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An atomic power nuclear reactor can deliver `300 MW`. The energy released due to fission of each nucleus of uranium atom `U^238` is `170 MeV`. The number of uranium atoms fissioned per hour will be.A. `30 xx 10^(25)`B. `4 xx 10^(22)`C. `10 xx 10^(2)`D. `5 xx 10^(15)` |
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Answer» Correct Answer - B b) Power = energy/time = `300 xx 10^(8)J//s` energy released due to fission `=170 MeV = 170 xx 10^(6)xx 1.6 xx 10^(-19)` `=27.2 xx 10^(-12)` J Number of atoms fissioned per second. `=(3 xx 10^(8))/(27.2 xx 10^(12)) = (3 xx 10^(20))/(27.2)` Number of atoms fissioned per hour `=(3 xx 10^(20)xx 3600)/(27.2)=(3 xx 36)/(27.2) xx 10^(22) = 4 xx 10^(22)`m |
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