Energy released by fission of one atom of U _ 12 ^ 325 is 200 MeV. The number of fission…
Physics · JEE Main · NTA Exams — Atoms and Nuclei
Energy released by fission of one atom of \(\mathrm{U}_{12}^{325}\) is 200 MeV. The number of fission required per sec to produce a power of 1 kW is
- \(3.125 \times 10^{11}\)
- \(3.125 \times 10^{11}\)
- \(3.125 \times 10^{12}\)
- \(3.125 \times 10^{13}\)
Answer
(C) 3.125 10^ 12
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