Question

A star initially has $10^{40}$ deuterons. It produces energy via the
processes ${}_1{H}^2{+}_1{H}^2{\to }_1{H}^3+p$ and ${}_1{H}^2{+}_1{H}^3{\to }_2{He}^3+n$.
If the average power radiated by the star is $10^{16}$ W, the
deuteron supply of the star is exhausted in a time o f the order of
$M(H{)}^2=2.014amu;$ $M(n)=1.008amu;$
$M(p)=1.007amu;$ $M(He{)}^2=4.001amu;$

• A. ${10}^6$s
• B. ${10}^8$s
• C. ${10}^{12}$s
• D. ${10}^{16}$s

Answer 1

Answer: (C)

The given reactions are :
${}_1{H}^2{+}_1{H}^2{\to }_1{H}^3+p$
${}_1{H}^2{+}_1{H}^3{\to }_2{He}^4+n$.
$3_1{H}^2{\to }_2{He}^4+n+p$.
$△m=(3\times 2.014-4.001-1.007-1.008)amu$
= 0.026 amu
Energy released =$=0.026\times 931MeV$
$=0.026\times 931\times 1.6\times {10}^{-13}J$
$=3.87\times {10}^{-12}J$
This is the energy produced by the consumption of three
deuteron atoms.
$\therefore$ Total energy released by ${10}^{40}$ deuterons