Nuclear Physics And Radioactivity Question 192
Question: If 200 MeV energy is released in the fission of a single $ U^{235} $ nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given $ 1\ eV=1.6\times {{10}^{-19}}J $ ) [AMU 1995; MP PMT 1999]
Options:
A) $ 3.125\times 10^{13} $
B) $ 3.125\times 10^{14} $
C) $ 3.125\times 10^{15} $
D) $ 3.125\times 10^{16} $
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Answer:
Correct Answer: A
Solution:
$ P=n,( \frac{E}{t} )\Rightarrow 1000=\frac{n\times 200\times 10^{6}\times 1.6\times {{10}^{-19}}}{t} $ $ $
$ \Rightarrow \frac{n}{t}=3.125\times 10^{13}. $
BETA
