A star initially has `10^(40) ` deuterons. It produces energy via the process `_(1)H^(2) + _(1)H^(2) + rarr _(1) H^(3) + p. `and `_(1)H^(2) + _(1)H^(3) + rarr _(2) He^(4) + n` .If the average power radiated by the state is `10^(16) W` , the deuteron supply of the star is exhausted in a time of the order of .
The masses of the nuclei are as follows:
`M(H^(2)) = 2.014 amu,`
`M(p) = 1.007 amu, M(n) = 1.008 amu, M(He^(4)) = 4.001 amu`.

Options

A. `10^(6)s`
B. `10^(8)s`
C. `10^(12)s`
D. `10^(16)s`

Answer

C

Explanation

As `._(1)^(2)H+._(1)^(2)Hto._(1)^(3)H+p `
`._(1)^(2)H+._(1)^(3)Hto._(2)^(4)He+n`
`3._(1)^(2)Hto._(2)^(4)He+n+p`
Mass defect, `Deltam=3xx2.14-(4.001+1.008+1.007)`
`Deltam=0.026u`
Energy released by 3 deutrons
`=(0.026u)(931.5MeV//u)=24.22MeV`
Energy released by `10^(40)` deutrons
`=24.22/3xx10^(40)xx1.6xx10^(-13)J=12.9xx10^(27)J`
`"Time (t)" =("energy")/("power") =(12.9xx10^(27)J)/(10^(16)J//s)`
`~~1.29xx10^(12)S~~10^(12)sec`