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^(18)` s
B. `10^(28)` s
C. `10^(12)` s
D. `10^(16)` s

Answer

C

Explanation

( c) Net process is , `3""_(1)H^(2) to ""_(2)He^(4)+p+n`
Mass defect , `Delta m,=(3xx2.014)-0.007-1.008- 4.001`
`=0.26 am u`
Energy released by the fusion deuterous
`=Delta mxx931.48=0.026xx931.48MeV `
`=0.026xx931.481.6zz10^(-13)J=3.87xx10^(-12)J`
`therefore ` total energy released `=(10^(40))/(3)xx3.87xx10^(-12)J=1.29xx10^(28)J`
total time `=(1.29xx10^(28))/(10^(16))=1.29xx10^(12)s( :' "power" =("energy ")/("time "))`