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New Bohr's worlds (atoms, molecules,bonds.) (13/6/6 new!)
♦ End of Quantum mechanics ! ( Fine structure, Sommerfeld, General relativity ) (12/ 12/30 new!)
♦ Bell inequality violation is an illusion. ( Delayed choice, Quantum computer is real ? ) (13/ 1/29 new!)
♦ End of Quantum field theory (QFT). (Dirac equation, Quantum electrodynamics, Higgs ) (13/ 1/20 new!)
Why they never tell the truth ?
Journal Nature's Bohr model. (13/6/18)
The quantum mechanical world is very strange.
The wave functions can only show "probability density" of the particle.
So the moment we try to observe the particle, the parobability wave spreading all over the space reduces to a point particle (= wavefunction collapse ).
Of course, this process of wavefunction "collapse" is faster-than-light.
(Fig.1) A single C60 buckyball interferes with itself !
Even a single C60 buckyball molecule can interfere with itself.
To explain this interference of a single particle in the double-slit experiment, the quantum mechanics needs "many-world" interpretation (= "superposition" ) to avoid faster-than-light collapse.
( See also Dismissing Bohr model means parallel worlds.)
Bohr model, which is based on real de Broglie wave, can explain this interference naturally, as shown in Fig.1 left.
And a single accelerating Bohr's electron does NOT radiate energy, as shown in this page.
(Fig.2) Dirac's hydrogen = Bohr-Sommerfeld model.
The important point is that the fine structure of the hydrogen-like atoms was first obtaied by Sommerfeld using Bohr's orbit ( ).
Later, Dirac equation could get the same solutions as Bohr-Sommerfeld model. ( See this page. )
The problem is that Dirac's hydrogen contains many accidental coincidenses such as 2s1/2 = 2p1/2, 3s1/2 = 3p1/2, 3p3/2 = 3d3/2 , .....
And most serious thing is that Dirac's hydrogen includes many fictional states such as 1p1/2, 2d3/2, 2f5/2.... (Eq.1). ( See this page. )
For example, 1P1/2 state is n=1 energy state and l=1 angular momentum, so 1P1/2 state does NOT exist in the quantum mechanical hydrogen atom.
Though Dirac's hydrogen clearly aimed at getting the same solutions as Bohr-Sommerfeld fine structure, ordinary textbooks do NOT say about this fact except for a few books .
This is strange.
(Fig.3) Schrodinger's hydrogen = Bohr Sommerfeld model.
Furthermore, Schrodinger's hydrogen also obeys an integer times de Broglie wavelength of Bohr-Sommerfeld quantization, as shown in this page.
In Schrodinger's hydrogen atom, the probability density of the electron always spreads into all over the space from zero to infinity.
(Fig.4) Schrodinger's 2P "radial" wave function (angular momentum = 1).
For example, in Schrodinger's 2P states, the radial kinetic energies (= 1/2mv2 ) become minus in both sides as shown in Fig.4.
( In 2P state, the energy level is n = 2 and the angular momentum is L = 1. )
This strange property has NO relation with tunnel effect.
Because in the region from 0 to a1 of Fig,4, the potential energy is lower than the total energy.
To cancel the increased tangential kinetic energy, radial kinetic energy has to be minus. ( See this page. )
These unreasonable states clearly show Schrodinger's hydrogen atom is wrong.
(Fig.5) Angular momentum zero = penetrating or rebounding ?
If Schrodiner's hydrogen includes circular orbit (= radial kinetic energy is zero ), the radial wavefunction becomes "constant".
If so, the normalization of the radial wavefunction diverges to infinity, unless this constant is zero.
To avoid this infinity, Schrodinger's radial kinetic energy must NOT be zero, which leads to the "zero" angular momentum in "S" state.
But as shown in Fig.5, this angular momentum zero states are impossible in this real world.
The electron always penetrates or rebounds at the nucleus, which makes the helium atom very unstable.
And the 3s electron of sodium has to pass through the n=2, n=1 inner electrons and nucleus.
( In this very hard circumstance, delicate anomalous Zeeman effect is impossible. )
(Fig.6) Spinning speed of electron ?
The ground state (= 1s ) of the hydrogen atom includes the magnetic moment (= Bohr magneton ).
So to explain this magnetic moment in the zero angular momentum state, they introduced a very strange concept of electron spin.
