|San José State University|
& Tornado Alley
in a Rotating Ngon of Particles
One of the major problems of the physics of nuclei is the explanation of the high binding energy of Helium 4 nucleus, the alpha particle. A deuteron (proton-neutron pair) has a binding energy of 2.225 million electron volts (MeV) but an alpha particle has a binding energy of 28.3 MeV. (The higher the binding energy of a particle the more energy is required to break it up into its constituent parts.) The binding energy of tritium, H3,a nuclide with one proton and two neutrons is 8.481 MeV, but that of He3, which also contains three nucleons but two protons instead of two neutrons, is 7.718 MeV. Thus the electrostatic repulsion of the two protons apparently reduces the binding energy of He3 by 0.763 MeV or 9 percent.
Binding energy is like, and perhaps is identical with, potential energy. In a previous study of deuteron-like nuclei it was found that the kinetic energy is proportional to the fourth power of the geometric mean of the cluster size. Thus the alpha particle with clusters twice the size of those of a deuteron should have kinetic energy 24=16 times that of a deuteron. On the other hand a H3 or He3 nucleus with mean cluster size √2 times that of a deuteron should have kinetic energy 24/2=4 times that of a deuteron. The relative sizes of the potential energies and binding energies should also be roughly 16 and 4 times that of a deuteron. The binding energies due to the formation of nucleonic pairs also have be taken into account but the binding energies of the deuteron of approximately 2 million electron volts (MeV), that of H3 and He3 at about 8 MeV and that of the alpha particle at 28 MeV are in roughly those proportions.
Another study established that the same analysis applies for particles which are linked together but not necessarily close together in clusters. That study considered a model of the type shown below.
The purpose of this material is to analyze polygonal rings of nucleons that rotate as a unit. Since like nucleons repel each other the ring has to be composed of alternations of neutrons and protons. Consider n neutrons and n protons arranged alternately in a 2n-gon. Neutrons are attracted to protons through the nuclear strong force but repelled from each other and the same holds for the protons. For now assume all of the particles have a mass of m. The deuteron is the simpliest case of n=1. The next case, that of n=2, is shown below.
If s is the distance between the centers of nucleons around the square then the distance between two like nucleons across the square is √2s and the distance from the center of the square to the nucleons is s/√2.
The 2n-gon rotates as a unit at an angular rate of ω around the nucleons' center of mass. The distance from the center of mass to the particles is denoted as r. The angular momentum L of the system is then nmωr²
The angular momentum must be an integral multiple of
h, Planck's constant divided by 2π; i.e.,
where k is a positive integer. For use later note that
The so-called centrifugal force on a nucleon is mω²r. For equilibrium it must be balanced by the net force acting on the nucleon due to the other nucleon. Suppose the strong force between two nucleons separated by a distance s is given by
where H is a constant, q1 and q2 are the nucleonic (strong force) charges of the nucleons and s0 is a scale parameter. The nucleonic charge of a proton is taken to be +1. The nucleonic charge of a neutron is a negative number less than 1 in magnitude. Let the nucleonic charge of a neutron be denoted as q.
For the arrangement of neutrons and protons in a square, shown previously, the formula for the net radial force on a neutron is
Since r=s/√2, s in the above formula can be replaced by √2r to give
The general formula for the net force would be quite complicated but it can be expressed as a function of n and r. If the force is carried by particles there will an inverse distance squared dependence. The form would be, say Hqf(r/s0,n,q)/r². Thus
The two expressions for ω² can be equated to give
This is the quantization condition for r. The term f(r/s0,n,q)r is zero if r is zero. It is reasonable to assume it can be approximated as γr. Thus
This value of r can be indexed by the quantum number k; i.e.,
From the quantization of angular momentum it follows that the tangential velocity v of a nucleon is
Thus the tangential velocity of the nucleons is inversely proportional to the quantum number k.
It is notable that vk is independent of the particle mass m.
The angular rotation rate ω is v/r and therefore
Thus the angular rotation rate is inversely proportional to the cube of the quantum number k.
The kinetic energy K of the system is ½nmv² which works out to be
Thus the quantized kinetic energy is inversely proportional to the square of the quantum number k. Also it is proportional to the cube of the number of particles in the system.
The potential energy V can be some complicated function of r. For an inverse distance squared force like the gravity or the electrostatic force it is simply inversely proportional to separation distance. Assume the potential energy over some relevant range of r can be approximated as
where ν is a quantity greater than or equal to 1.
Given the quantization condition found for r then
The ratio of Vk for n=2 to the value for n=1 is 2ν. If ν=1.2 then that ratio would be 5.28. A deuteron corresponds to n=1. The binding energy of a deuteron is about 2.22 MeV. This would make the binding energy for the case of n=2 equal to 11.7 MeV.
(To be continued.)
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