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 sec`
B. `10^8 sec`
C. `10^12 sec`
D. `10^16 sec`

Answer

C

Explanation

( c) Mass defect `= 3 xx 2.014 - 4.001 - 1.007 - 1.008 = 0.026 amu = 0.026 xx 931 xx 10^6 xx 1.6 xx 10^-19 J`
=`3.82 xx 10^-12 J`
Power of star `= 10^16 W`
Number of deuterons used `= (10^16)/(Delta M) = 0.26 xx 10^28`
Deuteron supply exhausts in `(10^40)/(0.26 xx 10^28) = 10^12 s`.