AP EAMCET · PHYSICS · Nuclear Physics
Each nuclear fission of \({ }^{235} \mathrm{U}\) releases 200 MeV of energy. If a reactor generates 1 MW power, then the rate of fission in the reactor is
- A \(3.125 \times 10^6\)
- B \(3.125 \times 10^8\)
- C \(3.125 \times 10^{10}\)
- D \(3.125 \times 10^{16}\)
Answer & Solution
Correct Answer
(D) \(3.125 \times 10^{16}\)
Step-by-step Solution
Detailed explanation
\(N = \frac{P}{E}\) \(N = \frac{1 \times 10^6 \text{ W}}{200 \times 10^6 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV}}\) \(N = \frac{1 \times 10^6}{3.2 \times 10^{-11}}\) \(N = 3.125 \times 10^{16}\) fissions/s
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