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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?
  • a)
    3.12 x 1013
  • b)
    3.12 x 1013
  • c)
    3.1 x 1017
  • d)
    3.12 x 1019
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
If 200 Mev energy is released in the fission of a single nucleus of 92...
Given: Energy released in the fission of a single nucleus of 92U235 = 200 MeV
Power required = 1 kW = 1000 J/s

To find: Number of fissions per second required to produce a power of 1 kW

Formula: Energy released per fission = 200 MeV = 3.2 × 10^-11 J

1. Calculation:
Number of fissions required per second = Power required / Energy released per fission
= 1000 J/s / 3.2 × 10^-11 J/fission
= 3.12 × 10^13 fissions/s

2. Answer:
The number of fissions per second required to produce a power of 1 kW is 3.12 × 10^13 fissions/s, which is option A.
Community Answer
If 200 Mev energy is released in the fission of a single nucleus of 92...
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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?a)3.12 x 1013b)3.12 x 1013c)3.1 x 1017d)3.12 x 1019Correct answer is option 'A'. Can you explain this answer?
Question Description
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