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THIS ARTICLE IS NOT YET FINALIZED—08-2015
Johann Marinsek
[email protected]
Standard atomic model refuted
Atoms don’t comprise nuclei and electron shells
Nuclei don’t comprise Z protons and (A – Z) neutrons
Violation of charge and angular momenta conservation for
transmutations and fissions of atoms refutes the nuclear atomic
model with its perpetually spinning protons, neutrons and electrons
Abstract
Rutherford’s scattering experiments and Moseley’s X-ray experiments are not crucial
for the nuclear atomic model. The same set of data can be explained by another
atomic model.
The main characteristics of the Rutherford-Bohr-Chadwick nuclear atomic model are:
1. A nucleus, which is made up of Z protons and (A – Z) neutrons, where A is mass
number and Z is the atomic number of the periodic table;
2. An electron cloud is located around the nucleus. The Z extra nuclear electrons are
placed in shells. The electron shells are arranged according to aufbau rules that
determine spins and angular momenta of the electrons. Similar rules exist for the
arrangement of nuclear shells.
Both, scattering experiments with metal foils and X-ray experiments for nearly all
elements were able to introduce the atomic number Z of target atoms as the variable
of the corresponding metal foils or of the corresponding elements, respectively. This
supposition is not justified because
1. Rutherford’s scattering theory considers electrostatic scattering only and disregards
therefore the magnetic scattering effect.
2. Most elements possess isotopes, which have obviously the same atomic number Z.
Rutherford left out of consideration isotope scattering experiments. According to
Rutherford, the scattering of
Sn-112 and Sn-124 (for example) is identical, but this is improbable, rather an isotope
shift can be expected. Presumably, the isotope shift is due to magnetic scattering too.
Consequently, it is not possible to infer that the distinctive property for scattering is
atomic number Z.
Rutherford’s scattering formula allows not in the least to infer that the atom consists
of a nucleus with Z protons, (A-Z) neutrons and Z extra nuclear electrons.
Mass number A is probably the predominant distinctive property for scattering. But
there are isobars (same A) that should yield different scattering results. Then
allotropes are known, for instance two for K-40, which have different atomic
structures and therefore should have different scattering features.
Conclusion: Scattering experiments cannot be crucial for the nuclear atomic model.
Scattering experiments can be compatible with other atomic models.
Nuclear reactions refute the nuclear and electronic shell model with its perpetually
spinning electrons, neutrons and protons: Charges, particle number, total atomic
angular momenta are not always conserved for atomic transmutations due to fission
and fusion.
Abbreviation: QM = Quantum Mechanics
A possible alternative interpretation:
Hydrogen consists of a proton and an electron that are magnetically coupled. The
proton consists of unknown sub particles. This atom is an oscillator with
eigenfrequencies.
Elements that are conceived as magnetically coupled hydrogen structures (Prout’s
idea in 1815) and molecules that possess real hydrogen lattice bars can show the same
experimental outcomes like the Rutherford-Bohr-Chadwick nuclear atomic model.
Rutherford’s scattering experiments not crucial
for the nuclear atomic model
The father of the atomic nucleus is Rutherford.
The Physics 2000 Colorado.edu textbook explains:
Rutherford's experiments consisted of shooting alpha particles at thin sheets of metal.
He then measured the angles at which they came sailing out.
Most of the alpha particles went right through the metal without changing course at
all, but a few turned a full 180 degrees and went shooting back the way they'd come.
It is not possible to infer from the outcome of this experiment that there are a dense
nucleus and extra nuclear electrons in orbitals around it.
The error made is a classical non sequitur!
Every beginner of the philosophy of science has learned that a set of data is
interpretable by more than one theory. So the interpretation of Rutherford is not the
unique one and therefore his experiment is not a crucial one!
The simplest objection to Rutherford’s claim that scattering depends on Z is that we
can set for Z = A/2! Then scattering depends on the mass number and
isotopes (same A) probably show an isotope shift. Rutherford did not notice this
possibility.
Below I will show otherwise that the outcome of the experiment does not imply that
nuclei have charge Ze where Z is the atomic number and e is the charge of the
electron.
If we conceive the metal as a 3D lattice, most alpha particles can pass the lattice,
some collide with the lattice components.
Consider the lattice crossing points (the nodal points) and the lattice bars as material!
The balls at the nodal points and the bars represent material hydrogen structures.
Alpha particles can pass between the bars. So it is not necessary to conceive the atom
consisting of a nucleus and orbiting electrons (or orbitals).
http://www.marless.de/natur/st_bernsteen.htm
is the source of the figure of such a lattice.
<======>
α can pass the lattice-----Collision with the lattice
Can the Rutherford-Bohr atomic model be confirmed by force microscopy?
University of Augsburg physicists Giessibl, Hembacher and Mannhart
“have imaged an individual tungsten atom by atomic force microscopy and found four distinct peaks that are attributed to highly located electron clouds. A printed image width of 5 cm corresponds to a magnification of two hundred million. A world-­‐record resolution of 77 pm is demonstrated. The electron structure originates from the quantum-­‐mechanical nature of tungsten bonding. Tungsten develops a body centered cubic crystal structure such that every tungsten atom is surrounded by eight nearest neighbor atoms, causing "arms" of increased charge density which point to the next neighbors. Four of these highly localized electron clouds are visible on surface atoms.” "Force microscopy with light atom probes", www.scienceexpress.org, 10 June 2004).
http://www.chemlin.de/news/jun04/20040610e01.htm Picture is from:
http://www.sciencemag.org/sciencexpress/recent.shtml
By no means atoms with nuclei and sourounding electron clouds can
be derived from force microscopy! The picture suggests rather a
lattice structure…
Not any evidence for nuclei that comprise Z protons
Rutherford obtained the scattering angle by assuming a
repulsive Coulomb force that corresponds to nuclear charges of magnitude Ze (Z =
atomic number). Some textbooks argue that Rutherford’s experiment implies
therefore a nucleus with Ze protons, because Ze can be found in Rutherford’s
scattering formula for the scattering angle.
