A star has $100\%$ helium composition. It starts to convert three $^4\text{He}$ into one $^{12}\text{C}$ via the triple alpha process as $^4\text{He} + {}^4\text{He} + {}^4\text{He} \to {}^{12}\text{C} + Q$. The mass of the star is $2.0\times10^{32}\ \text{kg}$ and it generates energy at the rate of $5.808\times10^{30}\ \text{W}$. The rate of converting these $^4\text{He}$ to $^{12}\text{C}$ is $n\times10^{42}\ \text{s}^{-1}$, where $n$ is ______. [Take, mass of $^4\text{He} = 4.0026\ \text{u}$, mass of $^{12}\text{C} = 12\ \text{u}$]
Numerical value type. Enter your answer.
Answer: 15
Energy per reaction:
$$Q = (3\times4.0026 - 12)\times931.5\ \text{MeV} = 0.0078\times931.5 \approx 7.27\ \text{MeV} \approx 1.163\times10^{-12}\ \text{J}$$
Reactions per second: $\dfrac{5.808\times10^{30}}{1.163\times10^{-12}} \approx 5\times10^{42}\ \text{s}^{-1}$.
Each reaction converts three $^4\text{He}$ nuclei, so the rate of converting $^4\text{He}$ is $3\times5\times10^{42} = 15\times10^{42}\ \text{s}^{-1}$, giving $n = 15$.
Solution by Sreeraj P, M.Sc Physics