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(C) The reaction is $3({ }^4 He) ightarrow { }^{12} C + Q$.The energy released per reaction $Q$ is given by $Q = (3 	imes m_{He} - m_C)c^2$.$Q = (3

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$15$

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(C) The reaction is $3({ }^4 He) ightarrow { }^{12} C + Q$.The energy released per reaction $Q$ is given by $Q = (3 	imes m_{He} - m_C)c^2$.$Q = (3 	imes 4.0026 \ u - 12 \ u)c^2 = (12.0078 - 12) \ u 	imes c^2 = 0.0078 \ u 	imes c^2$.Converting $u$ to $kg$: $Q = 0.0078 	imes 1.66 	imes 10^{-27} \ kg 	imes (3 	imes 10^8 \ m/s)^2$.$Q = 0.0078 	imes 1.66 	imes 10^{-27} 	imes 9 	imes 10^{16} \ J = 1.16568 	imes 10^{-12} \ J$.The power generated is $P = R 	imes Q$,where $R$ is the rate of reaction (number of reactions per second).$R = \frac{P}{Q} = \frac{5.808 	imes 10^{30} \ W}{1.16568 	imes 10^{-12} \ J} \approx 4.9825 	imes 10^{42} \ s^{-1} \approx 5 	imes 10^{42} \ s^{-1}$.Since each reaction consumes three ${ }^4 He$ nuclei,the rate of conversion of ${ }^4 He$ is $3 	imes R = 3 	imes 5 	imes 10^{42} = 15 	imes 10^{42} \ s^{-1}$.Thus,$n = 15$.

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