This argument can be refuted:
1. Rutherford’s scattering theory considers electrostatic scattering only and disregards
therefore the magnetic scattering effect.
Remember that Felix Bloch obtained the experimental proof for the magnetic
moment of the free neutron due to the magnetic scattering of neutrons in iron. Atomic
magnetic fields affected a passing neutron trough its magnetic moment.
(See Bloch’s Nobel lecture)
This experiment certainly cannot yield a scattering formula with atomic number Z as
the distinctive scattering property and by no means an evidence for the nuclear
atomic model.
One may object that Rutherford experimented with alpha particles that have no
resultant magnetic moment. Counter-arguments: The possibility of ortho-He particles
is not excluded, which have obviously a magnetic moment. But even when there
would be only a zero resultant magnetic moment for He, during collisions the
magnetic moments of protons and neutrons can have interactions with the magnetic
moments of the target atom.
2. Most elements possess isotopes (= same atomic number Z). Rutherford left out of
consideration isotope scattering experiments. According to Rutherford, the scattering
of
Sn-112 and Sn-124 (for example) is identical, but this is improbable, rather an
isotope shift can be expected. Presumably, the isotope shift is due to magnetic
scattering too.
Recipe:
Execute two scattering experiments with tin metal foils. The first experiment with
Sn-112, the second with Sn-124 (or with other isotopes, for example with
Li-6 and Li-7)).
The results of the experiments should be different.
You cannot explain the different results with Z = 50 for Sn, because both, Sn-112 and
Sn-124 have the atomic number Z = 50!
The unique possible variables are the mass numbers (A = 112, A = 124) and the
magnetic properties of the target atoms. For Z there is no implication.
Recall here that Rutherford made scattering experiments only on gold and aluminium.
M. Fowler [fow] reports:
“On replacing the gold foil by aluminium foil (some years later), it turned out that
small angle scattering obeyed the above law (for the scattering angle, JM), but large
angle scattering didn't. Rutherford correctly deduced that in the large angle
scattering, which corresponded to closer approach to the nucleus, the alpha was
actually hitting the nucleus. This meant that the size of the nucleus could be worked
out by finding the maximum angle for which the inverse square scattering formula
worked, and finding how close to the centre of the nucleus such an alpha came.
Rutherford estimated the radius of the aluminium nucleus to be about 10-14 meters.”
Obviously, Rutherford assumed that scattering came from single encounters with
nuclei due to electrostatic repulsion only. We see, the explanation of the observed
scattering patterns is “theory laden”. Therefore, Rutherford’s experiment is not an
experimentum crucis for the existence of a nucleus with Ze charges.
Recall collisions in macrophysics: the decisive parameters for the repulsion are the
masses of the colliding bodies, which means the number of atoms. When atoms
collide, the same occurs. The repulsive forces are of electromagnetic origin. The
cause is that atoms represent clusters of negative and positive charges and magnetic
moments. It does not matter that the atom as the sum of this charges is neutral.
Regarding the X-ray experiments of Moseley,
in the formula for the characteristic frequencies of the elements appear also the
atomic number Z. But Moseley did not research isotopes. If one investigates all
isotopes it is evident that mass number A is the appropriate variable. Therefore also
Moseley’s X-ray experiments are not crucial for the assertion that nuclei have charges
Ze. The occurring frequencies are frequencies of the dielectric medium (aether) due
to eigenfrequencies of an atomic oscillators. Observed frequencies are difference
frequencies. Of the one side the Eiegenfrequencies of the atoms, on the other side ther
Eigenfrequencies of the dielectric medium that is the carrier of thte vaves.. See below
for details
There is no rationale for spinning electrons in electron shells.
The spin was ad hoc invented to solve the problem of
doublets in atomic spectra.
Uhlenbeck and Goudsmit proposed the electron spin in 1925. A severe objection to
the possibility of spinning electrons Uhlenbeck and Goudsmit declared themselves:
Due to spin, peripheral parts of the electron would have velocities greater than c, the
velocity of light. Their objection was ignored… [ulg]
Before going further, a remark to the concept of spin in QM:
The prevailing opinion is that QM spins are not the same as spins in classical
mechanics. The QM spin is a “half-integer” property of atomic sub-particles.
But for Finkelnburg [fin] (in 1956!) the spin of an electron is a spin like in
mechanics:
s = (1/2) h/2π = 5,27 10-28 g cm2 sec-1
Spin is defined as moment of inertia Θ = ∫m dr times angular velocity dϕ/dt.
Let us calculate the peripheral velocity v of a spinning electron if we assume that its
shape is a ring with radius r. The mass is m.
Then Θ = mr2 and the spin is mr2 dϕ/dt = mr2 v/r = mv r
QM states that the spin is h/2 = mv r .
Data: h = 1,05 10 -34 kgm2/sec, m = 9,11 10 -31 kg, the radius may be about 1/100 of
the classical electron radius: r = 3. 10 -17 m.
Then => v = 6000 c.
For a proton radius of 0,865 10 -15 m we get v = 0,12 c
Uncertainties of the concept of a spinning electron
Classical physics is deterministic: natural laws specify positions and velocities of all
particles in the past and in the future, say for example for an electron.
