Q. 173.7( 6 Votes )

The fission properties of 23994Pu are very similar to those of 23592U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 23994Pu undergo fission?

Answer :

The atomic mass of 239Pu94 is 239.

Hence there are 6.023 × 1023 atoms in 239 gm.

So, there are = 2.52 × 1024 atoms

The energy released will be = average E × No. of atoms

Hence, E = 180 × 2.52 × 1024 Mev = 4.536 × 1026MeV

if all the atoms in 1 kg of pure 239Pu undergo fission then 4.536 × 1026MeV

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