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Q.

A star initially has 1040 deuterons. It produces energy via the process  12H+12H13H+p and  12H+13H24He+n Where the masses of the nuclei are m(2H)=2.014 amu, m(p)=1.007 amu, m(n)=1.008 amu and m(4He)=4.001 amu. If the average power radiated by the star is 1016W, the deuteron supply of the star is exhausted in a time of the order of 

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a

106s

b

108s

c

1012s

d

1016s

answer is C.

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Detailed Solution

Combining two given equations 

We have 31H2=2He4+p+n

Δm=3×2.0144.0011.0071.008=0.026u

Energy released by 3 deutrons =0.026×931.5×1.6×1013J=3.9×1012J

Now, (1016×t)=(10403)(3.9kt1012)

Solving we get, t1.3×1012s

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A star initially has 1040 deuterons. It produces energy via the process  12H+12H→13H+p and  12H+13H→24He+n Where the masses of the nuclei are m(2H)=2.014 amu, m(p)=1.007 amu, m(n)=1.008 amu and m(4He)=4.001 amu. If the average power radiated by the star is 1016W, the deuteron supply of the star is exhausted in a time of the order of