# Fermi energy of nucleus

How is it that the calculated Fermi energy of neutrons and protons in a heavy nucleus are different, but it's observed that they have the same energy? And how much is the Coulumb repulsion of the protons play a factor in that?

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Simon Bridge
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How is it that the calculated Fermi energy of neutrons and protons in a heavy nucleus are different, but it's observed that they have the same energy? And how much is the Coulumb repulsion of the protons play a factor in that?
Not sure I understand you - do you have a reference?

If calculated values differ from observed values than the model used for the calculation is wrong in some important way.

It was part of a question in a textbook: the back of the book gave their Fermi energies, with about 5 MeV difference between them, but the question said that their observational energies were the same. And later the energy of the proton repulsion was found to be 10 MeV.

Bill_K
The Fermi energy depends only on the particle density. Since the nucleons are distributed uniformly throughout the nuclear volume, an equivalent statement is that the Fermi energy for protons depends only on the number of protons Z present in the nucleus, and the Fermi energy for neutrons depends only on the neutron number N. If Z = N the two Fermi energies are equal. If we depart from Z = N the Fermi energies are unequal and the total energy increases. The difference is known as the symmetry energy.

The Coulomb repulsion favors lowering Z, to a point where the Coulomb energy and the symmetry energy balance one another.

That's kind of what I was thinking, with the Coulomb repulsion energy equaling the difference of the two Fermi energies. However, the very simple calculation for the Coulomb energy gave an energy of twice the difference of the two Fermi energies. Given that it was off by almost exactly 2, I was thinking that I was missing some physics somewhere.

Bill_K