Nuclear Physics And Radioactivity Question 76
Question: An atomic power nuclear reactor can deliver 300 MW. The energy released due to fission of each nucleus of uranium atom $ U^{238} $ is 170 MeV. The number of uranium atoms fissioned per hour will be [Pb. PET 2004; Kerala PET 2005]
Options:
A) $ 30\times 10^{25} $
B) $ 4\times 10^{22} $
C) $ 10\times 10^{20} $
D) $ 5\times 10^{15} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ P=\frac{nE}{t} $
$ \Rightarrow 300\times 10^{6} $ $ =\frac{n\times 170\times 10^{6}\times 1.6\times {{10}^{-19}}}{t} $ \ Number of atoms per sec $ \frac{n}{t}=1.102\times 10^{19} $ Number of atoms per hour = 1.02 ´ 1019 ´ 3600 = 3.97 ´ 1022.
BETA
