Each fission of a ²³⁵₉₂U nucleus releases 200 MeV. How many fissions per second are required to produce a power of 1 kW?
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Each fission of a ²³⁵₉₂U nucleus releases 200 MeV. How many fissions per second are required to produce a power of 1 kW?

G1 Physics · Accepted Answer

Energy released per fission = 200 MeV = 200 × 10⁶ × 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹¹ J. For 1 kW = 1000 J s⁻¹, fission rate = 1000/(3.2 × 10⁻¹¹) = 3.125 × 10¹³ s⁻¹. Correct answer: B.