Geometric Model for Nuclear Structure.

[Willem F Esterhuyse]

Abstract

  We first prove that nuclear structure is not randomly arranged nucleons. We then develop a geometric model for nuclear structure using JP data from the Internet. The model shows why Be(p = 4, n = 4) is unstable while Be(4, 5) is stable. It predicts correctly the mode of decay of unstable isotopes and it predicts the daughter isotope. It also predicts why Tc(43, n = 54 or 53) is unstable and why Tc(43, 52) is not allowed and why all other isotopes of Tc are unstable. There is also a clear indication of why nuclei of p > 26 require energy to form. Knowing the relative position of the particles may make field computations easier. The model also explains why an isotope can have zero Orbital Angular Momentum. Also shown is why the magic number 8 yields a threshold of minimal binding energy. The model predicts that 16 should also be a magic number for stability. It also predicts that for nuclei with A = 22 or more has the formula for radius as follows: R = R0 (2*Z)^(1/3) and not R = R0 (A)^(1/3). Thus the model predicts smaller radiuses for heavy nuclei. It is predicted that for F(9,9) decaying to O(8,10) there would be released 4 photons per nucleus. The transition energies i.e. frequencies for this decay is computed. It is shown how to calculate transition energies. It is predicted that Na(11, 11) with JP = 3+, will behave anomalously w.r.t. the Electromagnetic Interaction (i.e. it will not have the same chemistry as other isotopes of Na), as will the stable Ge(32, 41) at JP = 9/2+, Kr(36, 47) at JP = 9/2+, Rh(45, 58) at JP = 1/2- , Pd(46, 59) at JP = 5/2+ and Ag(47, 60) at JP = 1/2-. There are a few other properties one would not see without the model.

See: 
https://ijasr.org/2021 volume 4 issue 2 March - April.html

and scroll to article number 3.

[studiot]

Although not as well developed as the electron shell-orbital theory for atoms, Nuclear Shell Structure is still the most comprehensive model around for the nucleus.

 

https://en.wikipedia.org/wiki/Nuclear_shell_model

[swansont]

!

Moderator Note

Material for discussion is supposed to be posted here

 

 

Quote

Uranium fission would have produced variable daughter nuclei if the nucleons were arranged at random.

But uranium fission does produce variable daughter nuclei

[Willem F Esterhuyse]

12 hours ago, swansont said:

But uranium fission does produce variable daughter nuclei

They don't decay to any isotope whatsoever.

[exchemist]

19 hours ago, Willem F Esterhuyse said:

Abstract

  We first prove that nuclear structure is not randomly arranged nucleons. We then develop a geometric model for nuclear structure using JP data from the Internet. The model shows why Be(p = 4, n = 4) is unstable while Be(4, 5) is stable. It predicts correctly the mode of decay of unstable isotopes and it predicts the daughter isotope. It also predicts why Tc(43, n = 54 or 53) is unstable and why Tc(43, 52) is not allowed and why all other isotopes of Tc are unstable. There is also a clear indication of why nuclei of p > 26 require energy to form. Knowing the relative position of the particles may make field computations easier. The model also explains why an isotope can have zero Orbital Angular Momentum. Also shown is why the magic number 8 yields a threshold of minimal binding energy. The model predicts that 16 should also be a magic number for stability. It also predicts that for nuclei with A = 22 or more has the formula for radius as follows: R = R0 (2*Z)^(1/3) and not R = R0 (A)^(1/3). Thus the model predicts smaller radiuses for heavy nuclei. It is predicted that for F(9,9) decaying to O(8,10) there would be released 4 photons per nucleus. The transition energies i.e. frequencies for this decay is computed. It is shown how to calculate transition energies. It is predicted that Na(11, 11) with JP = 3+, will behave anomalously w.r.t. the Electromagnetic Interaction (i.e. it will not have the same chemistry as other isotopes of Na), as will the stable Ge(32, 41) at JP = 9/2+, Kr(36, 47) at JP = 9/2+, Rh(45, 58) at JP = 1/2- , Pd(46, 59) at JP = 5/2+ and Ag(47, 60) at JP = 1/2-. There are a few other properties one would not see without the model.

See: 
https://ijasr.org/2021 volume 4 issue 2 March - April.html

and scroll to article number 3.

