EXPLANATION OF KRYPTON IONIZATIONS

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

May 27 ,  2015

Krypton is a chemical element with symbol Kr and atomic number 36. 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 image of Krypton including the following ground state electron configuration: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶.

According to the “Ionization energies  of the elements-WIKIPEDIA” the ionization energies (eV) of krypton (from (E₁ to E₂₆) are the following:

E₁ = 13.99,  E₂ = 24.36,  E₃ = 36.95 , E₄ = 52.5 ,  E₅ = 64.7,  E₆ = 78.5,  E₇ = 111,  E₈= 125.8,  E₉ = 230.85, E₁₀ = 268.2, E₁₁ = 308 , E₁₂ = 350,  E₁₃ = 391, E₁₄ = 447 , E₁₅ = 492,  E₁₆ = 541,  E₁₇ = 592,  E₁₈ = 641,  E₁₉ = 786,  E₂₀ = 833 , E₂₁ = 884  E₂₂ = 937, E₂₃ = 998 , E₂₄ = 1051,  E₂₅ = 1151,  E₂₆ = 1205.3 

Firstly we examine the -( E₁ + E₂ + Ε₃ + Ε₄ + Ε₅ + Ε₆ ) = -271  =  E(4p⁶)

Here the E(4p⁶) represents the binding energy of the six outermost electrons.

Then, we observe that

-( Ε₇ + E₈ ) = - 236.8  = E(4s²)

- ( E₉  + … + E₁₈ )  = - 4261 = E(3d¹⁰).

-(E₁₉ +…+ E₂₄)  = - 5489 = E(3p⁶)

-(E₂₅ + E₂₆ ) =  -2356.3  =  E(3s²)

It is of interest to note that in the absence of data (from E₂₇ to E₃₆) one can write  

-( E₂₇ +…+ E₃₂ ) = E(2p⁶)

-(E₃₃ + E₃₄) = E(2s²)

-(E₃₅ + E₃₆) = E(1s²)

Such situations are analogous to those of copper. ( See my EXPLANATION OF COPPER IONIZATIONS ).  For understanding better the 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 +…+ E₆ ) = -271 =  E(4p⁶)

Here the binding energy E(4p⁶)  of the six outermost electrons  is given by applying  my formula of 2008.  The charges (-30e) of the  electrons (1s²2s²2p⁶3s²3p⁶3d¹⁰4s²) screen the nuclear charge (+36e) and for a perfect screening we would have an effective ζ = 6. However the electrons 4p⁶ of the three orbitals  repel the 4s² leading to the deformation of 4s² . Thus ζ  > 6. Under this condition we may write

(E₁ +… + E₆)  = 271 eV = -E(4p⁶)  =  -3[(- 27.21 )ζ² + (16.95) ζ - 4.1 ] / n²

Since  n = 4 the above equation could be written as   

5.1 ζ² – 3.178 ζ  - 270.23  =  0

Then solving for ζ we get ζ = 7.6 > 6

 

EXPLANATION OF -( Ε₇ + E₈ ) = - 236.8  = E(4s²)   

Here the E(4s²) represents the binding energy of the two electrons with opposite spin given by applying my formula of 2008.   The charges (-28e) of the inner electrons (1s²2s²2p⁶3s²3p⁶3d¹⁰) screen the nuclear charge (+36e) and for a perfect screening we would have ζ = 8. However according to the quantum mechanics the two electrons of 4s² penetrate the 3d¹⁰ and lead to the deformation of 3d¹⁰ . Thus we must observe that ζ > 8. Under this condition we may write

(Ε₇ + E₈ ) =  236.8  = -E(4s²)  = - [(-27.21)ζ² + (16.95)ζ - 4.1] / n²

Since n = 4 we may rewrite

1.7ζ²  -1.059ζ  -236.54 = 0

Then solving for ζ we get ζ =  12.11  > 8

 

EXPLANATION OF  ( E₉  + … + E₁₈ )  =  4261 eV = -E(3d¹⁰)

Here the E(3d¹⁰) represents the binding energy of the 10 electrons (3d¹⁰) . Note that the 10 electrons create five orbitals with electrons of opposite spin given by applying my formula of 2008. The charges (-18e) of the inner electrons (1s²2s²2p⁶3s²3p⁶) screen the nuclear charge (+36e) and for a perfect screening we would have ζ = 18. Under this condition we may write

 E₉ +..+ E₁₈ ) = 4261 eV  = - E(3d¹⁰) =   -5[(-27.21)ζ² + (16.95)ζ - 4.1] / n²

So  using  n = 3 we may rewrite

15.1167ζ²-  9.4167ζ  - 4258.72 = 0

Surprisingly solving for ζ we get ζ = 17.01  < 18 , which cannot exist . In fact, the 10 electrons of the five orbitals make a complete spherical sub- shell leading to a perfect screening with ζ = 18. Thus using ζ = 18 we expect to determine n > 3. Under this condition we may write

E₉ +..+ E₁₈ ) = 4261  eV = -E(3d¹⁰) =  -5[(- 27.21)18² - (16.95)18 + (4.1)] / n²

 Then solving for n we get n = 3.16  > 3 . 

 

EXPLANATION OF ( E₁₉ + … + E₂₄ ) = 5489 eV  =  -E(3p⁶)

Here the E(3p⁶) represents the binding energy of the 6 paired electrons given by applying my formula of 2008.  The charges (-12e) of the twelve inner electrons like (1s²2s²2p⁶3s² ) screen the nuclear charge (+36e) and for a perfect screening we would have an effective Z_eff  =  ζ = 24. However the 6 paired electrons  ( 3p⁶ ) repel the 3s² electrons and lead to the deformation of shells with ζ  > 24. Under this condition we may write

( E₁₉ +…+ E₂₄)  = 5489  eV = -E(3p⁶) =  -3[(-27.21)ζ² +(16.95)ζ - 4.1] / n²  

 Since n = 3   the above equation can be written as  

9.07ζ²  - 5.65ζ  - 5487.63 = 0

Then, solving for ζ we get ζ = 24.91 > 24  .

 

EXPLANATION OF ( E₂₅  +  E₂₆ ) = 2356.3  eV = -E(3s²)

Here the E(3s²) represents the binding energy of the two paired electrons (3s²) given by applying my formula of 2008.  The charges (-10e) of the inner 10 electrons of 1s².2s².2p⁶ screen the nuclear charge (+36e) and for a perfect screening we would have an effective  ζ = 26. However the  two electrons of 3s² penetrate the 2p⁶ leading to the deformation of shells with  ζ  > 26. Under this condition we write the following equation as

( E₂₅ + E₂₆ ) = 2356.3  eV  = - E(3s²) =  - [(- 27.21)ζ²  + (16.95) ζ  - 4.1 ] / n²

Since n = 3 we may write

3.0233ζ²  - 1.8833ζ  - 2355.84  = 0

Then solving for ζ we get ζ  = 28.23  > 26 .

Note that the two electrons of opposite spin (4s²) 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.

However 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.