As shown in this page, if this electron spin means real rotating, its spinning speed must be more than 100 times speed of light.
And the spinning electron can NOT return to its original configuration by one revolution. ( It needs two revolutions. )
(Fig.7) Electron spin (= Bohr magneton ) = circular movement (= Bohr magneton ) ?
As shown in this page, the magnetic moment means the current I × the area of the current loop (= πr2 ).
Surprizingly, the quantum mechanical spin also has the same magnetic moment (= Bohr magneton ) as Bohr's ground state circular orbit.
This trick exists in the electron spin g factor = "2".
As a result, the magnetic moment = g-factor × angular momentum = 2 × 1/2 = 1 (= Bohr magneton ).
The important point is that this Bohr magneton originates in Bohr's orbit, and can NOT be gotten from the quantum mechanics itself.
(Fig.8) Probability density = Many-world theory.
Schrodinger's wavefunction can NOT give the clear motion of the particle.
If we try to get the concrete motion using Coulomb force and de Broglie relation, it becomes Bohr-Sommerfeld model.
( Caution: the quantum mechanics ALSO uses de Broglie relation, without which the electron can come close to nucleus as it likes. )
More important thing is that the quantum machanics including pilot-wave theory can NEVER show the concrete probability density and motion of the two elections of helium atom.
Because the three-body helium tends to be chaotic and cause autoionization easily, as shown this page.
So the quantum mechanics escaped from reality into the "superpostion" probability concept, which leads to strange many-world interpretation.
(Fig.9) Delayed choice experiment = Many-worlds = photon.
Photon particle is strange concept.
Even though photon has NO mass and charge, the quantum mechanics and QED try to admit this photon.
But as shown in Fig.9, if we admit photon particle, we have to accept strange many-worlds.
Because even when a single photon enters one of the two paths after the beam splitter 1 (BS1), it shows interference from both two paths (Fig.9).
As shown in this page, this experiment clearly demostrates the existence of the electromagnetic waves in both two paths.
So above some threshold of the electromagnetic wave's intensity, the photon is recognized at the single photon detector.
(Fig.10) Bell inequality violation is impossible in this real world.
You may often hear that Bell inequality violation denied hidden variable proposed by Einstein.
But unfortunately, there are not many books, which explain this Bell inequality simply.
This Bell inequality violation was found using a "photon" pair.
The moment a photon (= A ) passes through the filter A and turns its polarization in the direction of filter A, the polarization of photon B also becomes the same as filter A, no matter how far they are apart from each other ( See this page ).
This process includes three impossible and convenient assumptions.
First, the existence of filter is too special, though the filter is only one of infinite things in the universe.
Second, if the photon is a rigid and point particle, it can NOT polarize and satisfy Malus law (= cosθ ).
Third, this clearly contradicts relativistic theory including relativistic quantum field thoery.
As shown in this page, if we use the classical electromagnetic waves, we can explain this violation as local and realistic phenomenon.
( We should NOT get away from "reality" so easily. )
(Fig.11) Electron transport system shows electron is "really" moving.
Our human body needs oxygen (O2) in the respiratory system.
Inside the mitochondria, coenzyme NADH is divided into H+ and an electron.
This electron is transferred through various complex enzymes and and coenzyme Q to the oxygen molecules.
Of course, in this process, the electron really exists and is actually moving, NOT a unrealistic electron cloud.
If electrons are NOT moving, we can NOT live.
(Fig.A) Which is the true circumference of disc ??
In Fig.A, the observer A is in the hole at the center of the disc.
N short sticks, which each length is l0, are put along the circumference of the disc.
When the disc is at rest (= Fig.A, left ), the circumference of the disc is 2πR = Nl0.
But when the disc is rotating like Fig.A, right , each stick is Lorentz-contracted, which means 2πR is NOT consistent with Nl0.
( The redius "R" is NOT contracted, because the moving direction is tangential. )
This fatal paradox (= Ehrenfest paradox ) has NOT been solved.
As shown in this page, special relativity can NOT deal with two directions, which causes fatal paradoxes.
Various fatal paradoxes of special relativity indicates relativistic quantum field theory, QED and standard model are wrong.
As shown in this page, virtual particles violate relativity.
And the anomalous magnetic moment (= g-factor ) of particles can be manipulated artificially ( see this page. )
As shown in this page ( Bohr model hydrogen ), the orbital length of Bohr model is just an integer times de Broglie wavelength.
