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If 200 MeV energy is released in the fis...

If `200 MeV` energy is released in the fission of a single `U^(235)` nucleus, the number of fissious required per second to produce 1 kilowatt power shall be ( Given 1 eV `=1.6xx10^(-19)J)`

A

10kg

B

1kg

C

1g

D

10g

Text Solution

Verified by Experts

The correct Answer is:
C
As, `n=m/AN_(A)impliesE=m/AN_(A)xx200xx1.6xx10^(-13)J`
Power `=E/t=(mN_(A)xx3.2xx10^(-11))/(235xx24xx3600)=10^(6)`
Mass `=(1xx10^(6)xx24xx3600xx235)/(200xx1.6xx10^(-13)xx6.02xx10^(23))~~1g`
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