By Prof. L. Kaliambos (Natural Philosopher in New Energy)
May 15, 2015
Phosphorus is a chemical element with symbol P and atomic number 15. 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 Phosphorus including the following ground state electron configuration: 1s2.2s2.2px2.2py2.2pz2.3s2.3px1.3py1 .3pz1.
According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (in eV) are the following: E1 = 10.48669 , E2 = 19.7694 , E3 = 30.2027 , E4 = 51.4439 , E5 = 65.0251, E6 = 220.421, E7 = 263.57 , E8 = 309.60, E9 = 372.13 , E10 = 424.4 , E11 = 479.46 . E12 = 560.8. E13 = 611.74, E14 = 2816.91, and E15 = 3069.842 . Here the -( E1 + E2 + E3 ) is equal to the binding energy E( 3px1 + 3py1 + 3pz1) of the three outer electrons. Then the - ( E4 + E5) is equal to the binding energy E(3s2). Also, the -( E6 + E7 + E8 + E9 + E10 + E11 ) is equal to the binding energy E( 2px2 + 2py2 + 2pz2) . On the other hand the -( E12 + E13 ) is equal to the binding energy E(2s2) , while the -( E14 + E15) equals the binding energy E(1s2). 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 + E2 + E3 ) = -E(3px1 + 3py1 + 3pz1)
Here the E(3px1 + 3py1 + 3pz1) represents the binding energy of the three outer electrons with parallel spin (S = 1). The charges (-12e) of the electrons 1s2.2s2.2px2.2py2.2pz2.3ps2 screen the nuclear charge (+15e) and for a perfect screening we would have an effective Zeff = ζ = 3. However the three outer electrons (3px1+ 3py1 + 3pz1 ) repel the 3ps2 electrons and lead to the deformation of shells with ζ > 3. Under this condition I apply the Bohr formula in order to write the binding energy of the three outer electrons E(3px1 + 3py1 + 3pz1) as
( E1 + Ε2 + E3 ) = 60.459 eV = -E(3px1 + 3py1 + Epz1 ) = - 3( -13.6057)ζ2 / n2
Since n = 3 we get ζ = 3.65 > 3
EXPLANATION OF ( E4 + E5 ) = 78.6348 eV = -E(3s2)
Here the E(3s2) represents the binding energy of the two electrons (3s2) . The charges (-10e) of the electrons of 1s2.2s2.2px2.2py2.2pz2 screen the nuclear charge (+15e) and for a perfect screening we would have an effective ζ = 5. However the two electrons of 3s2 penetrate the 2px2.2py2.2pz2 leading to the deformation of shells with ζ > 5. Under this condition I apply my formula of 2008 to write the equation of the binding energy E(3s2) = - (E4 + E5 ) of the two electrons as
( E4 + E5 ) = 116.469 eV = - E(3s2) = - [(- 27.21)ζ2 + (16.95) ζ - 4.1 ]/n2
Since n = 3 we may write
3.0233ζ2 - 1.8833ζ - 116 = 0
Then, solving for ζ we get ζ = 6.51 > 5
EXPLANATION OF -(E6 + E7 + E8 + E9 + E10 + E11 ) = - 2069.58 = E(2px2 + 2py2 + 2pz2)
Here E(2px2 + 2py2 + 2pz2) represent the binding energy of the six paired electrons. The charges (-4e) of the four electrons 1s2.2s2 screen the nuclear charge (+15e) and for e perfect screening we would have ζ = 11, because +15e - 4e = + 11e. Thus for n = 2 we should determine the effective ζ = 11 or ζ > 11 of the total binding energy of 6 paired electrons by applying my formula of 2008 as
E (2px2 + 2py2 + 2pz2) = 3[(-27.21)ζ2 + (16.95)ζ - 4.1)] / n2 = - 2069.58
Surprisingly, solving for ζ we get ζ < 11 , which cannot exist. In fact, since 2px2 , 2py2 , and 2pz2 make a complete spherical shell they provide a perfect screening with ζ = 11. So using ζ = 11 and solving for n we expect to find n > 2, because a perfect screening after the experiments of ionizations means that the quantum number n = 2 becomes n > 2. Under this condition for determining here the quantum number n the above equation could be written as
E(2px2 + 2py2 + 2pz2) = 3[(-27.21)112 + (16.95)11 - 4.1 )] / n2 = - 2069.58
Then ,solving for n we get n = 2.12 > 2
In other words the three orbitals of paired electrons do not lead to the deformations of 1s2 and 2s2 but differ from the symmetry of (2px1+ 2py1 +2pz1) . In that case because of the parallel spin ( S=1) the three electrons exert mutual elecric and magnetic repulsions. On the other hand the electric repulsions between the paired electrons of 2px2, 2py2 and 2pz2 make a complete spherical shell and lead to a perfect screening with ζ = 11. Under this condition the quantum number n = 2 becomes n = 2.12 .
Note that the two electrons of opposite spin (say the 2px2) 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 the 2px2 or 2py2 or 2pz2 , today many physicists believe incorrectly that it is due to the Coulomb repulsion between the two electrons of opposite spin. Under such fallacious ideas I published my paper of 2008.
EXPLANATION OF ( E12 + E13 ) = 1172.54 eV = - E (2s2)
Here the E(2s2) represents the binding energy of the two paired electrons ( 2s2). The charges (-2e) of 1s2 screen the nuclear charge (+15e) and for a perfect screening we would have an effective ζ = 13. However according to the quantum mechanics the two electrons (2s2) penetrate the 1s2 shell. Thus they lead to the deformations of both 1s2 and 2s2 spherical shells giving an effective ζ > 13. Since n = 2 we apply my formula of 2008 to write
( E12 + E13 ) = - E(2s2) = - [(-27.21) )ζ2 + ( 16.95) ζ - 4.1 ] / 22
Since ( E12 + E13 ) = 1172.54 eV, we may rewrite
6.8025ζ2 - 4.2375ζ - 1171.5 = 0
Then, solving for ζ we get ζ = 13.44 > 13.
Here ζ = 13.44 > 13 means that the repulsιοns (2s2-1s2 ) lead to the deformation of shells, because the two electrons (2s2) or (1s2) of opposite spin behave like one particle. Note that in both cases the repulsions are due to only electric forces of the Coulomb law. Whereas in the case of the three electrons of 3px1, 3py1, and 3pz1 of parallel spin (S = 1) the three electrons interact with both electric and magnetic repulsions from symmetrical positions.
EXPLANATION OF -( Ε14 + E15 ) = - 5886.752 = E( 1s2)
As in the case of helium the binding energy E(1s2) is due to the two remaining electrons of 1s2 with n = 1. Thus we may calculate the binding energy by applying my formula of 2008 for Z = 15 as
E(1s2) = [(-27.21) )152 (+ 16.95)15 - 4.1 ] /12 = - 5872.1
However the experiments of ionizations give - (E14 + E15 ) = - 5886.752 . In other words one sees here that after the ionizations my formula of 2008 gives the value of 5872.1 eV which is smaller than the experimental value of 5886.752 eV. Under this condition of ionizations I suggest that n = 1 becomes n < 1 due to the fact that the ionizations reduce the electron charges and now the nuclear charge is much greater than the electron charge of the two remaining electrons. So for Z =15 we determine the n by writing
(E14 + E15) = 5886.752 eV = - E(1s2) = - [(-27.21) )152 + ( 16.95)15 - 4.1 ] /n2
Then solving for n we get n = 0.9988.