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Question

# In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N=6.023×1026per kilo mole and 1 eV=1.6×10–19 J. The power output of the reactor is close to:

A
35 MW
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B
60 MW
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C
125 MW
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D
54 MW
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Solution

## The correct option is B 60 MWPower output of the reactor, P=energytime=Et Given that, Energy released in one fission reaction is, E1=200 MeV=200×1.6×10−19 MJ The number of atoms present in 2 kg of 92U235 is given by, N=2235×6.023×1026 ∴Total energy,E=NE1 Here, t=30 days=30×24×60×60 s ⇒P=2235×6.023×1026×200×1.6×10−1930×24×60×60 ⇒P≈60 MW. Hence, option (B) is correct.

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