A star initially has `10^40` deuterons. It produces energy via the processes `_1H^2+_1H^2rarr_1H^3+p` `_1H^2+_1H^3rarr_2He^4+n` 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 if the average power radiated by the star is `10^16W`, 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

Mass defect = 3 `("Mass of"H^(2)) - M(p) - M(n) -= M(He^(4))`
`= 3 xx 2.014 -1.007 -1.008 -4.001`
`= 0.026`
Energy = (Mass defect ) `xx 931.5` MeV
`= 0.026 xx 931.5 xx 1.6 xx 10^(-19) xx 10^(6)`
` = 3.82 xx 10^(-12)` J
Power = `10^(16)W`
As each reaction involves 3 deutrons , the total number of reaction involved in the process = `(10^(40))/3` , if each reaction produces and energy `DeltaE` , then
`"Energy" _("total") = (10^(40))/3 xx DeltaE = 1.29 xx 10^(28)`J
`"Energy"_("total") = "Power " xx "Time" rArr "Time (t)" =(1.29 xx10^(28))/(10^(16))`
`= 1.29 xx 10^(12) approx 10^(12)` sec .