$(a)\;10^{11} s\qquad(b)\;10^{12} s\qquad(c)\;10^{13} s\qquad(d)\;10^{14} s$

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Answer : (b) $\;10^{14} s$

Explanation :

The given reactions are

$\;_{1}H^{2}+_{1}H^{2} \to _{1}H^{3}+P\;$

$\;_{1}H^{2}+_{1}H^{3} \to _{2}He^{4}+n\;$

$\; 3 _{1}H^{2} \to _{2}He^{4}+n+P\;$

Mass defect

$\bigtriangleup m=(3\times2.014-4.001-1.007-1,008)amu$

$=0.026 amu$

Energy released $\;=0.026\times931 MeV$

$=0.026\times931\times1.6\times10^{-13} J$

$=3.87 \times10^{-12} J$

This energy produced by consumption of three deuteron atoms .

Total energy released by $\;10^{40}\;$ deuterons .

$=\large\frac{10^{40}}{3}\times3.87\times10^{-12} J$

$=1.29\times10^{28} J$

The average power radiated is $\;P=10^{16} W$

or $\;10^{16} J/s$

Total time to exhaust all deuterons of stars will be

$t=\large\frac{1.29 \times10^{28}}{10^{16}}=1.29 \times 10^{12} s$

$\approx 10^{12} s\;.$

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