COMEDK · Physics · 29. Nuclear Physics
On bombarding \(\mathrm{U}^{235}\) by slow neutron, \(200 \mathrm{MeV}\) energy is release(d)If the power output of atomic reactor is \(1.6 \mathrm{MW}\), then the rate of fission will be
- A \(5 \times 10^{22} / \mathrm{s}\)
- B \(5 \times 10^{16} / \mathrm{s}\)
- C \(8 \times 10^{16} / \mathrm{s}\)
- D \(20 \times 10^{16} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(5 \times 10^{16} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Given, output power, \(P_{o}=1.6 \mathrm{MW}\) \[ =1.6 \times 10^{6} \mathrm{~W} \] Energy released per fission,…
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