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In a nuclear reactor, the number of `U^(235)` nuclei undergoing fissions per second is `4xx10^(20).` If the energy releases per fission is 250 MeV, then the total energy released in 10 h is `(1 eV= 1.6xx10^(-19)J)`A. `576xx10^(6)J`B. `576xx10^(12)J`C. `576xx10^(15)J`D. `576xx10^(12)J` |
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Answer» Correct Answer - B (B) the fission per second `=4xx10^(20)` So , the energy relased per second `=4xx10^(20) xx250MeV ` `=4xx10^(20) xx250xx10^(6) xx1.6xx10^(-19) J` therefore , energy relased in 10 h`= 36xx10^(3) ` s is `E=36xx10^(3) xx4xx10^(20) xx250xx10^(6) xx1.6xx10^(-19) J` `=36xx4xx250xx1.6xx10^(10) J= 57600xx10^(10)` `=576xx10^(12)J` |
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