On bombarding U²³⁵ by slow neutron, 200 MeV energy is released. If the power output

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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.

Related Questions:

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Topics: Atoms and Nuclei (136)
Subject: Physics (2479)

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