Lasted edited by Andrew Munsey, updated on June 15, 2016 at 2:02 am.
Herein we analyse the stability of the isotopes of helium, lithium, and boron (because the structure of the boron isotopes has some similarities with the structure of the isotopes of lithium).
Theoretical calculations for the magnetic moments and nuclear spins of those light isotopes are corroborated by experimental data, suggesting that the present nuclear model with a central helium-4 is indeed the structure existing in the nature.
Earlier the discovery of the hydrogen atom model by Bohr, the theorist Voigt had developed a theory so that to explain the expectrum of light detected in the experiments. His theory was fenomenological, and by addopting a suitable mathematical formalism, his theory was successful to explain the experiments, in spite of it was wrong and did not consider any physical principle. The attempt was purely mathematical.
When Bohr discovered the structure of the hydrogen atom he changed the way addopted by Voigt, because in the Bohr theory he used the mathematics as a tool for the confirmation of the model, by making calculations and showing that the theoretical results were agree to the experimental data. In his model Bohr used a model with physical principles, and the mathematics was used for confirming the model and the principles. Unlike, Voigt used the pure mathematics, without any phyiscal principle, because he did not discover the structure of the hydrogen atom, and therefore the mathematics was a tool used so that to fit his theory to the experimental data.
It seems the current standard Nuclear Physics as a theory is similar to that attempt made by Voigt. The theorists did not discover yet the true structure of the nucleus. Standard Nuclear Physics is a phenomenological theory. Along more than 80 years they are using the mathematics so that to fit the several nuclear models proposed to the experimental data. And because the true structure of the nucleus was not discovered, of course along the 80 years the mathematics used by the nuclear theorists became more and more sofisticated and complex. So, the theory is very complex not because the model existing in the nature is so complex (and therefore would requires a complex mathematis). Instead of, it is very complex because the nuclear models are wrong''', and some principles proposed do not correspond to the principles by which the nature makes the nuclei to work.
Herein we will see that from the new nuclear model proposed in Quantum Ring Theory it is possible to explain the stability of light nuclei and to calculate the magnetic moments and nuclear spins of all light nuclei, starting from helium up to oxygen (after the oxigen the structure of the nuclei change - because with the oxygen it is complete the first hexagonal floor - and so the laws of interactions are different of those shown herein).
It's well to remember that in the book Quantum Ring Theory several other nuclear properties were already calculated, as binding eneries for light and haavy nuclei, magnetic moments for 2He3, electric quadrupole moments, etc., and the results obtained are close to the experimental data.
So, as the new nuclear model proposed in Quantum Ring Theory is successful to explain all the nuclear properties of nuclei, by using a simple mathematics, such fact suggest that this present theory is in the right way, as occurred in the case of the Bohr theory, because he had discovered the true structure of the atom.
In spite of the Bohr hydrogen atom is not correct, however his model has some correct fundamental principles, kept in the atom model of Quantum Mechanics, as for instance the electron orbiting the nucleus, the electron’s jumping between energy levels in the electrosphere responsible for the photons emissions, etc. In another words: The structure of the atom existing in the nature works through some fundamental principles considered in the model proposed by Bohr
And thanks to such a similarity of the Bohr model with the structure existing in the nature it was possible for Schroedinger to develop his theory, discovering his equation, and to find the Hamiltonian for the atom. So, the success of the Quantum Mechanics in the atomic level was possible because Bohr had discovered the initial partially correct model of the atom.
The situation of the standard Nuclear Physics is similar to that situation when Voigt proposed his theory for explaining the phenomena in the atomic level. Because as occurred in that age (earlier the Bhor’s discovery), nowadays there is not yet a nuclear model which structure is similar to the nuclear structure existing in the nature.
The nuclear model, in the nuclear level similar to Bohr model in the atomic level, is proposed in Quantum Ring Theory: it is the Hexagonal Floors Model.
Several experimental data are corroborating the predictions of this present new nuclear model. We can mention, for instance, that in Quantum Ring Theory was predicted that light isotopes with Z=N=pair have no spherical shape, and now in 2012 the experiments showed that the prediction is correct.
In the paper How atomic nuclei cluster:
published in the journal Nature in 2012, it is used the same argument proposed in the page 137 of my book (published in 2006) so that to explain why the light isotopes with Z=N=pair have no spherical shape, in spite of the experiments detected that they have null electric quadrupole moment (from the standard Nuclear Theory it's expected that nuclei with null electric quadrupole moment must have spherical shape).
