Guido Kinet
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QUANTUM CHEMISTRY
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Abstract
Because logic excludes that protons and neutrons have an outer layer to hold quarks together, there must by definition be a structural relationship between the quarks themselves to avoid jeopardizing the stability of the nucleus. To substantiate this, it is necessary to move away from the beaten track in an attempt to better understand the functioning of the nucleus and the atom as a whole. The structure of the nucleus of the atom has been re-mapped and spatially rearranged and, in its simplicity, explains completely the inner workings of that atomic nucleus. It became immediately clear that it was only a small glimpse into a whole new world that has remained hidden so far and undoubtedly will open new fields of research in both physics and chemistry.
Figures (11)
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Journal of Chemical Physics, 2022
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QUANTUM CHEMISTRY
A NEW APPRAOCH
REDESIGNING THE ATOMIC NUCLEUS
Author G. KINET – August 15, 2024
guido.kinet@telenet.be
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1
Because logic excludes that protons and neutrons have an outer layer to hold
quarks together, there must by definition be a structural relationship between the
quarks themselves to avoid jeopardizing the stability of the nucleus. To
substantiate this, it is necessary to move away from the beaten track in an attempt
to better understand the functioning of the nucleus and the atom as a whole.
The structure of the nucleus of the atom has been re-mapped and spatially
rearranged and, in its simplicity, explains completely the inner workings of that
atomic nucleus. It became immediately clear that it was only a small glimpse into
a whole new world that has remained hidden so far and undoubtedly will open
new fields of research in both physics and chemistry.
Alpha decay, Beta plus decay, Beta minus decay, electron capture and nuclear
fission are presented with detailed examples based on the new composition of the
atomic nucleus. Although it seems rather technical at first, it is in reality a very
elegant and easy-to-understand way how the mechanism of transformation of the
structures of the atomic nucleus works.
Important to know is that two UP quarks with each an electric charge of +2/3 and
one DOWN quark with an electric charge of – 1/3 make up a PROTON with a net
electrical charge of one, two DOWN quarks with each an electric charge of – 1/3
and one up quark with an electric charge of + 2/3 make up a NEUTRON with a
net electric charge equal to zero.
As mentioned before protons and neutrons each consist of three quarks. Two UP
quarks and one DOWN quark make up a proton with a net electric charge equal
to one. Two DOWN quarks and one UP quark make up a neutron with a net
electric charge equal to zero.
Here I allow myself some artistic freedom and give the quarks a visual calling
card that is without foundation but gives a foothold to the reasoning below. Let it
be clear that it is the design that is an artistic freedom, but that the parameters
with which they are endowed are scientifically substantiated.
This is the intellectual property of G. KINET. Free use of parts of text without modifications to those parts of the text tha t are
used/cited is allowed, with the understanding that the author is always mentioned together with the title of the work that
has been used. Every commercial use is excluded with the exception of the permission of the author.
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2
UP quark representation:
DOWN quark representation:
electric charge + 2/3 equal to 1.0681173 … x 10 -19 C
electric charge – 1/3 equal to 5.340586 ... x 10-20 C
Let’s start with some simple examples like Hydrogen-1, Deuterium, Tritium and
Helium.
Hydrogen-1 has only one proton in its nucleus with an internal structure of two UP
quarks and one DOWN quark.
the nucleus of Deuterium contains one proton and one neutron, which means that three UP
quarks and three DOWN quarks are needed to build these two nucleons.
So, let's deconstruct the proton and the neutron and forget for a moment the image that has been
presented up to now, take the constituent parts and put them together in a box. Having three
negative and three positive parts, two scenarios could develop … parts will arrange themselves
in a straight line in a sequence of positive and negative parts, but as natural forces are trying to
approximate circles and since those combining quarks naturally tend towards the most efficient
enclosure of space, they also tend towards that circularity, but as those quarks have opposite
electrical charges a hexagonal pattern appears spontaneously.
3
It is striking that when one draws a line from the center of
any side to the center of the opposite side, there is always
a proton and a neutron in the picture.
When all electrical charges of the quarks are added
together, a net electrical charge of +1 remains.
The nucleus of Tritium consists of one Proton and two neutrons. Again, deconstructing the
nucleons, we have a total of four UP quarks and five DOWN quarks. In the same scenario as
before quarks will arrange themselves in a hexagon and the remaining UP quark, for clear
reasons, will present it selves at one of the DOWN quarks on the hexagon followed by the two
DOWN quarks so that when all electrical charges of the quarks are added together, a net
electrical charge of +1 remains, as expected.
This is the intellectual property of G. KINET. Free use of parts of text without modifications to those parts of the text tha t are
used/cited is allowed, with the understanding that the author is always mentioned together with the title of the work that
has been used. Every commercial use is excluded with the exception of the permission of the author.
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On each hexagon there are three anchor points (DOWN quarks) for neutrons.
And then there is Helium, atomic number “two”, atomic mass “four” so there are two Protons
and two Neutrons in the nucleus.
Deconstruction yields six UP quarks and six DOWN quarks. Because there are only three UP
and DOWN quarks that can take their place in a single hexagon, the other three UP and DOWN
quarks form a second hexagon rotated 60 degrees to the first.
(a dimensional representation of the hexagonal alignment – hexagonal shapes are equal in size)
This second hexagon will naturally align with the first (by attraction). The Coulomb force
(attraction or repulsion of particles or objects due to their electrical charge) between the
positive UP quark and the negative DOWN quark is obviously a positive one ... the
strength of the attraction depends on how far they are separated from each other.
As in the example of the composition of the nucleus of Tritium where a UP quark anchors itself
to an DOWN quark on the hexagonal pattern, it is also possible for an DOWN quark to anchor
itself onto a UP quark on a hexagonal pattern.
A good example of this is Helium-3 (an isotope) with two Protons and one Neutron in its
nucleus.
Deconstructing yields five UP quarks and four DOWN quarks. Only three UP quarks and
three DOWN quarks (one proton and one neutron) can take their place in a hexagonal
arrangement, while the remaining DOWN quark, will present it selves at one of the UP quarks
on the hexagon followed by the two remaining UP quarks.
This is the intellectual property of G. KINET. Free use of parts of text without modifications to those parts of the text tha t are
used/cited is allowed, with the understanding that the author is always mentioned together with the title of the work that
has been used. Every commercial use is excluded with the exception of the permission of the author.
