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Graph of potential energy versus internucleon distance in an atom

  1. Feb 12, 2010 #1
    The graph of potential energy of a pair of nucleons as a function of their separation shows a minimum potential energy at a value r (approx. = 0.8 femtometers). Below r the PE is positive (actually rises sharply from a negative to a positive value). Above r the PE is negative, and becomes zero beyond a certain value of r. For a separation greater than r the force on the nucleons is attractive, while for separations less than r the force is strongly repulsive.

    I would appreciate if someone could explain the above phenomenon to me. Thanks.
  2. jcsd
  3. Feb 12, 2010 #2
    Have you seen the explanation for the similar shape of the Lennard-Jones potential between atoms? If not, you might find it interesting.


    You could try to apply this reasoning by treating the nucleons as composite particles (3xquark), and considering similar effects in the context of the strong nuclear force. In both cases you have a composite state. For the nucleons you have R,G,B quarks in each of the nucleons.

  4. Feb 12, 2010 #3
    Thanks. Let me have a look at your reference.
  5. Feb 12, 2010 #4
    I think that this is also worth reading:

    http://www.cartage.org.lb/en/themes/sciences/physics/NuclearPhysics/WhatisNuclear/Forces/Forces.htm" [Broken]

    It seems to explain the short range repulsion as coming from the Pauli explusion principle, and the medium range attraction as a Yukawa force, which appears as a consequence of the exchange of massive mesons.

    Last edited by a moderator: May 4, 2017
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