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On bombardment of `U^235` by slow neutrons, `200 MeV` energy is released. If the power output of atomic reactor is `1.6 MW`, then the rate of fission will beA. `5xx10^(10)`B. `5xx10^(12)`C. `5xx10^(4)`D. `5xx10^(16)` |
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Answer» Correct Answer - D Output power `1.6MW=1.6xx10^(6)J//s` Energy released/fission =200MeV `=200xx1.6xx10^(-13)J` `:.` Nuclear of fission/sec `=(1.6xx10^(6))/(200xx1.6xx19^(-13))=5xx10^(16)` |
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