The link seems to be to a (probably predatory) pay-to-publish, junk journal. It can't even print grammatical and complete sentences in English. This is from its home page:-

International Journal of Applied Science and Research [IJASR] is multidisciplinary double-blind peer-reviewed, open-access journal intended to publish original research papers in all main branches of science (All scientific disciplines) a peer reviewed refereed bimonthly journal that publishes empirical, conceptual and review papers of exceptional quality that contribute to enrich business administration thinking .The objective of the Journal is to disseminate knowledge, which ensures good practice of professional management and its focal point is on research and reflections relevant to academicians and practicing managers/Administrators for sustainable Applied Science and Research changes.

IJASR guides it to map new frontiers in emerging and developing areas in research, industry and governance as well as to link with centers of excellence worldwide to stimulate young minds for creating knowledge based community.

Our continued success lies in bringing together and establishing channels of communication between leading policy-makers and prominent experts in industry, commerce and related business as well as renowned academic, education and research based institutions to provide solutions for addressing the key issues of the contemporary society.

We see the need for synergy and collaboration between these fields rather than segmentation and isolation. Hence, our objectives are to build new links, networks and collaborations between communities of thinkers, scholars, managerial experts and practioners in order to stimulate and enhance creative and application-oriented solutions for society.

In order to foster and promote innovative thinking in the management studies and social sciences research, itself by introducing its Journal at global platform in ensuring the high quality and professional research standards.

Seems to be a fairly random collection of buzzwords, thrown together with little attempt at punctuation or at bothering to write complete sentences. What a farce.

 

[Willem F Esterhuyse]

I don't read it as a farce. It is expected to contain buzzwords. Anyway, they peer reviewed my article and accepted it for publishing, therefore it has at least some prestige.

[studiot]

2 hours ago, Willem F Esterhuyse said:

They don't decay to any isotope whatsoever.

Would you please explain that statement in the light of this diagram ?
(After Semat.)

 

[Willem F Esterhuyse]

I mean the direct daughter isotopes (one decay away). According to the diagram U decays directly just to Io with 90 protons and an isotope with p=91 though I don't know why this isotope is not graphed halfway between two horizontal lines.

[Sensei]

20 hours ago, Willem F Esterhuyse said:

It predicts correctly the mode of decay of unstable isotopes and it predicts the daughter isotope.
It also predicts why Tc(43, n = 54 or 53) is unstable and why Tc(43, 52) is not allowed and why all other isotopes of Tc are unstable.

43+52=95, so you meant Tc-95, right?

https://en.wikipedia.org/wiki/Isotopes_of_technetium

95Tc 	43 	52 	94.907657(6) 	20.0(1) h 	β+ 	95Mo 	9/2+ 

Its half-life is 20 hours..

[studiot]

23 minutes ago, Willem F Esterhuyse said:

I mean the direct daughter isotopes (one decay away). According to the diagram U decays directly just to Io with 90 protons and an isotope with p=91 though I don't know why this isotope is not graphed halfway between two horizontal lines.

Thank you for the reply.

I wonder what you mean by daughter isotope ?

Are you using this definition ?

Quote

https://www.thoughtco.com/definition-of-daughter-isotope-in-chemistry-605861

Example

For example, uranium-238 decays along what is called a decay chain:

 

²³⁸U → ²³⁴Th → ^234mPa → ... → ²⁰⁶Pb

 

Uranium-238 is the parent isotope, while thorium-234, protactinium-234m, and lead-206 are all daughter isotopes.

or are you meaning that the 'daughter' must be an isotope of the same element as the starter ?

That is the only way I can make sense of this absolute claim

3 hours ago, Willem F Esterhuyse said:

They don't decay to any isotope whatsoever.

Because clearly all decay is to an isotope of some element or another.

Edited February 4, 2023 by studiot

[Willem F Esterhuyse]

10 minutes ago, Sensei said:

43+52=95, so you meant Tc-95, right?

Right.

10 minutes ago, studiot said:

or are you meaning that the 'daughter' must be an isotope of the same element as the starter ?

No, I mean with daughter isotopes the isotopes when only one reaction happened.

"

I wonder what you mean by daughter isotope ?

or are you meaning that the 'daughter' must be an isotope of the same element as the starter ?

That is the only way I can make sense of this absolute claim

  3 hours ago, Willem F Esterhuyse said:

They don't decay to any isotope whatsoever.

Because clearly all decay is to an isotope of some element or another."

I mean they don't decay in one step to any isotope.

[studiot]

26 minutes ago, Willem F Esterhuyse said:

I mean they don't decay in one step to any isotope.

Sorry, that makes no more sense than the original statement.

