Physics, asked by amulyachowdary51, 8 months ago

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​

Answers

Answered by knjroopa
0

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

Similar questions