• # question_answer 39) If the binding energy per nucleon in $\text{L}{{\text{i}}^{\text{7}}}$and$\text{H}{{\text{e}}^{\text{4}}}$ nuclei is respectively 5.60 MeV and 7.06 MeV, then the energy of the reaction $L{{i}^{7}}+p\to {{2}_{2}}H{{e}^{4}}$ A)  19.6 MeV                   B)  2.4 MeVC)  8.4 MeV                     D)  17.3 MeV

Binding energy of $L{{i}^{7}}=$(binding energy per nucleon) $\times$(number of nucleons) $=\text{ }5.60\times 7=39.20\text{ }MeV.$ Similarly, Binding energy of $H{{e}^{4}}=7.06\times 4=28.24\,MeV.$ $\therefore$Binding energy of two $H{{e}^{4}}$nuclei $=28.24\times 2$ = 56.48 MeV Hence, the energy of reaction is (56.48 - 39.20) = 17.28 MeV Hence, the correction option is [d].