A star has 100% helium composition. It starts to convert three into one via triple alpha process as . The mass of the star is kg and it generates energy at the rate of W. The rate of converting these is , where n is

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Solution

Q.
A star has 100% helium composition. It starts to convert three ${}^{4}H e$ into one ${}^{12}C$ via triple alpha process as ${}^{4}H e + {}^{4}H e + {}^{4}H e \to {}^{12}C + Q$ . 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 ${}^{4}H e$ is $n \times 10^{42}$ $s^{- 1}$ , where $n$ is _______ .

[Take, mass of ${}^{4}H e$ = 4.0026 u, mass of ${}^{12}C$ = 12 u] [2024]

Ans.
(5)    ${}_{2}^{4}H e + {}_{2}^{4}H e + {}_{2}^{4}H e \to {}_{6}^{12}C + Q \Rightarrow 3 {}_{2}^{4}H e \to {}_{6}^{12}C + Q$

   $Q = (3 m_{{}^{4}H e} - m_{{}^{12}C}) c^{2}$

   $Q = (3 \times 4.0026 - 12) {(3 \times 10^{8})}^{2} = 7.266 MeV$

  Power generated $P = \frac{N}{t} Q$ (N = No. of reactions/sec)

   $\frac{N}{t} = \frac{P}{Q} = \frac{5.808 \times 10^{30}}{7.266 \times 10^{6} \times 1.6 \times 10^{- 19}} = 5 \times 10^{42}$

Rate of conversion of ${}^{4}H e$ in ${}^{12}C = 5 \times 10^{42}$

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