The first step always produces an isotope of something.

[Willem F Esterhuyse]

They don't decay in one step to Pb for example.

[swansont]

4 hours ago, Willem F Esterhuyse said:

They don't decay to any isotope whatsoever.

They don’t fission to form the same isotopes. The shell model explains why the fission yield curve has two peaks

[NTuft]

On 2/3/2023 at 7:44 AM, Willem F Esterhuyse said:

Abstract

[...]

"The model predicts that 16 should also be a magic number for stability."

"We will later prove that some nucleons go into the extra dimensions."[.pdf pg#1]

"Figure 1.4: Li(3,3)'s structure. Since the orbitals in layer Q have a unit of OAM each we have: J = 1 + 0 = 1, and the isotope has odd neutrons, so P = (-1) so parity agrees. I count only unsymmetrical filled orbitals since the rest cancel. The next level (labeled as R) has 6x2 orbitals so they are arranged in a hexagon."[pg#4]

"The neutron in brackets resides in the extra dimensions and does not contribute OAM or parity, it's spin pairs.
Theorem 3
Some neutrons go into the extra dimensions.
Proof: From Figure 1.6 we see Li(3, 4) has JP = 3/2+ without the extra dimensions (ED) filling. The second diagram shows that one neutron of (Li(3, 4) JP = 3/2- ) goes to the ED. If the theorem wasn't so we have that Li(3,4) with JP = 3/2+ would be stable and it is not. QED.
We will see that neutrons fill so that Z = N in the ordinary dimensions, with more neutrons filling ED orbitals."[pg#5]

"The stability rules are as follows: 1. A nucleus is unstable if one or more protons in a ring are unbalanced. 2. A nucleus is unstable if two or more neutrons are unbalanced. 3. A nucleus is unstable if its OAM is 6 or more."[pg#6]

"Figure 1.19: Structure of Na(11, 12). J = 1 + 1 - 1 + 1/2 = 3/2 and parity is positive: agrees with the data. The next layer (T) is again a hexagon with its L values as previously (Figure 1.5 of ref. [13])." [pg#10]

"Ar(18, 22) is also stable with JP = 0+, therefore there must be two more extra-dimensional neutron orbitals (see figure 1.28). These two orbitals are just activated if all the orbitals of level T are filled. The next layer is two decagons plus two orbitals in the center of each. See figure 1.29."[pg#14]

Can you explain more about the extra dimensions? It seems as though some results did not fit, so you postulate these ED orbitals.

Does your geometric model remain planar? You mention that P level is stacked on the z-axis(coming out of the page), P_L over P_R, one over another -- so is it more like 
      PL - Q_L
      P_L \ Q_L \
        \  P_R \  Q_R
        P_R       Q_L

once you get to P+Q level? Not sure I understand your diagrams...
Then you on to hexagrams and decagons (plus two orbitals in the center). Is the geometry planar?? 

I think this is interesting. I think the shell model can accomodate geometric nucleus structure. You start with a 4 (2 proton-2 neutron), add 6s or 8s, then 10s or 12s.. I think you could really gain something (and maybe lose the extra dimensions) from reading this write-up on another geometrical model for nuclear structure:
 

Quote

 

...
Professor Moon's Hypothesis
The existing dogma of nuclear physics requires us to believe that protons, being all of positive charge, will repel each other up to a certain very close distance corresponding to the approximat size of the nucleus. At that point, the theory goes, a binding force takes hold, and forces the little particles to stick together, until they get too close, at which point they repel again. Thus is the holding together of the protons in the nucelus accounted for.
  Disdaining such arbitrary notions of "forces", and preferring to view the cause of such phenomenon as resulting from a certain characteristic of physical space-time, Moon and the author demanded a different view. Considerations of "least-action" suggested to Moon a symmetric arrangements of the charges on a sphere, while the number of such charges (protons), and the existence of shells and orbitals beyond the nucleus (electrons) suggested a nested arrangement of such spheres. Our belief that the universe must be organized according to one set of laws, applying as well to the very large and the very small, suggested that the harmonic proportions which the astronomer Johannes Kepler found in the ordering of the solar system would also be evident in the microcosmic realm, so we looked for this also in the arrangement of the nucleus.
  We were led immediately to the five regular or Platonic solids, the tetrahedron, cube, octahedron, icosahedron, and dodecahedron. Moon developed a nested model, using the Platonic solids to deifne the atomic nucleus in much the same way that Kepler determined the orbits of the planets of the solar system. In Moon's "Keplerian atom", the 92 protons of the naturally occuring elements are determined by the vertices of two identical pairs of nested solids. Before elaborating the construction of this model, let us review the properties of the Platonic solids.

