Nuclear Physics and RadioactivityHard
Question
A star initially has 1040 deuteron. It energy via the processes 1H2 + 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 the order of :
Options
A.106s
B.108s
C.1012s
D.1016s
Solution
Add equation 312H → 12He + p + n
(A) Q = - [m (24He) + m (p) + m (n) + 3m (12H)] × 931 MeV
(B) 1016 W =
(A) Q = - [m (24He) + m (p) + m (n) + 3m (12H)] × 931 MeV
(B) 1016 W =
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