For a nuclear reactor, 48 kJ of energy is produced per minute. If the energy released per fission is 3.2 × 10^-11 J, then the number of fission

Question

For a nuclear reactor, 48 kJ of energy is produced per minute. If the energy released per fission is 3.2 x 10^-11 J, then the number of fission which would be taking place in a reactor per second is 

A. 5 × 10¹⁴ B. 2 × 10¹⁴ C. 52 × 10¹³ D. 2.5 × 10¹³

Answer 1

Given:

Total energy released per minute = 48 kJ 

= 48 × 10³ J 

Energy produced per second = \(\frac{(48\,\times\,10^2)}{60}\)

= 0.8 × 10³ J 

Energy release per fission = 3.2 x 10^-11 J

Solution:

Number of fission per second = \(\frac{[Energy\,produced\,per\,second]}{[energy\,released\,per\,fission]}\)

= \(\frac{[0.8\,\times\,10^3]}{[3.2\,\times\,10^{-11}]}\)

= 2.5x10¹³ 

Therefore number of fission per second is 2.5x10¹³ . 

So the correct answer is D.