$(a)\;121 gm/day\qquad(b)\;112 gm/day\qquad(c)\;128 gm/day\qquad(d)\;100 gm/day$

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Answer : (a) $\;121 gm/day$

Explanation :

Let us first calculate Q value of nuclear function

$Q=\bigtriangleup mC^2=\bigtriangleup m\;(931) MeV$

$Q=(2\times2.0141-4.0026)\times931MeV$

$=23.834MeV=23.834\times10^{6}eV$

Now efficiency of reactor is $\;25 \%$

$=\large\frac{25}{100}\times23.834\times10^{6}\times1.6\times10^{-19} J$

$=9.534 \times10^{-13} J$

Now $\;9.534\times10^{-13} J\;$ energy is released by fusion of 2 deuterium

$\large\frac{9.534\times10^{-13}}{2} J/deuterium\;$ is released Requirement is $\;200MW=200 \times10^{6} J/s\times86400\;$ for 1 day

no . of deuterium nuclei required = $\;\large\frac{200\times10^{6}\times86400}{\large\frac{9.534}{2}\times10^{-13}}$

$=3.624\times10^{25}$

No.of deucterium nuclei

$=\large\frac{g}{A}\times6\times10^{23}$

$3.624\times10^{25}=\large\frac{g}{2}\times6\times10^{23}$

$g=2\times\large\frac{3.624\times10^{25}}{6\times10^{23}}=120.83 gm/day\;.$

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