ and $c$ (in Me V), then the energy (in $\mathrm{MeV}$ ) released in this reaction is:
$a+b+c$
$a+b-c$
$c-a-b$
$c+a-b$
Solution
Energy in given reaction $=B E$ of products $-B . E$. of reactants
$\begin{aligned}
& =C-(a+b) \\
& =C-a-b
\end{aligned}$