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The energy released by fission of one U^...

The energy released by fission of one `U^(235)` atom is 200 MeV. Calculate the energy released in kWh, when one gram of uranium undergoes fission.

A

`2.278 xx 10^(8)`

B

`2.278 xx 10^(6)`

C

`2.278 xx 10^(4)`

D

`2.278 xx 10^(10)`

Text Solution

Verified by Experts

The correct Answer is:
C
Number of atoms in 1 gm of `U^(235)`
`= ("Avogadro number")/("Atomic weight of U"^(235)) = (6.023 xx 10^(23))/(235)`
Energy released per fission = 200 MeV
Therefore, energy released on fission of 1 g of `U^(235)`
`= (6.023 xx 10^(23) xx 200 xx 1.6 xx 10^(-13))/(235) J (Ws)`
`= (6.023 xx 10^(23) xx 200 xx 1.6 xx 10^(-13))/(235 xx 1000 xx 3600) k Wh`
`= 2.278 xx 10^(4)` kWh
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