The Binding energy per nucleon of ${}_{3}^{7}{Li}$ and ${}_{2}^{4}{He}$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction ${}_{3}^{7}{Li} + {}_{1}^{1}H \to {}_{2}^{4}{He} + {}_{2}^{4}{He} + Q$ the value of energy Q released is

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-2.4 MeV

8.4 MeV

17.3 MeV

19.6 MeV

answer is C.

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Detailed Solution

$\begin{array}{rcl} B . E of {}_{3}{Li}^{7} & = & 7 x 5.60 Mev = 39.2 Mev \\ B . E of {}_{2}{He}^{4} & = & 4 x 7.06 Mev = 28.24 Mev \\ Q & = & 2 (B . E of {}_{2}{He}^{4}) - B . E of {}_{3}{Li}^{7} \\ = & (2 x 28.24) - 39.2 = 17.28 Mev \end{array}$

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