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A star initially has $\;10^{40}\;$ deutrons . If produces energy via the process $\;_{1}H^{1}+_{1}H^{2} \to _{1}H^{3}+P\;$ and $\;_{1}H^{2}+_{1}H^{3} \to _{2}He^{4}\;.$ If the average power radiated by star is $\;10^{16}W\;,$ the deutron supply of star is exhausted in a time of order of  (The masses of nuclei are $\;m(H^2)=2.014\;amu\;,m(p)=1.007 amu , \; m(n)=1.0084 amu ,\; m(H)=4.001 amu))$

$(a)\;10^{6} s\qquad(b)\;10^{8} s\qquad(c)\;10^{12} s\qquad(d)\;10^{16} s$

Answer : (c) $\;10^{12} s$
Explanation :
$_{1}H^{2}+_{1}H^{2} \to _{1}H^{3}+P$
$_{1}H^{2}+_{1}H^{3} \to _{2}He^{4}+n$
$3\;_{1}H^{2} \to _{2}He^{4}+P+n$
$\bigtriangleup \varepsilon=m\;(_{2}He^{4})m(p)+m(n)-3m\;(_{1}H^{2})$
$\bigtriangleup m=[4.001+1.007+1.008-3(2.014)] amu$
$\bigtriangleup m=-0.026 amu$
$|\bigtriangleup \varepsilon|=C^2\;|\bigtriangleup m|$
$|\bigtriangleup \varepsilon|=(9 \times10^{16})\;(0.026\times1.67\times10^{-27})$
$\bigtriangleup \varepsilon=(931.5)\;(0.026)\;MeV$
$\bigtriangleup \varepsilon=3.87 \times10^{-12}\;J$
As each reaction involves 3 deutrons , so total number of reactions involved in process = $\;\large\frac{10^{40}}{3}\;$if each reaction produces an energy $\;\bigtriangleup \varepsilon\;,$ then
$\varepsilon_{total}=\large\frac{10^{40}}{3}\; \bigtriangleup \varepsilon =1.29\times10^{28} J$
$\varepsilon_{total}=Pt$
Time of exhaustion of star
$t=\large\frac{1.29\times10^{28}}{10^{16}}$
$t=1.29\times10^{12} s\;.$