By Prof. L. Kaliambos (Natural Philosopher in New Energy)
25 February 2019
After the discovery of the assumed uncharged neutron (1932) which led to the abandonment of the well-established laws of Coulomb (1785) and Ampere (1820) theoretical physicists under the influence of the invalid relativity (EXPERIMENTS REJECT RELATIVITY) developed various nuclear structure models which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 extra charged quarks in proton and 12 ones in neutron able to give the nuclear binding and nuclear structure by applying the electromagnetic laws. The paper was also presented at a nuclear conference held at NCSR "Demokritos" (See in photo).
For understanding better the nucleon-nucleon electromagnetic interaction you can see the new structure of protons and neutrons given by
proton = [93(dud) + 5d + 4u ] = 288 quarks = mass of 1836.15 electrons
neutron = [92(dud) + 4u + 8d ] = 288 quarks = mass of 1838,68 electrons
Nevertheless today physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. ( See the “Models of nuclear structure -San Jose State University ”). Under such fallacious theories the nuclear force was believed to exist equally between protons and neutrons and between neutrons as well as between protons. This led 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.
In order to overcome such difficulties 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 for understanding the pn bonds and the repulsions of the pp and nn systems. In the following diagrams of Li6 and Li5 you see that the spinning nucleons in Li6 form a simple rectangle with S =1 which explains the stability of Li6 , while the structure of Li5 with S = -3/2 is unstable.
Stable Lithium - 6 Unstable Lithium - 5
p3(+1/2)..n3 (+ 1/2) p2 (+1/2)
n2 (-1/2)..p2 (- 1/2) n2(-1/2)… p1(-1/2).. n1 (- 1/2)…p3(-1/2)
NUCLEAR STRUCTURE OF Li-6 WITH S = +1
In Li6 you see that the simplest p1n1 , p2n2, and p3n3 systems ( deuterons ) are characterized by the following weak binding energy
B(p1n1) = B(p2n2) = B(p3n3) = -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(p1p2) = U(p2p3) = 0.863 MeV and U(n1n2 ) = U(n2n3) = 0.099 MeV
Then from the binding energy of Li6
B(Li6) = - 31.98 MeV
one concludes that the three deuterons p1n1, p2n2, and p3n3 with a total binding energy
3B(p1n1) = 3(-2.2246) = 6.6738 MeV
are coupled along the spin axis or along the z axis with the following strong binding energies
B(p1n2) = B(n1p2) = B(n2p3) = B(p2n3)
Note that this strong nuclear binding energy was derived after a large number of integral equations. Here it should be weaker than that of Helium-4 because the strong axial repulsions of the p1p3 and n1n3 contribute to the reduction of axial binding energies. After many integral equations we found that U(p1p3) = 7.3 MeV and U(n1n3) = 4.7 MeV
Under this condition and writing
B(Li6) = 3B(p1n1) + 4B(p1n2) + 2U(p1p2) +2 U(n1n2) + U(p1p3) + U (n1n3)
-31.98 = 3( -2.2246) +4 B(p1n2) + 2(0.863) +2(0.099) + 7.3 + 4.7
That is, B(p1n2) = - 9.8 MeV.
This value differs from the B(p1n2) = -12.4 MeV of He-4, because here the p1p3 and n1n3 strong axial repulsions contribute to the reduction of the pn bonds along the spin axis.
NUCLEAR STRUCTURE OF Li-5 WITH S = -3/2In the unstable Lithium-5 the great 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 = -3/2 and the decay of Li5 . Here we observe three radial bonds