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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` |
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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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