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]
[Take, mass of $^4$He = 4.0026 u, mass of $^{12}$C = 12 u]
Correct Answer: 5
Approach Solution - 1
To solve the problem, we need to determine the number of triple alpha processes per second, denoted as \( n \times 10^{42} \, \text{s}^{-1} \).
Firstly, calculate the energy released in a single triple alpha process. The reaction is:
\[ ^4\text{He} + ^4\text{He} + ^4\text{He} \rightarrow ^{12}\text{C} + Q \]
The masses of the participating isotopes are:
- Mass of \( ^4\text{He} \) = 4.0026 u
- Mass of \( ^{12}\text{C} \) = 12 u
The initial total mass of the three \( ^4\text{He} \) nuclei:
\[ 3 \times 4.0026 \, \text{u} = 12.0078 \, \text{u} \]
The mass defect \(\Delta m\) is the difference in mass before and after the reaction:
\[ \Delta m = 12.0078 \, \text{u} - 12 \, \text{u} = 0.0078 \, \text{u} \]
The energy released \( Q \) can be calculated using \( E= \Delta m \times 931.5 \, \text{MeV/u} \):
\[ Q = 0.0078 \times 931.5 \, \text{MeV} = 7.2617 \, \text{MeV} \]
Convert the energy to joules (1 MeV = \( 1.60218 \times 10^{-13} \) J):
\[ Q = 7.2617 \times 1.60218 \times 10^{-13} \, \text{J} = 1.163 \times 10^{-12} \, \text{J} \]
Now, find the rate \( n \) using the given power output of the star:
Power \( P = 5.808 \times 10^{30} \, \text{W} \). Since power is energy per unit time,
\[ P = n \times 10^{42} \times 1.163 \times 10^{-12} \, \text{J/s} \]
\[ 5.808 \times 10^{30} = n \times 10^{42} \times 1.163 \times 10^{-12} \]
Solve for \( n \):
\[ n = \frac{5.808 \times 10^{30}}{1.163 \times 10^{-12} \times 10^{42}} \]
\[ n \approx \frac{5.808}{1.163} \times 10^{24} \]
\[ n \approx 5.0 \]
Thus, the value of \( n \) is 5. The computed value fits within the expected range of 5,5, confirming the solution is correct.
Approach Solution -2
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:
\( n = 5 \, (\text{where } \frac{N}{t} = n \times 10^{42} \, \text{s}^{-1}) \).
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