JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2024
- Hard
- Numerical
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 \({ }^{12} \mathrm{C}=12 \mathrm{u}\)]
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(15\)
Step-by-step Solution
Detailed explanation
\({ }^4 \mathrm{He}+{ }^4 \mathrm{He}+{ }^4 \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+\mathrm{Q}\)
\(\text { power generated }=\frac{\mathrm{N}}{\mathrm{t}} \mathrm{Q}\)
\(\text { where, } \mathrm{N} \rightarrow \text { No. of reaction/sec. }\)
\(\mathrm{Q}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right) \mathrm{C}^2\)
\(\mathrm{Q}=(3 \times 4.0026-12)\left(3 \times 10^5\right)^2\)
\(\mathrm{Q}=7.266 \mathrm{MeV}\)
\(\frac{\mathrm{N}}{\mathrm{t}}=\frac{\text { power }}{\mathrm{Q}}=\frac{5.808 \times 10^{30}}{7.266 \times 10^6 \times 1.6 \times 10^{-19}}\)
\(\frac{\mathrm{N}}{\mathrm{t}}=5 \times 10^{42}\)
rate of conversion of \({ }^4 \mathrm{He}\) into \({ }^{12} \mathrm{C}=15 \times 10^{42}\)
Hence, \(n=15\)
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