EXPLANATION OF INDIUM IONIZATIONS

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

June 5,  2015

Indium  is a chemical element with symbol In and atomic number 49. 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 Indium including the following ground state electron configuration:

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p¹

According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of  indium (from (E₁ to E₄ ) are the following:

E₁ = 5.79 ,  E₂ = 18.87,  E₃ = 28, and  E₄ = 54  .

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 = 5.79 eV = - E(5p¹)

Here the E(5p¹) represents the binding energy of 5p¹. The charges (-48e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰ 4s² 4p⁶4d¹⁰ 5s²) screen the nuclear charge (+49e) and for a perfect screening we would have ζ = 1. However the electron of 5p¹ repels the 4d¹⁰ and 5s² and leads to the deformations of electron clouds. Thus ζ > 1.  

Since 5p¹ consists of one electron, we apply the Bohr formula to write

E₁ = - E(5p¹)  = 5.79 eV =  - (-13.6057)ζ²/n²

Therefore   E₁ = 5.79 eV  = (13.6057)ζ² / 5²

Then  we get ζ = 3.26  > 1 .

 EXPLANATION OF E₂ = 18.87 eV = - E(5s²) + E(5s¹)

Here the E(5s²) represents the binding energy of 5s², while the E(5s¹) represents the binding energy of 5s¹. The charges  (-46e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰4s² 4p⁶4d¹⁰ ) screen the nuclear charge (+49e) and for a perfect screening we would have ζ = 3. However the electrons of 5s penetrate the 4d¹⁰ and lead to the deformations of  electron clouds. Thus ζ > 3. 

Note that 5s² consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008  we write

-E(5s²) = -[(-27.21)ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand, since 5s¹ consists of one electron, we apply the Bohr formula to write

E(5s¹)  =  (-13.6057)ζ²/n²

Therefore   E₂ = 18.87 eV = -E(5s²) + E(5s¹)  = [(13.6057)ζ² - (16.95)ζ + 4.1)] / n²

Since n = 5  the above equation can be written as

(13.6057)ζ² - (16.95)ζ  - 467.65  =  0

Then  solving for ζ we get ζ = 6.52  > 3 .

Of course the two electrons of opposite spin (5s²) 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. This situation 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 E₃ = 28  eV = -E(5s¹)  

 Here the E(5s¹) represents the binding energy of 5s¹. As in the case of E₂ the charges  (-46e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰4s² 4p⁶4d¹⁰ ) screen the nuclear charge (+49e) and for a perfect screening we would have ζ = 3. However the electron of 5s¹ penetrates the 4d¹⁰ and leads to the deformations of electron clouds. Thus ζ > 3.  

Since 5s¹ consists of one electron, we apply the Bohr formula to write

E₃ = - E(5s¹)  = 28 eV =  - (-13.6057)ζ²/n²

Therefore   E₃ = 28 eV  = (13.6057)ζ² / 5²

Then  we get ζ = 7.17  > 3 .

Here the 7.17  > 6.52 > 3 means that the one electron (5s¹) breaks more the symmetry of spherical shells than the two electrons (5s²) and leads to  greater deformations of spherical electron clouds. 

 

EXPLANATION OF   Ε₄  = 54  eV =  -E(4d¹⁰) + E(4d⁹)

Here the E(4d¹⁰) represents the binding energy of the 10 electrons (4d¹⁰), while the E(4d⁹) represents the binding energy of 9 electrons (4d⁹) .The charges (-36e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶ ) screen the nuclear charge (+49e) and for a perfect screening we would have an effective ζ = 13. However, in this case, the experiments of ionizations show that each pair of 4d repels not only the electrons of 4p⁶ but also the electrons of 4d. Surprisingly such repulsions occur in such a way that ζ < 13.

Note that 4d¹⁰ consists of five pairs (10 electrons with opposite spin)  Thus for the five pairs of opposite spin we apply my formula of 2008.  Under this condition one may write

-E(4d¹⁰) = -5[(-27.21) ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand the 4d⁹ consists of four pairs (8 electrons with opposite spin ) whose the binding energy is given by applying my formula of 2008, and of one electron whose the binding energy is given by applying  the Bohr formula. So we may write

E(4d⁹)  = 4[(-27.21)ζ² + (16.95)ζ - 4.1] / n²  + (13.6057)ζ² / n²

Therefore  Ε₄  = 54  eV =  -E(4d¹⁰) + E(4d⁹ )  = [(13.6057)ζ² - (16.95)ζ + 4.1] / n²

It is of interest to note that in this sub- shell of 4d the experiments of ionizations showed that  4 < n  < 5.  (See my paper EXPLANATION OF TECHNETIUM IONIZATIONS).

Thus using  n = 4 one writes

(13.6057)ζ² - (16.95)ζ  - 859.9  = 0

and solving for ζ we get ζ = 8.6 <  13

Whereas  using n = 5 one writes

(13.6057)ζ² - (16.95)ζ  - 1345.9 = 0

and solving for ζ we get ζ = 10.59  < 13 .