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Bonds, Pairing and Binding Energies
Among the Small Nuclides

It is thought that each nucleus is composed of neutrons and protons bound together by a hypothetical strong force and by the formation of nucleonic pairs. The degree of the binding of a nuclide is measured its mass deficit; i.e., the amount by which the mass of the nuclide falls short of the combined masses of its component parts. The mass deficit expressed in terms of its energy equivalent is called binding energy. The binding energy of a nuclide seems to be the negative of the loss of potential energy involved in its formation.

The Nuclear Strong Force

According to Hideki Yukawa the strong force is carried by π mesons. Any force carried by particles will have an inverse-distance-squared dependence. However the π mesons decay with time and therefore the number surviving at any distance is a negative exponential function of distance. This means that the strong force between two nucleons has the following formula.

F(s) = −H*exp(−s/s0)/s²

where s is the separation distance of two particles, H is a constant and s0 is a scale factor that depends upon the mass of the force-carrying particle. From Yukawa's relation and the mass of the π meson the parameter s0 has the value 1.522 fermi.

The Potential Function
for the Strong Force

The potential function V(s) for the force formula above is

V(s) = ∫s(H*exp(−z/s0)/z²)dz

By a change in the variable of integration from z to ζ=z/s0 the potential function can be converted into

V(σ) = (H/s0)∫σ(exp(−ζ)/ζ²)dζ

where σ=s/s0. Thus the crucial aspect of the separation distance is its value relative to the scale parameter s0.

For convenience the potential energy will be represented as

V(σ) = (H/s0)W(σ)

where W(σ)=∫σ(exp(−ζ)/ζ²)dζ.

The Physical Dimensions of the Small Nuclides

A group of physicists under the editorship of Savely G. Karshenboin published in 2008 a book devoted to the compilation of the best estimates of physical properties of particles, simple atoms and simple molecules (Precision Physics of Simple Atoms and Molecules, Springer-Verlag).

The best estimate of the root-mean-square(rms)-charge radii of the small nuclides are in the article "Precise Radii of Light Nuclei from Electron Scattering," by I. Sick from pages 57 to 77 of the above mentioned work. For example, the root-mean-square radius of the deuteron is 2.13 fermi with a margin of error of ±0.01 fermi and thus the diameter is 4.260 fermi. To get the separation distance of the centers of the nucleons the sum of their radii must be deducted from the rms-charge diameter.

The recommended estimate of the rms-charge radius of the proton, given on page 49 of the above work, is 0.895 fermi. Precision Physics of Simple Atoms and Molecules does not give an estimate for the radius of the neutron. Another source gives the rms-radius of the neutron as 1.113 fermi.

Thus the separation distance of the centers of the nucleons in a deuteron is

s = 4.260−0.895−1.113=2.252 fermi.

The separation distance s1 in the physical deuteron is thus about 2.252 fermi (2.252×10-15 m), Thus the ratio s1/s0 is equal to 1.4796, a pure number.

The Geometry of Triangular
Arrangements of Nucleons

The root-mean-square (rms) charge radius of the triteron is 1.755 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)=1.49 fermi. Relative to the scale parameter of 1.522 fermi for nuclei this is 0.98. For the deuteron this figure is 1.4796. Thus the potential energy involved in each of the nucleon bonds in the triteron should be larger than that in the deuteron. The conventional estimate of the binding energy of the deuteron is 2.22457 MeV whereas the binding energy per bond for the triteron is 8.482/3=2.827 MeV.

Nucleon Separation Distance in the
Alpha Particle, the He4 Nuclide

The rms-charge radius of the alpha particle given in I. Sick's article is 1.68 fermi. The structure of the alpha particle is most likely a tetrahedron. With the rms=charge radius of the proton being 0.895 fermi the distance from the centroid of the tetrahedron to the center of a nucleon is 0.785 fermi. This means that the separation distnce of the centers of the nucleons is √3(0.785)-1.36 fermi. Relative to the scale parameter of 1.522 fermi this is 1.36.

The Separation Distances of the Nucleon in Small Nuclides

The separation distances of the nucleons computed from the radii of the small nuclides are:

s (fermi)
Helium 31.921.250.172321
Helium 41.360.890.493693

The Statistical Explanation of Binding Energy
in Nuclides by the Number of Bonds and Pairs

Statistically most of the variation in binding energy of nuclides can be explained by variations in the number of bonds and pairs of various sorts. For a nuclide with #p protons and #n neutrons the number of proton-proton bonds is #p(#p-1)/2, the number of neutron-neutron bonds is #n(#n-1)/2 and the number of neutron-proton bonds is #p*#n. The number of proton-proton pairings is #p%2 (#p divided by 2 with any remainder thrown away). The number of neutron-neutron pairings is likewise #n%2. The number of neutron-proton pairings is the minimum of #p and #n, min(#p,#n).

A regression of binding energies (BE) for all nuclides with #p<28 on the above numbers of bonds and pairs yields the following results.


There are 511 nuclides involved in this regression. The negative values for #pp and #nn are indications that neutrons repel each other and protons repel each other. The positive value for #np indicates that neutrons and protons are attracted to each other. All three types of pairings result in increased binding energy but a neutron-proton pairing has twice the effect of a neutron-neutron or a proton-proton pairing.