Heisenberg invented the uncertainty principle, which states that there is no certainty
for both the position q and the momentum p of an electron.
Heisenberg’s uncertainty principle asserts for the uncertainties of position Δq and
momentum Δp the relation
(Δq) Í (Δp) ≥ h
For the example above the spin uncertainty of the electron would be h = 2π h,
whereas the spin itself has the magnitude h/2!
In order to explain his uncertainty principle, Heisenberg argued that the electron is
not a particle but a wave packet.
Because a wave packet does not possess fixed borders and because the constituent
waves of the packet have different velocities, both the position and the momentum
are properties with uncertainty.
Regarding the electron as a wave packet the first question is: What is waving?
For quantum mechanics a light-bearing medium, formerly known as æther, does not
exist. Space itself cannot vibrate, that would be a category-mistake.
Secondly, if a wave packet takes the place of the electron, then dispersion must occur.
To make the difficulties complete, QM asserts that the electron is a permanently
spinning one. Therefore, when the wave packet takes the place of a flying rotating
particle, the wave packet has to rotate! How is that possible? Is there a vortex of a
wave packet imaginable?
Permanently fast spinning massive particles or unintelligible spinning wave packets
make QM untenable.
Conclusion
The concept of an atom with a nucleus that consists of Z protons whereas Z electrons
surround the nucleus in shells is improbable.
Below we investigate some nuclear transmutations. When we suppose nuclei with Z
protons and (A - Z) neutrons we can show that conservation laws are violated. So we
get other indications that the basic atomic model is untenable…
Nuclear reactions refute the nuclear and electronic shell model
with its perpetually spinning electrons, neutrons and protons:
Charges, particle number, total atomic angular momenta are not always
conserved for atomic transmutations due to fission and fusion.
U-235 + n
→
Ce-140 + Zr-94 + 2n + Q
The electron shell and the particle transformations of the fission process are:
U-235
Ce-140
Zr-94
→
3
1
2
1
1
2
[Rn] 5f 6d 7s
[Xe] 4f 5d 6s
[Kr] 4d25s2
→
6p+6e
6 n (for Ce + Zr)
→
Due to fission, all electrons of the U-235 shell structure [Rn] 5f36d17s2 tumble at the
nucleus.
The nuclei of Ce and Zr cannot manage the resurrection of the electron shells of
Ce: [Xe] 4f15d16s2 and Zr: [Kr] 4d25s2!
The standard model cannot explain how 6 protons and 6 electrons of Uranium can be
transformed into 6 neutrons of Ce + Zr!
NUCLEAR SHELL MODEL IS A FAILURE
Nuclear reactions show that
neutrons and protons cannot have spin ½
Let us investigate some examples where neutrons are produced by nuclear reactions.
I is the resultant of nuclear spins and nuclear angular momenta. But of importance
for our purpose is only the conservation of spins and angular momentum of nuclides
in nuclear reactions:
1: N-14 (I = 1) + α ( I = 0 ) → F-17 ( I = 5/2) + n ( I = ½)
For this nuclear reaction the imbalance of spins and angular momenta is obvious.
2: F-19 ( I = ½) + α ( I = 0)→ Na -22 ( I = 3) + n ( I = ½)
Also for this reaction the spin and angular momenta imbalance is obvious.
(There are at least 10 known reactions that produce neutrons with different
calculated masses! Please see Clarence Dulaney’s Physics Page. What is a “Neutron”
http://mywebpage.netscape.com/clarencedulaney)
Next let us consider:
C-12 ( I = 2) + p ( I = ½) => N-13 ( I = ½) + γ ( I = 1).
The imbalance is evident, protons and neutrons cannot have spin 1/2!
Conclusion:
The QM nuclear shell model with spin h/2 for neutrons and protons is a failure.
α-Decays
1th example: Am-241 => Np-237 + He2+
Written in this form, charge is not conserved. Therefore we must write:
Am-241 => Np2—-237 + He2+
When the Am nucleus ejects an alpha particle, whereas the electronic shell remains
unchanged, then of course the total angular electronic momenta are conserved:
[Rn] 5f77s2 => [Rn] 5f77s2 : J = 7/2 = conserved .
But the unchanged electronic shell of Am is such as a deus ex machina!
For quantum mechanics theorists there are no troubles: 2p + 2n (= α) from the
outermost nuclear shell are ejected. The remainder is an Np nucleus. But the
electronic shell remains unchanged.
Every electron retains its “location”. All electrons retain their total angular
momentum. This is highly artificial…
Then recall that the electronic shell determines the chemical properties and not the
nucleus.
The Np2—- ion has the electronic shell of Am, therefore it is chemically Am and not
Np!
Note that total nuclear angular momentum for Am-241 => Np-237 + He is conserved:
5/2 (Am) => 5/2 (Np) + 0 (He)
When we regard the α decay as a fission with neutral daughter elements:
Am-241 => Np-237 + He then the total electronic angular momentum is not
conserved:
[Rn] 5f77s2 (J = 7/2) => [Rn] 5f4 6d17s2 (J = 11/2) + He (J = 0)
Note that one electron of 5f must alter completely its status in order to be now a 6d1
electron.
The He nucleus obviously has captured two f electrons… Is it possible that it is losing
the electrons afterwards? This alternative process is not more mysterious than the
first explained above…
Solution according to the Prout atomic model: Am-241 consists of 241 H-atoms. One
He building block ( = 4 H’s) splits off, the remainder is Np-237.
