TS EAMCET · Physics · Nuclear Physics
If the number of uranium nuclei required per hour to produce a power of 64 kW is \(7.2 \times 10^{18}\), then the energy released per fission is
- A \(0.64 \times 10^{-10} \mathrm{~J}\)
- B \(3.2 \times 10^{-13} \mathrm{~J}\)
- C \(0.32 \times 10^{-10} \mathrm{~J}\)
- D \(3.2 \times 10^{-10} \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(0.32 \times 10^{-10} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\( E_{fission} = \frac{P \times t}{N_f} = \frac{64 \times 10^3 \, \mathrm{W} \times 3600 \, \mathrm{s}}{7.2 \times 10^{18}} = 0.32 \times 10^{-10} \, \mathrm{J} \)
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