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`A U^(235)` reactor generated power at a rate of P producting `2xx10^(18)` fussion per second. The energy released per fission is 185 MeV. The value of P isA. `59.2` MWB. `370xx10^(18)MW`C. `0.59 MW`D. 370 MW |
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Answer» Correct Answer - A Power, `P=(nE)/(t)` Given, `n=2xx10^(18)` fission per second E=185MeV Here, `P=(2xx10^(18)xx185xx10^(6)eV)/(1)` `implies=2xx10^(18)xx185xx10^(6)xx16xx10^(-19)=59.2MW` |
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