A) 63.28 MW
B) 3.28 MW
C) 0.6 MW
D) 50.12 MW
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 \[=3.2\times {{10}^{-11}}\times 1.978\times {{10}^{18}}\] \[=63.28\times {{10}^{6}}W=63.28\,MW\]You need to login to perform this action.
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