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If the energy released in the fission of the nucleus is `200 MeV`. Then the number of nuclei required per second in a power plant of `6 kW` will be.A. `0.5 xx 10^14`B. `0.5 xx 10^12`C. `5 xx 10^12`D. `5 xx 10^14` |
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Answer» Correct Answer - D (d) Energy released in the fission of one nucleus =`200 MeV` =`200 xx 10^6 xx 1.6 xx 10^-19 J = 3.2 xx 10^-11 J` `P = 16 KW = 16 xx 10^3 Watt` Now, number of nuclei required per second `n = (P)/( E) = (16 xx 10^3)/(3.2 xx 10^-11) = 5 xx 10^14`. |
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