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`.

Explanation

`{:((._1H^(2)+._1H^(2),rarr,._1H^(3)+._1p^(3)),(._1H^(2)+._1H^(3),rarr,._2H^(4)+._0n^(1)))/(3._1H^(2)rarr._2He^(4)+._1p^(1)+._0n^(1):}`
Mass defect:
`Deltam = [3xx2.0141 -(4.0026+1.0078+1.0086)]`
`= 0.0233u`
Equivalent energy `= 0.0233xx931.5 = 21.7MeV`
`= 21.7 xx 1.6 xx 10^(-13)`
`= 34.7xx10^(-13)J`
Number of `._(1)H^(2)` atoms involved in reactions `= 3`
Total enegry released `= (10^(40))/(3) xx34.7xx10^(-13) = 11.6xx10^(27)`
`p = (E)/(t)`
`t = (E)/(P) = (11.6 xx 10^(27))/(10^(16)) = 11.6 xx 10^(11)~~10^(12) sec`