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

November 4, 2015

Zinc is an atom of the chemical element zinc with symbol Zn and atomic number 30. However unlike for hydrogen atom, a closed-form solution to the Schrödinger equation for the many-electron atoms like the zinc atom has not been found. So, under the invalid relativity (EXPERIMENTS REJECT RELATIVITY) various approximations, such as the Hartree–Fock method, could be used to estimate the ground state energies. Under these difficulties I analyzed carefully the electromagnetic interactions of two spinning electrons of opposite spin which give a simple formula for the solution of such ground state energies. You can see it in my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008). Hence the electronic structure of the zinc ground state should be given by this correct image as

1s^{2}.2s^{2}.2p_{x}^{2}.2p_{y}^{2}.2p_{z}^{2}.3s^{2}.3p_{x}^{2}.3p_{y}^{2}.3p_{z}^{2}.3d^{10}.
4s^{2}

According to the "Ionization energies of the elements-WIKIPEDIA" the ionization energies (from E_{1} to E_{20}) of zinc (in eV)
are the following:

E_{1} = 9.39,
E_{2} = 17.96, E_{3} = 39.7 , E_{4} =
59.4 , E_{5} = 82.6, E_{6}= 108, E_{7} =
134, E_{8}= 174, E_{9} = 203, E_{10} =
238, E_{11 }= 274 , E_{12 }= 310.8, E_{13 }=
419.7, E_{14 }= 454 , E_{15} = 490, E_{16}=
542, E_{17} = 579, E_{18} = 619, E_{19} =
698, and E_{20 }= 738.

Firstly we examine the -
( E_{1} + E_{2} ) = E(4s^{2}). Here the
E(4s^{2}) represents the ground state energy of the two outermost
electrons. Secondly, we observe that the -( E_{3 }+ E_{4} +
E_{5 }+ Ε_{6} + Ε_{7} + E_{8} + E_{9 } +
E_{10} + Ε_{11} +
E_{12}) = E(3d^{10}). The E(3d^{10}) represents
the ground state energy of 10 paired electrons (3d^{10}) of five
orbitals. Then, the - ( E_{13 }+ E_{14 }+
E_{15}+ E_{16} + E_{17 }+ E_{18} )
equals the ground state energy E(3p^{6}) of the 6 paired
electrons. Whereas the - ( E_{19} + E_{20}) equals
the ground state energy E( 3s^{2}).

It is of interest to note
that in the absence of data (from E_{21} to E_{30}) one
cannot calculate the ground state energies of - ( E_{21} +…+
E_{30} ).

For understanding better the ground state energies see also my papers about the ground state energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008.

** **

**GROUND STATE ENERGY OF ** **- ( E _{1} + E_{2 }) = -27.35 eV = E(4s^{2})**

Here
the ground state energy E(4s^{2}) of the two outermost
electrons is given by applying my formula of 2008. The
charges (-28e) of the inner electrons (1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10})
screen the nuclear charge (+30e) and for a perfect screening we would have an
effective ζ = 2.
However according to the quantum mechanics the electrons 4s^{2}penetrate
the ( 3d^{10}) leading to ζ > 2. Under this condition we may write

(E_{1} +
E_{2}) = 27.35 eV = -E(4s^{2}) = -[
(-27.21 )ζ^{2} +
(16.95) ζ - 4.1 ] /
n^{2}

Since n = 4 the above equation could be written as

1.7
ζ^{2} -
1.06 ζ - 27.09 = 0

Then solving for ζ we get ζ = 4.3

Here
the 4.3 > 2 means that the two outermost electrons (4s^{2})
lead to the deformation of 3d^{10} .

**GROUND STATE ENERGY OF** **- ( E _{3} +…. + E_{12} ) = - 1623.5 = E(3d^{10})**

Here
the E(3d^{10}) represents the ground state energy of the 10 electrons
(3d^{10}) . Note that here 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 (1s^{2}2s^{2}2p^{6}3s^{2}3p^{6})
screen the nuclear charge (+30e) and for a perfect screening we would have ζ = 12. Under this condition we may write

E_{3 }+..+
E_{12} ) = 1623.5 eV = - E(3d^{10}) = ^{ }-5[(-27.21)ζ^{2} +
(16.95)ζ - 4.1] / n^{2}

So using n = 3 we may rewrite

15.1167ζ^{2}-
9.4167ζ - 1621.2222 = 0

Surprisingly solving for ζ we get ζ = 10.67 < 12 , 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 = 12. Thus using ζ = 12 we expect to determine n > 3. Under this condition we may write

E_{3 }+..+
E_{12 }) = 1623.5 eV = -E(3d^{10}) = -5[(-
27.21)12^{2} - (16.95)12 + (4.1)] / n^{2}

Then solving for n we get n = 3.46 > 3 .