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4
Adhering neutrons are each bound to a DOWN quark on a successive HEX in such a way
that they will never face each other directly. Spatially, they will spiral over the number of
successive HEX. As for the protons the same structural build applies with the understanding
that each proton ties itself down on a UP quark on a successive HEX in the case of multiple
protons present.
Adhering protons and neutrons cannot be present on the same crystalline core at the same
time. When a neutron is converted to a proton, and at least one adhering neutron is present,
a new HEX is formed. It goes without saying that this also applies when a proton is converted
to a neutron and there is at least one adhering proton present a new HEX will be formed.
All nucleus settings of the different atom configurations can be compiled with all five previous
approaches and those two rules about how neutrons or protons combine or not.
GRAPHIC …
Carbon - atomic number 6 - atomic mass 12,011 – so here are six protons and six neutrons
present in the nucleus, which are distributed as follows:
5
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has been used. Every commercial use is excluded with the exception of the permission of the author.
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Beryllium: - atomic number 4 – atomic mass 9,0122 – counting four protons and five neutrons,
which are distributed as follows:
This is an example where the number of neutrons is larger than the number of
protons, and where that fifth neutron is anchored on a DOWN quark.
The composition of the nucleus consists of periodic structures having exactly the
same arrangement, rotated 60° in relation to each other. On each structure (or unit
cell) additional (free) protons and/or neutrons can be attached.
The nucleus of an (any) atom has a crystalline structure consisting of
hexagonal unit cells where attraction and repulsion is governed by the
individual quarks and their structural binding.
The regular, repeating arrangement of quarks, in a non-symmetrical way in a
three-dimensional setting adds strength to this structure.
(free neutrons are unstable and have a mean lifetime of 879.6±0.8 seconds.
When attached to a HEX they will take advantage of the fact that there is always
a proton configuration available to attach to, which makes it stable)
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used/cited is allowed, with the understanding that the author is always mentioned together with the title of the work that
has been used. Every commercial use is excluded with the exception of the permission of the author.
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6
HOW DOES THAT TRANSLATE TO THE CHEMICAL ELEMENTS?
A chemical element is a chemical substance that cannot be broken down into other
substances by chemical reactions. The basic particle that constitutes a chemical
element is the atom. Elements are identified by the number of protons in their
nucleus, known as the element's atomic number. For example - Barium 56 56
137
BA
has an atomic number of 56, indicating that each Barium atom has 56 protons in
its nucleus.
The atomic number indicates the number of protons in the nucleus of an atom and
is usually located to the left below the symbol of the element. The mass number
indicates the total number of protons and neutrons in the nucleus and is usually
located to the left above the symbol of the element
The amount of neutrons is equal to the atomic mass to the nearest integer, from
which the amount of protons is subtracted.
Barium has 56 protons and (137 -56) 81 neutrons in the nucleus.
In view of the above preconceived structures we have all the elements to visualize
the real picture of the atomic nucleus of 56137Ba :
When 56 protons and 56 neutrons combine 56 HEX emerge with the
remaining (81 – 56) = 25 neutrons spiraling over successive HEX.
The new chemical notation for the element
56
137Ba
looks like this :
BA56 25
the number of HEX (equaling the atomic number) written in the upper right corner
of the symbol of the chemical element and the number of neutrons at the bottom
on the right of the symbol of the chemical element
Further on one will find the same kind of notation for the regular
chemical notation, figuring the atomic number in the upper right
corner of the symbol of the chemical element and the mass number
at the bottom on the right of the symbol of the chemical element.
This is the intellectual property of G. KINET. Free use of parts of text without modifications to those parts of the text tha t are
used/cited is allowed, with the understanding that the author is always mentioned together with the title of the work that
has been used. Every commercial use is excluded with the exception of the permission of the author.
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7
THE NUCLEAR BINDING ENERGY - FORCES BETWEEN QUARKS
What is being put forward ….
The strong force, is a very short-range force (less than 0.8 fm, which
means that the distance between nucleons must be less than 10 -15m ),
that acts directly between quarks. This force holds quarks together to
form protons, neutrons, and other hadron particles. The strong
interaction is mediated by the exchange of massless particles called
gluons that act between quarks, antiquarks, and other gluons.
As two quarks are pulled apart, there comes a point where the amount
of energy required to separate them exceeds their own mass-energy.
With that amount of energy stored in the strong force field, a new pair
of quarks literally pops into existence. The reason for this is that when
one tries to separate two quarks the potential energy between them
increases and at some point, the potential energy is high enough that it
becomes energetically favorable to create two new quarks that quickly
pair up with the ones one was trying to pull apart.
…. but in the light of the new hypothesis …
A two-part story. On the one hand there are the up and down quarks
that together form hexagonal structures, and on the other hand these
hexagonal structures are the building blocks of the atomic nucleus.
In essence, there are forces present on two different levels that each
create a different bond.
FIRST of all there is the bond between the up and down quarks in the
hexagonal setting. This is a force that acts directly between quarks
themselves to form protons and neutrons. As the average distance
between nucleons has been set at 10 -15 m (or less), the Coulomb force
between each opposite UP and DOWN quark in the hexagonal setting
is equal to:
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8
F = K x (q1 x q2 / r 2)
Where “r” the distance between nucleons = 10 -15 m, assuming that quarks are
separated by minimal that distance
q1 = 2/3 of the electric charge of the electron = 1.0681177×10 −19 coulombs
q2 = 1/3 of the electric charge of the electron = 5.340586 x 10 -20 coulombs
K = 8.99 x 10 9 N m2 / C2 (Newton meter 2 / Coulombs 2 )
F = 8.99 x 10 9 Nm2 / C2 x (1.0681177 x 10 -19 C) x (5.340586 x 10 -20 C)
(10-16 m)2
= 5128 N or 5.128 kN of force acting between each pair of those (stationary)
point particles in that hexagonal setting.
(No negative or positive signs where plugged in, to not complicate the calculation,
knowing that positive and negative charges attract each other anyway)
At this point, some may claim that Coulomb’s law becomes invalid at distances
of the order of the Compton wavelength (10 -12 m) and smaller, due to vacuum
polarization, whereas in quantum electrodynamics the Coulomb force applies to
distances in the order of 10 -18 m … but on such small distances it is just a question
where one draws the line, the equation above is valid for any distance, and forces
can go from zero to infinity, condictio sine qua non, particles must be point
charges and particles must be stationary with respect to one another.
It is indeed a tremendous force acting between nucleons in that hexagonal setting,
question is whether this bond (strong or less strong) can be broken?
