By Prof. L. Kaliambos (Natural Philosopher in New Energy)
When we write the configuration we'll put all 14 electrons in orbitals around the nucleus of the Silicon atom. In writing the electron configuration for Silicon the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Silicon go in the 2s orbital. As 14nm is the present standard for the smallest components in a microprocessor and most of them are made with silicon. The van der val radius of silicon atom is 210pm which is equal to 0.21nm so in the smallest region of a microprocessor chip it may contain roughly 14/0.21 atoms of silicon that's almost 67 atoms.
The correct frequency of light (or heat) moves some of the bonded silicon electrons into the conduction band. An applied potential difference with move them.
Diameter of silicon atom is 0.2nm and we need at least several atoms to create a transistor. I am sure Intel will be able to fulfill its roadmap to 4nm, maybe 2nm, but at this size quantum effects are so large, that everything should be changed. Astroduston July 6, 2015–. Bohr Model of Silicon.
October 29, 2015
A silicon atom is an atomof the chemical element silicon with symbol Si and atomic number 14. However unlike for hydrogen, a closed-form solution to the Schrödinger equation for the many-electron atomslike the silicon atom has not been found. So, various approximations, such asthe Hartree–Fock method, could be used to estimate the ground state energies.Under these difficulties I analyzed carefully the electromagnetic interactionsof two spinning electrons of opposite spin and I published in Ind. J. Th. Phys (2008), my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures”.Under this condition we may use this correct image with the following electron configuration:
1s2.2s2.2px2.2py2.2pz2.3s2.3px1.3py1
According to the “Ionization energies of the elements-WIKIPEDIA”the ionization energies (in eV) of the silicon are the following: E1 =8.15169 , E2 = 16.34585 , E3 =33.49302 , E4 = 45.14181 , E5 = 166.767 , E6 = 205.27, E7 = 246.5 , E8 =303.54, E9 = 351.12 , E10 =401.37, E11 = 476.36 , E12 = 523.42, E13 = 2437.63, and E14 = 2673.182 .
Here the -( E1 +E2) gives the binding energy E( 3px1 + 3py1)of the two outer electrons. Then the -(E3 + E4)gives the binding energy E(3s2). Also, the - ( E8 +E9 + E10 ) gives the binding energy E(2px1 +2py1 + 2pz1) of thethree electrons with parallel spin , ( S = 1), while the -( E5 +E6 + E7 + E8 + E9 +E10 ) gives the binding energy E( 2px2 +2py2 +2pz2). On the other handthe -( E11 + E12 ) gives the binding energyE(2s2) , while the -( E13 + E14)gives the binding energy E(1s2). See also my papers about theexplanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS based on my paper of 2008.
For understanding theionization energies E1, E2, E3 , E4 ,E5 , E6 , E7 , E8 , E9 ,E10 , E11 , and E12, which give the groundstate energies of (2s2.2px2.2py2.2pz2.3s2.3px1.3py1) you can see my “EXPLANATION OF SILICON IONIZATIONS”.
In the same way for theexplanationof the E13 and E14 which give the ground stateof the 1s2 electrons one must apply the basic formula of mypaper of 2008.
EXPLANATION OF - ( Ε13 + E14 ) = - 5110.812 = E( 1s2), WHICH GIVES THE GROUND STATE ENERGY OF THE TWO 1s2 ELECTRONS
Asin the case of helium the binding energy E(1s2) is due to thetwo remaining electrons of 1s2 with n = 1. Thus we expect tocalculate the binding energy by applying my formula of 2008 for Z = 14 as
E(1s2) = [(-27.21) )142 + (16.95)14 - 4.1 ] /12 =- 5099.96
Howeverthe experiments of ionizations give - (E13 + E14 )= - 5110.812 .In other words one sees here that after theionizations my formula of 2008 gives the value of 5099.96 eV which issmaller than the experimental value of 5110.812 eV. Under this conditionof ionizations I suggest that n = 1 becomes n < 1 due to thefact that the ionizations reduce the electron charges and now the nuclearcharge is much greater than the electron charge of the two remaining electrons.So for Z =14 we determine the n by writing
(E13 +E14) = 5110.812 eV = - E(1s2) = -[(-27.21) )142 + (16.95)14 - 4.1 ] /n2
Then,solving for n we get n = 0.9989.
However in the absence of adetailed knowledge about the electromagnetic force between the two spinningelectrons of opposite spin physicist today using wrong theories cannot explainthe ground state energy of the electrons 1s2. For example underwrong theories based on qualitative approaches many physicists believeincorrectly that the second electron of the 1s2 shell is less tightlybound because it could be interpreted as a shielding effect; the other electronpartly shields the second electron from the full charge of the nucleus. Anotherwrong way to view the energy is to say that the repulsion of the electronscontributes a positive potential energy which partially offsets the negativepotential energy contributed by the attractive electric force of the nuclearcharge.
Under such false ideas Ipublished my paper of 2008 . You can see the paper in “User Kaliambos”.