I sent an email to the Editor-in-Chief of the Journal Nature, asking him to publish a note in the next issue of the journal, so that to explain that the argument used in Nature in 2012 was already proposed and published in 2006, in my book.
The Senior Editor Dr Karen Howell sent me the following reply:
Subject: RE: Plagiarism in the Journal Nature
Date: Tue, 24 Jul 2012 14:04:03 +0000
''Dear Dr Guglinski,
''Thank you for your comment regarding the manuscript entitled “How atomic nuclei cluster” by Ebran et al, and the accompanying News and Views article. Regretfully, we are unable to offer to publish it however, you are welcome to make use of our online commenting facility. Please go to the original article on our website and enter your message in the box provided beneath it.
''Thank you again for writing to us.
''Dr Karen Howell
If the structure of the nucleus existing in the nature is indeed like the structure proposed in Quantum Ring Theory, and the nuclear physicists decide to investigate this new nuclear model, a new era will be oppened for the Nuclear Physics, as the Bohr model inaugurated the era of the theory of the atom.
In the pages 228 and 232 of the book Quantum Ring Theory there are two figures:
Fig, 1.1 : shows the rotaion of the proton and the flux n(o) crossing its ring structure
Fig. 6.2 : shows the rotation of the nucleus 3Li6, which is established as reference for the study of the nuclear spin and magnetic moments due to nucleons in the light nuclei.
The 3Li6 is formed by a central 2He4, which magnetic moment is null, and a deuteron which magnetic moment is +0,857mn. The magnetic moment of 3Li6 is +0,822mn. Therefore the rotation of the deuteron into the structure of 3Li6 yields a magnetic moment equal to:
0,822-0,857 = - 0,035nm
The central 2He4 yields a flux named flux n(o), which captures deuterons, and it is divided in two sides named ANA and DOUGLAS, where the flux n(o) takes the directions n(o)-up and n(o)-down, as shown in the Figure AA:
The flux n(o) always enters within de deuterons and the neutrons as shown in the Figure 4-CA:
Therefore the deuteron has mag. moment +0,857 and spin i=+1 in the band left of DOUGLAS, but in the band right they change for -0,857 and i=-1. So, the spin and the mag. mom. are positive or negative in the same side DOUGLAS depending on its relative position regarding to the black line which devides that side, as shown in the Figures 4-CB and 4-CD.
The same is applied to the neutron, which mag. mom. is 1,913 and its spin is 1/2.
So, the contribution of the deuteron and the neutron, for the production of magnetic moments due to the rotation of a nucleus, depends on their position regarding to the black line.
The stability of light nuclei with Z < 8 is due to the following causes:
1- Spin-interaction between neutrons and deuterons captured by the flux n(o) of the central 2He4
2- The Pauli’s Exclusion Principle applied to the spin-interaction
3- Three types of spin-interactions neutron-deuteron - it is consequence of the two principles above, as follows:
A) First type- The excess neutron interacts with one deuteron – they form a spin i= 3/2
B) Second type- The excess neutron interacts with two deuterons – the three nucleons form a spin i= 3/2
C) Third type- The excess neutron interacts with three deuterons – the four nucleons form a spin i=5/2
So, the nucleus is no stable when those causes are not satisfied. We will see ahead some examples.
Of course Pauli’s Principle is applied to stable isotopes only. The no stable isotopes are instable just because the principle is not followed by them.
The hexagonal floor begins with the lithium, and it is complete in the oxygen nucleus 8O16.
The STABLE light isotopes with 2 < Z < 8 are the following:
Lithium: 3Li6 and 3Li7
Boron: 5B10 and 5B11
Carbon: 6C12 and 6C13
Nitrogen: 7N14 and 7N15
So, we realize that there are STABLE isotopes with one excess neutron only (it is just the first excess neutron captured by the deuterons of the side DOUGLAS).
Therefore, concerning the isotopes with excess neutrons, they are no stable when:
a) they are formed by the second excess neutron interacting with deuterons at the side DOUGLAS
b) they are formed by any other neutron captured by any deuteron in the side ANA
OBS:(in 7N15 the first excess neutron goes to get spin-interaction with 3 deuterons in the side ANA, as we will see in the PART FOUR of this article).
