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Q.
Calculate power output of ${ }_{92}^{235} U$ reactor, if it takes 30 days to use up $2\, kg$ of fuel, and if each fission gives $185\, MeV$ of useable energy. Avogadro's number $=6 \times 10^{23} / mol$ ?
Mass of $^{235}_{92}U$ per second in the reactor is
$m=\frac{2\times10^{3}}{30\times24\times60\times60}=7.72\times10^{-4}$ g/sec
$\therefore $ Number of fissions reaction per second
$=\frac{6\times10^{23}}{235}\times m=\frac{6\times10^{23}\times7.72\times10^{-4}}{235}$
$=1.97\times10^{18}$/sec
Power of nuclear reacto $=1.97\times10^{18}\times185$ MeV/s
$=1.97\times10^{18}\times 10^{6}\times 1.6\times 10^{-19} J/s.$
$=58.3$ MW.