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Question

# Assuming that about 200 MeV of energy is released per fission of 92U235 nuclei, then the mass of 92U235 consumed per day in fission reactor of power 1 MW will be approximately

A
102 g
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B
1 g
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C
100 g
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D
10 g
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Solution

## The correct option is B 1 g Given, Power of fission reactor is, P=106 W=106 Js−1 Time required =t=1 day=24×36×102=8.64×104 s Energy produced, U=Pt=106×8.64×104=8.64×1010 J Energy released per fission of 92U235 is, 200 MeV=200×106×1.6×10−19=32×10−12 J The number of 92U235 atoms used is, =8.64×101032×10−12=27×1020 The mass of 92U235 required is, m=(2356×1023)(27×1020)=1.058 g≃1 g <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (B) is the correct answer.

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