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An Analysis of Quantization
in a Rotating Structure 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 28.296 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 8.482 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.

For the nuclides beyond He 4 there is not a corresponding increase in the binding energy. The berylium 8 nuclide with 4 protons and 4 neutrons has a binding energy of 56.500 MeV, roughly only twice that of a He 4 nuclide. The carbon 12 nuclide has a binding energ of 92.162 MeV, a little more than three times the binding energy of He 4.

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 a structure of nucleons that rotate as a unit. Since like nucleons repel each other the structure has to be composed of alternations of neutrons and protons. Consider a nucleonic structure consisting of k layers each containing p particles all located at a distance r from the axis of rotation. (The integer p is presumed to be even.) The structure rotates as a unit at an angular rate of ω radians per second.

The angular momentum L of the structure is given by

L = (pk)mωr²

Angular Momentum

Angular momentum is quantized as an integral multiple of h, Planck's constant divided by 2π. Therefore

(pk)mωs² = nh

where n is a positive integer.

Thus

ω = nh/(pkms²)
and, for later use,
ω² = n²h²/((pk)²m²s4)

Dynamic Balance

The so-called centrifugal force on a neutron is mω²r1. 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

F = Hq1q2*exp(−s/z0)/s²

where H is a constant, q1 and q2 are the nucleonic (strong force) charges of the nucleons and z0 is a scale parameter. The direction of the force is taken care of by the sign of q1q2. 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. 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 net force on a particle depends upon which layer it is in and on the number of particles p in the layer. Assume the p particles in a layer are arrayed in a regular polygon. The relationship between the radius r and the side length s is given by the cosine law for triangles; i.e.,

s² = r² + r² − 2r²cos(2π/p)
and thus
s = √2[1−cos(2π/p)}½r
or, equivalently
r = √2[1−cos(2π/p)}s

Let the above equation be expressed as

r = αps

Let s/z0 be denoted as σ. The net force on a particle in the i-th layer can then be expressed as

Fi = (H/(z0²σ²))[(−q)fi(σ) − gi(σ)]

Therefore a balance of centrifugal force with the net attractive force on a particle requires

mω²z0αpσ = (H/(z0²σ²))[(−q)fi(σ) − gi(σ)]
and hence
ω² = [H/(z0³mαpσ³)][(−q)fi(σ) − gi(σ)]

The two expressions for ω² can be equated to yield

[H/(z0³mαpσ³)][(−q)fi(σ) − gi(σ)] = n²h²/((pk)²m²z04σ4)

This can be reduced to

[(−q)fi(σ) − gi(σ)]σ = n²h²/((pk)²mz0αpH)

Let the LHS of the above equation be approximated by Ωiσ. Then

σi = n²h²/((pk)²mz0αpi)
and hence
si = z0σi = n²h²/((pk)²mαpi)
and
ri = αpsi = n²h²/((pk)²mHΩi)

From this point on the dependence on the layer number is ignored.

Since ω=nh/(ms²) the tangential velocity v is given by

v = ωr = αpωs = αpnh/(ms)
v = αpnh((pk)²αpHΩ) /n²h²
which reduces to
v = αp²((pk)²HΩ) /nh

The kinetic energy K of the system is then

K = (pk)(½mv²) = ½mαp4(pk)5H²Ω²/(nh

The dependence of potential energy on n and on (pk) is difficult to establish but the potential energy is of a magnitude comparable to that of kinetic energy.

Some Comparisons

Going from a deuteron (H 2) with p=2 and k=1 to an alpha particle (He 4) with p=4 and k=1 would increase the potential energy by a factor of 25=32. If Be 8 has the structure of a cube then the kinetic energy should increase over that of He 4 by a factor of 32. It doesn't. It appears that there are no structures within a nucleus rotating as a unit beyond an alpha particle. The conclusion is that the there are no rotating units within a nucleus other than alpha particles.


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