A star initially has `10^40` deuterons. It produces energy via the processes `._1^2H+_1^2Hrarr_1^3H+p` and `._1^2H+_1^3Hrarr_2^4He+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 `10^16 W`, the deuteron supply of the star is exhausted in a time of the order of

Options

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

Answer

C

Explanation

The net reaction is
`3(._(1)^(2)H) rarr(._(2)^(4)He) +n+p`
`Q=[3 xx m(.^(2)H) -m(.^(4)He) -m(n) -m(p)]xx931 MeV`
`3.87 xx 10^(-12) J`
This is the energy produced by the consumption of `3` deuteron atoms, So,m the total energy released by `10^(40)` deutron is
`(3.87 xx10^(-12))/(3) xx 10^(40) =1.29 xx 10^(28)J`
Let total supply of deutrons in star be exhausted in t seconds. Then,
`10^(16) xx t=1.29 xx 10^(28)`
`rArr t=1.29 xx 10^(12) s` .