This concept agrees with the Davission and Germer experiment .
(Fig.12) Davisson-Germer experiment showed an electron is de Broglie wave.
In Davisson and Germer experiment, an electron is accelerated by the electric field and reflected by nickel crystal.
de Brolgie wavelength of an electron is gotten from the electric field strength and interference pattern of a single electron independently.
They agreed with each other (= de Broglie relation ).
This experiment showed an single electron has wave-like property , in which the opposite wave phases cancel each other, and expel the electron (Fig.12).
(Fig.13) de Broglie wave ends fit.
Not to cancel de Broglie wave, both ends of an electron wave in hydrogen atom need to agree with each other with respect to wave phases.
(Fig.14) Old Bohr's helium.
In old Bohr's helium (Fig.14), two electrons are moving at the opposite sides of the nucleus on the same circular orbit (= one de Broglie wavelength ).
Considering Fig.12 and Fig.13, old Bohr's helium makes two electrons of helium unstable due to de Broglie wave destructive interference.
Actually old Bohr's helium of Fig.14 gives wrong ground state energy of helium.
In circular orbit of helium, equating the centrifugal force to the Coulomb force, we have
where r is the circular orbital radius (meter), e is the electron charge
The circular orbital length is supposed to be an integer (n) times de Broglie's wavelength of the electron, we have
where h/mv means the de Broglie's wavelength.
The total energy E of the helium is the sum of the kinetic and the Coulomb potential energy of the two electrons, so
Solving the above three equations (Eq.2-4), the ground state energy (n=1) becomes - 83.33 eV.
This value is lower than the experimental value -79.005 eV.
Furthermore when some atoms such as oxygen come close to old Bohr's helium, its wave phases easily become chaotic due to the external force of other atoms, which is inconsistent with strong stability and closed shell property of helium atoms.
(Fig.15) Various old Bohr's helium atom.
In 1910s - 1920s, Lande (= outer and inner orbits, Fig.15A), Langumuir (= two parallel orbits, Fig.15B, two linear oscillating orbits, Fig.15C), Kramers (= 120 degree angle crossed orbits, Fig.15D ), and Heisenberg (= coplanar and inclined orbits, Fig.15E,F ) were trying to get better Bohr's helium atom.
But No old helium models could explain the correct ground state energy, stability, and closed shell property of helium atom.
Because about that time, they did not have computers to calculate three-body helium atom.
(Fig.16) Autoionization in the classical helium atoms
As shown in this page, helium has two electrons and one nucleus, so almost all cases show unstable autoionization (= one of the two electrons goes to infinity ) in the classical helium atoms.
( This classical helium does not consider de Broglie waves.)
If the helium structure has not been defined, the development of all the physics and chemistry would have stopped at that point.
In 1928-1930, Hylleraas succeeded in getting the approximate value of the helium ground state energy using the Schroedinger equation.
The quantum mechanical method has an advantage to be able to calculate approximate values without computers.
But the quantum mechanics can NOT show concrete probability and particle's state of two electrons of helium.
(Eq.5) Quantum mechanical helium
When Z=2, Eq.5 means the Hamiltonian of the helium atom. ( Δ = ∇2 )
When we try to get the correct value of the ground state energy of the helium atom, we have to use more than a thousand- terms variational functions of
(Eq.6) Quantum mechanical helium.
Of course, we can NOT imagine the helium's actual configuration from these many mathematical terms of Eq.6.
And this latest calculation value is -79.015 eV , which is a little different from the experimental value of -79.0051 eV.
On the other hand, our new Bohr's helium atom as shown in the latter section can get more correct answer of -79.0035 eV.
( This little difference between 79.0035 eV and 79.0051 eV is what we call "relativistic" effect, which value is good. See also this page. )
Also in other atoms and molecules ( see this page ), quantum mechanical methods can not give clear state of electrons, which obstructs the development of nanotechnology, and is the main reason why many-world concept becomes dominant.
(Fig.17) Accelerating electron radiates energy ??
Ordinary textbooks often say an accelerating electron of classical Bohr's orbit radiates and loses energy, falling into the nucleus.
If you see this explanation carefully, you will find this insistence is based on wrong assumption.
It uses Poynting vector (= E × H ) as the energy flow.
This Poynting vector is equal to the change of the electric and magnetic energies in the vacuum.
