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

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(C) The net reaction is: $3(_1H^2) 	o _2He^4 + p + n$.Mass defect $\Delta m = 3 	imes M(H^2) - [M(He^4) + M(p) + M(n)]$$\Delta m = 3(2.014) - [4.001

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$10^{12} \, sec$

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(C) The net reaction is: $3(_1H^2) 	o _2He^4 + p + n$.Mass defect $\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$.Energy released per reaction $Q = 0.026 	imes 931.5 \, MeV \approx 24.22 \, MeV$.$Q = 24.22 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J \approx 3.875 	imes 10^{-12} \, J$.Since $3$ deuterons are used to produce $Q$ energy, energy per deuteron is $E_d = Q/3 = 1.29 	imes 10^{-12} \, J$.Total energy available $E_{total} = N 	imes E_d = 10^{40} 	imes 1.29 	imes 10^{-12} = 1.29 	imes 10^{28} \, J$.Time taken $t = E_{total} / P = (1.29 	imes 10^{28}) / 10^{16} = 1.29 	imes 10^{12} \, s$.Thus, the time is of the order of $10^{12} \, s$.

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