1. Tardigrade
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  3. Physics
  4. A star initially has 1040 deuterons. It produces energy via the processes 12 H + 12 H arrow 13 He +p and 12 H + 13 H arrow 24 He +n. If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of [Given: M( 2 H )=2.014 u , M(n)=1.008 u , M(p)=1.008 u, and M( 4 He )=4.001 u ]

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Q. A star initially has $10^{40}$ deuterons. It produces energy via the processes ${ }_{1}^{2} H +{ }_{1}^{2} H \rightarrow{ }_{1}^{3} He +p$ and ${ }_{1}^{2} H +{ }_{1}^{3} H \rightarrow{ }_{2}^{4} He +n$. 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: $M\left({ }^{2} H \right)=2.014\, u , M(n)=1.008\, u , M(p)=1.008\, u$, and $M\left({ }^{4} He \right)=4.001\, u$ ]

Nuclei

Solution:

The net reaction is
$3\left({ }_{1}^{2} H \right) \rightarrow{ }_{2}^{4} He +n+p$
$Q=\left[3 \times m\left({ }^{2} H \right)-m\left({ }^{4} He \right)-m(n)-m(p)\right] \times 931\, MeV$
$=3.87 \times 10^{-12} J$
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,
$10^{16} \times t=1.29 \times 10^{28}$
$\Rightarrow t=1.29 \times 10^{12} s$