An answer to that question will be given when the subject concerning the decay
configurations will be explored and the case of "electron capture" is on the menu,
a form of beta decay in which the proton-rich nucleus of an electrically neutral
atom absorbs an inner atomic electron, (captured by the nucleus) with the
emission of an electron neutrino which changes a proton that is part of the hexagon
(the unit cell) into a neutron, jeopardizing the stability of the hexagon causing it
to fall apart.
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used/cited is allowed, with the understanding that the author is always mentioned together with the title of the work that
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9
SECONDLY the hexagonal structure of the proton and the neutron - the unit cell
- determines the structure of the atomic nucleus, when fission or fusion occurs,
the action will take place between those (unit cells) hexagonal parts.
With the new proposed configuration of protons and neutrons, we can assume that
the aforementioned distances at which the strong force is active, by extension also
apply between quarks when the unit cells merge into a crystalline structure.
Because of the alignment of the hexagonal unit cells in the atomic nucleus the upand down quarks will attract each other in a similar way as they do in their
hexagonal setting, but as the electrical charge of that hexagonal unit cell is equal
to + one (+ 1), those cells will also repel each other and weaken the bond between
them.
It ought to be crystal clear that the structural strength of the atomic nucleus
depends above all on the distance between the unit cells. Strongly bound elements
will be less prone to fission where others with a less strong bond will undergo the
fission process more quickly (easily).
Calculating the Coulomb force between nucleons in the unit cells depending on
the distance to each other, and taking into account the repulsive factor due to the
repelling electromagnetic force which acts between the unit cells, gives an
indication how sensitive an atomic nucleus can become to fission.
Distance
nucleons
between Force in Newton
Between nucleons
10-12 m
10-13 m
10-14 m
10-15 m
10-16 m
5.128 x 10 -5 Newton
0.005128 Newton
0.5128 Newton
51.28 Newton
5128 Newton
Forces redefined to
Milli Newton / kilo
Newton
0.05128 milli Newton
5.128 milli Newton
512.8 milli Newton
0.05128 kilo Newton
5.128 kilo Newton
0.05128 kilo Newton = 2.9333124983199126.1036 eV = 2.9333124983.10 30 MeV
5.128 kilo Newton
= 2.933312498319913.10 38 eV = 2.9333124983.1032 MeV
Which gives an idea of the enormity of the forces at work compared to the sum
of the rest masses of the three valence quarks in a proton 9.8 MeV and the total
mass of the proton 938 MeV.
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10
The above numbers show that even the smallest difference in distance between
the quarks in the adjacent hexagonal unit cells has major consequences in terms
of mutual attraction/repulsion.
The fact that the maximum binding energy is found in medium-sized nuclei is a
consequence of the trade-off in the effects of the two opposing forces that have
different range characteristics. The attractive nuclear force, which binds protons
and neutrons to each other, has a limited range due to a rapid exponential decrease
in this force with distance. However, the repelling electromagnetic force, which
acts between unit cells, falls off with distance much more slowly (as the inverse
square of distance).
There is however a threshold. Isotopes with (very) high-mass-number are
unstable and when colliding with high energetic neutrons are subject to fission.
The induced distortion is so high that the bond between the HEX themselves
(which will retain their hexagonal structure) and the adhering neutrons will for a
very brief moment be impaired (not undone) making it possible for that crystalline
nucleus to be split into smaller parts and since these fragments are a more stable
configuration, the splitting of such crystalline nuclei will be accompanied by
energy release. (Spontaneous radioactive decay, not requiring a neutron, is also
referred to as fission, and occurs especially in very high-mass-number isotopes).
On the other hand, there is also the possibility where the unit cells of the nucleus
will shift back to their original place releasing the (large) stored potential energy
into a spray of (possible) elementary and exotic particles, photons, ….
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DECAY CONFIGURATIONS EXPLAINED
ALPHA DECAY
Certain radionuclides of high atomic mass decay by the emission of alpha particles. These
alpha particles are tightly bound units of two neutrons and two protons each (He4 nucleus) and
have a positive charge. Emission of an alpha particle from the nucleus results in a decrease of
two units of atomic number and four units of mass number. The principal alpha emitters are
found among the elements heavier than bismuth (atomic number 83) and also among the rareearth elements from neodymium (atomic number 60) to lutetium (atomic number 71).
✓ first example:
U92 235 decays to Th90 231 by emission of alpha particles
What happened?:
The nucleus of U 92 235 (92 protons and 143 neutrons) is made up of 92 HEX (each HEX
= 1 proton and 1neutron) and the remaining 51 neutrons*. Through alpha radiation it
transforms to Th90 231 (90 protons and 141 neutrons) which is made up of 90 HEX and
the remaining 51 neutrons**.
⁕
U92235 : mass number (235) minus proton number (92) equals the number of
neutrons = (143). The nucleus is made up of 92 HEX and because each HEX has a
neutron in its composition the total number of remaining “free” neutrons equals 143
– 92 = (51). Those 51 neutrons are each bound to a DOWN quark on a successive
HEX in such a way that they will never face each other directly. Spatially, they will
spiral over 51 consecutive HEX.
** Th90 231 : mass number (231) minus proton number (90) equals the number of
neutrons = (141). The nucleus is made up of 90 HEX and because each HEX has a
neutron in its composition the total number of remaining “free” neutrons equals 141
– 90 = 51. Those 51 neutrons are each bound to a DOWN quark on a successive
HEX in such a way that they will never face each other directly. Spatially, they will
spiral over 51 consecutive HEX.
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✓ second example:
The nucleus of Po84 210 (84 protons and 126 neutrons) is made up of 84 HEX (each HEX = 1
proton and 1neutron) and the remaining 42 neutrons*. Through alpha radiation it transforms
to Pb82 206 (82 protons and 124 neutrons) which is made up of 82 HEX and the remaining 42
neutrons **.
Po84 210 decays to Pb82 206 by emission of alpha particles
What happened? :
⁕
Po84 210 : mass number (210) minus proton number (84) equals the number of neutrons
= (126). The nucleus is made up of 84 HEX and because each HEX has a neutron in its
composition the total number of remaining “free” neutrons equals 126 – 84 = (42).
Those 42 neutrons are each bound to a DOWN quark on a successive HEX in such a
way that they will never face each other directly. Spatially, they will spiral over 42
consecutive HEX.
