Q84 / 120 · VITEEE 2012 · Physics

If the binding energy per nuclear in $\mathrm{Li}^{7}$ and $\mathrm{He}^{4}$ nuclei are respectively $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$, then energy of reactor $\mathrm{Li}^{7}+\mathrm{P} \rightarrow 2{ }_{2} \mathrm{He}^{4}$ is

  • A
    $19.6 \mathrm{MeV}$
  • B
    $2.4 \mathrm{MeV}$
  • C
    $8.4 \mathrm{MeV}$
  • Correct answer
    D
    $17.3 \mathrm{MeV}$

Solution

$\mathrm{BE}$ of $\mathrm{Li}^{7}=39.20 \mathrm{MeV}$ and $\mathrm{He}^{4}=28.24 \mathrm{MeV}$ Hence binding energy of ${ }_{2} \mathrm{He}^{4}=56.84 \mathrm{MeV}$ Energy of reaction $=56.84-39.20$ $=17.28 \mathrm{MeV}$

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

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