2nd example:
Be-8 => He-4 + He-4
Nuclear and electronic angular momenta are conserved. But this is not the rescue for
the nuclear atomic model. When the Be nucleus breaks into two, electrons spiral into
the nuclear fragments and lose their spins. The electrons must know that exactly two
electrons have to arrive at each He nucleus. Till now there is no causal explanation
for this process available…
What happens now? The electrons have to rise at their (probable) locations and must
begin to spin. The p electron has to “orbit” because it must have an angular
momentum.
It is an invalid argument that the electrons simply occupy again their state of
dynamical equilibrium. Without support they cannot rise from the nucleus at their
places in the electronic shells. Because no such support is comprehensible, QM must
suppose a resurrection of the electron shells.
But this is not physics…
232
228
4
3rd example:
90Th =>
88Ra + 2He
Imbalance for total electronic angular momenta:
2 ≠>
0 + 0
Conclusion: the aufbau rules for the electronic shells and their accompanied states of
angular momenta and spins violate the conservation law for total atomic angular
momenta. This fact brings the nuclear atomic model into discredit…
Electrons and positrons in β-decays cannot originate from nuclei.
The origin of the e+ and e—’s is an intergalactic dielectric medium.
Some neutral atoms undergo transmutations into other neutral atoms with the
same mass number A, this transmutation is accompanied by the occurrence
of either e— or e+:
A
XN => AZ+1XN-1 + e—
or
Charges: (0) => (0)
+ (—1) or
Z
A
XN => AZ-1XN+1 + e+
(0) => (0) + (+1)
Z
Violation of charge conservation indicates that the parent atoms cannot be the source
of either the electron e— or the positron e+.
Because if either e— or e+ originated from the parent atom, then the daughter atom
should be a charged one, which is not the case. The source of the β-particles therefore
cannot be the nuclei of parent atoms as it is the claim of QM.
Obviously, a vacuum does not contain electrons and positrons. The source of the
occurring electrons and positrons is an all-pervading intergalactic dielectric medium
that consists of electrons and positrons.
Transformations of atomic structures cause excitations of that medium. The results
are gamma radiation and release of beta particles.
The ongoing processes are not explainable in terms of the Rutherford-Bohr atomic
model that acts in a vacuum.
Introduction
A textbook [koe] explains:
Beta decay is the "transformation" of a neutron into a proton, with the emission of an
electron for charge conservation, and an antineutrino for energy and momentum
conservation. It occurs in those situations in which alpha decay would leave the
nucleus less stable than it was before. In Beta decay, the following parameters
change:
Z -> Z + 1
A -> A
N -> N – 1
Another form of Beta decay occurs in which a proton changes into a neutron plus a
positron and a neutrino, with parameter changes:
Z -> Z – 1
A -> A
N -> N + 1
Now, here is an example for an alleged positron decay:
C-11
à
B-11 + e+ + νe
Constituents: C-11: 6e + 6p + 5n
B-11: 5e + 5p + 6n
Allegedly it works as follows: 1p from C-11 transmutes according to the formula:
Energy + p => n + e+ + neutrino νe,
then B-11 has its 6th neutron, as required. An e+ is ejected.
=> 1st mistake: charges are not conserved, moreover a charge was created:
C-11 => B-11 + e+ + νe
(0) => (0) + (+1) + (0)
(obviously, atoms are neutral!)
=> 2nd mistake: total angular momenta of electrons and nucleons are not conserved!
For electron angular momenta and spins see any handbook for the ground states
listed. For nuclear reactions see the attached table (author N. J. Stone), „Column 4.
Gives the spin (I) and parity of the state.”
C-11 => B-11 + e+ + νe
Electron shell:
0 ≠> ½
Nuclear reactions:
3/2 => 3/2 + ±1/2 + ±1/2
Comment: For the nuclear reactions, total angular momenta can be conserved when
the neutrino spin is introduced. But because for the electrons total angular momenta
are not conserved, the conservation of overall angular momenta and spins of the
parent atom is violated.
The neutrino spin is the rescue only for the nuclear reaction, but not for both together,
the electron shell plus the nuclear reaction involved!
The clou is: The conservation of overall total angular momenta requires no neutrino!
This proof shows that the neutrino is ad hoc, namely in order to rescue the spin
conservation for the neutron decay only:
n => p + e—. But the necessity to introduce the neutrino spin depends on the assertion
that the neutron has spin h/2, which is a not justified by QM. See the article The
Neutrino is Stopgap of Flawed Quantum Mechanics.
=> 3rd mistake:
There is a 6th electron in the outermost shell of the resulting B-11 making it
negatively charged! It should be neutral. According to some authors (Bröcker B.,
dtv Atlas Atomphysik) the surplus electron breaks off. In the same book (table of
nuclei) for a C-11 decay only β+ is mentioned as the decay product...
=> The 4th mistake represents a magic transubstantiation like in religions:
If p => n + e+ + neutrino νe, then
(because the claim is that n => p + e— + anti-neutrino):
p => p + e— + e+ + neutrino + anti-neutrino.
This is obviously an ontological nonsense. This is magic, not physics.
An empirical verification of that mysterious transubstantiation is impossible.
Probably the formula E = mc2 led to the error that any transubstantiation is possible,
namely that with the addition of energy
p + energy => n + e+ + neutrino νe
... Anything goes!
Author of the formula:
p => n + e+ + neutrino νe is Fermi.
To distinguish between neutrino and anti neutrino meaning that one has
—½ spin the other + ½ is nonsense.
One and the same neutrino could have ± ½ spin.