**GROUND STATE ENERGY OF ( E _{13 }+ … + E_{18} ) = 3103.7 eV = -E(3p^{6})**

Here
the E(3p^{6}) represents the ground state energy of the 6 paired
electrons given by applying my formula of 2008. The charges (-12e)
of the twelve inner electrons like (1s^{2}2s^{2}2p^{6}3s^{2 })
screen the nuclear charge (+30e) and for a perfect screening we would have an
effective Z_{eff } = ζ = 18. However the 6 paired electrons ( 3p^{6 }) repel
the 3s^{2 }electrons and lead to the deformation of shells with ζ > 18. Under this condition we may
write

(
E_{13} +…+ E_{18}) = 3103.7
eV = -E(3p^{6}) = -3[(-27.21)ζ^{2} +
(16.95)ζ - 4.1] / n^{2} ^{ }

Since n = 3 the above equation can be written as

9.07ζ^{2} -
5.65ζ -
3091.4 = 0

Then, solving for ζ we get ζ = 18.776 > 18 .

**GROUND STATE ENERGY OF ( E _{19 } + E_{20 }) = 1436 eV = -E(3s^{2})**

Here
the E(3s^{2}) represents the ground state energy of the two paired
electrons (3s^{2}) given by applying my formula of 2008. The
charges (-10e) of the inner 10 electrons of 1s^{2}.2s^{2}.2p^{6} screen
the nuclear charge (+30e) and for a perfect screening we would have an
effective ζ = 20.
However the two electrons of 3s^{2} penetrate the 2p^{6} leading
to the deformation of shells with ζ > 20. Under this condition we write the following equation as

(
E_{19} + E_{20 }) = 1436 eV = - E(3s^{2})
= - [(- 27.21)ζ^{2} +
(16.95) ζ -
4.1 ] / n^{2}

Since n = 3 we may write

3.0233ζ^{2 } -
1.8833ζ - 1435.54
= 0

Then solving for ζ we get ζ = 22.1 > 20 .

**WHY EINSTEIN'S RELATIVITY LED TO THE ABANDONMENT OF LAWS IN FAVOR OF WRONG ATOMIC THEORIES**

It is of interest to note that in the absence of a detailed knowledge about the electromagnetic interaction of two electrons of opposite spin physicists so far under the influence of Einstein’s relativity could not calculate such ground state energies of many-electron atoms. Historically, despite the enormous success of the Bohr model and the quantum mechanics of the Schrodinger equation based on the well-established laws of electromagnetism in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far, under the influence of Einstein’s relativity which led to the abandonment of natural laws neither was able to provide a satisfactory explanation of the two-electron atoms.

In atomic physics a two-electron atom is a quantum mechanical system consisting of one nucleus with a charge Ze and just two electrons. This is the first case of many-electron systems. The first few two-electron atoms are:

Z =1 : H^{- }hydrogen
anion. Z = 2 : He helium atom. Z = 3 : Li^{+} lithium
atom anion. Z = 4 : Be^{2+} beryllium ion.

Prior to the development of quantum mechanics, an atom with many electrons was portrayed like the solar system, with the electrons representing the planets circulating about the nuclear “sun”. In the solar system, the gravitational interaction between planets is quite small compared with that between any planet and the very massive sun; interplanetary interactions can, therefore, be treated as small perturbations.

However, In the helium atom with two electrons, the interaction energy between the two spinning electrons and between an electron and the nucleus are almost of the same magnitude, and a perturbation approach is inapplicable.

In 1925 the two young Dutch physicists Uhlenbeck and Goudsmit discovered the electron spin according to which the peripheral velocity of a spinning electron is greater than the speed of light. Since this discovery invalidates Einstein’s relativity it met much opposition by physicists including Pauli. Under the influence of Einstein’s invalid relativity physicists believed that in nature cannot exist velocities faster than the speed of light.(See my FASTER THAN LIGHT).

So, great physicists like Pauli, Heisenberg, and Dirac abandoned the natural laws of electromagnetism in favor of wrong theories including qualitative approaches under an idea of symmetry properties between the two electrons of opposite spin which lead to many complications. Thus, in the “Helium atom-Wikipedia” one reads: “Unlike for hydrogen a closed form solution to the Schrodinger equation for the helium atom has not been found. However various approximations such as the Hartree-Fock method ,can be used to estimate the ground state energy and wave function of atoms”.

