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I Energy of pp, np, nn bonds

  1. Jun 1, 2017 #1
    I am curious as to why proton-proton, neutron-neutron bonds are preferable in the nucleus; the pairing term in the semi empirical mass formula for the nucleus comes from the fact that pp and nn bonds are more stable in the nucleus (higher binding energy or more negative potential energy, however you want to look at it) whereas when talking about the formation of simple, two nucleon particles, the only one allowable is np, i.e. deuterium because of the PEP. Why isn't the PEP applying in the nucleus? Why does the pairing term not favour np bonds?
     
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  3. Jun 1, 2017 #2

    Vanadium 50

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    Nuclear physics is not chemistry. You don't make nuclei by considering individual nucleon-nucleon bonds. Each nucleon feels the potential created by the other nucleons.
     
  4. Jun 1, 2017 #3
    Of course, but the pairing term of the SEMF is to be thought of as being to do with pp and nn pairs in the nucleus. Is this an effect of the shell model? Like how in atoms it's lower energy (high binding energy) for the spins to be aligned if the electrons are all in different orbitals?
     
  5. Jun 1, 2017 #4

    mfb

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    Nuclei can have spin up and down, every available energy level can get two neutrons or protons, respectively. That makes pairs favorable, the next (odd) nucleon has to occupy a higher energy level.
     
  6. Jun 1, 2017 #5

    ChrisVer

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    What is the PEP? I thought the combination pn was favorable because of isospin issues?
     
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