EXPLANATION OF THORIUM IONIZATIONS

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

June 29 , 2015

 

INTRODUCTION

Thorium is a chemical element with symbol Th and atomic number 90. However 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 many electon atoms related to the 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. (EXPERIMENTS REJECT 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. 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. Under this condition we may use this correct image of Thorium including the following ground state electron configuration:

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f ¹⁴5d¹⁰6s²6p_x²6p_y²6p_z²6d¹6d¹7s²

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

E₁ = 6.3 , E₂ = 11.9 ,  E₃ = 20 ,  E₄ = 28.8 , and E₅ = 65  .  

For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008.  

EXPLANATION OF E₁ = 6.3 eV = -E(7s²) + E(7s¹)

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

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹⁴5d¹⁰6s²6p_x²6p_y²6p_z²

screen the nuclear charge (+90e) and for a perfect screening the electrons of 7s²  and 6d² would provide an effective Z_eff  = ζ = 4. However the electrons of 7s² and 6d¹ and 6d¹ penetrate the electrons of 6p⁶ and lead to the deformation of electron clouds. Therefore  ζ  > 4.

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 = 6.3  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  - 304.6  =  0

and solving for ζ we get ζ = 5.4 > 4 .

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  = 11.9 eV = -E(7s¹)

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

(1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶ 4f¹⁴5d¹⁰6s²6p_x²6p_y²6p_z²)

screen the nuclear charge (+90e)  and for a perfect screening we would have the same effective  ζ = 4.  However the one electron of 7s breaks the symmetry  and leads to the  greater deformation of electron clouds. Consequently ζ > 5.4 > 4.

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

E₂ = 11.9  eV = E(7s¹)  =  (-13.6057)ζ²/n²

Then using  n = 7  we get ζ = 6.55  > 5.4 > 4.

Here the ζ = 6.55 > 5.4 > 4 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 4f ¹⁴5d¹⁰6s²6p_x²6p_y²6p_z²)

screen the nuclear charge (+90e)  and for a perfect screening we would have the same effective  ζ = 4.  However after the two ionizations the one electron of 6d breaks more the symmetry  and leads to the  greater deformation of electron clouds. Thus ζ > 6.55 > 5.4 > 4.

Here the E(6d¹) represents the binding energy of the first 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  > 6.55 > 5.4 >  4.

Here the ζ = 7.28 > 6.55 > 5.4 > 4 means that after the ionizations the electrons of 6d break more the symmetry and lead to a greater deformation of electron clouds.

 

EXPLANATION OF E₄ = 28.8 eV = -E(6d¹)

As in the cases of E₁ , E₂ and E₃ the electron charges (-86e) of the 86 electrons of the following electron configuration

(1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶ 4f¹⁴5d¹⁰6s²6p_x²6p_y²6p_z²)

screen the nuclear charge (+90e)  and for a perfect screening we would have the same effective  ζ = 4.  However the second electron of 6d breaks more the symmetry  and leads to the  greater deformation of electron clouds. Thus ζ  >7.28 > 6.55 >5.4 > 4.

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

E₄ = 28.8  eV = E(6d¹)  =  (-13.6057)ζ²/n²

Then using  n = 6  we get ζ = 8.73  > 7.28 > 6.55 > 5.4 > 4.

Here the ζ = 8.73 > 7.28 > 6.55 > 5.4 > 4 means that after the  ionizations the second electron of 6d breaks more the symmetry and leads to a greater deformation of electron clouds.

EXPLANATION OF E₅  = 65 eV = -E(6p_x²) + E(6p_x¹)

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

(1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶ 4f¹⁴5d¹⁰6s²)

screen the nuclear charge (+90e) and for a perfect screening the electrons of 6p⁶  would provide an effective ζ = 10. However the electrons of 6p⁶ repel the electrons  of 6s² and 5d¹⁰ and lead to the deformation of electron clouds. Thus ζ  > 10.

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   E₅ = 65  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  - 2335.9  =  0

and solving for ζ we get ζ = 13.7 > 10 .