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I understand that a nuclei with too many protons is unstable as the electromagnetic repulsion becomes too strong to be balanced by the strong nuclear force.

However in the case of there being too many neutrons, surely the addition of neutrons should be adding extra strong nuclear force without extra electromagnetic repulsion, which should make a nucleus more stable. Why is this not the case?

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marked as duplicate by John Rennie, Pieter, M. Enns, StephenG, ZeroTheHero Apr 13 '18 at 4:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Good question. See what Frank Wilczek said in Nature Hard-core revelations Vol 445|11 January 2007 Doi 10.1038/445156a. Sorry it's paywalled: “ironically from the perspective of QCD, the foundation of nuclear physics appear distinctly unsound”. $\endgroup$ – John Duffield Apr 12 '18 at 16:28
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In the simple yet remarkably accurate liquid drop model of the nucleus, there are five contributions to the nuclear binding energy. As a function of the mass number $A$ and the atomic number $Z$, the semi-empirical formula for the binding energy is

$$E_B = a_V A - a_S A^{\frac{2}{3}} - a_C \frac{Z(Z-1)}{A^{\frac{1}{3}}} - a_A \frac{(A-2Z)^2}{A} + \delta(A,Z)$$

  • The first term is the volume term, which accounts for the pairwise nucleon-nucleon binding energy provided by the residual strong force
  • The second is the surface term, which corrects the prior estimate because the nuclei near the "surface" of the nucleus are subject to smaller binding energies than those in the center
  • The third is the Coulomb term, which accounts for the mutual electrostatic repulsion of the protons embedded in the nucleus
  • The fourth is the asymmetry term, which is minimized when the number of protons and neutrons is equal. The theoretical motivation for this term is the Pauli exclusion principle - because no two protons (resp. neutrons) can occupy the same state, adding additional protons (resp. neutrons) to a nucleus necessarily means that the latter will exist in significantly higher energy states.

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  • The final term is the pairing term, which accounts for the pairwise interaction between nucleon spins. This function takes the form

$$\delta(A,Z) = \cases{\delta_0 & if N and Z are even \\ -\delta_0 & if N and Z are odd \\ 0 & otherwise}$$

The parameters $a_V,a_S,a_C,a_A,$ and $\delta_0$ can be experimentally measured. The resulting prediction for binding energy is remarkably good, especially considering the simplicity of the model. More sophisticated models take into account things like the nuclear shell structure, which provides extra stability for the "magic numbers."


In the context of this model, your question is answered by the asymmetry term. An excess of neutrons causes the neutron Fermi level to be too high - the nucleus would be more stable if some of extra neutrons underwent beta decay and became protons, so generically this is what happens.

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    $\begingroup$ your description of the different terms in the equation is very useful, thanks for taking the time to furnish them. $\endgroup$ – niels nielsen Apr 12 '18 at 18:25
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This is a qualitative answer:

Neutrons decay with a mean life of ~15 minutes. Within a nucleus the strong interaction can bind the neutron to a proton, with a binding energy analogous to the energy levels of the atomic binding of electrons around nuclei.

A neutron with a proton make a deuteron, a stable isotope of hydrogen. Tritium also an isotope of hydrogen decays with a half life ~12 years.

The qualitative reason is that the two neutrons are part time shielding each other from the attraction of the proton as neutron rich nuclei mostly decay by the beta decay , which is the decay of a neutron. There are various nuclear models that fit the behavior, for example this publication fits data using a shell model for the nucleus.

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