Question

A star initially has $10^{40}$ deuterons. It produces energy via the processes ${}_1^{2}H+{}_1^{2}H\to {}_1^{3}He+p$ and ${}_1^{2}H+{}_1^{3}H\to {}_2^{4}He+n$. 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 [Given: $M\left({}^{2}H\right)=2.014u,M(n)=1.008u,M(p)=1.008u$, and $M\left({}^{4}He\right)=4.001u$ ]

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

Answer 1

Answer: (C)

The net reaction is
$3\left({}_1^{2}H\right)\to {}_2^{4}He+n+p$
$Q=\left[3\times m\left({}^{2}H\right)-m\left({}^{4}He\right)-m(n)-m(p)\right]\times 931MeV$
$=3.87\times 10^{-12}J$
This is the energy produced by the consumption of $3$ deuteron atoms.
So, the total energy released by $10^{40}$ deuterons is
$\frac{3.87\times 10^{-12}}{3}\times 10^{40}=1.29\times 10^{28}J$
Let total supply of deuterons in star be exhausted in $t$ seconds.
Then,
$10^{16}\times t=1.29\times 10^{28}$
$\Rightarrow t=1.29\times 10^{12}s$