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A star initially has $\;10^{4}\;$ deuterons . It produces energy via the process \[\] $\;_{1}H^{2}+_{1}H^{2} \to _{1}H^{3}+P\;$ and \[\] $\;_{1}H^{2}+_{1}H^{3} \to _{2}He^{4}+n\;$\[\] If average power radiated by star is $\;10^{16} W\;$ the deutron supply of star is exhausted in a time of order of

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

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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
$t=\large\frac{1.29 \times10^{28}}{10^{16}}=1.29 \times 10^{12} s$
$\approx 10^{12} s\;.$
answered Feb 27, 2014 by yamini.v

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