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Fermi energy of nucleus

  1. Mar 16, 2013 #1
    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?
  2. jcsd
  3. Mar 17, 2013 #2

    Simon Bridge

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    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.
  4. Mar 17, 2013 #3
    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.
  5. Mar 17, 2013 #4


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    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.
  6. Mar 17, 2013 #5
    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.
  7. Mar 17, 2013 #6


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    We have to be careful what to compare. The Coulomb energy is usually given as a total energy, whereas the Fermi energy is energy per nucleon. Also it's the maximum energy - not the case that all the neutrons in the nucleus will have the same energy eF, there will be an average < eF.
  8. Mar 17, 2013 #7
    It was actually the average Coulomb energy per proton that I was comparing. That was twice the difference of the average Fermi energies per nucleon.
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