Atoms And Nuclei Question 235

Question: If the energy released in the fission of one nucleus is 200 MeV. Then the number of nuclei required per second in a power plant of 16 kW will be [KCET (Engg.) 2000; CPMT 2001; Pb. PET 2002]

Options:

A) 0.5×1014

B) 0.5×1012

C) 5×1012

D) 5×1014

Show Answer

Answer:

Correct Answer: D

Solution:

Energy released in the fission of one nucleus = 200 MeV
=200×106×1.6×1019J=3.2×1011J

P=16KW=16×103Watt Now, number of nuclei required per second

n=PE=16×1033.2×1011=5×1014 .



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