** Pb82 206 : mass number (206) minus proton number (82) equals the number of neutrons =
(124). The nucleus is made up of 82 HEX and because each HEX has a neutron in its
composition the total number of remaining “free” neutrons equals 124 – 82 = 42. Those
42 neutrons are each bound to a DOWN quark on a successive HEX in such a way that
they will never face each other directly. Spatially, they will spiral over 42 consecutive
HEX.
⸎ The attentive reader will have noticed that the number of “free” neutrons in
both the mother nucleus and in the decay product remains the same after
transition, which means that the mother nucleus has to shed two HEX (He4
nucleus) to make that transition possible.
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− DECAY
A nucleus with an unstable ratio of neutrons to protons may decay through the emission of a
high-speed electron called a beta particle. This results in a net change of one unit of atomic
number. Beta particles have a negative charge and the beta particles emitted by a specific
radionuclide will range in energy from near zero up to a maximum value, which is characteristic
of the particular transformation.
An example of electron emission
nitrogen-14:
(β− decay) is the decay of carbon-14 into
C614 → N714
✓ C6 14 has 6 protons and six + 2 neutrons in its nucleus.
To avoid having to draw every time “x” number of hexagons, I will refer to "HEX" preceded
by a number that indicates their quantity.
The crystalline core of the carbon atom consists of 6 HEX + 2 neutrons each connected to a
DOWN quark.
✓ N7 14 has 7 protons + 7 neutrons
The crystalline core of nitrogen has a total number of 7 HEX, which means there is no change
in the mass number but the atomic number has increased by “one”.
What happened? one of the neutrons is converted to a proton, and the process creates an electron
and an electron antineutrino via an intermediate W− boson, and a new HEX is added.
+ e− + ν-e
Proton + Neutron = additional 7th HEX
To make it visually clearer, I have limited myself to two HEX, which shows the
transformation of a neutron into a proton, and the formation of a new HEX.
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DETAILED EXAMPLE
-
DECAY
✓ First example
Co2760 decays by - decay to Ni2860
The nucleus of Co27 60 (27 protons and 33 neutrons) is made up of 27 HEX (each HEX = 1 proton
and 1neutron) and the remaining 6 neutrons. By beta decay it decays to Ni28 60 (28 protons and 32
neutrons) of which the nucleus is made up of 28 HEX and 4 neutrons.
What happened? :
One neutron of the remaining six neutrons in the cobalt nucleus will split into one
positively charged proton leaving 5 neutrons with no net electric charge. The proton will
interconnect with one of the five neutrons and a new HEX is added to the nucleus of
Co27 60 which will then be changed into Ni28 60 .
Co27 60 = 27 HEX + 6 neutrons
Ni28 60 = 28 HEX + 4 neutrons + γ + e-
5 neutrons + 1 Proton = 1 HEX + 4 neutrons
+ e− + ν-e
15
✓ Second example
Kr3693 decays by - decay to Rb3793
The nucleus of Kr36 93 (36 protons and 57 neutrons) is made up of 36 HEX (each HEX = 1
proton and 1neutron) and the remaining 21 neutrons. By beta decay it decays to Rb 37 93 (37
protons and 56 neutrons) of which the nucleus is made up of 37 HEX and 19 neutrons.
What happened? :
One neutron of the adhering 21 neutrons in the krypton nucleus will split into one positively
charged proton leaving 20 neutrons with no net electric charge. The proton will interconnect
with one of the 20 neutrons and a new HEX is added to the nucleus of Kr36 93 which will then
be changed into Rb37 93 .
Kr36 93 = 36 HEX + 21 neutrons
Rb37 93 = 37 HEX + 19 neutrons + γ + e-
20 neutrons + 1 Proton = 1 HEX + 19 neutrons
+ e− + ν-e
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β+ DECAY
An example of positron emission (β+ decay) is the decay of magnesium-23 into sodium-23:
Mg1223 → Na1123
✓ Mg12 23 has 12 protons and 11 neutrons in its nucleus
The crystalline core of the Magnesium atom consists of 11 HEX + 1 proton connected to an UP
quark.
✓ Na11 23 has 11 protons + 12 neutrons in its nucleus
The crystalline core of sodium has a total of 11 HEX + one neutron connected to a DOWN
quark, which means there is no change in the mass number but the atomic umber has decreased
by “one”.
What happened? That one bound proton in the crystalline structure of the magnesium atom that
is composed of two UP quarks and one DOWN quark loses one UP quark that is converted into
a DOWN quark, which turns that proton into a neutron.
Having a different structure this neutron cannot hold its position being attached to an UP quark
and will rotate (or jump) to the next available DOWN quark to take its new position.
Proton
Neutron
⸎ The reason that adhering neutrons or protons can change positions, lies in their weak point
of attachment. The coulomb force between the UP and DOWN quark is quite strong, but the
two adjacent UP quarks (in the case of a proton) and the two adjacent DOWN quarks (in the
case of the neutron) provide a repulsive force that weakens the attachment.
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DETAILED EXAMPLES
+
DECAY
Mg1223 decays by + decay to the stable element Na 1123
The nucleus of Mg1223 (12 protons and 11 neutrons) is made up of 11 HEX
(each HEX = 1 proton and 1neutron) and 1 proton. Through + decay
(emitting a neutrino and a positron) it decays to Na 1123 (11 protons and 12
neutrons) of which the nucleus is made up of 11 HEX and 1 neutron.
What happened? :
The one adhering proton in the crystalline nucleus by emitting a neutrino
and a positron, changes into a neutron.
Mg12 23 = 11 HEX + 1 proton
Na11 23 = 11 HEX + 1 neutron
e+ + νe
17
✓ Other examples of + decay
C611 decays by + decay to a stable element B 511
The nucleus of C 611 (6 protons and 5 neutrons) is made up of 5 HEX (each
HEX = 1 proton and 1neutron) and 1 proton. Through + decay (emitting
a neutrino and a positron) it decays to B 511 (5 protons and 6 neutrons) of
which the nucleus is made up of 5 HEX and 1 neutron.
What happened? :
The one adhering proton in the crystalline nucleus by emitting a neutrino
and a positron, changes into a neutron.
C6 11 = 5 HEX + 1 proton
B5 11 = 5 HEX + 1 neutron
e+ + νe
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ELECTRON CAPTURE
Electron capture is a form of beta decay in which the proton-rich nucleus
of an electrically neutral atom absorbs an inner atomic electron, (captured
by the nucleus) with the emission of an electron neutrino and a change in
the neutron and proton numbers in the nucleus.
