The fission properties of 23994 Pu are very similar to those of 23592 U . Theaverage energy released per fission is 180 MeV. How much energy,in MeV, is released if all the atoms in 1 kg of pure 23994 Pu undergofission?
Given: The average energy released per fission of P 94 239 u is 180 MeV and the amount of P 94 239 u is 1 kg .
239 g of P 94 239 u contains 6.023 × 10 23 atoms .
Let N be the number of atoms present in 1 kg of P 94 239 u .
The number of atoms contain by 1 kg of P 94 239 u is given by,
N = 6.023 × 10 23 239 × 1000 = 2.52 × 10 24 atoms
Let E t be the total energy released in the fission.
E t = E × N
Where, the average energy released per fission of P 94 239 u is E .
By substituting the given values in above equation, we get.
E t = 180 × 2.52 × 10 24 = 4.536 × 10 26 MeV
Thus, the energy released by the fission of 1 kg of P 94 239 u is 4.536 × 10 26 MeV .
Given: The average energy released per fission of
Let
The number of atoms contain by
Let
Where, the average energy released per fission of
By substituting the given values in above equation, we get.
Thus, the energy released by the fission of
.