1.00 kg of `.^(235)U` undergoes fission process. If energy released per event is `200 MeV`, then the total energy released is

Options

A. `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

c

Explanation

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`.