It is of interest to note
that in 1993 in Olympia of Greece I presented at the international conference
“Frontiers of fundamental physics” my paper “Impact of Maxwell’s equation of displacement current on electromagnetic laws and comparison of the Maxwellian waves with our model of dipolic particles ". In that paper I showed that LAWS AND EXPERIMENTS INVALIDATE FIELDS AND RELATIVITY .
At the same time I tried to find not only the nuclear force and structure
but also the coupling of two electrons under the application of the abandoned
electromagnetic laws. For example in the photoelectric effect the absorption of
light contributed not only to the increase of the electron energy but also to
the increase of the electron mass, because the particles of light have mass m =
hν/c^{2. }(
See my paper "DISCOVERY OF PHOTON MASS" ).

However the electron spin
which gives a peripheral velocity greater than the speed of light cannot be
affected by the photon absorption. Thus after 9 years I presented at the 12^{th} symposium
of the Hellenic nuclear physics society my paper “Nuclear structure is governed by the fundamental laws of electromagnetism". (NCSR “Demokritos, 2002). The paper
was published in Ind. J. Th. Phys (2003) in which I showed not only
my discovery of nuclear force and structure but also that the
peripheral velocity (u >> c) of two spinning electrons with opposite spin
gives an attractive magnetic force (F_{m}) stronger than the electric
repulsion (F_{e}) when the two electrons of mass m and charge (-e) are
at a very short separation (r < 578.8 /10^{15} m). Because of
the antiparallel spin along the radial direction the interaction of the
electron charges gives an electromagnetic force

F_{em }= F_{e}
- F_{m} .

Therefore in my research
the integration for calculating the mutual F_{em} led to the
following relation:

F_{em }= F_{e}
- F_{m} = Ke^{2}/r^{2} - (Ke^{2}/r^{4})(9h^{2}/16π^{2}m^{2}c^{2})

Of course for F_{e} =
F_{m} one gets the equilibrium separation r_{o }=
3h/4πmc = 578.8/10^{15} m.
That is, for an interelectron separation r < 578.8/10^{15} m
the two electrons of opposite spin exert an attractive electromagnetic force,
because the attractive F_{m} is stronger than the repulsive F_{e} .
Here F_{m }is a spin-dependent force of short range. As a
consequence this situation provides the physical basis for understanding the
pairing of two electrons described qualitatively by the Pauli principle, which
cannot be applied in the simplest case of the deuteron in nuclear physics,
because the binding energy between the two spinning nucleons occurs when the
spin is not opposite (S=0) but parallel (S=1). According to the experiments in
the case of two electrons with antiparallel spin the presence of a very strong
external magnetic field gives parallel spin (S=1) with electric and
magnetic repulsions given by

F_{em }=
F_{e} + F_{m}

So, according to the
well-established laws of electromagnetism after a detailed analysis of
paired electrons in two-electron atoms I concluded that at r <
578.8/10^{15} m a motional EMF produces vibrations
of paired electrons.

Unfortunately today many physicists in the absence of a detailed knowledge believe that the two electrons of two-electron atoms under the Coulomb repulsion between the electrons move not together as one particle but as separated particles possessing the two opposite points of the diameter of the orbit around the nucleus. In fact, the two electrons of opposite spin behave like one particle circulating about the nucleus under the rules of quantum mechanics forming two-electron orbitals in helium, beryllium etc. In my paper of 2008, I showed that the positive vibration energy (Ev) described in eV depends on the Ze charge of nucleus as

Ev = 16.95Z - 4.1

Of course in the absence of such a vibration energy Ev it is well-known that the ground state energy E described in eV for two orbiting electrons could be given by the Bohr model as

E = (-27.21) Z^{2}.

So the combination of the energies of the Bohr model and the vibration energies due to the opposite spin of two electrons led to my discovery of the ground state energy of two-electron atoms given by

-E = (-27.21) Z^{2} +
(16.95 )Z - 4.1

For example the laboratory
measurement of the ionization energy of H^{-} yields an energy of
the ground state -E = - 14.35 eV. In this case since Z = 1 we
get

-E = -27.21 + 16.95 - 4.1 = -14.35 eV. In the same way writing for the helium Z = 2 we get

-E = - 108.8 + 32.9 - 4.1 = -79.0 eV

The discovery of this simple formula based on the well-established laws of electromagnetism was the first fundamental equation for understanding the energies of many-electron atoms, while various theories based on qualitative symmetry properties lead to complications.