A star has 100 % helium composition. It… - JEE Main 2024 Physics PYQ

JEE Main 2024 — Physics Modern Physics

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A star has $100 \%$ helium composition. It starts to convert three ${ }^4 \mathrm{He}$ into one ${ }^{12} \mathrm{C}$ via triple alpha process as ${ }^4 \mathrm{He}+{ }^4 \mathrm{He}+{ }^4 \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+\mathrm{Q}$ . The mass of the star is $2.0 \times 10^{32} \mathrm{~kg}$ and it generates energy at the rate of $5.808 \times 10^{30} \mathrm{~W}$ . The rate of converting these ${ }^4 \mathrm{He}$ to ${ }^{12} \mathrm{C}$ is $\mathrm{n} \times 10^{42} \mathrm{~s}^{-1}$ , where $\mathrm{n}$ is _________ [ Take, mass of ${ }^4 \mathrm{He}=4.0026 \mathrm{u}$ , mass of ${ }^{12} \mathrm{C}=12 \mathrm{u}$ ]

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Held on 9 Apr 2024 · Verified 6 Jul 2026.

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Solution

${ }^4 \mathrm{He}+{ }^4 \mathrm{He}+{ }^4 \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+\mathrm{Q}$ power generated $=\frac{\mathrm{N}}{\mathrm{t}} \mathrm{Q}$ where, $\mathrm{N} \rightarrow$ No. of reaction $/ \mathrm{sec}$ . $\begin{aligned} & \mathrm{Q}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right) \mathrm{C}^2 \\ & \mathrm{Q}=(3 \times 4.0026-12)\left(3 \times 10^8\right)^2 \\ & \mathrm{Q}=7.266 \mathrm{MeV}\end{aligned}$ $\begin{aligned} & \frac{\mathrm{N}}{\mathrm{t}}=\frac{\text { power }}{\mathrm{Q}}=\frac{5.808 \times 10^{30}}{7.266 \times 10^6 \times 1.6 \times 10^{-19}} \\ & \frac{\mathrm{N}}{\mathrm{t}}=5 \times 10^{42}\end{aligned}$ rate of conversion of ${ }^4 \mathrm{He}$ into ${ }^{12} \mathrm{C}=15 \times 10^{42}$ Hence, $\mathrm{n}=15$

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