Atoms And Nuclei Question 350

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

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Answer:

Correct Answer: A

Solution:

  • 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 $



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