And the question obviously is: Why are stable only those isotopes which capture the first neutron ???
We will see the answer ahead.
The first deuteron captured by the central 2He4 occurs in the nucleus 3Li6, in the side DOUGLAS. Therefore such side is energetically lower than the side ANA [see Fig. 4-(aa)].
The first neutron captured by the nucleus 3Li6 aligns its spin with the deuteron, so that they form a spin-interaction into the structure of the 3Li7 [see Fig. 6-(aa)].
The next neutron captured tries to align its spin with the deuteron, so that to get a spin-interaction (see Fig. 7-(aa). However, as the deuteron already has a spin-interaction with the first neutron, then due to Pauli’s Principle the second neutron cannot get the spin-interaction. So, the second neutron cannot be kept by the flux n(o), and the 3Li8 decays.
The neutron No. 3 [see Fig. 8-(aa)] is constrained to go to the side ANA, because the neutrons No. 1 and No. 2 have occupied the available enegy levels in the side DOUGLAS. The instability of the neutron No.3 is worst than in the case of the neutron No. 2, and so 3Li9 has a half-life 178,3ms, shorter than 3Li8, with half-life 840ms.
The more neutrons are captured in the lithium isotopes, the instability obviously grows.
Look at the 4Be8 in the Figure 1-A.
There is repulsion between the two deuterons, in order that they both are forced to go to take a place close to the deuterons of the central 2He4. Due to a conflic of spins with the central 2He4, the 4Be8 is no stable, and it decays.
The two spin-interactions are shown in the Fig. 1-A. Note that the two spins i= +3/2 and i= -3/2 are contrary and, what is the worst, they are face to face. So their interaction weakens each deuteron-neutron spin-interaction, and the 4Be10 is subject to decay.
It is interesting that in 6C14 occurs the same phenomenon. Compare the two structures of the isotopes in the Figure 8-(ad), and the similarity of their half-life. In both structures the spins of the two interactions neutron-deuterons are oposite, and face to face.
The 11Be isotope has the same problem which causes the instability of the 10Be, but in 11Be the situation is worst, because as n-1 already has interaction with D-1, then n-3 cannot interact, and it leaves out the nucleus, causing its decay.
However n-3 has a weak interaction with D-1, and 4Be11 survives along 13,81 seconds only. It seems such weak interaction can explain the halo-neutron observed in the experiments made at the University of Lenz and published in 2009. They had detected that the halo-neutron in 4Be11 has a 7fm radius orbit.
Of course with the growth of neutrons in the Be isotopes their instability becomes worst, as shown in the Fig. 1-A.
As DOUGLAS is the side of lower energy, the first neutron gets spin-interaction with the deuteron of that side, as shown in the Fig. 18-(aa):
The second neutron takes place again in the side DOUGLAS, because it is the side of lower energy. But it cannot get spin-interaction with the deuteron D-1, because the first neutron already got it, and according to Pauli’s Exclusion Principle the second neutron cannot share the spin-interaction with the first neutron. So the second neutron takes a spin down, and as it did not get interaction with any deuteron, the 5B12 is not stable.
The two energy levels in the side DOUGLAS are filled by the neutrons n-1 and n-2. Then the third neutron in 5B13 will take place in the side ANA.
In the 5B12 there are two deuterons spin-up and one deuteron spin-down. Then the third neutron n-3 takes a spin-up into the 5B13, as shown in the Fig. 19-A-(aa):
Note that n-3 and D-3 have contrary spins, and therefore they cannot get spin-interaction. On another hand, D-2 and D-3 have contrary spins, and n-3 is actually getting spin-interaction with both them. But as D-2 and D-3 have contrary spins, then n-3 actually does not get any spin-interaction, and so 5B13 is no stable.
With the capture of the fourth neutron dramatic changes take place:
D-1 and n-1 keep the alignment of their spins toward the same direction
n-2 and D-1 have antiparallel spins
n-3 and D-3 have antiparallel spins
n-4 and D-2 have antiparallel spins
Fig. 1 shows why.
( A) - A proton will be captured by the flux n(o) of the central 2He4 of the isotope 3Li6, and they will form the isotope 4Be7. As the side Douglas is the side of lower energy, the proton will be captured at that side. As there is magnetic attraction between the right side of the cenral 2He4 and the proton, it will be captured in the region shown in the figure.