✓ Example
Xe54127 through electron capture decays to I53127
Xe54127 + eI53127 + νe
The nucleus of Xe54127 (54 protons and 72 neutrons) is made up of 54 HEX
(each HEX = 1 proton and 1neutron) and 19 neutrons. Through electron
capture it decays to I53127 (53 protons and 74neutrons) of which the nucleus
is made of 53 HEX and 21 neutrons.
What happened? :
The inclusion of an electron in a nucleus will influence one of the UP
quarks in a proton, the transformation proceeds by the exchange of a
positive W boson which changes the charge by one unit and therefore
modifies that proton into a neutron. That positive W boson is a virtual
particle that in turn immediately decays to a positron and an electron
neutrino. (a negative W boson decays to an electron and antineutrino)
Xe54127 + e- = 54 HEX + 19 neutrons
= (53 HEX + 1 HEX) + 19 neutrons
each HEX = 1 proton and 1neutron
= ((53 HEX + (1 proton + 1 neutron)) + 19 neutrons
Proton = 2 UP quarks + 1 Down quark
(uud)
Neutron = 1 UP quark + 2 DOWN quarks (udd)
u = UP Quark / d = DOWN quark
= ((53 HEX + ( uud + udd )) + 19 neutrons
Proton
neutron
1 HEX + 53 HEX = 54 HEX
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One UP quark in the proton composition changes to a DOWN quark and
changes that proton into a neutron.
e+ + νe
W+
= ((53 HEX + ( udd + udd )) + 19 neutrons
+
+
whereby Xe54 127 decays to I 53 127 which is made up of 53 HEX and 21 neutrons.
⸎ A nucleus which is in an excited state may emit one or more packets of
electromagnetic radiation (photons) of discrete energies. The emission of
gamma rays does not alter the number of protons or neutrons in the nucleus
but instead has the effect of moving the nucleus from a higher to a lower
energy state (unstable to stable). Gamma ray emission frequently follows
beta decay, alpha decay, and other nuclear decay processes.
19
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NUCLEAR FISSION
Breaking heavy nuclei apart (fission), light and heavy fractions will emerge
which oddly enough will have mostly the same ratio of neutrons to protons
as the heavy nucleus had.
In order for the fission process to take place, a sufficient amount of energy
must be added to the nucleus.
As an example: introducing a neutron into the nucleus of U92235 it gets
absorbed to produce U 92236 which then fissions and produces two
fragments, (for instance) Kr 3692 and Ba 56142
The fragments will have approximately the same Proton/Neutron ratio
number than THE ORIGINAL ATOMIC NUCLEUS . In this case the
Proton/Neutron ratio of U 92235 (92 protons & 143 neutrons) is equal to
1.55
The fractions will have adjacent numbers in the proton/neutron ratio of the
parent nucleus. The Proton/Neutron ratio of Kr 3692 (36 protons & 56
neutrons) equates to 1.55, and the Proton/neutron ratio for Ba 56142 (56
protons & 86 neutrons) equates to 1.53
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FISSION EXAMPLES AND TRANSLATION INTO HEX NOTATION
1
0n
+ U92235
U92236
Kr3692 + Ba56142 + 2 1 0 n
( 1 0 n = short for neutron)
HEX notation :
1
0n
+ U9251
U9252
Kr3620 + Ba5630 + 2 1 0 n
The number on the upper right (92) next to the uranium symbol indicates
the number of HEX that makes up the atom core, the number at the bottom
(51) indicates the number of adhering neutrons (or protons). All (decay)
elements can be written and read in the same way.
In the standard notation it is not immediately obvious what the correct
number of protons and neutrons are after the fission, whereas in the HEX
notation it is clear at a glance.
Krypton has 36 protons, 36 neutrons and 20 adhering neutrons in its
nucleus, Barium has 56 protons, 56 neutrons and 30 adhering neutrons in
its nucleus. Looking at the total of adhering neutrons in the Uranium
nucleus there is one adhering neutron short when counting adhering
neutrons in Krypton and Barium fragments, which is to be found in the
ejected neutron together with the neutron that triggered the fission (2 1 0 n).
Second example :
1
0n
+ U92235
U92236
Xe54140 + Sr3894 + 2 1 0 n
HEX notation :
1
0n
+ U9251
U9252
Xe5432 + Sr3818 + 2 1 0 n
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Third example :
1
0n
+ U92235
U92236
La57145 + Br3588 + 3 1 0 n
HEX notation :
1
0n
+ U9251
U9252
La5731 + Br3518 + 3 1 0 n
There are of course other combinations possible, but always with the same
ratio proton/neutron in mind, and that in itself narrows possibilities down.
Most newly formed elements are still unstable and are (can be) transformed
to (more) stable elements by some form of decay, for instance:
1
0n
+ U92235
U92236
Kr3693 + Xe54141 + 3 1 0 n
HEX notation :
1
0n
+
U92
51
22
U92
52
Kr36
21
+
Xe54
33
+3
1
0n
The first fragment Krypton will decay to Niobium through - decay.
Kr3693
Rb3793
Sr3893
Y3993
Zr4093 towards Nb4193
Sr3817
Y3915
Zr4013 towards Nb4111
(Niobium 41 stable)
HEX notation :
Kr3621
Rb3719
The second fragment Xenon will decay to Praseodymium also through
- decay.
Xe54141
Cs55141
Ba56141
La57141
Ce58141 towards Pr59141
(Praseodymium 59 stable)
HEX notation :
Xe54
33
Cs5531
Ba5629
La5727
Ce5825 towards Pr5923
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小出の質量公式
Koide's mass formula
The Koide formula is an unexplained empirical equation discovered by Yoshio
Koide in 1981. In its original form, it relates the masses of three charged leptons;
later authors have extended the relation to neutrinos, quarks, and other families of
particles.
In essence Koide's formula shows a mathematical relationship between physical
quantities.
In the following lines a slightly different approach will be taken and the formula
applied to the mass and charge numbers of the constituent parts of the proton and
neutron, extrapolated to the hexagonal setting of protons and neutrons proposed
as part of the atomic nucleus, with remarkably results.
The proton is composed out of two up quarks and one down quark, while the
neutron is composed out of two down quarks and one up quark.
The bare mass of the up quark is not well determined, but probably lies between
1.8 MeV/c2 and 3.0 MeV/c2.
The middle between 1.8 MeV/c2 and 3.0 MeV/c2 is 2.4 MeV/c2 , value
integrated into Koide’s formula.
Also the bare mass of the down quark is not well determined, but probably lies
between 4.5 MeV/c2 and 5.2 MeV/c2.
