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Minimum Energy Nuclear
Composition and Nuclear
Stability: Version 2

It is clear that the stability of nuclear isotopes is dependent upon their neutron/proton composition. A previous study successfully explained the stability of the isotopes of the heavier element as being due to their having a minimum energy for all nuclides having the same number of nucleons. The neutron/proton ratio for the stable isotopes heavier elements tends to be 1.5 whereas for the lighter elements it is about 1.0.

That previous study dealt only with the energy involved with the interaction of nucleons, but there is another major component of the energy of a nuclide. That is the energy due to the spin pairing of nucleons; i.e., protons with protons, neutrons with neutrons and neutrons with protons. The binding energies involved in each of these spin pairings are an order of magnitude higher than those involved in an interaction of nucleons. However the spin pairings are exclusive in the sense that one neutron can pair with one other neutron and with one proton but no more. The same applies for a proton.

The nucleonic interactions are not exclusive so in nuclides with larger numbers of nucleons the large numbers of smaller energy interactions may be more important than the small number of large energy spin pairings.

Let n and p be the numbers of neutrons and protons, respectively, in a nuclide. The energies in the three types of pairings are roughly equal. The total energy due to spin pairing is proportional to the sum of the numbers of the neutron pairs, proton pairs and neutron-proton pairs. The number of neutron-proton pairs is equal to the minimum of n and p.

Consider the variation in energy in all nuclides such that n+p is constant, say a. A shift in a nucleon pair from one category to the other would not alter the energy except that it changes the minimum of n and p. It is notable that the maximum energy from spin pairing occurs where n equals p.

But it is the total energy that is relevant rather than that due to spin pairing alone or interaction energy alone.

Let the nucleonic charge of a proton be taken to 1 and that of a neutron be denoted as q. Then the force due a neutron-neutron interaction is proportional to q². The forces between neutron-proton and proton-proton interactions are proportional to q·1 and 1·1, respectively.

If q is a negative value the force between a neutron and a proton is an attraction, whereas the forces for the other two interactions are necessarily repulsions. The energies involved for these interactions are in the same proportions: q², q and 1. Ignoring the electrostatic repulsion between protons for the total energy can be represented as proportional to

E = ½(p²−p)(1·1) + pn(q·1) + ½(n²−n)q² − K[ p%2 + n%2 + min(n, p)]

where p%2 and n%2 represent the number of pairs of protons and neutrons, respectively.

The previous study showed that the effect of the electrostatic repulsion between protons can be handled by treating the nucleonic charge of the proton as being (1+d) instead 0f 1. The total energy function is then

E = ½(p²−p)(1+d)²l + pnq + ½(n²−n)q² − K[ p%2 + n%2 + min(n, p)]

The parameter K represents the energy of a spin pairing relative to that of an interaction.

Here are some graphs of E versus n for a few values of a.

For a=30 the minimum energy occurs at n=16 so p=14 and the neutron/proton ratio is essentially unity. For a=40 the minimum energy is at n=21 and p=19, Again the n/p ratio is essentially unity.

The n/p ratio depends upon the parameter K, the relative magnitude of the spin pairing energy to the interaction energy. To demonstrate this here are graphs for the above cases with K=0.

Now for the a=30 case n=19 and thus p=11. The n/p ratio is essentially 1,5. For the a=40 case n=25 and p=15 so the n/p ratio is 1.67.

The spin pairing energy as a function of a is roughly (a+a²/4), The energy due to interactions as a function of a is roughly 2(a/2)(a/2-1)+a²/4=3a²−a. The ratio of the these is

(1+a/4)/(3a − 1) = (1/4 + 1/a)/(3 − 1/a)

The limit of this as a increases without bound is 1/12. The graph of the data for a=40 and K=1/12 is essentially the same as the one for K=0. Thus the n/p ratio tends toward 1.5.

The nuclides with energies near the minimum are likely to be nearly as stable as the one having minimum energy. This suggests there would a range of stable nuclides. The curves for K=0 are flatter on the bottom near the minimum than those for K=9. This suggests that the range of stable nuclides would be broader for the n/p ratios being 1.5 than for the ones with n/p ratios of 1.0,


The greater surplus in the stable isotopes of the heavier elements of neutrons compared to protons stems from the relative magnitude of the energy due to nucleon interactions in the formation of the nuclei of those elements compared to the energy involved in the pairing of nucleons. The ratio of neutrons to protons in the stable heavier nuclei is roughly 1.5 whereas for the lighter nuclei it is roughly 1.0.

Spin pairings involves relatively large energy amounts but are exclusive. Nucleon interaction involves relatively small energy amounts but is nonexclusive. Consequently for nuclei having large enough numbers of nucleons the energy involved in interactions exceeds that due to spin pairing. Thus for small nuclei a neutron/proton ratio of 1.0 prevails but for large nuclei neutron/proton ratio of 1.5 prevails.

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