Question:
In a reactor, $2 \mathrm{~kg}$ of ${ }_{92} \mathrm{U}^{235}$ fuel is fully used up in 30 days. The energy released per fission is $200 \mathrm{MeV}$. Given that the Avogadro number, $\mathrm{N}=6.023 \times 10^{26}$ per kilo mole and $1 \mathrm{eV}$ $=1.6 \times 10^{-19} \mathrm{~J}$. The power output of the reactor is close to:
Correct Option: , 2
Solution:
(2) Power output of the reactor,
$P=\frac{\text { energy }}{\text { time }}$
$=\frac{2}{235} \times \frac{6.023 \times 10^{26} \times 200 \times 1.6 \times 10^{-19}}{30 \times 24 \times 60 \times 60} \simeq 60 \mathrm{MW}$
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