A star initially has `10^40` deuterons. It produces energy via the processes `._1H^2+_1H^2rarr_1H^3+p` and `._1H^2+_1H^3rarr_2He^4+n`. 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
(a) `10^6s` (b) `10^8s` (c) `10^12s`
The masses of the nuclei are as follows
`M(H^2)=2.014` amu, `M(n)=1.008` amu,
`M(p)=1.007` amu,`M(He^4)=4.001`amu

Explanation

The two process are:
`._1H^2+._1H^2to._1H^3+p`,
`._1H^2+._1H^3to._2H^4+n`,
Adding the two equations, we get
`3._1H^2to._2H^4+p+n`,
Mass defect,`Deltam`
`=3xx2.014-4.001-1.007-1.008`
`=0.026a m u=0.026xx1.67xx10^(-27)kg`
Total energy released `=(Deltam)(c^2)`
`=0.026xx1.67xx10^(-27)(3xx10^8)^2`
Energy released per neutron
`=(0.026xx1.67xx10^(-27)xx9xx10^(16))/3J`
As average power radiated `=10^6W=10^(16)J//s`
`:.` Number of deutrons consumed/sec
`=(10^(16)xx3)/(0.026xx1.67xx10^(-27)xx9xx10^(16))`
`=(10^(27))/(0.078xx1.67)`
As total supply of deutrons`=10^(40)`
`:.` Time for which supply would last
`t=(10^(40)xx0.078xx1.67)/(10^(27))=1.3xx10^(12)"second"`