Q84 / 160 · JIPMER 2007 · Physics

In the reaction ${ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \longrightarrow{ }_2^4 \mathrm{He}+{ }_0^1 n$, 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 MeV ), then the energy (in MeV ) released in this reaction is

  • A
    $c+a-b$
  • Correct answer
    B
    $c-a-b$
  • C
    $a+b+c$
  • D
    $a+b-c$

Solution

The energy released per nuclear reaction is the resultant binding energy. Binding energy of $\left({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H}\right)=a+b$ Binding energy of ${ }_2^4 \mathrm{He}=c$ In a nuclear reaction the resultant nucleus is more stable than the reactants. Hence, binding energy of ${ }_2^4 \mathrm{He}$ will be more than that of $\left({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H}\right)$ Thus, energy released per nucleon $\begin{aligned}& =\text { resultant binding energy } \\& =c-(a+b)=c-a-b\end{aligned}$

Hence, the correct answer is option: 'B'.

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