CHAPTER 7: THE NUCLEAR FORCE THEORY

SECTION 7-0: INTRODUCTION

In this chapter, we will look for describing functions, which help us to understand the nuclear force. From the study of the neutron, we learned that the electron reaches a speed of 0.91859333C before it enters the proton to form the neutron. This super-electron energy is a clue to what happens with the nuclear force. For this solution we are only looking for an Engineering ballpark answer which will help us to understand the nuclear force. Thus this is only a short initial study of the nuclear force problem.

The Nuclear Force Theory proposes that the force which holds the atoms nuclei together is the result of the electron’s energy at 0.9186C, which is shared by neighboring neutrons and protons. However two protons, or a proton and a neutron can only share a high energy electron if both lose an equivalent amount of energy in the process. Thus it is the protons which lose the energy whether within the protons themselves or as part of the neutrons. To complete the energy requirements, we must also subtract the energy of proton/proton repulsion, which will cause the atom to unbind. In addition we must subtract the energy of each original electron. To see if this theory holds true let us look at some examples from my old physics book.

SECTION 7-1: THE NUCLEAR FORCE

We know that the nuclear force is a close action force. In Weidner & Sells Modern Physics 1965 by Allyn & Bacon, the nuclear force appears at approximately 1.4E-15 meters. Weidner & Sells state that the nuclear force is reduced by the repulsion of the protons. They state that the electric force applies to all nucleons while the nuclear force only applies to nearby nucleons. Also the Proton/Proton force has the same force as the proton/neutron force or the neutron/neutron force.

Let us now write the equation for the nuclear binding energy and calculate some examples.

Nuclear Binding Energy = (Einsteinian electron energy/pair). (#Active pairs) –

(# Electrons . energy) – (proton/proton energy per pair).

(# Active proton pairs) .......(7-1)

Equation 7-1 species that the nuclear binding energy of an atom equals the neutron’s Einsteinian energy per pair times the number of active pairs. From this we subtract the rest energy of the number of electrons in the neutron. Then we subtract the proton/proton electrical energy per pair times the number of active proton pairs.

From equation 3-34 the Einsteinian energy of the electron in the Neutron is:

Energy = 1.293MEV ......(7-2)

In order to share the high-energy electron with a neutron or a proton, the total energy of a pair of nuclei must be deficient by twice this amount. Therefore the energy per pair is:

Energy per pair = 2.586MEV .................(7-3)

The repulsive energy in electron volts for a pair of protons touching at a proton radius of 1.32142E-15 is

Energy per pair of protons = 2 (KQ/2Rp) = KQ/Rp = 1.08971 MEV..........(7-4)

Finally the energy per electron is:

Energy per electron = 0.511 MEV ..............(7-5)

We can now calculate the binding energy of the Deuteron:

Let us look at Deuteron. This has one proton and one neutron.

The proton/neutron electric repulsive force is zero. There is no proton/proton force.

The nuclear force is:

Nuclear Force = 2.586 MEV ...............(7-6)

The electron rest energy is:

Electron rest energy = 0.511 ..............(7-7)

The Binding energy is the difference:

BE or Deuteron = 2.075 MEV ................(7-8)

The Deuteron binding force is 2.22 MEV. There are many small forces at play. Additional pressure could cause the radius to shrink and this would increase the binding energy. Therefore the solution is within an Engineering ballpark.

Let us look at helium. According to Physics by Hausman & Slack Brooklyn Polytechnic June 1957, the binding energy of Helium is listed as 29 MEV.

Helium has two protons and two neutrons. Therefore it has 3 pathways for each nuclei times 4 nuclei. The total nuclear energy is:

Nuclear energy = 2.586 MEV x 3 x 4 = 31.032 MEV ...........(7-9)

The atom has two protons, thus the negative binding energy is:

Proton/Proton energy = 1.08971 MEV ...........(7-10)

The rest energies of 2 electrons is:

Electron energy = 1.022 MEV ..............(7-11)

The total binding energy is:

Helium binding energy = 28.92 MEV ...............(7-12)

This answer is extremely close to the measured results as per my old text books. Thus it appears that the Einsteinian energy of the electron is lost in the production of the atoms.

Let us now look at the binding energy of oxygen. Oxygen has 8 protons and 8 neutrons. In general not all protons and neutrons will bind to each other. In addition not all protons will repel each other because the electrostatic field will short out within the atom. We cannot say that every proton sees every other proton for this reason.

The atoms follow rules and regulations similar to the Bohr atom. For oxygen we know that the binding energy is 127.5 MEV. It is reasonable to understand that each pair of nuclei will only match up with at most 4 others. Sixteen is 4 squared. Therefore Let us take 4 as the number of electron pathways for each nuclei. The nuclear energy is:

NE = 2.586 x 4 x 16 = 165.504 MEV ................(7-13)

The electron rest energy is:

Electron energy = 8 x 0.511 = 4.088 MEV .................(7-14)

The proton/proton negative binding energy is:

Proton repulsive energy = 1.08971 x 4 x 8 = 34.8707 ...........(7-15)

Therefore the binding energy is:

Oxygen binding energy = 126.545 MEV ...............(7-16)

We see that the binding energy of the atoms is merely the Einsteinian energy of the electron, which is released from the protons and the neutrons during the compression process. We can now write the general formula for the binding energy of large atoms:

BE = 4 x 2.586MEV - 0.511/2 MEV – 2 x 1.08971MEV ...............(7-17)

Be = 10.344 – 0.2555 – 2.17942 = 7.90908 MEV ..............(7-18)

We see in Equation 7-18 that the binding energy of atoms larger than oxygen is equal to 4 times the nuclei binding energy minus half of the rest energy of the electron minus twice the proton/proton repulsive energy. Each atom will have some small differences. Some may have more neutrons. Some may have slightly different proton/proton interactions. In general Equation 7-18 describes the general solution for the binding energy of atoms. It is a very simple Einsteinian mass/energy force.

Equation 7-18 shows the energy formula for the binding of the atoms. The sharing of the electrons helps to bind the atoms. There will be additional factors in the binding of the atoms. Bipolar dots will convert into plus and minus photonic dots, which will surround the protons and neutrons. This will cause some surface binding.

The important thing is that the Einsteinian energy at 0.9186C is common to all the atoms. In the case of the neutron it provides the extra energy. However it does not bind the neutron. The neutron falls apart readily. The same amount of energy as a deficiency in the atoms causes a sharing of the electrons.

The above analysis is only a small study involving a few days work. It was not part of the original dot theory and is not part of the dot-wave theory. It only came into being as part of the proton thruster rocket theory. Thus once the neutron’s Einsteinian energy was discovered, this became self evident to me as the main reason for the binding energy of atoms.