By  Prof. Lefteris Kaliambos (Natural Philosopher in New Energy)

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NCSR Demokritos 2002 the discovery of nuclear structure leads to the photon mass

( June 2014)

Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories  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 which led to my discovery of 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

The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying  the laws of electromagnetism. You can see my papers of nuclear structure in  my FUNDAMENTAL PHYSICS CONCEPTS  . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications

Beryllium(Be) has 12 known isotopes, but only one of these isotopes (9Be) is stable and a primordial nuclide. As such, beryllium is considered a monoisotopic element. It is also a mononuclidic element, because its other isotopes have such short half-lives that none are primordial and their abundance is very low. Beryllium is unique as being the only monoisotopic element with both an even number of protons and an odd number of neutrons. Here we describe the ubstable Be8 and the stable Be9.

To simplify the problem you can use the following diagrams of Be-8 and Be-9, and for detailed knowledge you can see Fig. 5a and Fig. 5b of my published paper.


To compare the structure of Be-8 with the structures of He-4, O-16, and Pb-208 see the following figures.


structures of He-4, Be-8, O-16, and Pb-208

Also in the following first diagram of Be-8  you see that it consists of the two  simple rectangles (p1n1p2n2) and (p3n3p4n4) which represent the rectangles of two alpha particles having a total binding energy 2B(He4) = 2(-28.29) MeV with S = 0. However when the two alpha particles are closely packed for the formation of the unstable parallelepiped of Be-8 one observes that two extra pp  repulsions as p1p3 =  p2p4 appear. They are very strong because the protons have parallel spin and exert both electric and magnetic repulsions along the diagonals of the squares (p1n1p3n3) and (p2n2p4n4) respectively. Note that the nucleons cannot form a cube because of their oblate spheroid.  In the same way one sees the nn repulsions as n1n3 =  n2n4. Also between the two alpha particles along the diagonals of the rectangles (n1p3n4p2) and (p1n3p4n2) one observes the repulsions p2p3 =  p1p4 and  n1n4 =  n2n3 respectively. All these repulsions give a total extra repulsive energy Uex as

Uex = 2U(p1p3) + 2U(p2p3) + 2U (n1n3) +2U(n2n3)

 Of course such extra repulsions contribute to the reduction of the binding energy of the four extra   bonds having the structure of deuterons, as p1n3 = n1p3 = p2n4 = n2p4 with a total extra binding energy

Bex = 4B(p1n3) .


                                                                   Diagram of stable Be-9 with S =-3/2

                                                                   n4 (-1/2)..p4( -1/2)..n5(-1/2)      

                                                                   p2(+1/2)..n3 (+1/2).p3(+1/2)      

   '                                                               ' n1( -1/2)..p1(-1/2)..n2( -1/2) 


'                                                                                                                                                                                                                                                                                                                                          ' '             'DIAGRAM OF THE UNSTABLE Be-8 WITH S =0

Here the p1n1n2p2 make the first horizontal square with positive spins at the first horizontal plane (+HP1) , while the n3p3p4n4 make the second horizontal square at the second horizontal plane of negative spins (-HP2).All these nucleons form the first unstable parallelepiped of the structure of atomic nuclei.


                 n3.........p3            -HP2


             p1..........n1           +HP1                                                                                                                                                 


Using the binding energies  B(Be-8) = -56.5 MeV  and B( He-4) = -28.29 MeV  one can write

 B(Be-8) = 2B(He-4) + ( Uex + Bex)      Or    -56.5 = 2(-28.29) + ( UexBex)                                                     

That is ( Uex + Bex) = 0.08 MeV

It means that the large number of extra repulsions overcome the extra bonds and lead to the decay of Be8  which splits into two alpha particles. It is of interest to note that the fallacious nuclear structure model of the well-known Fermi gas leads to serious problems, because one might expect that Be8 with a great symmetry with  S = 0 and Z = N could be able to be one of the most stable nuclides. Actually the Be8 of a great symmetry is the collection of two very stable alpha particles (rectangles) which cannot form a stable nucleus in three dimensions for making the first parallelepiped in the nuclear structure.


 In the stable Be-9 with S = -3/2 as shown in the second diagram the structure occurs not as a parallelepiped in three dimensions but as a rectangle (n1n2n5n4) in two dimensions in which a smaller rectangle (n1p1p4n4)  could represent the nuclear structure of Li-6. In this case one observes that outside of Li-6 there are extra repulsions and extra pn bonds . The extra pp repulsions as p3p1, p3p2 and p3p4 have an extra U(pp) repulsive energy, while the extra nn repulsions as n2n1, n2n3, n2n4, and n2n5, along with the extra repulsions as n5n1, n5n3 and n5n4 have a total  extra U(nn) repulsive energy. Here one concludes that all these extra repulsive energies Uex = U(pp) + U(nn) cannot overcome the  extra total binding energy  Bex  of the  pn bonds like p1n2, p3n3, p4n5, p3n2, and p3n5 , because the p3n2 and p3n5 systems are very strong bonds acting along the spin axis.    

Using the binding energies B(Be-9) = -58.15 MeV and B(Li-6) = -31.98 MeV one can write

B(Be-9) = B(Li-6) + (Bex + Uex )     Or    -58.15 = -31.98 + (Bex +Uex)  

That is (Bex + Uex) = - 26.17 MeV

This means that the radial and the strong axial pn bonds overcome the pp and nn repulsions and lead to the stability of Be-9. Note that  nuclear physicist applying  not the electromagnetic laws but the fallacious nuclear structure models believe that Be-9 is composed of two α-particles separated to form a deformed Be8-core and an extra neutron strongly coupled to the motion of the Be8-core by the neutron core potential. ( See “Alpha-Particle Model for Be9 – Progress of Theoretical Physics ”).

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