Question: Q. 2. Calculate the energy in fusion reaction: ${ }{1} \mathrm{H}^{2}+{ }{1} \mathrm{H}^{2} \rightarrow{ }{2}^{3} \mathrm{He}+{ }{0} n^{1}$ where, BE of ${ }_{1} \mathrm{H}^{2}=2.23$
$\mathrm{MeV}$ and of ${ }_{2}^{3} \mathrm{He}=7.73 \mathrm{MeV}$.
A [Delhi Set I, II, III 2016]
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Solution:
Ans. Total Binding energy of Initial System
$$ \text { i.e., } \quad{ }{1} \mathrm{H}^{2}+{ }{1} \mathrm{H}^{2}=(2.23+2.23) \mathrm{MeV} $$
$$ =4.46 \mathrm{MeV} \quad 1 / 2 $$
Binding energy of final system i.e., ${ }_{2}^{3} \mathrm{He}$
$$ =7.73 \mathrm{MeV} $$
Hence energy released $=7.73 \mathrm{MeV}-4.46 \mathrm{MeV} \quad 1 / 2$
$$ =3.27 \mathrm{MeV} $$
[CBSE Marking Scheme 2016]
