By L. Kaliambos (Λευτέρης Καλιαμπός) Τ.Ε. Institute of Larissa Greece.( June 2014)

It is indeed unfortunate that the discovery of the assumed uncharged neutron along with the invalid relativity led to the abandonment of the well-established laws of electromagnetism. Thus the various nuclear theories and structure models could not lead to the nuclear structure . The problem became more complicated when Heisenberg assumed that the proton-proton and neutron-neutron repulsions create fallacious attractive nuclear forces.

For example in the “Evidence for Alpha Particle Substructures in nuclei-San Jose State University ” one reads: “The nuclear force was believed to exist equally between protons and neutrons and between neutrons as well as between protons. This led Werner Heisenberg to speculate that a proton and a neutron are merely different forms of a single particle which he called the nucleon. This nucleon hypothesis came to be widely accepted even though there is considerable evidence against it. It was just a convenient assumption that made theorizing simpler. If Heisenberg's hypothesis were really true there would exist He2 nuclei, two protons bound together by the overwhelming nuclear force. Such a nuclide does not exist. There would also be bound neutron complexes and the emission of gamma rays from neutron collections when such complexes form.”

Under this physics crisis in my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003) I revived the natural laws which led to my discovery of extra 9 charged quarks in proton and 12 ones in neutron responsible for the nuclear structure and binding. Note that such extra charged quarks ate a part of 288 quarks in nucleons.

Today it is well-known that the structures and binding energies of nuclei are based not on invalid nuclear theories but on the well-established laws of electromagnetism. You can see my DEUTERON STRUCTURE AND BINDING for the simplest explanation of the deuteron structure and binding able to tell us how the charges of two spinning nucleons interact electromagnetically with parallel spin ( S = 1) for giving the nuclear binding and force in the simplest nuclear structure. Also you can see my paper SRUCTURE AND BINDING OF H3 AND He3 in my FUNDAMENTAL PHYSICS CONCEPTS . To compare the structure of He-4 with the structures of Be-8, O-16, and Pb-208 see the following figures

Also in the following diagrams of He4 and He6 you see that the spinning nucleons in He4 form a simple rectangle with S=0 which explains the stability of alpha particles.

He-4

n_{2}(-1/2)..p_{2 }(- 1/2)

p_{1}(+1/2)..n_{1}(+1/2)

He-6

n_{2 }(-1/2)..p_{2}(- 1/2)..n_{4}(-1/2)

n_{3}(+1/2)..p_{1}(+1/2)..n_{1}(+1/2)

**NUCLEAR STRUCTURE OF He-4 WITH S = 0**

Here you see that the simplest p_{1}n_{1} and p_{2}n_{2 }systems ( deuterons ) are characterized by the following weak binding energy

B(p_{1}n_{1}) = B(p_{2}n_{2}) = -2.2246 MeV

According to electromagnetic laws they are coupled along the radial direction or along the x axis giving the total parallel spin along the spin axis or along the z axis as

S = +1/2 +1/2 = 1

which invalidates the so-called Pauli Principle. Then from the structure of the mirror nuclei H3 and He3 it was possible to find the repulsive energies

U(p_{1}p_{2}) = 0.863 MeV and U(n_{1}n_{2} ) = 0.099 MeV

Then from the strong binding energy of He4

B(He4) = - 28.29 MeV

one concludes that the two deuterons are coupled along the spin axis or along the z axis with very strong binding energies

B(p_{1}n_{2}) = B(n_{1}p_{2}) = -12.4 MeV

Note that this very strong nuclear binding energy is derived after a large number of integral equations when z = 0.41 R_{p} where R_{p} is the proton radius (R_{p } = 0.84 /10^{15 }m). This short separation justifies the oblate spheroid of nucleons.

Under this condition we write

B(He4) = B(p_{1}n_{1}) + B(p_{2}n_{2}) + B(p_{1}n_{2}) + B(n_{1}p_{2}) + U(p_{1}p_{2}) + U(n_{1}n_{2})

Therefore the above equation in terms of MeV can be written as

-28.29 = -2.2246 -2.2246 -12.4 -12.4 + 0.863 + 0.099

**NUCLEAR STRUCTURE OF He-6 WITH S = 0**

In the He6 the greater number of bonds and repulsions complicates the calculations of energies. However the shape as shown in the second diagram explains very well the total spin S=0 and the decay of He6 since the repulsive forces of the nn systems contribute to the reduction of the p_{1}n_{3} bond and the p_{2}n_{4} bond. Note that in the absence of nn repulsions the binding energy of them should be equal to the binding energy of deuteron.

For example the repulsions n_{3}n_{2} and n_{3}n_{1}, (the repulsion n_{3}n_{4 }is negligible) contribute to the reduction of the p_{1}n_{3} bond. In the same way the repulsions n_{4}n_{1} and n_{4}n_{2} _{ }contribute to the reduction of the p_{2}n_{4 }bond.

Although the distances n_{1}n_{3} and n_{2}n_{4} are greater than the distances n_{2}n_{3} and n_{1}n_{4} we assume that all the repulsions have the same value because in the systems n_{3}n_{1} and n_{4}n_{2} the neutrons have parallel spin with electric and magnetic repulsions.

Thus

U(nn) = U(n_{3}n_{2}) +U(n_{3}n_{1}) + U(n_{4}n_{1}) + U(n_{4}n_{2}) = 4U(n_{1}n_{2}) = 4(0.099) = 0.396 MeV

Then from the binding energy of He6

B(He6) = -29.27 MeV

one concludes that the weak bonds of p_{1}n_{3} and p_{2}n_{4 } have a binding energy weaker than the binding energy of -2.2246 MeV (deuteron). Such weak bonds along with the repulsions of the nn systems give a total energy

B(p_{1}n_{3}) + B((p_{2}n_{4}) + U(nn ) = B(He6)- B(He4) = (-29.27) – (-28.29) = -0.98 MeV

Because of symmetry we have B(p_{1}n_{3}) = B(p_{2}n_{4}). Thus we may write

2B(p_{1}n_{3}) +U(nn)) = -0.98 or 2B(p_{1}n_{3}) + 0.396 = - 0.98 MeV

That is B(p_{1}n_{3}) = Bp_{2}n_{4}) = - 0.344 MeV. Since this value is weaker than the - 1.29 MeV it leads to the decay. Note that a free neutron or a single np bond with a binding energy B(np) < 1.29 MeV leads to the decay, because the difference in energies between a down quark (3.69 MeV) and an up quark (2.4 MeV) is 1.29 MeV. For example for the decay of a neutron we discovered that the neutron with 288 quarks turns into the proton with 288 quarks as

n = p +e +ν or [92(dud) +4u +8d] = [(92(dud) +dud +4 +5d ] +e + ν οr d = u +e +ν.