San José State University

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Thayer Watkins
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Binding Energy Levels and
the Limits of Nuclear Stability

## Background

The conventional theory of nuclear structure explains the holding together of protons in a nucleuse by hypothesizing a force that drops off faster with separation distance than does the electrostatic force. Neutrons as well as protons are supposedly subject to this hypothetic force. All nucleons are assumed to be attracted attracted to each other due to this hypothetical force. ,

Let PP, NN and NP denote the number of proton-proton. neutron-neutron and neutron-proton spin pairs respectively. There is binding energy due to the interaction of the nucleons. Let pp, nn and np be the number of proton-proton. neutron-neutron and neutron-proton interactions, respectively. If n and p denote the numbers of neutrons and protons, respectively, then pp, nn and np are given by

#### pp = p(p-1)/2 nn = n(n-1)/2 np = n·p

The model presumes that Binding Energy BE is a linear function of these six variables.

The results of regressing binding energy on these six variables are:

Regression Results
Variable np nn pp NP NN PP
Reg Coeff 0.263433925 -0.184060235 -0.467933089 21.46899263 14.23721024 3.278476244
t-Ratio 33.55377125 -35.0687162 -38.63352697 32.84417693 79.84818847 5.732760146

The coefficient of determination (R²) is 0.9999. Below is a graph of the actual binding energies of nuclides plotted against their estimates based upon the regression eqation.

In order to see the details of the statistical fit the results for just the isotopes of Tin are plotted below.

The lines for the regression estimate total and the actual values are so close that they coincide and cannot be seen separately.

What is revealed is that all of the binding is due to the spin pair formation. There is attraction between neutrons and protons but it is more than overwhelmed by the repulsion between like nucleons. The net repulsion is less for a proper mixture of neutrons and protons, but it is still a repulsion.

The results for the isotopes of Krypton (p=36) are qualitatively the same as those for the isotopes of Tin.

Likewise the same pattern prevails for the isotopes of Uranium.

## The Maximum Net Binding EnergyDue to the Interaction of the Nucleons

The equation for the Interaction Binding Energy IBE is

#### IBE = cppp(p-1)/2 + cnnn(n-1)/2 + cnpcppnp

Its partial derivative with respect to n is

#### ∂IBE/∂n = cnn(n - ½) + cnpp

It is equal to zero where

#### n = −(cnp/cnn)p + ½

For the values of the regression coefficients this is

#### n = 1.444018977p + ½

For p=92 (Uranium) this evaluates to 133.3497459, and rounded off to n=133. This suggest that the isotope of Uranium U225 is the most stable.

For more on the ideal mixture of protons and neutrons see Nucleon composition.

## Conclusions

A nucleus is held together entirely by the spin pairing of the nucleons. The attraction between unlike nucleons offsets some but not all of the repulsion between like nucleons.

(To be continued.)