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The Upper and Lower
Limits of Nuclear Stability

Previous work established, contrary to the conventional theory, that like nucleons are repelled from each other and it is only unlike nucleons that are attracted to each other and hold a nucleus together. Thus a stable nucleus needs a proper mix of neutrons and protons. The stable isotopes of the heavier element have fifty percent more neutrons than protons. If the nucleonic composition of a nucleus deviates too much from the proper ratio the nucleus is unstable.

The Energies of Nucleonic Interactions

Let the nucleonic charge of a proton be taken to be 1 and that of a neutron be denoted as q. Then the force due a neutron-neutron interaction is proportional to q². The forces between neutron-proton and proton-proton interactions are proportional to q·1 and 1·1, respectively.

If q is a negative value the the force between a neutron and a proton is an attraction, whereas the forces for the other two interactions are necessarily repulsions. The energies involved for these interactions are in the same proportions: q², q and 1. Ignoring the electrostatic repulsion between protons for the moment, the energy for a nucleus of p protons and n neutrons is proportional to the sum of the products of the numbers of the various interactions and the proportionality factors. The number of proton-proton interactions is ½(p²−p). Those for neutron-neutron and neutron-proton interactions are ½(n²−n) and np, respectively. The Interaction Energy EI of the nucleons of a nucleus is therefore

EI = g[½(p²−p) + pnq + ½(n²−n)q²]

where g is a constant.

The Effect of the Electrostatic
Repulsion between Protons

The above ignores the electrostatic repulsion between protons. That can be represented by making the energy of the interaction between protons as (1+d) rather that 1, where d represents the relative magnitude of the electrostatic force compared to that of the nucleonic force at the distances involved in the nuclear interactions. This makes the energy of interactions equal to

EI = g[½(p²−p)(1+d) − pnq + ½(n²−n)q²]

The Binding Energy of
the Spin Pairing of Nucleons


The numbers of proton-proton, neutron-neutron and neutron-proton spin pairs are p%2, n%2 and min(n, p), respectively, where p%2 and n%2 are p and n divided by 2 and the results rounded down to a whole number. The binding energy due to spin pairings is then

ESP = h1(p%2) + h2(n%2) + h3min(n, p)

where the hj are constants.

Total Binding Energy

Nuclear stability occurs where

ET = EI + ESP < H

where H is a critical level of binding energy for nuclear stability and might be zero.

If the step functions p%2 and n%2 are replaced by their approximations as (p/2 − 1/2) and (n/2 − 1/2) then the equation ET=H is quadratic in n and p. It can be put into the form of a equation for an ellipse.

The Equations for Ellipses

The equation for an elipse of semimajor axes a and b centered on (0, 0) is

(x/a)² + (y/b)² = 1

If the ellipse is centered on (h, k) the equation is of the form

((x−h)/a)² + ((y−k)/b)² = 1

If the ellipse is tilted with respect to the horizontal there is an xy term. For a tilted ellipse centered on (0, 0) the equation is

(x/a)² + q(x/a)(y/b) + (y/b)² = 1

The Empirical Limits of Stability

For stability defined as "stable enough to have its mass measured" here are the graphs of the limits

The slope of the initial straight line for the maximums is 3/4. For the minimums it is 3/5. The average of these values is 2/3.

The shapes are remarkably similar. They are the reflections of each other about the line n=p. That of course is just the relationship of x plotted against y versus y plotted against x.

Because in the equation for ET=H there is the term for min(n, p) and min(n, p)=n if n< p and min(n, p)=p if n> p there could be a difference in shape between the upper edge and the lower edge of the ellipse.

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