Question

The fission properties of \[\ce{_94Pu^239}\] are very similar to those of \[\ce{_92U^235}\]. How much energy (in MeV), is released if all the atoms in 1 g of pure \[\ce{_94Pu^239}\] undergo fission? The average energy released per fission is 180 MeV.

Answer 1

Number of atoms in 1 g of pure \[\ce{Pu}\]

= `(6.023 xx 10^23)/239 xx 1000`

= 2.52 × 10²⁴

As average energy released in fission is 180 MeV, the total energy released

= 2.52 × 10²⁴ × 180 MeV

= 4.53 × 10²⁶ MeV