There is an objection to this line of arguments, meaning that the daughter atom is in
reality a ion:
C-11 => (B-11)— + e+ + νe
Written in this form, charge and total atomic angular moments are conserved. The
overwhelming majority treats the daughter atom as a neutral one. A (B-11)— ion
would be chemically carbon that has a nucleus of 5p and 6n instead of 6p and 5n.
+
Wikipedia: Fundamentally, an up quark is converted into a down quark, emitting a W boson,
which then decays into a positron and a neutrino.
For our purpose quarks are irrelevant…
+
Wiki: In all the cases where β decay is allowed energetically (...), it is accompanied by the
electron capture process, when an atomic electron is captured by a nucleus with the
emission of a neutrino:
Comment:
The electron capture (EC) is unbelievable... A proton of the nucleus allegedly
captures an electron of the innermost electron shell, incorporates it and creates this
way
a neutron!
In the example above, one p of C-11 would capture an electron, then instead of
6e + 6p + 5n of C-11,
5 e + 5p + 6n are the result, which is B-11! Magic...
γ-emitting excited atoms discredit the nuclear atomic model:
Nuclear γ-rays would destroy the fragile electronic shell structure
We are taught [lbl]
that gamma rays are a type of electromagnetic radiation that results from a
redistribution of electric charge within a nucleus. A γ ray is a high-energy photon.
Take for example the following γ-decay
60
60
28Ni* ––>
28Ni + 1.173 MeV + 1.332 MeV
60
where 28Ni* denotes an excited Ni atom. The two γ -photons with about 1 MeV
each are high-energy bullets emitted from the nucleus. They must pass on their path
the wonderfully arranged electron shell or electron wave structure made up of 28
electrons.
High-energy γ photons/waves would surely destroy the fragile quantum mechanics
complex of shells or orbitals. But according to current theory the Ni shell structure
remains unaffected by the γ bombardment. This survival of the atomic shell
structure could be possible only by a wonder. Converting γ-photons into
positronium is the next puzzle to solve in terms of QM:
Peripheral collisions of heavy ions that yield positronium e+ e—
For gold ions this process can be written:
Au+ Au+ ––> Au* Au* + e+ e—
Where Au* Au* mean excited Au atoms.
Morozov explains the occurring e+ e— pairs: „The e+ e− pairs are a product of purely
electromagnetic interaction of the virtual photon fields emitted by the Au ions. The process of
+ −
two photons converting into an e e pair has been studied experimentally with a good
precision in quantum electrodynamics (QED) for pair energies up to 100 GeV.
+ −
… The virtual photons can produce an e e pair in collisions with the photons from the ion in
the opposite beam. Additionally, photons can cause an excitation of the Au ions in the
opposite beam.“
The occurrence of positronium was therefore interpreted as the product of a
conversion of radiation into positronium due to a collision of γ-photons. But there are
some difficulties to apply this interpretation: the two radiation beams are represented
by two photons but photons are not charged particles, whereas both the produced
electron and positron have charge 1. The net charge is 0 because the electron charge
is 1 and the positron charge is +1. Nevertheless it cannot be argued that charge is
conserved:
0 —> (+1) + (-1)
The solely possible interpretatiton is that two charges are created out of nothing,
creatio ex nihilo.
Then regard spin conservation:
Photons would have spin 0 (see the article Photons Would Have Spin 0), positronium
can have spin ±1 (ortho positronium) or 0 (para positronium).
But there is a spin imbalance: 0 –> ±2 for ortho positronium, therefore spin
conservation is violated. By the way, allegedly one single photon with > 1.01 MeV
can be converted into positronium, but spin conservation is violated: 0 —> ±1.
Violated spin and charge conservation discredit the concept of a charge less photon
with spin 1.
Last regard magnetic moments. Obviously, γ rays don’t’ possess magnetic moments.
Both the electron and the positron represent tiny magnets. Again, this is creation out
of nothing, before the creation act, the photons are without magnetic moments.
The conversion of radiation or its energy into inert mass has no physical rationale...
Peripheral collisions of ions would slip of the electron shells. Also excited ions would
lose their electron jacket. The 156 slipped of electrons of
Au+ Au+ are not detectable...
The alternative:
The approaching ions excite each other. The ions oscillate and release positronium
out of the cosmic dielectric medium that consists of electrons and positrons. The atom
does not comprise a nucleus with surrounding electrons that can be slipped of.
Electron and positron of hydrogen are magnetically coupled. Hydrogen atoms are the
building blocks of the atom.
Anderson disregarded spin and charge conservation violation during the
alleged creation of electon-positron pairs
Anderson declared in his Nobel lecture in 1936 (www.nobel.se):
… "creation" of a positive-negative electron pair in the neighborhood of an atomic
nucleus. The energy corresponding to the proper mass of both of the particles, is, ....,
supplied by the incident radiation. Since the energy corresponding to the proper mass
of a pair of electrons is approximately one million electron-volts one should expect
gamma rays of energy greater than this amount to produce positrons in their passage
through matter...
Obviously, Anderson did not pay attention to charge and spin conservation.
Conclusion
The Rutherford-Bohr-Chadwick nuclear atomic model cannot explain beta plus (e+)
decays as decays! This is one of many indications that this model is a failure.
I repeat: The occurring positrons e+ are not explainable by the magic transmutation of
a proton into neutron and a positron and a neutrino. The magic is that all particles can
transmute into all other particles.
Note here a difference:
A neutron can decay because it consists of a proton and an electron.
A proton does not consist of a neutron and a positron (plus a neutrino) therefore it can
neither decay nor transmute into those parts...
