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Question: Answered & Verified by Expert
If 200 MeV energy is released in the fission of a single nucleus of  92U235. How many fissions must occur per second to produce a power of 1 kW?
PhysicsNuclear PhysicsJEE Main
Options:
  • A 3.125 × 1013
  • B 6.250× 1013
  • C 1.525 ×1013
  • D None of these
Solution:
2098 Upvotes Verified Answer
The correct answer is: 3.125 × 1013
We know that 1 kW = 1 × 103 Js-1

Also, 1.6 × 10-9 J = 1 eV

200 MeV= 200 × 1.6 × 10-19 × 106 J

Number of fissions = PowerEnergy released

= 103200 × 1.6 ×10-13= 3.125 × 1013

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