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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 1eV=1.6×10-19J. close to


A

a) 60 MW

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B

b) 54 MW

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C

c) 125 MW

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D

d) 35 MW

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Solution

The correct option is A

a) 60 MW


Step 1: Given terms:

Avogadro number, NA = 6.023×1026perkilomole

1eV=1.6×1019J

Massnumberofuranium,m=235gmTotalmass.M=2kg

Step 2: Formula used

n(moles)=Mm

Numberofnucleus(N)=NA×n

Power=Nt(sec)

Step 3: Find the number of moles

The number moles can be found out by using the formula:

n(moles)=Mm

Where M is the given mass and m is the molar mass

n(moles)=2kg235gm=2000235

Step 4: Find the number of nucleus

By substituting the given values in the formula to find number of nucleus, we get:

Numberofnucleus=6.022×1023×2000235=51.25×1023=102.5×1025MeV

Now, to covert eV into Joules:

N=102.5×1025×106×1.6×10-16J=164×106MJ

Step 5: Find the required value of power

By substituting the given values in the formula to find power, we get:

Power=Nt(sec)

Power=164×106MJ30×24×60×60sec=0.063×103MW=60MW

Hence the answer is option a ) 60 MW


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