Therefore the source of the occurring positrons cannot be the nucleus of the parent
atom because a proton cannot transmute to a neutron and a positron and a neutrino.
Equally the charge imbalance shows that the positron must come from outside. This
is an indication that outside of atoms and molecules there is not the absolute void or
the vacuum.
Instead of a vacuum we must assume a plenum. It should contain positrons and
electrons. This makes possible the polarizability, which is the characteristic of a
dielectric.
Furthermore, both the electron and the nuclear shell model are untenable because
conservation laws are violated. Also the assertion that the atom consists of Z protons
and electrons and (Z-A) neutrons is untenable.
Another indication for such an electron-positron dielectric plenum is the capacitor.
Between the plates of a capacitor is a dielectric. Because a capacitor containing only
a „vacuum“ between its plates also works like a capacitor with any other dielectric
inside, the erroneously so-called vacuum must be in reality also a dielectric.
Therefore, the „vacuum“ contains electrons and positrons. Erroneously so called beta
decays (see also below) indicate these aufbau particles…
The polarization “current” of a capacitor is an indication that the electrons and
positrons are existing building blocks of the cosmic plenum. They must not be created
for example as a spherical standing wave of that medium.
Nothing decays during „beta plus decay“
A transformation of the atom releases a positron out of the all-pervading cosmic
dielectric medium.
Prout’s idea that all elements are complex hydrogen structures
C-11 is a configuration of 11 hydrogen atoms. This configuration is not a stable one.
It must undergo a trans-formation that is also a trans-position of the hydrogen
building blocks. This transformation occurs not in the void but in a dielectric medium
that consists of e+ and e—.
The transformation process liberates positrons out of the all-pervading dielectric (the
old aether).
Why positrons and not electrons or both, e+ and e-? (Some decays liberate electrons
and positrons like Ag108, Ag 110).
..... I don’t know. Till now we don’t know the nature of electrons and positrons and
how they are built-in in the dielectric medium structure....
There is only one regularity:
With increasing mass number A of the isotopes of an element there is a group with a
positron decay product and then it follows the group with an electron decay product.
Take nitrogen: N-12, N-13 have beta plus decay, N-16, -17, -18 beta minus decay
„Beta minus decay“ is not a decay.
The source of the electron is the
all-pervading intergalactic dielectric
There are atomic transformation processes with electrons as the
by-products.
1st example
Co-60-à Ni*-60 + e— + neutrino
QM explanation: One neutron of Co decays into proton and
electron plus a neutrino.
The Co nucleus ejects an electron and a neutrino.
Zà Z+1, so Ni has one electron more in its shell:
Co: [Ar]3d74s2 => Ni:[Ar] 3d84s2
Note the creation ex nihilo of the 8th 3d electron of Ni!
Obviously, charge conservation is violated:
(0) => (0) + (—1).
Therefore the occurring electron cannot have its origin in the Co nucleus. It must be
released from the surrounding dielectric medium.
Are spins and angular momenta conserved? Co-60 à Ni*-60 + e— + neutrino
Shell electrons:
9/2 ≠> 4
Nuclear reaction:
5 ≠> 2 + ±1/2 + ±1/2
Conservation of spins and angular momenta for the electron shell is clearly violated.
Total angular momenta for the nuclear reaction are not conserved.
One proposal is, that Ni is a ion: Ni:[Ar] 3d74s2
Then charge is conserved. But chemically this atom is Co and not Ni! The nuclear
reaction shows again an imbalance for total angular momenta! The nuclear atomic
model is untenable…
Of course a transformation that produces a gamma radiation means an
excitement of the
dielectric electron-positron-medium by a transposition
process of the hydrogen building blocks....
In terms of QM and the Bohr atomic model the Co
nucleus emits gamma ray and gamma photons. This
process would destroy the electron orbital structure. Then
must assume a mysterious resurrection of that structure
plus a creation of a 3d electron in order to get Ni!
http://hyperphysics.phyastr.gsu.edu/hbase/nuclear/betaex.html#c1
The complete metamorphosis of Co-60 into the stable Ni-60 produced
one
2,5 MeV gamma radiation plus 0.31 MeV electrons.
Second example of a beta minus „decay“:
Citation (italics) of the colorado.edu Physics 2000 textbook
http://www.colorado.edu/physics/2000/isotopes/radioactive_decay.html
Figure: Karlsruhe Tritium Neutrino Experiment,
http://www-ik.fzk.de/tritium/overview/index.html
You can write out the nuclear reaction involved in the beta decay of tritium by giving
the electron a "mass number" of 0 and an "atomic number" of -1:
3
H1 => 3He2 + 0e-1
Notice that the mass numbers on each side add up to the same total (3 = 3 + 0), and
so do the charges (1 = 2 + -1). This must always be true in any nuclear reaction.
Comment: Textbook authors erroneously only compare the charge of the H-1 nucleus
(+1) with the charge of the nucleus of He-2 (+2) and the electron (-1). But first of all
the charge balance for the entire transformation process must be taken into
consideration!
Charge conservation is violated:
(0) => (0) + (-1) because both H-3 and
He-3 are neutral and there remains only the charge (-1) of the electron at the right
side.
Are spin and angular momentum conserved when we invent an anti neutrino with
spin 1/2 ?
3
H1 => 3He2 + 0e-1 + anti neutrino νe
Shell electrons:
1/2 ≠> 0
Nuclear reaction:
½ => ½ + ±1/2 + ±1/2
Spins of nuclear reaction don’t violate the conservation law. But overall spins and
angular momenta are not conserved because there is the electron spin of the tritium
atom that is not conserved! It was forgotten…
The neutrino and its spin was invented to fulfil the conservation of spins for the
neutron decay: n => proton + electron.