The middle between 4.5 MeV/c2 and 5.2 MeV/c2 is 4.8 MeV/c2, value integrated
into Koide’s formula.
First step, the Koide formula applied on the data of the proton which is composed
of two up quarks and one down quark :
Proton =
2.4+2.4+4.8
(√2.4+ √2.4+ √4.8)^2
= 9.6 / 27.976443 = 0.343145(8) (a dimensionless quantity)
≈ 1/3 (difference = 0,343145 – 0.333333= 0,0098)
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Second step, the Koide formula applied on the data of neutron which is composed
of two down quarks and one up quark :
Neutron =
4,8+4,8+2,4
(√4,8+ √4,8+ √2,4)^2
= 12 / 35.176444 = 0.341137(3) (a dimensionless quantity)
≈ 1/3 (difference 0.341137 – 0.333333= 0,0078)
Taking the two results together being each part of the hexagonal setting :
+ = 0.343145(8) +0.341137(3) =
0.684283 ≈ 2/3 (difference = 0.684283 – 0.666666 = 0.017617)
Applying the Koide formula on the electric charge numbers of the constituent
parts of the proton and neutron in the hexagonal setting.
The proton is composed out of two up quarks and one down quark, while the
neutron is composed of two down quarks and one up quark.
The up quark has an electric charge of + 2/3 and the down quark an electric charge
of – 1/3.
Using these electric charge values in the Koide formula without their charge
indication reveals a remarkably outcome :
First step, the Koide formula applied on the data of the proton which is composed
of two up quarks and one down quark :
Proton =
2/3+2/3+1/3
(√2/3+ √2/3+ √1/3)^2
= 5/3 divided by 4.885616 = 0.341137(4) (a dimensionless quantity)
≈ 1/3 (difference 0.341137 – 0.333333= 0,0078)
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Second step, the Koide formula applied on the data of the neutron which is
composed of two down quarks and one up quark :
Neutron =
1/3+1/3+2/3
(√1/3+ √1/3+ √2/3)^2
= 4/3 divided by 3.885618 = 0.343145(6) (a dimensionless quantity)
≈ 1/3 (difference = 0,343145 – 0.333333= 0,0098)
Taking the two results together being each part of the hexagonal setting :
+ = 0.3411374 + 0.3431456 =
0.684283 ≈ 2/3 (difference = 0.684283 – 0.666666 = 0.017617)
ANALYSING THE NUMBERS
Note: these numbers are not mass or charge numbers, they only refer
to the mutual relationship between the mass and charge numbers of two
leptons, the Up and down quark.
NUMBERS
INTRODUCED IN
KOIDE’S FORMULA
PROTON
NEUTRON
HEX
0.343145(8)
0.341137(3)
0.684283
0.341137(4)
0.343145(6)
0.684283
0.684283
0.684283
MASS
ELECTRIC CHARGE
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Filling in Koide's formula with the mass numbers of the constituent parts of the
proton , and the charge numbers of the constituent parts of the neutron ,
the use of these different properties of the nucleons, generates an identical result
(up to six decimals), 0.343145(8).
The same happens when filling in Koide’s formula with the mass numbers of the
constituent parts of the neutron and the charge numbers of the constituent
parts of the proton . The use of the different properties of the nucleons
generates identical results (up to six decimals), 0.341137(4).
Looking at the numbers in the table, there seems to be some sort of cross
pollination going on. Same numbers arise when the mass numbers of the proton
and the charge numbers of the neutron are inserted in the Koide formula, the
same happens when the charge numbers of the proton and the mass numbers
of the neutron are inserted in that same Koide formula.
Adding the mass and charge numbers of the proton + and doing the same for
the neutron + the numbers for both are exactly the same (up to six decimals):
0.684283.
Adding the mass numbers of the proton and neutron in the HEX configuration
+ results in the number 0.684283. The same goes for the proton and neutron in
the HEX configuration + when their charge numbers are added together, again
resulting in the number 0.684283 (numbers identical to six decimal places)
No matter how one turns it, the same numbers just keep popping up . This is too
unlikely to be coincidental … some additional observations :
1. The chosen mass values for the two constituent elementary particles of the
proton and the neutron are between their minimum and maximum
experimentally determined values. The results from Koide formula when
using those numbers are an indication that those values are (almost) correct.
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2. Fact is that protons are the only hadrons known to be stable in isolation,
neutrons are stable only when they are incorporated into atomic nuclei.
When a free neutron decays,
ordinarily one of the down quarks
becomes an up quark, changing
the neutron into a proton, with the
emission of a W boson. This W
boson, in turn, becomes an
electron plus an anti-neutrino. The
change from two down quarks and
one up quark in the neutron to one
down quark and two up quarks for
the proton balances electrically.
3. NUMBER pointing in a specific direction :
When a subatomic particle decays into a different particle new particles are
created, they are not merely separated out of the previous particle.
Because a neutron can decay into a proton releasing an electron and an antineutrino does not mean that it contains a proton and an electron and an antineutrino.
The result in subtracting the numbers 0,343145 and 0,341137 = 0,002008.
This number 0.002008 indicates the numerical representation of the
necessary decay energy for a neutron to decay into a proton. Adding
0.002008 to the neutron number in the table will transform a neutron into
a proton.
(Protons have not been observed to decay, so it is in fact a one way
operation.)
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THE BUCKINGHAM π THEOREM
The Buckingham π theorem indicates that validity of the laws of
physics does not depend on a specific unit system. A statement of
this theorem is that any physical law can be expressed as an
identity involving only dimensionless combinations (ratios or
products) of the variables linked by the law (e. g., pressure and
volume are linked by Boyle's Law – they are inversely
proportional)
(
A dimensionless quantity is a quantity to which no physical
dimension is assigned, also known as a bare, pure, or scalar quantity or a
quantity of dimension one.)
The number 0,343145 relates the mass numbers of the nucleons of the
proton with the electric charge numbers of the nucleons of the neutron and
the number 0,341137 relates the mass numbers of the nucleons of the
neutron with the electric charge numbers of the nucleons of the proton. The
number 0.002008 binds those two numbers energetically.
REASON WHY THOSE THREE NUMBERS CAN BE
CATALOGED AS “DIMENSIONLESS PHYSICAL CONSTANTS”
IN THE HEXAGONAL SETTING OF PROTONS AND NEUTRONS
IN THE ATOMIC NUCLEUS.
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… JUST FOOLING AROUND WITH NUMBERS.
