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

Triple alpha process: $$3 \times {}^4\text{He} \to {}^{12}\text{C} + Q$$

Mass defect = $$3 \times 4.0026 - 12 = 12.0078 - 12 = 0.0078$$ u.

Energy released per reaction: $$Q = 0.0078 \times 931.5$$ MeV = $$7.2657$$ MeV = $$7.2657 \times 1.6 \times 10^{-13}$$ J = $$1.1625 \times 10^{-12}$$ J.

Rate of energy generation = $$5.808 \times 10^{30}$$ W.

Number of reactions per second = $$\frac{5.808 \times 10^{30}}{1.1625 \times 10^{-12}} = 4.997 \times 10^{42} \approx 5 \times 10^{42}$$ s$$^{-1}$$.

Each reaction converts 3 He atoms to 1 C atom. Rate of conversion of He = $$3 \times 5 \times 10^{42} = 15 \times 10^{42}$$ s$$^{-1}$$.

So $$n = 15$$.

The answer is $$\boxed{15}$$.