The Helium 3 nuclide contains three nucleons, one neutron and two protons. The conventionally accepted figure for the binding energy of the Helium 3 nuclide is 7.718058 MeV. Because this is computed using a mass for the neutron that is too low by 0.98638 MeV the binding energy of the Helium 3 nuclide is 7.718058+0.98638)= 8.7044 MeV.

According to the article, "Precise Radii of Light Nuclei from Electron Scattering," by I. Sick published in Precision Physics of Simple Atoms and Molecules (edited by Savely G. Karshenboin), the root-mean-square (rms) charge radius of the Helium 3 nuclide is fermi; and that of a proton is 0.895 fermi. Thus the distance from the centroid of the triangle to the center of the proton is 0.860 fermi.

For an equilateral triangle the distance between adjacent vertices is √3*(0.86)=1.92 fermi. Relative to the scale parameter of 1.522 fermi for nuclei this is 1.25. The value of W( 1.25) is 0.172321.

The Helium 3 nuclide is also subject to the electrostatic repulsion between the two protons. The magnitude of the potential energy due to this repulsion can be measured by comparing the binding energies of the triteron and the Helium 3 nuclide. The difference in those binding energies is (8.481821−7.718058)=0.76376 MeV.

The equation to be sastified for the Helium 3 nuclide is, if its spatial arrangement is that of an equilateral triangle,

2P_np + P_pp + (4/3−1)(H/s₀)W(1.25) − 0.76376 = 7.718058 + 0.98638
or, upon substitution of known values
2(1.98671) + P_pp + (1/3)(18.3636)(0.172321) − = 8.7044 MeV
which evaluates to
3.9734 + P_pp + 1.05481 − 0.76376 = 8.7044 MeV
and hence
P_pp = 4.4400 MeV

This figure is surprisingly high but within the realm of plausibility. However, it is at variance with the evidence for the incremental binding energies of protons. For example, for the nuclides containing 50 neutrons here is the effect of additional protons on the binding energies.

The enhancement in binding energy when the additional proton can pair with an existing proton is on the order of 2 to 3 Mev. A more precise assessment of the effect of proton pairing can found by taking the absolute value of the difference between the incremental binding energy and the average of the two adjacent levels.

The average of the values plotted in the above graph is 2.67341 MeV.

Although the two approaches to estimating the effect of the formation of a proton-proton spin pair are different they are of the same order of magnitude.