Another set of ratio’s ….
Dividing 0,343145 by 0.002008 = 170,88894422310756972111553784861
Dividing 0,341137 by 0.002008 = 169,88894422310756972111553784861
The difference between the two results is EXACTLY “one” (a
dimensionless number) with the additional fact that the digits after the
decimal point all correspond to each other.
This is perfectly normal because the difference between the denominators
is 0.002008.
Let’s equate 170,88894422310756972111553784861 with “X”, and
169,88894422310756972111553784861 with “Y”, then the following is
true :
X–Y=1
X–Y–1=0
29
Let’s do the same with the numbers 0.343145 = X’, and 0.341137 = Y’
then the following is true :
X’ – Y’ = 0.002008
X’ – Y’ – 0.002008 = 0
So it is reasonable to say that that number “one” (that dimensionless
number) equals the number 0.002008.
X–Y=1
X=Y+1
Y=X–1
X . 1 = 0.343145
Y . 1 = 0.341137
(X . 1) + (Y . 1) = 0.684283
X = 0.343145 / 1
Y = 0.341137 / 1
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W-ELECTRONS ( e-W ) AND P-ELECTRONS ( e-P )
According to quantum theory, electrons orbit the nucleus of an atom in a series of
orbitals, ( regions around an atom’s nucleus where electrons are likely to be found. These
orbitals are defined by quantum mechanics and can be visualized as clouds where the
probability of finding an electron is highest. ) bound to specific shells which correspond
to the energy level of the electron. No two electrons of exactly the same properties
(every electron is identical to every other electron, they all have the same mass, the
same electric charge, and the same spin BUT while all electrons have the same
electrical charge, they do not always have the same energy. The energy of an electron
depends on its position relative to the nucleus and the energy level or orbital it occupies )
can inhabit the same space at the same time (the Pauli exclusion principle) thus, atomic
orbitals are characterized as "probability clouds".
An electron is a quantum object, it can act as a wave or as a particle. This dual nature of
electrons means that they do not fit neatly into classical categories of “wave” or
“particle.”, instead their behavior is context-dependent.
As mentioned before the way elementary particles come about makes them
indistinguishable between themselves, meaning that every up quark is indistinguishable
from another up quark, every down quark is the same as is the case for an electron …
We have to forget the image of the number of particle like electrons that match the
amount of protons "within" the so-called probability cloud around the nucleus ... there
are indeed electrons present in the electromagnetic field but not all with the same
appearance.
Since Louis de Broglie proposed the wave interpretation of a particle of momentum p,
as having wavelength λ = h/p it explains the discrete energy levels in atoms in terms of
standing waves, with electrons at specific energy levels (shells & subshells) consisting
of a standing wave with “n” cycles forming an electrical cloud around the nucleus, the
W-electron confined to dimensions on the order of the size of an atom, the Wave like
electron ( e-W ), it is never to be found as a free electron.
The twin electron on the other hand has a different occurrence, it is the electron that is
to be found in the valence band, hitchhiking on the electrical cloud around the nucleus,
a P-electron, the Particle like electron ( e-P).
It is the electron that hops in and out the conduction band in an attempt to establish a
bond between atoms.
It is also the electron that brings about the equilibrium of the electrical cloud around the
nucleus (ionization being the exception).
This electron which when not attached to an atom is a free electron that can move freely
when external energy is applied.
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The nucleus being an electrical charged particle is served by W- and P-electrons, it is
the natural way it gains its “neutral” status.
The quantum number “n” labels the energy level “E n” around the nucleus .
The lowest energy level n=1, so “n” is related to the number of wavelengths that fit into
the W-electrons orbit. All orbits with the same n form a (spherical!) shell. Since the
size of a shell increases with n, outer shells can hold more W- electrons in shells and
subshells. Together they form the cloud.
It is the energy from the W-electrons, which constitutes the so called “cloud”, around
the nucleus. At large distance from the nucleus this electromagnetic field is weak,
however closer to the nucleus, the field becomes stronger. Because the cloud wants to
reduce its kinetic energy by shrinking closer to the core, its potential energy rises more
than its kinetic energy goes down, which is why it regulates itself at an average size.
This electromagnetic field (the cloud) around the nucleus actually extend all the way
down to the nucleus itself where protons/neutrons are constantly interacting,
electromagnetically. In quantum field theory we would say that there is constant photon
exchange.
((every photon is the outcome of an electromagnetic wave, meaning that all frequencies
and therefore all energy, of the electromagnetic radiation are/is translated into photons
traveling at the speed of light (in vacuum), this also means that every particle moving
through the matrix is submerged in an environment saturated with photons …))
The equilibrium of the electromagnetic field around the nucleus will ultimately depend
on the amount of P-electrons present in the valence band and the total value of the
electric charges of the W-electrons around the nucleus.
Therefore when atoms combine, electromagnetic waves overlap, P-electrons will or
will not be exchanged, and (eventually) a new kind of pattern will be created with a
stable new electromagnetic “inner” cloud and a new electron configuration at the outer
rim of that electromagnetic cloud, a new “bond” that will balance the electrical charge
of the nuclei. The possible imbalance is smoothed out by the leakage (or absorption) of
photons with appropriate electrical values.
When atoms combine to form molecules, their P-electrons, or valence electrons, often
interact in various ways. This interaction depends on the type of chemical bond formed .
In covalent bonding, atoms share pairs of P-electrons. This sharing allows each atom
to achieve a more stable electron configuration. For example, in a water molecule (H₂O),
the oxygen atom shares P-electrons with two hydrogen atoms.
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In ionic bonding, one atom donates one or more P-electrons to another atom, resulting
in the formation of positively and negatively charged ions. These oppositely charged
ions attract each other. For instance, in sodium chloride (NaCl), sodium donates a Welectron to chlorine.
In metallic bonding, P-electrons are not associated with any specific atom and can move
freely throughout the metal lattice. This “sea of P-electrons” gives metals their
characteristic properties, such as conductivity and malleability.
The P-electrons present in the outer conduction bands of the atom contribute to the
balance of the electrical charge of the nucleus and allow possible bonding, they dictate
the state of matter, being a conductor, semiconductor or insulator… being a negative or
positive charged atom, ion … the number of P-electrons in the conduction band
determines many of the chemical properties of the atom …
W-electrons in the valence band & P-electrons in the conduction band
➢ CONDUCTOR
Conduction Band :
Full of P -electrons
Valence Band :
the valence band is either not fully occupied with Welectrons, or the filled valence band overlaps with the
empty conduction band. In general, both states occur at
the same time, the W-electrons can therefore move inside
the partially filled valence band or inside the two
overlapping bands.
