Question

A star initially has ${10}^{40}$ deuterons. It produces energy through the processes
${}_1{H}^2$ ${+}_1$ ${}_1{H}^2$ ${\to }_2$ $H^3$ + P and ${}_1{H}^2$ ${+}_1$ ${\to }_2$ ${He}^4$ + $n$
If the average power radiated by the star is ${10}^{16}$ $W$, the deutron supply of the star is exhausted in a time of the order of

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

Answer 1

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

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 ${10}^{16}W$
Therefore, number of deuterons lost per second is: $N\times \frac{3.73\times {10}^{-12}}{3}={10}^{16}\Rightarrow N=\frac{1}{1.24}\times {10}^{28}$
Deuteron supply will be exhausted in $\frac{{10}^{40}}{\frac{1}{1.24}\times {10}^{28}}=1.24\times {10}^{12}s$