EXPLANATION OF ACTINIUM IONIZATIONS

By Prof. L. Kaliambos (Natural Philosopher in New Energy)

March 11, 2023

Today it is well-known that the Nobel prize in physics (2022) confirmed Newton's third law of instantaneous interaction (faster than light) and rejected Maxwell's and Einstein's invalid fields, as I presented them at an international conference on physics in 1993 and at a nuclear conference at NCSR "Democritos" in 2002. (2022 NOBEL PRIZE WINNERS proved Einstein WRONG). Historically in 1925 the two young Dutch physicists Uhlenbeck and Goudsmit discovered the electron spin according to which the peripheral velocity of a spinning electron is faster than the speed of light. Since this discovery invalidates Einstein’s relativity it met much opposition by physicists including Pauli. Nevertheless surprisingly in my paper "Nuclear structure is governed by the fundamental laws of electromagnetism" (2003) I showed that the peripheral velocity (u) of the electron spin is greater than light ( u >> c ), which explains all atomic and molecular phenomena. Therefore I published my paper "SPIN-SPIN INTERACTIONS OF ELECTRONS AND ALSO OF NUCLEONS CREATE ATOMIC MOLECULAR AND NUCLEAR STRUCTURES (2008) for finding the ground state energies of all atoms. For example I have found a closed- form solution to the Schrodinger equation for the helium atom, based on the stronger magnetic attraction than the electric repulsion of two electrons of opposite spin at very small distances. Whereas in the dominant article "HELIUM ATOM-WIKIPEDIA" under the ACADEMIC ESTABLISHMENT OF WRONG THEORIES we read incorrectly the following various approximations: "Unlike for hydrogen, a closed-form solution to the Schrödinger equation for the helium atom has not been found. However, various approximations, such as the Hartree–Fock method, can be used to estimate the ground state energy and wavefunction of the atom."

 INTRODUCTION

Despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of elements related to the atoms of many electrons and of chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin.(Faster than light). So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008) which supplied the clue that resolved this puzzle.

Actinium is a chemical element with symbol Ac and atomic number 89. So after my paper of 2008 we may use this correct image of actinium with the following electron configuration:

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹⁰ 6s² 6p_x² 6p_y² 6p_z² 6d¹ 7s²

According to the “Ionization Energies (eV) of Atoms and Ions” the ionization energies  of actinium  ( from  E₁ to E₅ ) are the following:

E₁ = 5.17 , E₂ = 11.75 ,  E₃ = 20 ,  E₄ = 49 , and E₅ = 62  .  

For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS.

EXPLANATION OF E₁ = 5.17 eV = -E(7s²) + E(7s¹)

Since 7s² and 6d¹ appear with three valence electrons we suggest that the electron charges (-86e) of the 86 electrons of the following configuration

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f ¹⁴5d¹⁰6s²6p_x²6p_y²6p_z²

screen the nuclear charge (+89e) and for a perfect screening the electrons of 7s²  and 6d¹ would provide an effective Z_eff  = ζ = 3. However the electrons of 7s² and 6d¹ penetrate the electrons of 6p⁶ and lead to the deformation of electron clouds. Therefore  ζ  > 3.

Here the E(7s²) represents the binding energy of 7s², while the E(7s¹) represents the binding energy of 7s¹, which appears after the first ionization of 7s² .

Note that the 7s² consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write

-E(7s²) = -[(-27.21)ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand, since the 7s¹ consists of one electron, we apply the Bohr formula as

E(7s¹)  =  (-13.6057)ζ²/n²

Therefore   E₁ = 5.17  eV = -E(7s²) + E(7s¹)  = [(13.6057)ζ² - (16.95)ζ + 4.1)] / n²

Then using  n = 7  the above equation can be written as

(13.6057)ζ² - (16.95)ζ  - 249.23  =  0

and solving for ζ we get ζ = 4.95 > 3 .

Of course the two electrons of opposite spin (7s²) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT).

However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that two electrons with opposite spin exert a mutual Coulomb repulsion. Under such fallacious ideas I published my paper of 2008  .

 

EXPLANATION OF E₂  = 11.75 eV = -E(7s¹)

As in the case of E₁ the electron charges (-86e) of the 86 electrons of the following electron configuration

(1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶ 4f ¹⁴5d¹⁰6s²6p_x²6p_y²6p_z²)

screen the nuclear charge (+89e)  and for a perfect screening we would have the same effective  ζ = 3.  However the one electron of 7s breaks the symmetry  and leads to the greater deformation of electron clouds. Therefore ζ  > 4.95.

