The energy released in the fission of a 92U235 nucleus is 200 MeV. The fission rate of
92U235
fuelled reactor operating at a power level of 3.2 kW is
elect one:
Da, 1010g1
2b. 101281
I c. 1015 si
O d. 1014 g
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Explanation:
Given The energy released in the fission of a 92 U235 nucleus is 200 MeV. The fission rate of 92 U235 fueled reactor operating at a power level of 3.2 kW is
- So energy released in the fission is 200 Mev. We need to calculate the fission rate.
- Now energy E = 200 Mev = 200 x 10^6 x 1.6 x 10^-19 and power P = 3.2 KW = 3200 W
- Now fission rate / sec = P / E
- = 3200 / 200 x 10^6 x 1.6 x 10^-19
- = 16 / 1.6 x 10^-13
- = 10 x 10^13
- = 10^14 fission / sec
Reference link will be
https://brainly.in/question/12105988
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