EXPLANATION OF XENON IONIZATIONS

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

June 10 , 2015

Xenon is a chemical element with symbol Xe and atomic number 54.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 Xenon including the following ground state electron configuration:

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p_x²5p_y²5p_z²

According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of xenon (from (E₁ to E₃ ) are the following:

E₁ = 12.13 ,  E₂ = 21.21,  and  E₃ = 32.12  .

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 “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” published in Ind. J. Th. Phys. (2008).

 

EXPLANATION OF E₁ = 12.13  eV = -E(5p_x²) + E(p_x¹)

Here the E(5p_x²) represents the binding energy of 5p_x², while the E(5p_x¹) represents the binding energy of 5p_x¹.The charges (-48e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰4s² 4p⁶4d¹⁰5s²) screen the nuclear charge (+54e) and for a perfect screening we would have ζ = 6.

Note that the 5p_x² consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008  we write

-E(5p_x²) = -[(-27.21)ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand, since the 5p_x¹ consists of one electron, we apply the Bohr formula to write

E(5p_x¹)  =  (-13.6057)ζ²/n²

Therefore   E₁ = 12.13  eV = -E(5p_x²) + E(5p_x¹)  = [(13.6057)ζ² - (16.95)ζ + 4.1)] / n²

Then, using  n = 5  the above equation can be written as

(13.6057)ζ² - (16.95) ζ  - 299.15  =  0

However, solving for ζ we get ζ = 5.35  < 6, which cannot exist, because the three pairs of 5p make a complete spherical sub-shell leading to a perfect screening with ζ = 6. Under this condition we expect to determine n > 5 . Under this condition we write 

E₁ = 12.13  eV = -E(5p_x²) + E(5p_x¹)  = [(13.6057)6² - (16.95)6 + 4.1)] / n²

Then, solving for n we get n = 5.7 > 5

 Of course the two electrons of opposite spin (5p_x²) 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 it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008.

 

EXPLANATION OF E₂ = 21.21  eV = -E(5p_y²) + E(p_y¹)

Here the E(5p_y²) represents the binding energy of 5p_y², while the E(5p_y¹) represents the binding energy of 5p_y¹. As in the case of E₁ the charges (-48e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²) screen the nuclear charge (+54e) and for a perfect screening we would have ζ = 6.

Note that the 5p_y² consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008  we write

-E(5p_y²) = -[(-27.21)ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand, since the 5p_y¹ consists of one electron, we apply the Bohr formula to write

E(5p_y¹)  =   (-13.6057)ζ²/n²

Therefore   E₂ = 21.21  eV = -E(5p_y²) + E(5p_y¹)  = [(13.6057)ζ² - (16.95)ζ + 4.1)] / n²

Since n = 5.7  the above equation can be written as

(13.6057)ζ² - (16.95)ζ  - 685  =  0

Then solving for ζ we get ζ = 7.75  > 6

Here we see that in the absence of one electron the electrons of 5p break the symmetry and lead to the deformations of electron clouds. Thus ζ = 7.75  > 6.

EXPLANATION OF E₃  = 32.12 eV = -E(5_z²)  + E(5_z¹)

Here the E(5p_z²) represents the binding energy of 5p_z², while the E(5p_z¹) represents the binding energy of 5p_z¹. As in the case of E₂ the charges (-48e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²) screen the nuclear charge (+54e) and for a perfect screening we would have ζ = 6.

Note that the 5p_z² consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008  we write

-E(5p_z²) = -[(-27.21)ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand, since the 5p_z¹ consists of one electron, we apply the Bohr formula to write

E(5p_z¹)  =   (-13.6057)ζ²/n²

Therefore   E₃ = 32.12  eV = -E(5p_z²) + E(5p_z¹)  = [(13.6057)ζ² - (16.95)ζ + 4.1)] / n²

Since n = 5.7   the above equation can be written as

(13.6057)ζ² - (16.95)ζ  - 1039.48  =  0

Then solving for ζ we get ζ = 9.39  > 6

Here we see that in the absence of two electrons the electrons of 5p break more the symmetry and lead to the great deformations of electron clouds.