TS EAMCET · Physics · Nuclear Physics
The energy released by the fission of one uranium nucleus is 200 MeV . The number of fissions per second required to produce 128 W power is
- A \(6 \times 10^{12}\)
- B \(2 \times 10^{12}\)
- C \(8 \times 10^{12}\)
- D \(4 \times 10^{12}\)
Answer & Solution
Correct Answer
(D) \(4 \times 10^{12}\)
Step-by-step Solution
Detailed explanation
Energy per fission \(=200 \mathrm{MeV}\) \(\begin{aligned} & =.200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & \text { Fission rate }=\frac{128}{200 \times 10^6 \times 1.6 \times 10^{-19}}=0.4 \times 10^{13} \\ & =4 \times 10^{12}\end{aligned}\)
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