A star has 100% helium composition. It starts to convert three 4He into one 12C via triple alpha process as 4He + 4He + 4He → 12C + Q.

Question

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 \({ }^{12} \mathrm{C}=12 \ \mathrm{u}\)]

Answer 1

Correct answer is : 15 

\({ }^{4} \mathrm{He}+{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+\mathrm{Q}\) 

power generated \(=\frac{\mathrm{N}}{\mathrm{t}} \mathrm{Q}\)

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

\(\mathrm{Q}=\left(3 \mathrm{m}_{\mathrm{He}}-{m}_{\mathrm{C}}\right) \mathrm{C}^{2}\)

\(\mathrm{Q}=(3 \times 4.0026-12)\left(3 \times 10^{8}\right)^{2}\)

\(\mathrm{Q}=7.266\ \mathrm{MeV}\) 

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

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

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

Hence, \(\mathrm{n}=15\)