JEE Main 2024 — 9 April, Shift 1 — Question 52

$\quad{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+\mathrm{Q}$

power generated $=\frac{\mathrm{N}}{\mathrm{t}} \mathrm{Q}$

where, $\mathrm{N} \rightarrow$ No. of reaction $/ \mathrm{sec}$.

$\mathrm{Q}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right) \mathrm{C}^{2}$ $\mathrm{Q}=(3 \times 4.0026-12)\left(3 \times 10^{8}\right)^{2}$

$\mathrm{Q}=7.266 \mathrm{MeV}$

$\frac{\mathrm{N}}{\mathrm{t}}=\frac{\text { power }}{\mathrm{Q}}=\frac{5.808 \times 10^{30}}{7.266 \times 10^{6} \times 1.6 \times 10^{-19}}$

$\frac{\mathrm{N}}{\mathrm{t}}=5 \times 10^{42}$

rate of conversion of ${ }^{4} \mathrm{He}$ into ${ }^{12} \mathrm{C}=15 \times 10^{42}$

Hence, $\mathrm{n}=15$