# In a reactor, 2 kg of  fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, per kilo mole and  The power output  (in MW) of the reactor is close to: Option: 1 60 Option: 2 35 Option: 3 125 Option: 4 54

Number of uranium atoms in 2 kg=$\frac{2 \times 6.023 \times 10^{26}}{235}$

energy from one atom is $200 \times 10^{6} \ e.v.$

hence total energy from 2 kg uranium$=\frac{2 \times 6.023 \times 10^{26}}{235} \times 200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}$
And 2kg uranium is used in 30 days hence this energy is received in 30 days hence energy received per second or power is
$\text { Power }=\frac{2 \times 6.023 \times 10^{26} \times 200 \times 10^{6} \times 1.6 \times 10^{-19}}{235 \times 30 \times 24 \times 3600}$
$Power =63.2 \times 10^{6} \ Watt\ \ or \ \ 63.2 \ Mega \ \ Watt$

So  60 MW is the answer as it is closest to 63.2 MW

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