Nuclear Physics And Radioactivity Question 420
Question: Determine the power output of a $ _{92}U^{235} $ reactor if it takes 30 days to use 2kg of fuel. Energy released per fission is 200 MeV and $ N=6.023\times 10^{26} $ per kilo mole.
Options:
A) 63.28 MW
B) 3.28 MW
C) 0.6 MW
D) 50.12 MW
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Number of atoms in 2kg fuel $ =\frac{2}{235}\times 6.023\times 10^{26}=5.12\times 10^{24} $ number of fission per second $ =\frac{5.12\times 10^{24}}{30\times 24\times 60\times 60}=1.978\times 10^{18} $ Energy released per fission $ =200,MeV=200\times 1.6\times {{10}^{-13}}=3.2\times {{10}^{-11}}J $ Power output $ =3.2\times {{10}^{-11}}\times 1.978\times 10^{18} $ $ =63.28\times 10^{6}W=63.28,MW $
BETA
