The energy released when $\dfrac{7}{17.13}$ kg of $^7_3\text{Li}$ is converted into $^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32}$ eV. The value of $\alpha$ is ______. (Nearest integer) (Mass of $^7_3\text{Li} = 7.0183$ u, mass of $^4_2\text{He} = 4.004$ u, mass of proton $= 1.008$ u and $1\text{u} = 931\ \text{MeV/c}^2$ and Avogadro number $= 6.0 \times 10^{23}$)
Numerical value type. Enter your answer.
Answer: 6
Reaction: $^7_3\text{Li} + ^1_1\text{p} \to 2\,^4_2\text{He}$.
Mass defect: $7.0183 + 1.008 - 2 \times 4.004 = 0.0183$ u, so $Q = 0.0183 \times 931 \approx 17.04$ MeV per reaction.
Number of Li nuclei (molar mass $7$ g/mol): $\dfrac{7000/17.13}{7} \times 6 \times 10^{23} = \dfrac{1000}{17.13} \times 6 \times 10^{23} \approx 3.50 \times 10^{25}$.
Energy $= 3.50 \times 10^{25} \times 17.04 \times 10^6$ eV $\approx 5.97 \times 10^{32}$ eV, so $\alpha \approx 6$.
Solution by Sreeraj P, M.Sc Physics