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On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output of atomic reactor is 1.6 MW, then the rate of fission will be
Options
(a) 8×10¹⁶/s
(b) 20×10¹⁶/s
(c) 5×10²²/s
(d) 5×10¹⁶/s
Correct Answer:
5×10¹⁶/s
Explanation:
Energy released per fission of uranium = 200 × 10⁶ × 1 × 10⁻¹⁹ J
Power output = 1.6 × 10⁶ W
Number of fission /s = 1.6 × 10⁶ / 200 × 10⁶ × 1 × 10⁻¹⁹ = 5 × 10¹⁶ /s
This is the rate of fission.
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A proton and helium nucleus are shot into a magnetic field at right angles to the field
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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