... How the Model Works  
  In Moon's "Keplerian atom", the 92 protons of the naturally occuring elemnts are determined by two identical sets of nested solids each containing 46 vertices. Moon's proposed arrangement is as follows: Two pairs of regular Platonic solids, the cube-octahedron pair and the icosahedron-dodecahedron pair, may be called duals: one will fit inside the other such that its vertices fit centrally on the faces of the other, each fitting perfectly inside a sphere whose surface isthus perfectly and symmetrically divided by the vertices. The tetrahedron is dual unto itself and therefore plays a different role.

...The four dual solids my be arranged in a nested sequence--cube, octahedron, icoshedron, dodecahedron--such that the sum of the vertices is 46: cube=8 +Octahedron=6 +Icosahedron=12 +Dodecahedron=20 =46

...Building the Nucleus If we now take the vertices of the solids so arranged to be the singularities in space where the protons are found, a remarkable structure to the nucleus appears. First we see a sort of periodicity in the nucleus, formed by the completion of each of the "shells", as we might call the circumspheres of the cube, octahedron, icosahedron, and dodecahedron. Let us first look at which unique elements correspond to the completed "shells":  
Oxygen(8)=completed cube
Silicon(14)=completed octahedron
Iron(26)=completed icosahedron
Palladium(46)=completed dodecahedron
...
Uranium(92)=completed twin nested figures.

...Fission of the Nucleus
  Moon's model beautifully accounts for the process of fission. Filling out with protons the outermost figure, the dodecahedron, brings us to palladium, atomic number 46, an element that has an unusually symmetric character. First, a look at the table of electron configurations shows palladium to be the only element in which an outer electron shell, previously occupied, is completely abandoned by the extra-nuclear electrons. Second, palladium is a singularity in the fission process, falling at a minimum on the table of distribution of fission products. Palladium also marks the boundary point for the sort of fission that occurs with very high energy (for example, protons of billion-electron-volt energies), when nuclei are split up into two parts of similar size. Silver, atomic number 47, is the lightest of the elements that may split up this way.
  To go beyond palladium in our model, a twin structure joins at one of the faces of the dodecahedron and begins to fill up its vertex positions with protons, beginning on the outermost figure. (Silver, atomic number 47, is the first.) Six positions are unavailable to it--the five vertex positions on the binding face of the second figure and the one at the face center where a vertex of the inscribed icosahedron pokes through.
  Thus on the second nested dodecahedron figure, 15 out of 20 of the dodecahedral vertices are availabled, and 11 out of 12 of the icosahedral vertices. We now fill 11 of the available dodecahedral vertices, thus creating 47-silver and continuing through 57-lanthanum. At this point, one face of the dodecahedron remains open to allow filling of the inner figures. The cube and octahedron fill next, producing the 14 elements of the lanthanide, or rare earth series (58-cerium to 71-luterium). Placing the proton charges on the inner solids causes a corresponding inward pulling of the electron orbitals. Thus, the otherwise unaccounted for filling of the previously unfilled 4-f orbitals and the mystery of the period of 14 for the rare earths are explained.

... The structure begins with a helium nucleus, or alpha particle-- a tetrahedron conttaining two protons and two neutrons at its four vertices. To go on to the third element, lithium, the protons must move outward to start building up their first shell on the vertices of a cube. The two neutrons that were on the vertices of the alpha particle have no need to leave. However, any additional neutrons will place themselves at the center faces of the cube, which is the same place as the midpoints of the edges of the large tetrahedron. (The smaller tetrahedron is called the alpha particle.) Thus,at 6-carbon-12, there are two neutrons on the alpha particle an four on the faces of the cube. For clarity, here is another example: the proton structure of 8-oxygen-16. Of the eight neutrons, two are on the inner alpha particle and six on the midpoints of the six edges of the large tetrahedron(or, the same thing, the face centers of the cube), marking the completion of this shell. The eight protons locate on the eight vertices of the cube. Thus, not only is oxygen highly symmetrical with respect to proton configuration, but also one of its neutron shells is complete. ...

emphasis added. Attached .pdf: 13pgs.

 

I'll try to get through the rest.

  

MOON-HECT-Keplerian-ATOM-Periodicity.pdf