This electric energy in the vacuum (= 1/2εE2 ) means the potential energies needed to gather minus (or plus) small charges to the sphere (= Fig.17 left ).
But a single electron is NOT made from smaller charges. ( A single electron is the smallest charge. )
It means the vacuum electric energy in a single electron is NOT energy.
( Magnetic field is induced by the electric field, so it's NOT energy, either. )
So the Poyinting vector does NOT mean energy flow in a single electron case.
As a result, unless the interactions between other charged particle change, it does NOT radiate.
( See also this page and this page. )
(Fig.18) Stable and independent two de Broglie waves. (= perpendicular to each other.)
To avoid the problems of vanishing de Broglie's wave in the upper section, we suppose another model as shown in Fig.18.
As shown in Fig.18, this new Bohr's helium consists of two electron orbits which are perpendicular to each other.
If the two orbits are perpendicular to each other, their wave phases are independent from each other and stable.
If the electron tries to obey repulsive Coulomb force completely and lay down its orbit to Fig.14, the destructive interference of their de Broglie waves expel the electron. ( Of course, these wave phases are chaotic, and unstable in this case. )
( See also this page. )
(Fig.19) New Bohr's helium (= A.) is not electrically polarized.
As you know, helium atom does NOT form compounds with other atoms, and has the lowest boiling point in all atoms.
Unfortunately, the quantum mechanical electron spin has NO power to stop forming compounds, because the magnetic moment of spin is very weak in comparison with Coulomb force.
( Spin interaction = very tiny fine structure level. )
So ONLY de Broglie waves is left to explain this important property of helium.
As shown in Fig.19 left, when the two electron orbits are perpendicular to each other, the space around 2e+ nucleus becomes just neutral.
( In this case, two negative electrons are equally distributed around the 2e+ nucleus both in vertical and horizontal directions. )
And due to the destructive interference of de Broglie wave, these states continue even when other atoms come close to helium.
In other helium models, the space is electrically polarized, and their wave phases easily become chaotic when other atoms are close to them.
Of course, there is no space for the third electron to enter in Fig.18 model (= Pauli exclusion principle ).
(Fig.20) Two same-shaped orbital planes are perpendicular to each other.
Next we calculate the new helium using simple computer program.
Fig.20 shows one quarter of the orbits.
Electron 1 starts at (r1, 0, 0), while electron 2 starts at (-r1, 0, 0).
(Fig.21) The two electrons have moved one quarter of their orbitals.
In Fig.21, the electron 1 is crossing y axis perpendicularly, while electron 2 is crossing z axis.
When the two orbits are crossing perpendicularly, the motion pattern as shown in Fig.20 and Fig.21 is the most stable one.
( If the electrons are moving like Fig.20 and 21, the potential energy becomes the lowest. )
Here we investigate how the electrons of the helium are moving by calculating the Coulomb force among the two electrons and the nucleus at short time intervals.
The computer programs of JAVA ( version 1.5.0 ), simple C languages and Python ( 2.7 ) to compute the electron orbit of the helium are shown in the link below.
Sample JAVA program
C language program
As shown in Fig.20 and Fig.21, the helium nucleus is at the origin.
The electron 1 initially at ( r1, 0, 0 ) moves one quarter of its orbital to ( 0, r2, 0 ), while the electron 2 initially at ( -r1, 0, 0 ) moves to ( 0, 0, r2 ).
As meter and second are rather large units for measurement of atomic behavior, here we use new convenient units
(Eq.7) New units of time and length.
From Eq.7, the accelaration is
If you copy and paste the above program source code into a text editor, you can easily compile and run this.
First we input the initial x-coordinate r1 (in MM) of electron 1 (see Fig.20 ), and the absolute value of the total energy E (in eV) of helium.
From the inputted values, we calculate the initial velocity of the electron 1.
And at intervals of 1 SS, we compute the Coulomb force among the two electrons and the nucleus.
When the electron 1 is at ( xx, yy, 0 ), the electron 2 is at ( -xx, 0, yy ) (in MM). (See Fig.20 and Fig.21.)
Change MM to meter as follows; x (m) = xx × 10-14. y (m) = yy × 10-14.
So the x component of the acceleration ( m/sec2 ) of the electron 1 is,
(Eq.9) x component of the acceleration.
where the first term is by the Coulomb force between the nucleus and the electron 1, and the second term is by the force between the two electrons.
