Assume that just after the fission of according to the reaction

the resulting and nuclei are just touching at their surfaces. (a) Assuming the nuclei to be spherical, we have to calculate the Coulomb potential energy (in MeV) of repulsion between the two fragments. (b) We have to compare this energy with the energy released in a typical fission process. We have to answer in what form this energy will ultimately appear in the laboratory.

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(a)

The fission of is assumed to place according to the reaction

We consider the situation when the resulting and nuclei are just touching at their surfaces. We will calculate the radii of and nuclides using the empirical relation

The radius of the nuclide will therefore be

And, the radius of the nuclide will be

The atomic number of nuclide is 54 and the atomic number of nuclide is 38. Therefore, the Coulomb potential energy between and nuclides, assuming that these nuclides are spherical and are just touching at their surfaces, will be

(b)

The energy released in a typical fission process is about 200 MeV. This energy appears mainly in the laboratory as the kinetic energies of the fission neutrons.