A major mystery of the physics of nuclei is why the Helium 4 nuclide, the alpha particle, has such a high level of binding energy, 28.3 million electron volts (MeV), compared to the binding energies of the deuteron (Hydrogen 2) at 2.2 MeV, the triteron (Hydrogen 3) at 8.5 Mev, and the Helium 3 at 7.7 MeV. The alpha particle does have more bonds and pairings: 6 strong force bonds (4 neutron-proton, 1 neutron-neutron and 1 proton-proton) and 4 possible spin pairs (2 neutron-proton, 1 neutron-neutron and 1 proton-proton). In contrast, the Helium 3 has 3 strong force bonds (2 neutron-proton and 1 proton-proton) and 3 spin pairs (2 neutron-proton and 1 proton-proton). The triteron has the same number as the He 3 except neutron-neutron replaces the proton-proton combinations. It does not seem possible to find values for the effects of the various bonds and pairs that would account for the differences in binding energies.

What follows below is a construction which shows that the binding energy of an alpha particle can be accounted for in terms of its constituent bonds and pairings. First consider all of the nuclides which could and probably are composed entirely of alpha particles. These will be referred to as the alpha (α) nuclides. There are 25 of them. Then consider the nuclides that could consist of alpha particle plus two neutrons (a neutron pair). These are the α+nn nuclides. The difference between the binding energies of the α+nn nuclides and the α nuclides is the effect of a neutron pair (nn) on binding energy. The graph of this binding energy as a function of the number of alpha particles in the nuclide is shown below.

There are shell phenomena involved in the relationship. The sharp drops come at points where the numbers of neutrons (and the numbers of protons) are equal to a magic number. These sharp drops reflect the filling of shells.

The same construction can be carried out for proton-proton pairs, as shown below:

Likewise the effect of the addition of a deuteron (neutron-proton pair) to alpha nuclides can be determined.

The effect of the addition of an alpha particle to the alpha nuclides is as follows.

This is the pattern to be explained. The values are generally higher than the binding energy of a single alpha particle. That value may be error due to an error in the mass of the neutron. In contrast the incremental binding energies in the above graph are independent of any error in the mass of the neutron. The average incremental binding energy for the last shell in the above display is 30.82 MeV. This would indicate that the mass of the neutron is underestimated by 1.27 MeV. A previous study estimated that the mass of the neutron is underestimated by about 0.98638 MeV and hence the binding energy of the alpha particle is 30.27 MeV.

The effects of one neutron pair, one proton pair and two neutron-proton pairs can be combined and compared with the binding energy of an additional alpha particle for each of the alpha nuclides.

The match is quite close for the larger nuclides. The fit is not perfect and it is useful to look at the differences on an expanded scale, as below.

For the larger nuclides the difference is essentially zero. The effect is nonzero and generally positive because the effect of the interaction of the neutron pair, proton pair and neutron-proton pairs is not captured by using the effects of the separate addition of the pairs to the alpha nuclides. For small nuclides where these components are close together the effect of their interaction is substantial but for the large nuclides where the components are widely separated the effects are negligible.

The dependence of the above differences D on the number of alpha particles (#a) in the nuclide seems to be of the form

D = B/(#a)^β

The value of B and β can be determined from a regression of the form

ln(D) = ln(B) − βln(#a)

However in order to carry out this regression the negative values of the differences have to be deleted. Also the values for #a equal to 3 and 11 appear to be connected with nuclear magic numbers and unrelated to the pattern for the other points. Therefore the data points for #a equal to 3 and 11 were eliminated. The regression equation then obtained is

ln(D) = 2.96809 − 1.00611ln(#a)

with a coefficient of determination of 0.952 and a t-ratio for the regression coefficient of −14.1. The regression coefficient is not statistically significantly different from 1.0 at the 95 percent level of confidence. This means that the product of D and #a should be constant. This is confirmed in the plot of the values of these products as a function of the number of alpha particles in the nuclide.

Thus the relationship between the difference in binding energy and the number of alpha particles is given approximately by

D = 19.52871/(#a)

The distance between particles in a nucleus is proportional to the cube root of the number of particles. This also applies to the radius of the nucleus. Thus the difference in binding energy D is approximately equal to the cube of the radius. The volume of the nucleus is proportional to the cube of the radius, but there would be no logical connection of the energies with the volume of the nucleus. However if r represents the radius of the nucleus the relationship to the energy difference D would be

D = K/r³

The potential energy of the interaction of particles would be dependent upon their distances which would be closely related to the radius of the nucleus. The generally cited formula for the radius of a nucleus in fermi is

r = 1.25A^1/3

where A is the number of nucleons in the nucleus.

There is a scale factor for the nuclear strong force which was derived by Hideki Yukawa from the mass of the particle which carries the strong force, the π meson. Its value is 1.522 fermi. The fermi values of r are divided by this scale factor to get dimensionless values for the nuclear radii.

If D=K/f(r) then the plot of D*f(r) would be a constant. Below are the graphs of D*r and D*r².

Neither of these are constant but the pattern for f(r)=r might be a negative exponential. If D=K*exp(−r)/r then D*r*exp(r) would have a constant level. Below is the plot.

It just happens that K*exp(−r)/r is the Yukawa potential. The regression of D*r*exp(r) on r has a regression coefficient which is not significantly different from zero at the 95 percent level of confidence. The coefficient of determination for the regression equation is 0.02. Thus D*r*exp(r) is a constant whose value is 88.09 MeV.

Conclusion

The binding energy of the alpha particle is the sum of the effects of the neutron pair, the proton pair and the two neutron-proton pairs that it contains. The difference between the incremental binding energy of an alpha particle and the above indicated sum is of the form that can be explained by interactions involving a negative exponentially weighted inverse square law of force.

The alternative to this explanation of the binding energy of the alpha particle in terms of its two-particle interactions would be to invoke some special three-particle and four-particle interactions. The quantum mechanical explanation proceeds in that way by considering the overlap of the wave functions of three and four particles. What the above material shows is that the large effect on binding energy of an additional alpha particle in heavy nuclides is entirely the result of the interactions of the component bonds and pairings of the alpha particle.

(To be continued.)