The clou of the tritium decay is that there is no need for a stopgap neutrino because
overall total angular momenta are conserved without the neutrino spin. Conversely, if
one introduces a neutrino spin, overall angular momenta of the atom are not
conserved!!
Now consider the alleged metamorphosis of the neutral tritium into the neutral He-3:
One neutron of tritium decays into proton and electron. The electron is emitted. Then
the He-3 nucleus is complete: 2p + n, but one 1s electron of He is missing! (See
above)
The alternative to the emission of the electron, namely that the electron of the
decaying neutron n => p + e—
comes to be the accompanist of the single electron of H-3 on the 1s shell and forms
this way the He-3 neutral atom cannot resolve the trouble because then the observed
free electron (0e-1 at the right side) would be a creation out of nothing...
I propose a possible interpretation for the occurrence of the electron:
H-3 and He-3 are neutral isotopes; therefore the origin of the occurring free electron
with its charge (-1) can be neither the nucleus of H-3 nor the vacuum but an allpervading dielectric medium.
See above the same argument fort he occurring positron in a beta plus „decay“.
Like for the occurrence of positrons during the metamorphosis of an atom, the
electrons are released out of the all-pervading dielectric medium because the
transformation process of the atom excites this medium. The transformation process
is simply a re-arrangement of the building blocks of the atom.
The structure becomes more stable. In terms of QM (which happens in the vacuum)
this process is not explainable. See above the missing electron in the daughter atom
and the evidence that the electron cannot be emitted but must be released from the
background that is not a void.
Conclusion
The atomic model that consists of
• nuclear shells of perpetually spinning protons and neutrons
• plus electron shells of perpetually spinning electrons around the nucleus
is untenable.
The scene for all ongoing physical processes of ruling QM is the vacuum, in this void
particles interact. This ruling paradigma is also untenable.
The scene for physical processes is the intergalactic dielectric medium that represent
the privileged frame of the cosmos. The impossibility of a perpetuum mobile was a
firm principle of physics. Quantum mechanics violates this principle and the principle
of causality by the ad hoc assumption of perpetually spinning atomic electrons,
neutrons and protons. There is neither theoretical nor empirical evidence for spinning
electrons and nucleons.
The recent spin crisis concerns the problem how the spin ½ of the proton can be
compatible with the alleged quark structure of the proton.
No rationale is given for perpetually spinning nucleons and electrons…
Uncertainties of the concept of a spinning electron
Classical physics is deterministic: natural laws specify positions and velocities of all
particles in the past and in the future, say for example for an electron. Heisenberg
invented the uncertainty principle, which states that there is no certainty for both the
position q and the momentum p of an electron.
Heisenberg’s uncertainty principle asserts for the uncertainties of position Δq and
momentum Δp the relation
(Δq) Í (Δp) ≥ h
For the electron Heisenberg argued that it is not a particle but a wave packet. Because
a wave packet does not possess fixed borders and because the constituent waves of
the packet have different velocities, both the position and the momentum are
properties with uncertainty.
Regarding the electron as a wave packet the first question is: What is waving?
For quantum mechanics a light-bearing medium, formerly known as æther, does not
exist. Space itself cannot vibrate, that would be a category-mistake.
Secondly, if a wave packet takes the place of the electron bullet, then dispersion must
occur. To make the difficulties complete, QM asserts that the electron is a permanent
spinning one. Therefore, when the wave packet takes the place of a flying rotating
particle, the wave packet has to rotate! How is that possible?
Spin is defined as moment of inertia Θ = ∫ mdr times angular velocity dϕ/dt.
Let us calculate the periferal velocity v of a spinning electron if we assume that its
shape is a ring with radius r and mass r. then Θ = mr2 and the
spin is mr2dϕ/dt = mr2 v/r = mv r
QM states that the spin is mv r = h/2 .
h = 1,05 10 -34 kgm2/sec, m = 9,11 10 -31 kg, the radius may be about 1/100 of the
classical electron radius: r = 3 10 -17 m.
Then v = 6000 c.
For a proton radius of 0,865 10 -15 m we get v = 0,12 c
Permanent fast spinning massive particles or unintelligible spinning wave pakets
make QM untenable.
Historical review:
The conceptual development of the nuclear atom
GEIGER’S AND MARSDEN’S SCATTERING EXPERIMENTS
The α particle bombardment of ultra thin metal foils tended to detect the structure of
the atom.
Geiger wrote the first paper on this topic in 1910. [gei]
An important fact for the history of the nuclear atomic model is that Rutherford
communicated this paper. Geiger desired to express my thanks to Prof. Rutherford.
Geiger’s conclusion was that the most probable angle through which a α-particle is
turned when passing through an atom is proportional to its atomic weight. It is
striking that at that times the proportionality factor of scattering was atomic weight
and not nuclear charge!
In the next paper on this topic Geiger and Marsden wrote [geim]:
We are indebted to Prof. Rutherford for his kind interest in these experiments, and
for placing at our disposal the large quantities of radium emanation necessary. We
are also indebted to the Government Grant Committee of the Royal society for a
grant to one of us, out of which part of the expenses has been paid.
Now Geiger and Marsden (an assistant of Rutherford) had to be grateful to their
mentor and wrote:
Professor Rutherford has recently developed a theory to account for the scattering of
α-particles … the assumption being that the deflections are the result of an intimate
encounter of an α-particle with a single atom of the matter traversed. In this theory
an atom is supposed to consist of a strong positive or negative central charge…and
surrounded by electricity of the opposite sign…
It was found out that some of the α particles were deflected, some of them were
scattered back toward the α particle source and most of them went straight through
the Au- foil that was used.
