By Prof. Lefteris Kaliambos (Λευτέρης Καλιαμπός) Τ.Ε. Institute of Larissa Greece.
( June 2014)
After the discovery of the assumed uncharged neutron (1932) and the invalid relativity (1905) which led to the abandonment of the well-established electromagnetic laws, theoretical physicists developed contradicting theories of the nuclear force and 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 charged quarks in proton and 12 ones in neutron able to give the nuclear binding and nuclear structure by applying the laws of electromagnetism.(See my papers of nuclear structure 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.
Carbon (C) has 15 known isotopes, from 8C to 22C, 2 of which (12C and 13C) are stable. The longest-lived radioisotope is 14C, with a half-life of 5,700 years. This is also the only carbon radioisotope found in nature - trace quantities are formed cosmogenically by the reaction 14N + 1n → 14C + 1H.
NUCLEAR STRUCTURE OF C-12 WITH S = 0
Comparing the shapes of Fig. 6a and Fig.5a of my published paper one sees that C-12 consists of a parallelepiped similar to the unstable Be8 with a binding energy B(Be8) = -56.5 MeV and a rectangle similar to the stable He-4 of binding energy B(He4 ) = -28.29 MeV. In the following diagram of the structure of C12 and N14 you see that indeed the C12 consists of the unstable parallelepiped of Be8 in which we add the rectangle (p5n5p6n6). In fact we observe the three rectangles (p1n1p2n2) , (n3p3n4p4) and (p5n5p6n6) which represent the rectangles of three alpha particles having a total binding energy 3B(He4) = 3(-28.29) = -84.87 MeV with S = 0. However when the three alpha particles are closely packed for the formation of the stable parallelepiped of C12 the experiments showed that the new parallelepiped of C12 has a binding energy B(C12) = -92.14 MeV which means that the pn bonds outside the alpha particles have an extra binding energy Bex stronger than the extra pp and nn repulsions of an extra repulsive energy Uex.
That is (Bex + Uex) = B(C12) – 3B(He4) = (-92.14) -(- 84.87) = - 7.27 MeV
Whereas for the structure of the unstable Be8 we have
(Bex +Uex) = B(Be8) -2B(He4) = (-56.50) – 2(-28.29) = 0.08 MeV
This means that the extra pp and nn repulsions outside the alpha particles overcome the extra pn bonds and lead to the instability of Be8. (See my STRUCTURE OF Be8 and Be9 ).
This situation is not surprising because in the structure of C12 between the α particle and the Be8 one sees that there are four pn bonds per nucleon, while for the formation of the unstable Be8 there are always three pn bonds per nucleon. For example at the point p1 of the structure of Be8 the p1 makes three pn bonds, like the p1n1, p1n2, and p1n3 , while in the structure of C12 at point n3 which is between the Be8 and the additional He4 the n3 makes four pn bonds per nucleon, like the n3p1, n3p3, n3p4, and n3p5. This situation is very important because in all parallelepipeds of the so-called α-particle nuclei the structure is stable because it contains points with four pn bonds per nucleon.
C-12 (from p1to n6) and N-14 (from p1 to n7)
n7(+1/2) Third horizontal plane
n2 (-1/2)..........p4 (- 1/2)……….n6(-1/2) Second horizontal plane
p1(+1/2)..........n3(+1/2)……….p5(+1/2) First horizontal plane
NUCLEAR STRUCTURE OF C-14 WITH S = 0
In the case of the radioactive C-14 we assume that an extra n8 with spin +1/2 makes the single p3n8 bond like the structure of deuteron, while an extra n7 neutron with spin -1/2 makes the single p4n7 bond. Thus the total spin is S = 0. (On purpose such single bonds which give the structure of C-14 are not shown in the above structure). Of course such bonds favor the binding energy of the parallelepiped of C12 because at points p3 and p4 there are five pn bonds per nucleon. However the nn repulsions contribute to the reduction of the binding energies of these single pn bonds and lead to the decay.
Note that a free neutron or a single np bond with a binding energy B(np) weaker than the binding energy of 1.29 MeV leads to the decay because the difference in energy 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 write
n = p +e +ν or [92(dud) +4u +8d] = [(92(dud) +dud +4 +5d ] +e + ν οr d = u +e +ν.
Here C-14 goes through radioactive beta decay
C14 = N14 + e + ν
NUCLEAR STRUCTURE OF N-14 WITH S = +1
For the structure of N-14 you can see also Fig 8c of my published paper. In this procedure the neutron n8 turns into the proton p7 which goes over the neutron n4 for making a very strong axial n4p7 bond . At the same time the n7 of the single radial p4n7 bond goes over the p4 with spin +1/2 for making the very strong axial p4n7 bond and the radial n7p7 bond. That is, in this case the n7 cannot decay because it makes two bonds with the p4 and the p7. Instead, both the p7 and the n7 make a rectangle having also the nucleons p4 and n4 . This rectangle contributes to the structure of the stable nucleus N14 with spin S =1 having a binding energy B(N14) = -104.63 MeV. That is, the new rectangle formed by the nucleons p7 and n7 contributes to the increase of the binding energy from B(C12) = -92.14 MeV to the binding energy B(N14) = -104.63 MeV. In other words the single pn bonds of the radioactive C-14 with S=0 which are not shown in the diagram make a stable rectangle with S = +1/2 +1/2 = +1 . The rectangle is over the structure of C-12 for the formation of the stable N-14 with S =+1.