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} $ .
BETA