Today’s concepts of the nucleus differ in one important point from Rutherford’s
nucleus: Rutherford obviously conceived his neutron not as a decaying neutral
compound of proton and electron (and neutrino). Rutherford’s neutron was a neutral
hydrogen atom, H. Therefore Rutherford conceived two different types of H: the
nuclear H is a proton with a stationary electron, obviously near the nucleus. The
singular hydrogen atom is a proton with an exterior electron, far away from the tiny
nucleus.
The starting points for Rutherford were the data of scattering on particles that
occurred in experiments begun by Geiger and Marsden. Rutherford argues [rua]:
We have seen that from an examination of the scattering of α particles by matter, it
has been found out that the positive charge on the nucleus is approximately equal to
Ae/2, when A is the atomic weight and e is the unit charge. This is equivalent to the
statement that the number of electrons in the external distribution is about half the
atomic weight in terms of hydrogen…
It seems improbable that the electrons within the nucleus would contribute to this
scattering, for they are packed together with positive nuclei und must be held in
equilibrium by forces of a different order of magnitude from those who bind the
external electrons. It is obvious… that the number of electrons cannot be exactly half
the atomic weight in all cases. This has led to an interesting suggestion by van den
Broek that the number of units of charge on the nucleus, and consequently the
number of the external electrons, may be equal to the number of the elements when
arranged in order of increasing atomic weight. On this view, the nucleus charges of
hydrogen, helium and carbon are 1, 2, and 6 respectively, and so on for the other
elements, provided there is no gap due to a missing element.
Bohr has taken this view in his theory of the constitution of simple atoms and
molecules.
This apt quotation involves almost the nucleus of the atomic nucleus theory and its
implications for Bohr’s shell model and the periodic table of elements.
Indeed, Rutherford was convinced that he discovered the major feature of atomic
structure in 1911. Without Rutherford there would not be nuclear physics!
1903: Watts discovered the proportionality of square root of
frequency and atomic weight
10 years before Rutherford and Moseley did research the X-ray spectra of the
elements, the British physicist Watts published his research results in the same
journal as Rutherford and Moseley later did, namely in the Philosophical Magazine in
1903 ((6) 5, 203).
Watt’s findings had a broad resonance in the scientific community, Chwolson
appreciated the remarkable findings of Watts and similar of Ramage and Hartley in
his [chw], vol. II, p. 459. In the chapter Spektralanalyse Chwolson listed the
following chemical groups:
I. Li, Na, K, Rb, Cs; II. Cu, Ag; III. Mg, Ca, Sr, Ba; IV. Zn, Cd, Hg; V. Al,
In, Tl, Ga.
Chwolson noticed some remarkable characteristics: For every group with increasing
atomic weight the spectral lines move over to the red end of the spectrum. In the
exchange from one group into the following, a violet-shift of the spectral series was
observed.
And for any group the difference of the frequency between two double-lines or
between the two first lines of triplet-lines is proportional to the square of the atomic
weight.
Rutherford and Moseley did not mention these reports. Rutherford’s objective was to
find empirical correlations that support his dogma, namely that there is a nucleus with
Z protons, Z being the atomic number, and that physical properties depend on Z and
not on atomic weights.
MOSELEY’S X-RAY SPECTRA of the ELEMENTS
See article on Moseley’slaw
References
[koe] Koehler ,http://www.rwc.uc.edu/koehler/biophys/7c.html
[chw] Chwolson O. D., Lehrbuch der Physik, Braunschweig 1904
[gei] Geiger, H., The Scattering of the α-Particles, Proc. Roy. Soc.,vol A83, p. 492
(1910)
[geim] Geiger/Marsden, The Laws of Deflection of α-Particles trough Large
Angles, Philos.Mag. series 6, vol. 25, no 148 (1913)
[fow] Fowler, M.,
(http://galileo.phys.virginia.edu/classes/252/Rutherford_Scattering/Rutherford_Scatt
ering.html
[ma] Marinsek, J. Non-convertibility of Inertial Mass into Energy and Vice Versa.
Conjectures Regarding an e+e–-Pair–Cluster Atom Model in Connection with an
e+e–-Cosmic Lattice. Proceedgs 2nd Cologne Workshop “Physics as a Science”
2000,
Journal of New energy, vol. 5, no. 3 (2001)
[pra] Prout, W., On the Relation between the Specific Gravities of Bodies in their
Gaseous State and the Weights of their Atoms, Ann. Philosophy, 6, p. 321, 1815,
Also in:
[prb] Prout, W., Correction of a Mistake in [17], Ann. Phil. 7, p 111, 1816
[rua] Rutherford, E., The Structure of the Atom, Phil. Mag., series 6, vol. 27, 1914,
p.48
[rub] Rutherford, E. Bakerian Lecture: Nuclear Structure of Atoms, Proc. Roy. Soc.
A, 97, 374, 1920
[giu] Giunta C., http://webserver.lemoyne.edu/faculty/giunta/ Here Prof. Giunta has
placed the classic papers of: Aston, Avogadro, Cannizzaro, Dalton, Doebereiner,
Gay-Lussac, Petit/Dulong, Marignac, Mendeleev, Newlands, Prout,
Ramsay, Rutherford, Stas,…
[nur] Nuclear Reactions, www.nidlink.com/~jfromm/history/nuclear2
[lbl] www.lbl.gov/abc/basic