Here the E(7s¹) represents the binding energy of 7s¹, given by applying the Bohr formula as

E₂ = 11.75  eV = E(7s¹)  =  (-13.6057)ζ²/n²

Then using  n = 7  we get ζ = 6.5  > 4.95 > 3.

Here the ζ = 6.5 > 4.95 > 3 means that after the first ionization the one electron of 7s breaks the symmetry and leads to a greater deformation of electron clouds.

 

EXPLANATION OF E₃ = 20 eV = -E(6d¹)

As in the cases of E₁ and E₂ the electron charges (-86e) of the 86 electrons of the following electron configuration

(1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶ 4f¹⁴5d¹⁰ 6s²6p_x²6p_y²6p_z² )

screen the nuclear charge (+89e)  and for a perfect screening we would have the same effective  ζ = 3.  However the one electron of 6d breaks more the symmetry  and leads to the  greater deformation of electron clouds. Thus ζ  > 6.5 > 4.95 > 3.

Here the E(6d¹) represents the binding energy of 6d¹, given by applying the Bohr formula as

E₃ = 20  eV = E(6d¹)  =  (-13.6057)ζ²/n²

Then using  n = 6  we get ζ = 7.28  > 6.5 > 4.95 > 3.

Here the ζ =  7.28 > 6.5 > 4.95 > 3 means that after the  ionizations the one electron of 6d  breaks more the symmetry and leads to a greater deformation of electron clouds.

 

EXPLANATION OF E₄  = 49 eV = -E(6p_x²) + E(6p_x¹)

The electron charges (-80e) of the 80 electrons of the following configuration

(1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶ 4f¹⁴5d¹⁰ 6s²)

screen the nuclear charge (+89e) and for a perfect screening the electrons of 6p⁶  would provide an effective ζ = 9. However the electrons of 6p⁶ repel the electrons  of 6s² and 5d¹⁰ and lead to the deformation of electron clouds. Thus ζ  > 9.

Here the E(6p_x²) represents the binding energy of 6p_x², while the E(6p_x¹) represents the binding energy of 6p_x¹, which appears after the first ionization of 6p_x² .

Note that the 6p_x² consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write

-E(6p_x²) = -[(-27.21)ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand, since the 6p_x¹ consists of one electron, we apply the Bohr formula as

E(6p_x¹)  =  (-13.6057)ζ²/n²

Therefore   E₄ = 49  eV = -E(6p_x²) + E(6p_x¹)  = [(13.6057)ζ² - (16.95)ζ + 4.1)] / n²

Then using  n = 6  the above equation can be written as

(13.6057)ζ² - (16.95)ζ  - 1759.9  =  0

and solving for ζ we get ζ = 12 > 9 .

 

 EXPLANATION OF E₅ = 62  eV = -E(6p_y²) + E(6p_y¹)

As in the case of E₄ the electron charges (-80e) of the 80 electrons of the following configuration

(1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶ 4f¹⁴5d¹⁰6s²)

screen the nuclear charge (+89e) and for a perfect screening after the ionization the five electrons of 6p  would provide the same effective  ζ = 9. But the electrons of 6p after the first ionization of 6p_x² break the symmetry and provide an effective ζ  > 12 > 9.

Here the E(6p_y²) represents the binding energy of 6p_y², while the E(6p_y¹) represents the binding energy of 6p_y¹, which appears after the first ionization of 6p_y² .

Note that the 6p_y² consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write

-E(6p_y²) = -[(-27.21)ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand, since the 6p_y¹ consists of one electron, we apply the Bohr formula as

E(6p_y¹)  =  (-13.6057)ζ²/n²

Therefore   E₅ = 62 eV = -E(6p_y²) + E(6p_y¹)  = [(13.6057)ζ² - (16.95)ζ + 4.1)] / n²

Then using  n = 6  the above equation can be written as

(13.6057)ζ² - (16.95)ζ  - 2227.9  =  0

and solving for ζ we get ζ = 13.44  > 12 >  9.

Here the ζ = 13.44 > 12 > 9 means that after the ionization  the five electrons of 6p break more the symmetry and lead  to a greater deformation of electron clouds.