A star initially has ${10}^{40}$ deuterons. It produces energy via the processes  ${ }_1{H}^2{+}_1{H}^2{\to }_1{H}^3+p{\&}_1{H}^2{+}_1{H}^3\to 2H{e}^4+n$. 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 
(mass of  ${ }_1{H}^2=2.014amu,$ mass of ${ }_{2}H{e}^4=4.001amu,$ $m_p=1.007amu,$ $m_n=1.008amu$ )

$$\begin{gathered} { }_1{H}^2{+}_1{H}^2{\to }_1{H}^3+p { \\ }_1{H}^2{+}_1{H}^3{\to }_{1}H{e}^4+n \\ {3}_1{H}^2{\leftarrow }_{2}H{e}^4+n+p \end{gathered}$$

Mass defect
$\Delta m$= (3 x 2.014 – 4.001 – 1.007 – 1.008)amu = 0.026amu
Energy released = 0.26 x 931 MeV
$=0.26x931x1.6x{10}^{-12}J=3.87x{10}^{-13}J$

This is the energy produced by the consumption of three deuteron atoms.
Therefore total energy released by ${10}^{40}$ deuterons
$=\frac{{10}^{40}}{2}\times 3.87\times {10}^{-12}J=1.29\times 1028J$

the average power radiated is P = 10¹⁶W or 10¹⁶Js. Therefore, the total time to exhaust all deuterons of the star, time taken will be

$t=\frac{1.29\times {10}^{28}}{{10}^{16}}=1.29\times {10}^{12}\approx {10}^{12}$