( B ) – Left side of Douglas – There is Coulomb repulsion between the proton and the deuteron, and the proton is pushed away of the deuteron.
( C ) – Right side of Douglas – The proton changes its spin. As its spin is contrary of the deuteron’s spin, they cannot get spin-interaction. By emission of positron the proton transmutes to neutron and 4Be7 transmutes to 3Li7.
As there is not any deuteron kept in the flux n(o) of the helium-4, one neutron cannot get spin-interaction so that to be kept by the nucleus. That’s why the neutron tries to get spin interaction with one of the two deuterons that form the structure of the 2He4, as shown in the Figure 4-A. But then the structure of the 2He4 is broken, and the nucleus has decay.
The same happens with the other isotopes 2He6, 2He7, etc. The neutrons do not have a deuteron available so that to get spin-interaction, and so the nucleus decays. This is shown in the Figure 4-B:
There is no information on magnetic moment of the isotope 3Li5 in nuclear tables.
The magnetic moments induced by the rotation of the nucleus is shown in the Figure 6-A:
As the nucleons are submitted to the rotation of the nucleus, due to such rotation they induce magnetic fields.
The intensity of the induced magnetic field is increased when two (or more nucleons) have their spins aligned toward the same direction.
But such growth in the intensity of the induced magnetic field depends on the following conditions:
1- The two (or more nucleons) must be kept in the same side (either ANA of DOUGLAS).
2- If there are two nucleons with aligned spins in the same side either ANA or DOUGLAS, but there is another nucleon with contrary spin, there is no growth in the intensity of the induced magnetic moments.
3- The phenomenon is calculated by the multiplication factor Mf.
There are two sort of Mf:
a) Internal Mfi - is concerning the nucleons situated in the region between the two black lines of the sides ANA and DOUGLAS
b) External MfE - is concerning the nucleons situated in the region ouside the region limited by the two black lines of the sides ANA and DOUGLAS
In the 3Li8 a very interesting phenomenon occurs, thanks to the alignment of the 3 spins toward the same direction.
In the superior part of the Fig.7-A we see the positions occupied when there is no strong spin-interaction yet, because the deuteron is in the side Douglas and the neutrons are in the side Ana. So, each neutron induces positive magnetic moment Mm= +0,078.
Note the position of each neutron regarding the line center of the nucleus 3Li8, and the direction of the flux n(o)-up regarding to the the line center of rotation.
Now look at the inferior part of the Fig. 7-A. It shows that due to the strong spin-interaction of the two neutrons with the deuteron, the deuteron attracts them, and they take a new position close to the deuteron. In the new position the flux n(o)-up has changed with regard to the line center of rotation.
So, thanks such changigs due to the spin-interaction of the deuteorn with the two neutrons, the sign of the magnetic moment induced by the two neutrons changes: in the new position each one indeces a NEGATIVE Mm= -0,078.
Thanks to such change in the sign of the magnetic moment induced by the neutrons rotation about the line center of the nucleus rotation, the phenomenon also contributes in the total magnetic moment of the nucleus. That's why there is need to introduce the multiplication factor Mf, calculated in the Fig. 7.
So, now let's calculate the magnetic moment of 3Li8, by considering these changes.
The calculation of the external multiplication factor MfE = 3,94, used here, was obtained from the isotope 5B8, in the Fig. 15.
The structure of 5Be10 is shown in the page 231 of the book Quantum Ring Theory, where it is shown that, due to repulsions and the flux n(o) crossing within the three deuteriuns, the structure of 10Be thakes that shape shown in the Figure 5.2(A) of the book:
As seen in the page 231, in the structure of 5B10 all the three deuteriuns are captured by the flux Ana, no one is captured by Douglas.
Like the 3Li8, also 5B8 has spin 2.
But we cannot use the internal multiplication factor Mfi = 15,86 , obtained from the magnetic moment of 3Li8, because the two protons of 5B8 have an orbit outside of the black line.
Let's see what happens if we do not use any multiplication factor.
The 5B9 decays by emitting the alone proton, and the nucleus transmutes to 8Be.
The nucleus is stable.
Stability of Light Nuclei – PART ONE
Stability of Light Nuclei – PART THREE
Stability of Light Nuclei – PART FOUR
Stability of Light Nuclei – PART FIVE
Guglisnki, W. , Quantum Ring Theory-foundations for cold fusion, 2006, Bäuu Institute Press