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

Lk4

nuclear binding rejects relativity N.C.S.R. Demokritos (2002)

After the important discovery of the assumed uncharged neutron (1932) which led to the abandonment of the well-established laws of Coulomb (1785) and Ampere (1920) theoretical physicists 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 structureby applying the electromagnetic laws. (See my papers in my FUNDAMENTAL PHYSICS CONCEPTS ).

 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)

           p1(+1/2)..n1(+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)

one gets

-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 = -3/2

 In 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 of n2p1 , p1n1 , and n1p3 systems, while along the spin axis there exists only the strong n1p2 bond. On the other hand the repulsions of the p1p2, p1p3, and p2p3 systems contribute to the reduction of the pn bonds. In the same way the repulsion n1n2  contributes to the reduction of the radial bonds  .  This condition of many repulsions leads to the decay of Li5. Therefore the p3 leaves the system because the repulsions of p3p2  and p3p1 overcome the p3n1  bond. Since the axial n1p2 bond is very strong ,  at the same time the neutron n2 goes over the p1 in order to make the n2p2 radial bond and the n2p1 axial bond for the formation of the very stable He4 which will consist of p1, p2, n1, and n2 nucleons.