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Nuclear and elctronic binding energy

  1. Jun 19, 2012 #1
    "For almost all nuclei, the binding energy/nucleon is roughly the same. Hence the total binding energy of a nucleus is roughly proportional to the number A of the constituents. Every constituent of the nucleus is more or less equally strongly bound unlike electrons in atoms where the mean binding energy is higher for higher Z atoms.

    This implies that the nuclear force has a short range of the order of inter-nucleon separation. If the force had a much longer range then the binding energy/nucleon would not be constant but would increase as ~A2.

    I don't get the part in italics:

    1. I understand that every constituent of the nucleus is more or less equally strongly bound because the binding per nucleon is roughly the same. But how is the mean binding energy of electrons in atoms higher for higher Z atoms? Is it because the Coloumb interaction (at the same radial separation) is stronger for bigger Z and so more energy is needed to free the electron from the evil grasp of the nucleus?

    2. I have no idea how the implication in the second paragraph comes about! I'd appreciate any comments!
  2. jcsd
  3. Jun 20, 2012 #2


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    Staff: Mentor

    Correct. In addition, the higher charge of the nucleus reduces the size of the orbitals, which is an additional increase in binding energy.

    If the range of the interaction is very short, the binding energy for a nucleon just depends on its neighborhood (a very classical view, but fine here) - which is similar in all large nuclei: Nucleons are surrounded by other nucleons.
    If the range would be large compared to the size of a nucleus, the binding energy for a nucleon would depend on the total amount of nucleons in it.
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