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in a Rotating Rhombus 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 2^{4}=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 2^{4/2}=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 a rhomboidal structure of nucleons that rotate as a unit. Since like nucleons repel each other the rhombus has to be composed of alternations of neutrons and protons. 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 structure being considered here is that shown below with the axis of rotation perpendicular to the plane of the particles.
Let r_{1} be the distance between a neutron and the center of rotation of the rhombus. Likewise r_{2} is the distance between a proton and the center of rotation. Let φ denote the angle shown in the above diagram. If s is the distance between the centers of nucleons around the rhombus then
If the system rotates as a unit at an angular rate of ω then the angular momentum of the system L is given by
Angular momentum is quantized as an integral multiple of h, Planck's constant divided by 2π. Therefore
where n is a positive integer.
Thus
The so-called centrifugal force on a neutron is mω²r_{1}. For equilibrium it must be balanced by the net force acting on the neutron due to the other nucleons. Suppose the strong force between two nucleons separated by a distance s is given by
where H is a constant, q_{1} and q_{2} are the nucleonic (strong force) charges of the nucleons and z_{0} is a scale parameter. The direction of the force is taken care of by the sign of q_{1}q_{2}. The nucleonic charge of a proton is taken arbitrarily to be +1. There is empirical evidence that the nucleonic charge of the neutron is −3/4. For now, however, let the nucleonic charge of a neutron be denoted as q.
For the arrangement of neutrons and protons in a rhombus, shown previously, the formula for the net radial force on a neutron is
Dynamic balance requires that the net attractive force on a neutron and the centrifugal force add up to zero; i.e.,
Thus
The above formula can be simplified somewhat by letting s/z_{0} be denoted as σ; i.e.,
The situation for a proton differs from that of a neutron only in that the repulsion between two neutrons is proportional to q² whereas the repulsion between two protons is proportional to 1². Thus for a proton
The two expressions for mω² can be equated to give
Subtracting 2(−q)*exp(−σ) from both sides of the above equation yields
This is one condition which has to be satisfied by σ and φ. Another condition arises from equating the expression for ω² which comes from the quantization of angular momentum with one that comes from dynamic balance. The expression for a proton is simpler since it does not involve q². Thus
Thus there are two equations in the two unknowns σ and φ which are functions of the two parameters q and ν=n²h²/(4mHz_{0}).
Once σ and φ are determined as functions of the quantum number n then s is determined as z_{0}σ. Then ω is determined from
From s and φ the radii of the orbits of the neutrons r_{1} and the protons r_{2} are determined from r_{1}=s*cos(φ) and r_{2}=s*sin(φ). The tangential velocities are given by
The tangential velocities determine the kinetic energy K of the system as
However
But ωs is equal to nh/(ms). Thus
Thus if s is proportional to n² then K is inversely proportional to n². The dependence of potential energy on the value of n is not simple to describe, but potential energy should be roughly comparable to the level of kinetic energy. In the case of an inverse distance squared force kinetic energy is equal to one half of the magnitude of potential energy. For the case of a force which is not strictly dependent only on the inverse distance squared see the Virial Theorem and the Nuclear Force.
(To be continued.)
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