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?

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