Considering the helium nuclear mass (= alpha particle), we use here the reduced mass (= rm ) except when the center of mass is at the origin.
See also reduced mass of three-body helium.
In the same way, the y component of the acceleration (m/sec2) is,
(Eq.11) y component of the acceleration.
Based on that calculation value, we change the velocity vector and the position of the electrons.
We suppose electron 1 moves only on the XY-plane, so the z component of the acceleration of the electron 1 is not considered.
If we consider all components of the Coulomb force against the electrons, the electron's motion becomes as shown in Fig.14 (= old Bohr's helium ).
But in Fig.14, the two electrons are packed in one orbit of one de Broglie's wavelength.
So their de Broglie wave phases are cancelled.
We also calculate de Broglie's wavelength of the electron from the velocity ( λ = h/mv ) at intervals of 1 SS.
The number of that wave ( λ in length ) contained in that short movement section ( the sum of them is WN ) is,
where (VX, VY) are the velocity of the electron 1 (in MM/SS), the numerator is the movement distance (in meter) for 1 SS. the denominator is de Broglie's wavelength (in meter).
Here, the estimated electron's orbit is divided into more than one million short segments for the calculation.
When the electron 1 has moved one quarter of its orbit and its x-coordinate is zero (Fig.21), this program checked the y-component of the electron 1 velocity ( last VY ).
When the last VY is zero, two electrons are periodically moving around the nucleus on the same orbitals as shown in Fig.20 and Fig.21.
So, only when -0.0001 < last VY < 0.0001 (MM/SS) is satisfied, the program displays the following values on the screen,
r1, VY, preVY ( VY 1 SS ago ), and (mid)WN ( the total number of de Broglie's waves contained in one quarter of the orbit ).
Table 1 shows the results in which the last VY is the closest to zero.
|E (eV)||r1 (MM)||WN||WN x 4|
WN × 4 is the total number of de Broglie's waves contained in one round of the orbital.
The experimental value of helium ground state energy is -79.005147 eV ( Nist, CRC 2012 ).
This result shows the relativistic correction to the energy = -79.005147 - (-79.0037) = -0.001447 eV.
The theoretical ground state energy value of the helium ion (He+) can be gotten from usual Bohr model or Schrodinger equation using the reduced mass.
This value is -54.41531 eV.
And the experimental value of He+ ground state energy is -54.41776 eV (Nist).
So the relativistic correction to the energy in He+ ion is -54.41776-(-54.41531) = -0.00245 eV.
The theoretical ground state energy value of the hydrogen atom (H) can be gotten from usual Bohr model or Schrodinger equation using the reduced mass, too.
This value is -13.5983 eV.
And the experimental value of H ground state energy is -13.59844 eV (Nist).
So the relativistic correction to the energy in hydrogen atom is -13.59844-(-13.5983) = -0.00014 eV.
The electron's velocity of the neutral helium atom is slower than helium ion, but faster than hydrogen atom.
So the relativistic correction in neutral helium atom should be between -0.00245 eV and -0.00014 eV.
The above calculation value of -0.001447 eV is just between them !
As a control program, we show the program of hydrogen-like atoms ( H and He+ ) using the same computing method as above. Try these, too.
JAVA program ( H or He+ )
C language ( H or He+ )
Here we use the new unit ( 1 SS = 1 × 10-23 second ) and compute each value at the intervals of 1 SS.
If we change this definition of 1 SS, the calculation results of the total energy (E) in which the orbital length is just one de Broglie's wavelength change as follows,
|1 SS = ? sec||Result of E(eV)|
|1 × 10-22||-79.00540|
|1 × 10-23||-79.00370|
|1 × 10-24||-79.00355|
|1 × 10-25||-79.00350|
This means that as the orbit becomes more smooth, the calculation values converge to -79.00350 eV.
The programs based on other 1 SS definition is as follows,
Sample JAVA program 1 SS = 1 × 10-25 sec, calculation takes much time.
Old sample JAVA program 1 SS = 1 × 10-22 sec--fast but the result and Eq.no are a little different
The standard helium model of the quantum mechanics has the spin-up and spin-down electrons.
So it seems to generate no magnetic field.
But to be precise, in all areas except in the part at just the same distance from the two electrons, magnetic fields are theoretically produced by the electrons even in the standard helium model.
