$(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

$t=\large\frac{1.29\times10^{28}}{10^{16}}$

$t=1.29\times10^{12} s\;.$

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