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The Role of the Electrostatic Repulsion of Protons in the
Estimate of the Ratio of the Nucleon Charge of a Neutron
to that of a Proton Based on the Limits of Nuclear Stability

Nuclei are held together by the mutual attraction between neutrons and protons. Neutrons are repelled from each other through the strong force. Protons are also repelled from each other not only through the electrostatic force but also through the strong force. Therefore there has to be some balance between the number of neutrons and the number of protons for a nucleus to hold together. If there are too many protons compared to the number of neutrons the repulsion between the protons overwhelms the attraction between neutrons and protons. Likewise if there are too few protons the repulsion between the neutrons overwhelms the neutron-proton attraction. There is an asymmetry between the numbers of neutrons and protons that indicates the strength of the repulsion between protons due to the strong force is greater than that between neutrons. The strong force drops off faster with distance than the electrostatic force so the electrostatic repulsion between protons becomes relatively stronger in larger nuclides where the average distance between protons becomes greater. The situation is made more complicated by the fact that neutrons form spin pairs with each despite their mutual repulsion and protons do likewise. Spin pair formation is relatively more important for the smaller nuclides.

There are 2931 nuclides stable enough to have had their masses measured and their binding energies computed. For each number of neutrons the minimum number and the maximum number of protons were compiled. The results are displayed in the following graph.

In the graph there is some piecewise linearity displayed.

A Previous study developed evidence that the nucleonic (strong force) charge of a neutron is of the opposite sign and smaller in magnitude from that of a proton. Let ν denote the ratio of the nucleonic charge of a neutron to that of a proton. The actual value of ν is undoubtedly a simple fraction. Previous work indicated that the relative magnitude of the neutron charge could be 2/3 or 3/4. Furthermore such a difference in charge of the nucleons can account for the limits to the values of the proton numbers of the known nuclides, shown above.

Another study demonstrated that the binding energy increments experienced by additional nucleons to a nuclide is a function of two components. One is simply the difference in the number of protons and neutrons in the nuclide. This component has to do with the formation of a neutron-proton spin pair. The other component has to do with the interaction of nucleons through the strong force and it is a function of the net nucleonic charge of the nuclide. If p and n are the numbers of protons and neutrons, respectively, of the nuclide then the net nucleonic charge ζ is

ζ = p − νn

where ν is the magnitude of the nucleonic charge of the neutron relative to that of a proton.

The binding energy associated with the interaction nucleons through the strong force is a nonlinear function of ζ, but for small values of ζ to a reasonable approximation it is kζ, where k is a constant. Nucleons also interact through the formation of spin pairs. For example, the addition of another neutron to a nuclide with an odd number of neutrons would result in the formation of a neutron-neutron spin. Let Enn be the binding energy associated with the formation of a neutron-neutron spin pair. If there are unpaired protons in the nuclide the addition of another neutron would result in the formation of a neutron-proton spin pair with a binding energy of Enp. The binding energies associated with the formation of spin pairs are not really constants independent of the levels of n and p but for the present they are assumed to be constants.

An Additional Neutron

The incremental binding energy of a neutron, IBEn, is the binding energy change associated with the addition of another neutron to a nuclide. For a nuclide with p protons and n neutrons in which n is odd and less than p it is

IBEn = kζ + Enn + Enp
or, expanded
IBEn = k(p−νn) + Enn + Enp

For stability the values of n and p must be such that IBEn≤0. The minimum number of protons for a nuclide with p protons is reached when IBEn is driven to zero. This means that

k(pmin − νn) + Enn + Enp = 0
and hence
pmin = νn − Enn/k − Enp/k

The maximum number of neutrons is also where IBEn=0 and hence

k(p − νnmax) + Enn + Enp = 0
and hence
nmax = (1/ν)p + Enn/(kν) + Enp/(kν)

An Additional Proton

In addition to the factors which affect neutrons, protons are subject to electrostatic repulsion by other protons. This factor is thought to only effective in large nuclides where the average distance between proton becomes large. The effect of this repulsion on the binding energy for an additional proton is negative and proportional to the number of protons in the nuclide, say −qp. The binding energy of an additional proton to a nuclide with p protons and n neutrons in which p is odd and less than n is

IBEp = −kζ −qp + Epp + Enp
and thus
IBEp = kνn - (k+q)p + Epp + Enp = 0

For IBEp to be greater than or equal to zero requires a minimum n of

nmin = (1/ν)((k+q)/k)p − Epp/(kν) − Enp/(kν)
or, equivalently
nmin = (1/ν)((1+q/k)p − Epp/(kν) − Enp/(kν)

The maximum p is the value such that IBEp=0; i.e.,

pmax = (k/(k+q))νn + Epp/(k+q) + Enp/(k+q)

Now the equations for the maximum and minimum number of neutrons can be displayed together; i.e.,

nmax = (1/ν)n + Enn/(kν) + Enp/(kν)
 
nmin = (1/ν)((1+q/k)p − Epp/(kν) − Enp/(kν)

If these two equations are added together and the result divided by 2, the result is

nmid = (1/ν)(1+½q/k)p + (Enn−Epp)/(2kν)

The equations for the maximum and minimum number of protons are

pmax = (k/(k+q))νn + Epp/(k+q) + Enp/(k+q)
 
pmin = νn − Enn/k − Enp/k

The average of these two equations is

pmid = (2k+q)/(2(k+q))νn + (Epp/2(k+q) − Enn/2k − Enp(1/k − 1/(k+q))/2)
or, equivalently
pmid = (1+½q/k)/(1+q/k))νn + (Epp/2(k+q) − Enn/2k − Enp(1/k − 1/(k+q))/2)

In order to illustrate the effect of the electrostatic force suppose the true value of ν were 0.75 and q/k=0.25. Then the slope of the relationship between pmid and n would be (1.125/1.25)(.75)=0.675. The slope of the relationship between nmid would be (1/0.75)(1.125)=1.5.

The regression coefficient for pmid on n is 0.67264 and that of nmid on p is 1.37559.* Thus

(1+½q/k)/(1+q/k))ν = 0.67264
(1/ν)(1+½q/k) = 1.37559

If these two equations are multiplied together the result is

(1+½q/k)²/(1+q/k) = 0.92528
or, equivalently
1 + q/k + (q/k)²/4 = 0.92528 + 0.92528(q/k)
0.074723 + 0.074723(q/k) + 0.25(q/k)² = 0

The solutions to this quadratic equation are

q/k = (−0.074723 ± (0.0055835 − 0.074723)½)/0.5

These roots are not real. Therefore the electrostatic repulsion of the protons cannot explain the discrepancy of the results for the slopes of the equations for nmid and pmid. This is a negative result but nevertheless an important one.


*For the analyses in terms of the maximum and minimum numbers of neutrons and protons see Nuclear Charge Ratio and Nuclear Charge Ratio 2.


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