The important result is that the size of the effect of neutron-neutron bonds is about one half the magnitude of the effect of a neutron-proton bond. For a proton-proton bond the effect is about (4/3) the effect of a neutron-proton bond.

The coefficient of determination (R²) reported by the regression program (Excel) is 0.999562152, but this is misleading. Because the regression equation does not have a constant term the regression program compares the explained variation with the average of the squares of the dependent variable instead of the average of the squared deviation from the mean, the variance. The variance of the binding energies of the 511 nuclides used in the regression is 22494 (MeV)². The unexplained variation was 49.89 (MeV)² so the explained variation was 22444.11 (MeV)². The ratio of the explained variation to the variance is therefore 0.99778; a good value but not the 0.99956 of the reported figure.

Another regression was run using only the 32 nuclides such that #p<5. The results were


For this data set the effect of the neutron-neutron bond is again negative and about 45 percent of the magnitude of the neutron-proton bond. The effect of the proton-proton bonds is also negative and about 122 percent of the magnitude of th neutron-proton bond.

When all of the composite nuclides are included here are the results.


The effect of a neutron-neutron bond is negative and in magnitude 69 percent of the effect of a neutron-proton bond. The effect of a proton-proton bond is also negative and 176 percent of the magnitude of a neutron-proton bond.

The Explanation of the Binding Energies of
the Four Smallest Composite Nuclides

The binding energy could depend upon four parameters: the binding energies for the three types of pairs and the force constant H. Let Pnp, Pnn and Ppp denote the binding energy created by the formation of a neutron-proton pair, a neutron-neutron pair and a proton-proton pair, respectively.

For a deuteron the equation to be satisfied is

Pnp + (H/s0)(0.100714) = 2.224573

The triteron can have two neutron-proton pairs, one neutron pair and three bonds. Two of those three bonds are between neutrons and protons and one between neutrons. The equation to be sastified for the triteron is

2Pnp + Pnn + (2−1/2)(H/s0)(0.37357) = 8.481821

For the He3 nuclide the equation is

2Pnp + Ppp + (2−4/3)(H/s0)(0.172321) = 7.718058

The alpha particle (He4 nuclide) with two protons and two neutrons can have four neutron-proton bonds and one each of neutron bonds and proton bonds. The possible pairs are two neutron-proton and one proton-proton. The equation to be satisfied is then

4Pnp + Pnn + Ppp + (4−4/3−1/2)(H/s0)(0.493693) = 28.295674

The four equations in the four unknowns brought together are:

Pnp + 0.100714(H/s0) = 2.224573
2Pnp + Pnn + 0.560355(H/s0) = 8.481821
2Pnp + Ppp + 0.114886(H/s0) = 7.718058
4Pnp + Pnn + Ppp + 1.069685(H/s0) = 28.295674

First the first equation is multiplied by two and subtracted from the second and from the third. the variable Pnp is eliminated. Likewise the first equation is multiplied by four and subtracted from the fourth. Thus the variable Pnp is eliminated.

Pnn + 0.35893(H/s0) = 4.03268
Ppp − 0.086536(H/s0) = 3.26891
Pnn + Ppp + 0.66912(H/s0) = 19.393352

Now the first two equations can be added together and the result subtracted from the third to eliminate Pnn and Ppp to give

0.396726(H/s0) = 12.091792
and thus
(H/s0) = 18.996626 MeV

From this it follows that

Pnn = −2.785779 MeV
Ppp = 4.912802 MeV
Pnp = 0.311334 MeV

These values for the binding energies for the three types of pairings are not plausible in terms of signs and magnitudes. The exercise cannot be considered a success in explaining the binding energies of the small composite nuclides.

Sensitivity of Results to Errors
in the Mass of the Neutron

The conventional estimates of the binding energies of the nuclides are based upon the mass deficit of the deuteron being equal to the energy of the gamma photon associated with its formation or dissolution. If there is an error of ΔN in the mass of the neutron the error in the binding energy of a nuclide would be equal to the number of neutrons times the error in the mass of the neutron; i.e., #n*ΔN. The equations to be solved would then become:

Pnp + 0.100714(H/s0) = 2.224573 + ΔN
2Pnp + Pnn + 0.560355(H/s0) = 8.481821 + 2ΔN
2Pnp + Ppp + 0.114886(H/s0) = 7.718058 + ΔN
4Pnp + Pnn + Ppp + 1.069685(H/s0) = 28.295674 + 2ΔN

The second set of equations would then become

Pnn + 0.35893(H/s0) = 4.03268
Ppp − 0.086536(H/s0) = 3.2689 −ΔN
Pnn + Ppp + 0.66912(H/s0) = 19.393352

Adding together the first two equations in the above set and subtracting the result from the third would then give

0.396726(H/s0) = 12.091792 + ΔN
and thus
(H/s0) = 18.996626 + 2.52063ΔN

This value for (H/s0) would result in

Pnn = −2.785779 − 0.90473ΔN
Ppp = 4.912802 −1.218125ΔN
Pnp = 0.311334 + (1-0.25386)ΔN

For example, if ΔN=1 MeV then

Pnn = −3.69051 MeV
Ppp = 3.69468 MeV
Pnp = 1.05747 Mev

The value for Pnp is more plausible but the values for Pnn and Ppp are still not plausible.

(To be continued.)

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