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\times 10^{14} $

B) $ 0.5\times 10^{12} $

C) $ 5\times 10^{12} $

D) $ 5\times 10^{14} $

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Answer:

Correct Answer: D

Solution:

Energy released in the fission of one nucleus = 200 MeV
$=200\times 10^{6}\times 1.6\times {{10}^{-19}}J=3.2\times {{10}^{-11}}J $

$ P=16KW=16\times 10^{3}Watt $ Now, number of nuclei required per second

$ n=\frac{P}{E}=\frac{16\times 10^{3}}{3.2\times {{10}^{-11}}}=5\times 10^{14} $ .



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