EXPLANATION OF CERIUM IONIZATIONS

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

June 14 , 2015

Cerium is a chemical element with symbol Ce and atomic number 58.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 REJECTING EINSTEIN). 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 Cerium including the following ground state electron configuration:

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p_x²5p_y²5p_z² 4f¹5d¹6s²

According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of cerium (from (E₁ to E₆ ) are the following:

E₁ = 5.53 , E₂ = 10.85,  E₃ = 20.2,  E₄ = 36.76,  E₅ = 65.55, and  E₆ = 77.6 .

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 = 5.53  eV = -E(6s²) + E(6s¹)

Here the E(6s²) represents the binding energy of 6s², while the E(6s¹) represents the binding energy of 6s¹.

According to the  “ Cerium- WIKIPEDIA” the energy levels of the inner 5d¹ and 4f¹ are nearly the same as that of the outer of valence electrons. Thus the charges (-54e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶) screen the nuclear charge (+58e) and for a perfect screening we would have ζ = 4.

Note that the 6s² consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008  we write

-E(6s²) = -[(-27.21)ζ² + (16.95)ζ - 4.1 ] / n²

On the other hand, since the 6s¹ consists of one electron, we apply the Bohr formula to write

E(6s¹)  =  (-13.6057)ζ²/n²

Therefore   E₁ = 5.53  eV = -E(6s²) + E(6s¹)  = [(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  - 194.98  =  0

and solving for ζ we get ζ = 4.46  >  4.

This  means that the two electrons of 6s² along with the two electrons of 5d¹ and 4f¹ penetrate the  5p⁶ leading to the deformation of electron clouds.

Of course the two electrons of opposite spin (6s²) 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 = 10.85 eV =  -E(6s¹)

As in the case of E₁ the charges (-54e) of the 54 electrons of the following electron configuration   (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p_x²5p_y²5p_z²) screen the nuclear charge (+58e) and for a perfect screening we would have an effective ζ = 4. However the one electron of 6s¹ along with the two electrons of 5d¹ and 4f¹ penetrate the 5p⁶ and lead to the deformation of electron clouds. Thus we would have the same  ζ  > 4.

Here the E(6s¹) represents the binding energy of (6s¹) given by applying the Bohr formula as

E₂ = 10.85 eV = -Ε(6s¹) =  - ( -13.6057)ζ² /6²

Then solving for ζ we get ζ = 4.9  > 4 . Here the 4.9 > 4.46  > 4 means that the one electron breaks the symmetry and leads to a greater deformation of electron clouds.

EXPLANATION of E₃ = 20.2 eV = -E(5d¹)

Here the E(5d¹) represents the binding energy of the one electron (5d¹) given by applying the Bohr formula as

E₃ = 20.2 = -E(5d¹) = -(-13.6057)ζ² /n²

As in the two cases of E₁ and E₂ the charges  (-54e) of  the 54 electrons of the following configuration (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p_x²5p_y²5p_z²) screen the nuclear charge (+58e) and for a perfect screening we would have an effective ζ = 4. However in the absence of 6s² the one electron of 5d¹ along with the one electron of 4f¹ repel the 5p⁶ and lead to the deformation of electron clouds under n = 5. Thus ζ  > 4.

Therefore using n = 5 we get ζ =  6 > 4.  

EXPLANATION of E₄ = 36.76 eV = -E(4f¹)

Here the E(4f¹) represents the binding energy of the one electron (4f¹) given by applying the Bohr formula as

E₄ = 36.76 = -E(4f¹) = -(-13.6057)ζ² /n²

As in the three cases of E₁ , E₂ , and E₃ the charges  (-54e) of the 54 electrons of the following configuration  (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p_x²5p_y²5p_z²) screen the nuclear charge (+58e) and for a perfect screening we would have an effective ζ = 4. However (under n = 4) after the ionizations the one electron of  4f¹ brakes more the symmetry and leads to a greater deformation of electron clouds. Thus ζ  >> 4.

Then using n = 4 we get ζ = 6.57  > > 4. 

Here we observe a strange phenomenon, because the 4f¹ seems to be in inner orbital than that of the 5p electrons. In fact, the deformed electron cloud of one electron (4f¹) under n = 4 repels the electrons of 5p in such a way that the electron cloud of 4f¹ is always at the opposite position with respect to the deformed electron cloud of 5p.

 

EXPLANATION OF E₅ = 65.55 = -E(5p_x²)  +  E(5p_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 (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²) screen the nuclear charge (+58e) and for a perfect screening we would have ζ = 10.

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   E₅ = 65.55  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  - 1634.65 =  0

and solving for ζ we get ζ = 11.6 >  10.

Here ζ = 11.6  > 10  means that the electrons of 5p  repel the electrons of 5s² from symmetrical positions in order to provide an effective ζ = 11.6 > 10 which is a little greater than the perfect screening with ζ = 10 .

EXPLANATION OF E₆ = 77.6  eV = -E(5p_y²)  +  E(5p_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 E₅ the charges (-48e) of (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²) screen the nuclear charge (+58e) and for a perfect screening we would have ζ = 10.

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   E₆ = 77.6  eV = -E(5p_y²) + E(5p_y¹)  = [(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  - 1935.9  =  0

and solving for ζ we get ζ = 12.57  >  10.

Here ζ = 12.57  > 11.6  > 10 means that the electrons of 5p after the ionizations break  the symmetry for providing an effective ζ = 12.57 > 11.6 .