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

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N.C.S.R. "Demokritos" (2002)


After my discovery of extra 9 charged  quarks in proton and 12 ones in neutron which led to my published paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003)  today it is well-known that the structures and binding energies of nuclei are based not on invalid nuclear theories or on Einstein's contradicting theories of relativity, but on the well-established laws of electromagnetism.  In my DEUTERON STRUCTURE AND BINDING  one sees the simple 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.

 Nevertheless today nuclear theorists continue to believe that the structure and binding of Tritium and He3 is based not on the well-established laws of electromagnetism but on the false ideas of invalid nuclear theories. So in “The Structure and Binding Energy of the H3 (Tritium) and He3 Nuclides- San Jose State University ”  one reads:

“The binding energy of a triteron is 8.482 million electron volts (Mev) but that of the Helium 3 nucleus is only 7.718 Mev.... It might be expected that the distance between two repelling protons in He3 would be larger than the distance between a proton and a neutron in a deuteron which are only subject to the attractive strong nuclear force. However if the nucleons in H3 and He3 are arranged in a triangle each nucleon is subject to a greater net attraction than those in a deuteron. The Nuclear Strong Force.  According to Hideki Yukawa the strong force is carried by π mesons. Any force carried by particles will have an inverse distance squared dependence. However the π mesons decay with time and there the number surviving at any distance is a negative exponential function of distance.”

 It is  of interest to note that such fallacious ideas cannot lead to the nuclear structure and binding energy and in the absence of the detail knowledge of our DISCOVERY OF NUCLEAR FORCE AND STRUCTURE today theoretical physicist in order to find the difference between the H3 (Tritium) and He3  use the p-p repulsion of two protons treated as point charges, because they believe that the nuclear strong force is due to the fallacious meson theory of Yukawa. Under this physics crisis due to the abandonment of electromagnetic laws  I  revived the natural laws which led to our discovery of the nuclear structure. This discovery  is understandable by using the following diagrams of the four lightest nuclides as

Deuteron                  Tritium  (H3)                Helium3                         Helium4

p(+1/2)..n(+1/2)        n2(-1/2)                                       p2(-1/2)            n2(-1/2)..p2(-1/2)

                                  p1(+1/2)..n1(+1/2)       p1(+1/2)..n1(+1/2)            p1(+1/2..n1(+1/2)

Here you see that the simplest pn system of nuclear structure ( Deuteron ) is characterized by the following weak binding energy

 B(pn) = -2.2246 MeV

According to electromagnetic laws it is  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 very stable He4 with S =0  it is possible to describe the structures of Tritium and He3 which have some disorder as a result of a missing nucleon.  In both cases of H3 and He3 the strong  bonds along the spin axis like p1n2 and n1p2 give a zero spin S=0 . So the total spin S = ½  is due to the spin  s = 1/2  of  n1 or  p1.

Because of  the repulsions of nn and pp systems along the diagonals we see that in Tritium and in He3  the same weak binding energy along the x axis is reduced from

 B(pn) = -2.2246 MeV to the following value

B(p1n1) = -1.3 MeV

which is responsible for the decay of Tritium or for the transformation of n1  to proton . In Helium3  the n1 cannot turn into proton, because it makes two bonds with p1 and p2.  Note that the deuteron cannot decay because its binding energy of -2.2246  is stronger than the binding energy of -1.29 MeV.  Whereas a free neutron or a neutron in a single neutron-proton bond with a binding energy

 B(pn)  < 1.29 MeV

leads to the decays, because the value of 1.29 MeV is the difference in energies between the down quark (3.69 MeV ) and the up quark (2.4 MeV). (See my papers of my FUNDAMENTAL PHYSICS CONCEPTS ).    

 Note that after a detailed analysis of the pn bonds and the pp repulsions and the nn repulsions of the structure of He4 we found that the repulsive energies U(n1n2) and Up1p2) along the diagonals are

U(n1n2 ) = 0.099 MeV and U(p1p2) = 0.863 MeV

Under this conditions for the Tritum we may write the energies in MeV as

B(Tritium) = B(p1n1) + B(p1n2) +U(n1n2)    or - 8.482 = -1.3  - 7.281 + 0.099  

 In the same way for the He3 we write

B(He3) = B(p1n1) + B(n1p2) + U(p1p2)  or   -7.718 = -1.3 – 7.281 + 0.863