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

Answer

C

Explanation

( c) Mass defect,
`Delta m = 3 xx 2.014 - 4.001 -1.007 -1.008`
=`0.026 amu`
=` 0.026 xx 931 xx 1.6 xx 10^-13 J`
Power of star `= 10^16 W`
Number of deuterons required
=`(10)/(Delta m) = 7.75 xx 10^27`
Now, deuteron supply exhausts in
=`(10^40)/(7.75 xx 10^27) = 10^12 sec`.