TS EAMCET · Physics · Nuclear Physics
113. Energy released in the fission of a single uranium nucleus is \(200 \mathrm{MeV}\). Then the number of fissions per second to produce \(5 \mathrm{~mW}\) power is
- A \(1.56 \times 10^8\)
- B \(1.56 \times 10^{13}\)
- C \(3.12 \times 10^8\)
- D \(3.12 \times 10^{13}\)
Answer & Solution
Correct Answer
(A) \(1.56 \times 10^8\)
Step-by-step Solution
Detailed explanation
Number of fissions per second, \(\begin{aligned} & \mathrm{n}=\frac{5 \times 10^{-3}}{200 \times 10^6 \times 1.6 \times 10^{-19}} \\ & =\frac{50}{32} \times 10^8=1.56 \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}\)
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