Triple alpha reaction

The given nuclear reaction is $3\,{}^4\text{He} \rightarrow {}^{12}\text{C} + Q$

We first find the energy released in one such reaction.

Mass defect per reaction

Mass of three helium nuclei: $3 \times 4.0026\,u = 12.0078\,u$

Mass of carbon nucleus produced: $12.0000\,u$

Hence mass defect is $\Delta m = 12.0078 - 12.0000 = 0.0078\,u$

Energy released per reaction

Using $1\,u = 931.5\,\text{MeV}$ the energy released is $Q = 0.0078 \times 931.5\,\text{MeV}$ $Q \approx 7.2657\,\text{MeV}$

Converting to joules: $1\,\text{MeV} = 1.6 \times 10^{-13}\,\text{J}$ so $Q \approx 7.2657 \times 1.6 \times 10^{-13}$ $Q \approx 1.1625 \times 10^{-12}\,\text{J}$

Rate of reactions

The star generates energy at the rate $P = 5.808 \times 10^{30}\,\text{W}$

If the reaction rate is $R$, then $P = RQ$ Therefore, $R = \frac{P}{Q}$ $R = \frac{5.808 \times 10^{30}}{1.1625 \times 10^{-12}}$ $R \approx 4.996 \times 10^{42}\,\text{s}^{-1}$

Thus, $R \approx 5 \times 10^{42}\,\text{s}^{-1}$

Value of $n$

Comparing with $R = n \times 10^{42}\,\text{s}^{-1}$ we get $n = 5$

Comparison with stored answer

Stored correct answer: $5$

Our derived answer matches it.