Question

A star has 100\% helium composition. It starts to convert three $^4$He into one $^{12}$C via the triple alpha process as \[ ^4\text{He} + ^4\text{He} + ^4\text{He} \rightarrow ^{12}\text{C} + Q. \] 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]

Answer 1

Correct Answer: 5

The given triple alpha process is:

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

The power generated by the star is given as:

\( \text{Power} = \frac{N}{t} Q \),

where \( \frac{N}{t} \) is the number of reactions per second.

The energy released per reaction (\( Q \)) is:

\( Q = (3m_{\text{He}} - m_{\text{C}})c^2 \)

Substituting the given values:

\( Q = (3 \times 4.0026 - 12) \times (3 \times 10^8)^2 \)

\( Q = 7.266 \, \text{MeV} \)

Convert \( Q \) into joules:

\( Q = 7.266 \times 1.602 \times 10^{-13} \, \text{J} \)

\( Q = 1.163 \times 10^{-12} \, \text{J} \)

Rearranging the power equation:

\( \frac{N}{t} = \frac{\text{Power}}{Q} \)

Substitute the values:

\( \frac{N}{t} = \frac{5.808 \times 10^{30}}{1.163 \times 10^{-12}} \)

Simplify:

\( \frac{N}{t} = 5 \times 10^{42} \, \text{s}^{-1} \)

Final Answer: The rate of conversion of \( ^4\text{He} \) to \( ^{12}\text{C} \) is:

\( n = 5 \, (\text{where } \frac{N}{t} = n \times 10^{42} \, \text{s}^{-1}) \).