AP EAMCET · PHYSICS · Nuclear Physics
If the energy released per fission of a \({ }_{92}^{235} \mathrm{U}\) nucleus is 200 Me V . the energy released in the fission of 0.1 kg of \({ }_{92}^{235} \mathrm{U}\) in kilowatt - hour is.
- A \(22.8 \times 10^5\)
- B \(22.8 \times 10^7\)
- C \(11.4 \times 10^5\)
- D \(850 \times 10^{10}\)
Answer & Solution
Correct Answer
(A) \(22.8 \times 10^5\)
Step-by-step Solution
Detailed explanation
Energy released per fission per atom, \(\mathrm{E}=200 \mathrm{MeV}\) Number of atoms in 0.1 kg of \({ }_{92}^{235} \mathrm{U}\) is…
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