STANDARD Physics Nuclear Fission MCQ Question
Assuming that 200 MeV of energy is released per fission of ₉₂U²³⁵ atom. Find the number of fission per second required to release 1 kW power :-
Option 1
Option 2
Option 3
Option 4
Correct Answer
Detailed Explanation
To find the number of fissions, convert 1 kW to MeV/s: 1 kW = 10⁶ J/s = 6.242 × 10¹² MeV/s. Divide by energy per fission: 6.242 × 10¹² MeV/s ÷ 200 MeV/fission = 3.121 × 10¹⁰ fissions/s.
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