# Nuclear Physics - Analysis of Stability in Atomic Nuclei

A simplified example can be presented which gives the basic premise. It will be clear that the concept would apply to all nuclei that are more complex, although there are some slight variations that will be discussed. We will consider a standard Helium atom nucleus, which is described as containing two protons and two neutrons. For the sake of this discussion, we will consider the nucleons to not move and to be in a formation of a tetrahedron, where the four nucleons are all equally distant from each other. Known parameters of atomic nuclei provide a guideline regarding how rapid such migration would have to occur. We know that atomic nuclei are on the order of 10-13 cm in diameter. A proton has a mass of 1.65 * 10-24 gm. We calculated above the force of electrostatic repulsion, at 2.3 * 10+7 dynes. Assuming non-relativistic motions, and for minimal variations in the distances involved, F = m * a or a = F / m, will give an approximation of the acceleration of the proton. This solves to an acceleration of 1.4 * 10+31 cm/sec2.

We might consider an absolute limit of an individual proton's movement due to Coulomb repulsion from another proton, to be half the nucleus diameter, or 5 * 10-14 cm, if nuclear stability is to be maintained. Again assuming non-relativistic velocities, then d = 1/2 * a * t2 or t2 = 2 * d / a or t2 = 7 * 10-45 or t = 8.5 * 10-23 seconds. This value is not particularly precise because we did not consider the variable force due to the variable distances that exist, but it is only meant to give a ball-park idea of the time involved for the repulsion portion of this cycle. The entire cycle would then be on the order of 1.5 times that long or 1.2 * 10-22 seconds.

This reasoning represents the longest interval that the proton-proton repulsion could be in effect without the electrostatic repulsion destroying the integrity of the nucleus. Therefore, it represents a guide to the longest possible cycle time for the process described above. As long as the electrons complete their entire migration pate in a shorter time than this, then the protons would not be de-stabilized, although they would likely experience a cyclic oscillation at that rate. If the cycle occurred more rapidly than that, the movements of the protons would be smaller and stability would be greater.

This situation suggests that the electrons, if described as moving, would need to traverse a cycle of three segments, or around 2.5 * 10-13 cm in a period no longer than 1.2 * 10-22 seconds, which implies a minimum velocity of around 2 * 10+9 cm/sec, about 1/15 the speed of light. This is interesting in that, should it be a higher velocity, then relativistic velocities of the electrons would increase their mass and possibly affect the reasoning regarding the mass defect and many other effects.

There are a variety of ways the electrons might actually move within the nucleus. A more generalized form of the above argument involves taking the time integral of the attraction between each proton and the electron during whatever path is followed. For most geometries of electron movement, the resulting effect is a slight reduction of the net attractive force. The initial (migrate) argument above would have had too much attractive force for stability, and effects such as this might ensure that the time-average of the attraction exactly equals the time-average of the proton-proton repulsion, in making a stable nucleus. A discussion below will consider the situations where there are more or less than an optimal number of electrons within the nucleus, and the effects on stability, on the half-life and the radioactive decay schemes.

## Preference for Beta- or Beta+ or EC Decay in Light Nuclei Isotopes

There has been recognized that the "single atomic weight" graphs discussed and displayed below tend to have much steeper sloped curves for lower atomic weights than the same graphs for higher atomic weights. Since one conclusion of this analysis is that the slope of those curves represent an indication of stability regarding beta decay, there is the implication that heavier nuclei would more rarely emit beta- or beta+ particles, which is borne out by experimental evidence.

Regarding beta- and beta+ decay, the graphs displayed below suggest that the half-life of any isotope might be predicted from the slope of the appropriate graph, and that the total energy (kinetic and radiation) might also be predicted for any isotopic beta decay.

## Preference for Alpha Decay in Heavy Nuclei Isotopes

A similar analysis of actual atomic weight graphs shows an entirely different pattern regarding decay involving alpha particles. For light nuclei, the curve is inverted which suggests that alpha particle emission is not possible. For the heaviest isotopes, the slope of such curves exceeds the slope of the curves for beta decay, which suggests a natural preference for alpha decay for such isotopes.

In addition to this, the universal extreme symmetry preference for isotopes with even numbers of protons and also even numbers of neutrons seems to suggest a special stability of 4He nuclei, which is the alpha particle. This symmetry-based stability might suggest that such structures exist within heavy nuclei, which might explain why they leave the nucleus as a bundle as an alpha particle. This might imply that within heavy nuclei there are distinct organized structures, the simplest of which would be the alpha particle.

This presentation was first placed on the Internet in November 2003.

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