You are given that the mass of ${}^7_3\text{Li} = 7.0160$ u, mass of ${}^4_2\text{He} = 4.0026$ u and mass of ${}^1_1\text{H} = 1.0079$ u. When $20$ g of ${}^7_3\text{Li}$ is converted into ${}^4_2\text{He}$ by proton capture, the energy liberated (in kWh) is: [Mass of nucleon $= 1$ GeV/$c^2$]
Answer: (D) $1.33\times10^6$
Reaction: ${}^7_3\text{Li} + {}^1_1\text{H} \rightarrow 2\,{}^4_2\text{He}$
Mass defect: $\Delta m = 7.0160 + 1.0079 - 2(4.0026) = 0.0187$ u
Energy per reaction: $0.0187\times931.5 \approx 17.4$ MeV $\approx 2.79\times10^{-12}$ J
Number of Li nuclei in $20$ g: $N = \dfrac{20}{7}\times6.02\times10^{23} \approx 1.72\times10^{24}$
Total energy: $1.72\times10^{24}\times2.79\times10^{-12} \approx 4.8\times10^{12}$ J
$$\frac{4.8\times10^{12}}{3.6\times10^6} \approx 1.33\times10^6\ \text{kWh}$$
Solution by Sreeraj P, M.Sc Physics