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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.PowerEasy2 minQB

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In a reactor, $2 kg$ of ${}_{92}U^{235}$ fuel is fully used up in $30$ days. The energy released fission is $200 MeV$ . Given that the Avogadro number, $N = 6.023 \times 10^{26}$ per kilo mole and $1 eV = 1.6 \times 10^{- 19} J$ . The power output of the reactor is close to:

Options:

Answer:

Solution:

$P = \frac{E}{t}$

$= \frac{2}{235} \times \frac{6.023 \times 10^{26} \times 200 \times 1.6 \times 10^{- 19}}{30 \times 24 \times 60 \times 60} = 60 MW$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Power

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