1.

1.00 kg of `.^(235)U` undergoes fission process. If energy released per event is `200 MeV`, then the total energy released isA. `5.12 xx 10^(24)MeV`B. `6.02 xx 10^(23)MeV`C. `5.12 xx 10^(16)MeV`D. `6.02 xx 10^(6)MeV`

Answer» Correct Answer - c
The number of nuclei in `1 kg .^(235)U` is
`N=(N_(A))/(235)xx(1xx10^(23))`
`N=(6.023 xx 10^(23))/(235) xx 10^(3) =2.56 xx 10^(24) "nuclei"`
Total energy released is
`E =N xx 200 MeV`
`=5.12 xx10^(26)MeV`.


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