A star initially has 1040 deuterons. It produces energy via the processes,1H2+1H2→ 1H3+p1H2+1H3→ 2He4+nThe masses of the nuclei are as follows,m(1H2)=2.014 amu ; m(1H3)=3.016 amum(p)=1.007 amu ; m(n)=1.008 amu m(2He4)=4.001 amuIf the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time period of the order of[Take 1 amu=931.5 MeVc2 ]

A star initially has 1040 deuterons. It produces energy via the proces

1.	A star initially has 1040 deuterons. It produces energy via the processes,1H2+1H2→ 1H3+p1H2+1H3→ 2He4+nThe masses of the nuclei are as follows,m(1H2)=2.014 amu ; m(1H3)=3.016 amum(p)=1.007 amu ; m(n)=1.008 amu m(2He4)=4.001 amuIf the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time period of the order of[Take 1 amu=931.5 MeVc2 ]
Answer» A star initially has $10^{40}$ deuterons. It produces energy via the processes, $_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} + p$ $_{1} H^{2} +_{1} H^{3} \to_{2} {He}^{4} + n$ The masses of the nuclei are as follows, $m (_{1} H^{2}) = 2.014 amu; m (_{1} H^{3}) = 3.016 amu m (p) = 1.007 amu; m (n) = 1.008 amu m (_{2} {He}^{4}) = 4.001 amu$ If the average power radiated by the star is $10^{16} W$ , the deuteron supply of the star is exhausted in a time period of the order of [Take $1 amu = 931.5 \frac{MeV}{c^{2}}$ ]

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