A star initially has 1040 deuterons. It produces energy via the processes1H2+1H2→1H3+p and 1H2+1H3→2He4+n. If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of 11.6×10x s. Then the value of x is -Take m(1H2)=2.0141u, m(2He4)=4.0026u, m(1p1)=1.0078u, m(0n1)=1.0086u1u c2=931.5 MeV

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

1.	A star initially has 1040 deuterons. It produces energy via the processes1H2+1H2→1H3+p and 1H2+1H3→2He4+n. If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of 11.6×10x s. Then the value of x is -Take m(1H2)=2.0141u, m(2He4)=4.0026u, m(1p1)=1.0078u, m(0n1)=1.0086u1u c2=931.5 MeV
Answer» A star initially has $10^{40}$ deuterons. It produces energy via the processes $_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} + p and_{1} H^{2} +_{1} H^{3} \to_{2} {He}^{4} + n$ . If the average power radiated by the star is $10^{16} W$ , the deuteron supply of the star is exhausted in a time of the order of $11.6 \times 10^{x} s$ . Then the value of $x$ is - $Take m (_{1} H^{2}) = 2.0141 u, m (_{2} {He}^{4}) = 4.0026 u, m (_{1} p^{1}) = 1.0078 u, m (_{0} n^{1}) = 1.0086 u 1 u c^{2} = 931.5 MeV$

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