( Because they do not stick to each other. )
So as the electrons move to cancel the magnetic field out, they lose energy by emitting electromagnetic waves.
Actually, one-electron atom hydrogen has the magnetic moment, two-electron atom helium has no magnetic moment.
So the standard quantum-mechanical helium model has self-contradiction.
In this new helium, the two symmetrical orbits crossing perpendicularly are wrapping the whole helium atom completely.
See also Why "solitary" helium doesn't form compounds ?.
( The Bohr model hydrogen which has only one orbit, can not wrap the direction of the magnetic moment completely. )
It is just consistent with the fact of the strong stability and the closed shell property of helium.
(Fig.22) Hydrogen and Helium atoms.
These orbits are all just one de Broglie's wavelength.
Surprisingly, this new atomic structure of Bohr's helium is applicable to all other two and three electron atoms ( ions ).
|Atoms||r1 (MM)||WN x 4||Circular orbit||Result (eV)||Experiment||Error (eV)|
Table 4 shows three electron atoms such as lithium.
|Atoms||r1 (MM)||WN x 4||Result (eV)||Experiment||Error (eV)|
About the calculation method, see this page.
In the standard model of the quantum mechanics (QM), it is said that the electrons are stable as electron clouds, which are not actually moving.
They say this is the reason why the electrons don't fall into the nucleus radiating energy in QM.
But if so, how do we explain about the relativistic corrections to the energy (caused by the high electron's velocity) and the use of the reduced mass ?
If we use the reduced mass of an electron, the calculation results of the hydrogen energy levels becomes more accurate.
Does this mean that the electron and the nucleus are actually moving around the center of mass ? So the quantum-mechanical model contains self-contradiction also in this subject.
And if Dirac's hydrogen is wrong the intepretation of Lamb shift needs to change.
This very little Lamb shift relies on the assupmtion of metastable 2S1/2 by the photon, which Willis Lamb himself did not believe.
See also de Broglie wave, Virial theorem, and Zeeman effect.
In the neutral neon (Ne), how are eight outer shell (valence) electrons moving ?
Again the quantum mechanics cannot show any clear images at all.
The neutral neon has ten electrons, in which two 1S electrons are very close to the nucleus.
So "approximately" these eight outer electrons are moving around one +8e central nucleus (+10e-2e = +8e).
(Fig.23) Neutral Neon model
Considering repulsive Coulomb forces among eight valence electrons ( e0 - e7 ), these outer electrons are distributed like regular hexahedron.
In this hexahedron configuration, the outer 8 electrons are most stable.
But different from other atoms such as carbon, in Fig.23 model, the outer electrons of upper and lower crash into each other, while they are orbiting around the nucleus.
(Fig.24) Upper and lower electrons' crash.
As explained above, mysterious Pauli exclusion principle is caused by two 1s orbits, which are perpendicular to each other.
So also in noble gases such as Neon, this de Broglie wave's nature is naturally related.
To avoid this crash, using helium perpendicular de Broglie waves, the eight outer electrons' orbits are thought to be
(Fig.25) Neutral Neon's eight orbits of valence electrons.
In Fig.25, for example, the orbits of the electron 3 and 5 ( e3 and e5 ) have two common vertices, and perpendicular to each other, like
(Fig.26) Orbits perpendicular to each other.
In Fig.25 model, the eight valence electrons don't crash into each other.
And the eight valence electrons don't need to enter into a single common plane, when they are passing each other.
And each two electrons having two common vertices are perperndicular to each other like helium model above.
( Pairs are e1-e7, e3-e5, e2-e4, e0-e6. )
So the eight outer electrons have their own different orbits.
Considering the symmetric structure of polyhedron, and de Broglie wave's destructive interference, this eight outer electrons is the limit of n = 2 (= two de Broglie wavelength ) energy levels. (= Pauli exclusion principle. )
( Different from n = 1, the orbit can be elliptical due to radial de Broglie wave. )
Of course, about the Neon, this model is still an estimation.
So I'm glad if someone could find better and realistic Neon model.
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 P.A.M. Dirac, The Principles of Quantum Mechanics (Fourth Edition).
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PDF old version(2009/2) (The reduced mass isn't used in this old paper. And here, 1SS = 1 × 10-21 sec. So the calculation values are a little rough and different from the above correct values.)
2011/3/9 updated. Feel free to link to this site.