A star initially has 10^{40} deuterons. It pr | vedclass

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(C) The net reaction is obtained by adding the two given processes:$3({}_1H^2) 	o {}_2He^4 + p + n$Calculate the mass defect $(\Delta m)$:$\Delta m =

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$10^{12} \, sec$

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(C) The net reaction is obtained by adding the two given processes:$3({}_1H^2) 	o {}_2He^4 + p + n$Calculate the mass defect $(\Delta m)$:$\Delta m = 3 	imes M(H^2) - [M(He^4) + M(p) + M(n)]$$\Delta m = 3(2.014) - [4.001 + 1.007 + 1.008] = 6.042 - 6.016 = 0.026 \, amu$Calculate the energy released $(Q)$ per net reaction:$Q = 0.026 	imes 931.5 \, MeV \approx 24.22 \, MeV$$Q = 24.22 	imes 1.6 	imes 10^{-13} \, J \approx 3.875 	imes 10^{-12} \, J$Total energy produced by $10^{40}$ deuterons:Since $3$ deuterons are used per reaction,total reactions $N = \frac{10^{40}}{3}$.Total Energy $E_{total} = N 	imes Q = \frac{10^{40}}{3} 	imes 3.875 	imes 10^{-12} \approx 1.29 	imes 10^{28} \, J$Time taken to exhaust the supply $(t)$:$t = \frac{E_{total}}{Power} = \frac{1.29 	imes 10^{28}}{10^{16}} = 1.29 	imes 10^{12} \, sec$Thus,the time is of the order of $10^{12} \, sec$.

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