Band Gap :
No band gap
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➢ INSULATOR
Conduction Band :
Remains empty
Valence Band :
the valence band is fully occupied with electrons. The Welectrons cannot move because they're "locked up"
between the atoms. To achieve a conductivity, Welectrons from the valence band have to move into the
conduction band. This is prevented by the band gap, which
lies in-between the valence band and conduction band.
Band Gap :
Large band gap
➢ SEMI CONDUCTOR
In semiconductors, there is a band gap, but compared to insulators it is so small that
even at room temperature W-electrons from the valence band can be lifted into the
conduction band.
The W-electrons can move freely and act as charge carriers. For each W-electron that
jumps to the conduction band there is a hole left in the valence band, that can be filled
by other W-electrons.
Quantum tunneling …
…. occurs when electrons pass through an energy barrier that they classically shouldn’t
be able to cross. This happens because, according to quantum mechanics, particles like
electrons have besides particle like properties also wave-like properties and can as such
exist in a “superposition” of states. Imagine a barrier that an electron doesn’t have
enough energy to overcome. Classically, the electron would just bounce back.
However, due to its wave properties, there’s a small probability that the electron can
“tunnel” through the barrier and appear on the other side.
The tunneling process is influenced by the energy level(s) and the wave function
amplitude(s) of the electron.
At first, the electron’s energy is less than the height of the potential barrier. This means
that classically, the electron doesn’t have enough energy to overcome the barrier.
Before the barrier the electron’s wave function has a certain amplitude on the side where
it starts.
Inside the Barrier the wave function doesn’t abruptly stop, instead, it decays
exponentially within the barrier. The wave function’s amplitude inside the barrier is
smaller compared to outside, but it never goes to zero. This decay is due to the fact that
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the electron’s energy is less than the potential energy of the barrier. The non-zero
amplitude inside the barrier is what allows for the possibility of tunneling.
So the tunneling effect is a result of the wave function’s ability to adapt its amplitude
across different regions, including the barrier. The energy levels and the continuity of
the wave function play crucial roles in making tunneling possible.
After the Barrier on the other side of the barrier, the wave function re-emerges, but with
a reduced amplitude but the wave function’s continuity and smoothness requires that it
adapts its amplitude to match the conditions on both sides of the barrier. This adaptation
is a natural consequence of the Schrödinger equation, which governs the behavior of
quantum systems.
The wave function, which describes the quantum state of the electron, must satisfy this
equation throughout the entire region, including the barrier.
When the electron’s wave function enters the barrier, it doesn’t disappear but rather
decays exponentially. This is because the potential energy inside the barrier is higher
than the electron’s total energy. Mathematically, if the barrier is located between
(x = 0) and (x = a), the wave function inside the barrier can be expressed as:
ψ(x) = ψ(0)e−κx for 0 ≤ x ≤ a
where (kappa) is a constant related to the height and width of the barrier.
At the boundaries of the barrier (at (x = 0) and (x = a)), the wave function and its first
derivative must be continuous. This means:
𝜓left (0) = 𝜓barrier (0) and
𝑑𝜓left
𝑑𝜓barrier
(0) =
(0)
𝑑𝑥
𝑑𝑥
𝜓barrier (𝑎) = 𝜓right (𝑎) and
𝑑𝜓right
𝑑𝜓barrier
(𝑎) =
(𝑎)
𝑑𝑥
𝑑𝑥
Similarly, at (x = a):
Once the wave function emerges on the other side of the barrier, it resumes its oscillatory
nature, but with a reduced amplitude. This reduction reflects the probability of the
electron having tunneled through the barrier. The wave function on the right side of the
barrier can be written as:
𝜓(𝑥) = 𝜓(𝑎)𝑒 𝑖𝑘(𝑥−𝑎) for 𝑥 ≥ 𝑎
where (k) is related to the electron’s energy outside the barrier.
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The key point is that the wave function’s continuity and smoothness ensure that it adapts
its amplitude and phase to match the conditions on both sides of the barrier.
The total energy of the electron remains constant throughout the tunneling process. The
electron doesn’t gain or lose energy while tunneling instead the wave function’s
behavior allows it to “borrow” energy temporarily in a way that doesn’t violate the
overall conservation of energy. This is a manifestation of the Heisenberg Uncertainty
Principle, which allows for temporary fluctuations in energy.
So the likelihood of an electron tunneling through a barrier depends on factors like the
height and width of the barrier and the energy of the electron. The probability decreases
exponentially with increasing barrier height and width.
In semiconductors, the band gap is the energy difference between the valence band
(where W-electrons are bound to atoms) and the conduction band (where P-electrons
are free to move and conduct electricity). When a W-electron tunnels from the valence
band to the conduction band, it effectively moves through this band gap without needing
the energy to jump over it directly. Compared to insulators the band gap is so small that
even at room temperature P-electrons from the valence band can be lifted into the
conduction band.
35
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IN ANTICIPATION OF FURTHER RESEARCH
Redrawing the atomic nucleus into a crystalline structure made of
hexagonal proton-neutron relationships where the elegance, the structural
simplicity of that crystalline structure of the atomic nucleus, gave new and
promising insights into its inner workings.
The structure of the nucleus of the atom has been re-mapped and spatially
rearranged and, in its simplicity, explains completely the inner workings of
that atomic nucleus. It became immediately clear that it was only a small
glimpse into a whole new world that has remained hidden so far and
undoubtedly will open new fields of research in both physics and chemistry.
Alpha decay, Beta plus decay, Beta minus decay, electron capture and
nuclear fission are presented with detailed examples based on the new
composition of the atomic nucleus. Although it seems rather technical at
first, it is in reality a very elegant and easy-to-understand way how the
mechanism of transformation of the structures of the atomic nucleus works.
After analysis of the input of the numerical data of the constituent parts of
the protons and the neutrons in the Koide formula, it seems a reasonable
conclusion that there is a clear and real physical connection between the
hexagonal structure of the atomic nucleus and that Koide formula, which in
turn strengthens the theorem of the hexagonal structure of the atomic
nucleus and at the same time also gives a different weight to the value of
the Koide formula itself, away from the claim by some that it is p urely
numerology.
The twin electron approach (W- an P electrons) is a new approach on how
electrons are distributed around an atomic nucleus which gives a better
insight in the inner workings of that electron cloud.
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