A star initially has 1040 deuterons. It produces energy via the processes 21H+21H→31H+10pand21H+31H→42He+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
[Given M (2H)=2.014u,M(n)=1.008u,(p)=1.008u, and M(4He)=4.001u]
A
106s
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B
108s
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C
1012s
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D
1016s
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Solution
The correct option is C1012s Adding two equations gives, 321H⟶42He+n+p