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 ${}_1{H}^2{+}_1{H}^3{\to }_{2}H{e}^4+n$. 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

[The mass of nuclei are as follows: $M\left(H^2\right)=2.014amu,M\left(n\right)=1.008amu,$

$M\left(p\right)=1.007amu,M\left(H{e}^4\right)=4.001amu$ }

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

Answer 1

The correct option is C ${10}^{12}s$

Adding two equations gives:
$3_1^{2}H⟶{}_2^{4}He+n+p$

Mass-Energy conversion:
$(3\times 2.014-4.001-2\times 1.008){3\times {10}^8}^2$

Therefore, energy released by $3$ deuterons is $3.73\times {10}^{-12}J$
$1u=1.66\times {10}^{-27}kg$

Given, average power of a star is ${10}^{16}W$.

If number of deuterons lost per second is $N$ then:

$N\times \frac{3.73\times {10}^{-12}}{3}={10}^{16}\Rightarrow N=\frac{1}{1.24}\times {10}^{28}$
Deuteron supply will be exhausted in $\frac{{10}^{40}}{\frac{1}{1.24}\times {10}^{28}}=1.24\times {10}^{12}s$