A star has 10^{40} deuterons. It produces ene | vedclass

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

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$10^{12} \ s$

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(C) The net reaction is obtained by adding the two processes:$3(_1H^2) 	o _2He^4 + p + n$Calculate the mass defect $\Delta m$:$\Delta m = 3 	imes m(_1H^2) - [m(_2He^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 net reaction $(Q)$:$Q = 0.026 	imes 931.5 \ MeV = 24.219 \ MeV = 24.219 	imes 1.6 	imes 10^{-13} \ J \approx 3.875 	imes 10^{-12} \ J$Since $3$ deuterons are consumed to release $Q$ energy, energy per deuteron is $Q/3 \approx 1.29 	imes 10^{-12} \ J$.Total energy available $E_{total} = (10^{40} / 3) 	imes Q = 1.29 	imes 10^{28} \ J$.Time taken $t = E_{total} / P = (1.29 	imes 10^{28}) / 10^{16} = 1.29 	imes 10^{12} \ s$.Thus, the order of time is $10^{12} \ s$.

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