Shortly after the neutron was discovered Werner Heisenberg conjectured that the neutron and proton could be considered to be the same particle; one with the charge turned on and the other with the charge turned off. This conjectured particle was named the nucleon. While there were insights gained from Heisenberg's conjecture it proved not to be literally true. Now the proton and neutron are considered to be different triplets of quarks. The proton is thought to be composed of two up quarks and a down quark. The neutron is thought to consist of two down quarks and one up quark.

A weakened version of Heisenberg's conjecture is that the strong force between any pair of nucleons is the same. The net force has to take into account the electrostatic repulsion between two protons. This conjecture seems to be borne out. Consider the nuclides which could contain an integral number of alpha particles. Hereafter these will be called alpha nuclides. Now consider the nuclides that contain an integral number of alpha particles plus one neutron. The difference between the binding energies of these nuclides and the corresponding alpha nuclide is the effect of an additional neutron on binding energy. Also consider the same computation for the effect of an additional proton. Then take the difference of the two effects. The results are shown in the following graph.

The effect of an additional neutron is shown as the upper edge of the yellow area. The green area shows the effect of an additional proton; very much smaller than the effect for a neutron. The red rectangles show the neutron effect minus the proton effect. This difference as a function of the number of alpha particles is remarkably smooth. It appears to be nearly linear with a slight quadratic effect. The regression of the difference in binding energy as a function of the number of alpha particles is

ΔBE = 0.356677 + 0.824096#α − 0.01189(#α)²
[1.75]    [33.38]    [-10.97]
R² = 0.99745

This difference represents the negative of the effect on binding energy of a positive charge separate from the effect of the strong force due to the nucleon.

The results for the corresponding computation of the effect of two additional neutrons and two additional protons, along with their difference, are shown below.

The regression equation is

ΔBE = 0.854576 + 1.761323#α − 0.02981(#α)²
[2.20]    [28.28]    [-8.93]
R² = 0.997215

The effect of two positive charges on binding energy is remarkably close to twice the effect of one positive charge. The comparison is shown below in terms of the negative of the effects of the positive charges. The red rectangles are for the effect of two charges, the edge of the green for the effect of one charge and double the effect of one charge is the edge of the yellow area.

The fit is not perfect, but it is very close. The correlation between the two is 0.999058.

Although the statistical performance of the above regression equations is impressive the relations could not actually be quadratic. The extrapolation of the quadratic relations would involve the relationships reaching a maximum and then declining. If the relationships were quadratic the increments in the values for an increase in the number of alpha particles would be linear and declining toward zero. The incremental values are shown in the graph below.

The increments are generally declining but not linearly. Instead the increments are generally asymtotically approaching some level. This is as it should be. For larger nuclides the separation distances are larger and the electrostatic force corresponding smaller.

The form of the functional relationship would be

δBE = β + γ/(#α+ε)

This would mean that

ΔBE = β#α + γln(#α+ε)

The value of β could be zero.

It is notable that the two patterns in the above graph display the similar irregularities. Some of the relative extreme points correspond to nuclear magic numbers.

The parameter ε is the only one that cannot be determined as a regression coefficient. Its value can be chosen to maximize the coefficient of determination R².

The regression equation for the two positive charge effect is:

δBE = 0.552562 + 6.903068/(#α+3.615)
[2.60]    [8.42]
R² = 0.81578

For comparison the coefficient of determination for the linear equation is 0.653274.

The regression equation for the one positive charge effect is:

δBE = 0.273837 + 4.056938/(#α+5.15)
[1.51]    [4.40]
R² = 0.491732

For comparison the coefficient of determination for the linear equation is 0.384618.

For a comparison with the quadratic regression equation a regression equation of the form

ΔBE = c₀ + c₁#α + c₂ln(#α+ε)

was estimated. The results for the one positive charge effects were

ΔBE = −4.67495 + 0.157143#α + 2.763195ln(#α+5.15)
[-32.92]    [19.93]    [16.41]
R² = 0.998763

For the two positive charge effect the results were

ΔBE = −5.58986 + 0.364415#α + 4.293751ln(#α+3.615)
[-34.96]    [32.87]    [23.46]
R² = 0.99953

Even though the statistical performances of the quadratic equations were high, these results are even better.

(To be continued.)