Historically, despite theenormous success of the Bohr model and the quantum mechanics of the Schrodingerequation based on the well-established laws of electromagnetism in explainingthe principal features of the hydrogen spectrum and of other one-electronatomic systems, so far, under the abandonment of natural laws neither was ableto provide a satisfactory explanation of the two-electron atoms. In atomicphysics a two-electron atom is a quantum mechanical system consisting of onenucleus with a charge Ze and just two electrons. This is the first case of many-electronsystems. The first few two-electron atoms are:
Z =1 : H- hydrogenanion. Z = 2 : He helium atom. Z = 3 : Li+ lithiumatom anion. Z = 4 : Be2+ beryllium ion.
Prior to the development ofquantum mechanics, an atom with many electrons was portrayed like the solarsystem, with the electrons representing the planets circulating about thenuclear “sun”. In the solar system, the gravitational interaction betweenplanets is quite small compared with that between any planet and the verymassive sun; interplanetary interactions can, therefore, be treated as smallperturbations.
However, In the helium atomwith two electrons, the interaction energy between the two spinning electronsand between an electron and the nucleus are almost of the same magnitude, and aperturbation approach is inapplicable.
In 1925 the two young Dutchphysicists Uhlenbeck and Goudsmit discovered the electron spin according towhich the peripheral velocity of a spinning electron is greater than the speedof light. Since this discovery invalidates Einstein’s relativity it met muchopposition by physicists including Pauli. Under the influence of Einstein’sinvalid relativity physicists believed that in nature cannot existvelocities faster than the speed of light.(See my FASTER THAN LIGHT).
So, great physicists likePauli, Heisenberg, and Dirac abandoned the natural laws of electromagnetism infavor of wrong theories including qualitative approaches under an idea ofsymmetry properties between the two electrons of opposite spin which lead tomany complications. Thus, in the “Helium atom-Wikipedia” one reads: “Unlike for hydrogen aclosed form solution to the Schrodinger equation for the helium atom has notbeen found. However various approximations such as the Hartree-Fock method ,canbe used to estimate the ground state energy and wave function of atoms”.
It is of interest to notethat 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 '. The conference was organised by the natural philosophers M. Barone and F. Selleri ,who gave me an award including a disc of the atomic philosopher Democritus, because 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 structurebut also the coupling of two electrons under the application of the abandonedelectromagnetic laws. For example in the photoelectric effect the absorption oflight contributed not only to the increase of the electron energy but also tothe increase of the electron mass, because the particles of light have mass m =hν/c2. (See my paper 'DISCOVERY OF PHOTON MASS' ).
However the electron spinwhich gives a peripheral velocity greater than the speed of light cannot beaffected by the photon absorption. Thus after 10 years I published my paper 'Nuclear structure...electromagnetism' (2003), in which I showed not onlymy DISCOVERY OF NUCLEAR FORCE AND STRUCTURE but also that theperipheral velocity (u >> c) of two spinning electrons with opposite spingives an attractive magnetic force (Fm) stronger than the electricrepulsion (Fe) when the two electrons of mass m and charge (-e) areat a very short separation (r < 578.8 /1015 m). Because ofthe antiparallel spin along the radial direction the interaction of theelectron charges gives an electromagnetic force
Fem = Fe - Fm .
Therefore in my researchthe integration for calculating the mutual Fem led to thefollowing relation:
Fem = Fe - Fm = Ke2/r2 - (Ke2/r4)(9h2/16π2m2c2)
Of course for Fe =Fm one gets the equilibrium separation ro =3h/4πmc = 578.8/1015 m.
Silicon Atom Image
That is, for aninterelectron separation r < 578.8/1015 m the two electronsof opposite spin exert an attractive electromagnetic force, because theattractive Fm is stronger than the repulsive Fe . Here Fm is a spin-dependent force of short range. As aconsequence this situation provides the physical basis for understanding thepairing of two electrons described qualitatively by the Pauli principle, whichcannot be applied in the simplest case of the deuteron in nuclear physics,because the binding energy between the two spinning nucleons occurs when thespin is not opposite (S=0) but parallel (S=1). According to the experiments inthe case of two electrons with antiparallel spin the presence of a very strongexternal magnetic field gives parallel spin (S=1) with electric andmagnetic repulsions given by
Silicon Atom
Fem = Fe + Fm
So, according to thewell-established laws of electromagnetism after a detailed analysis of paired electrons in two-electron atoms I concluded that at r < 578.8/1015 m a motional EMF produces vibrationsof paired electrons.
Unfortunately today manyphysicists in the absence of a detailed knowledge believe that the twoelectrons of two-electron atoms under the Coulomb repulsion between theelectrons move not together as one particle but as separated particlespossessing the two opposite points of the diameter of the orbit aroundthe nucleus. In fact, the two electrons of opposite spin behave like oneparticle circulating about the nucleus under the rules of quantum mechanicsforming two-electron orbitals in helium, beryllium etc. In my paper of 2008, I showed that the positive vibration energy (Ev) described in eV dependson the Ze charge of nucleus as
Ev = 16.95Z -4.1
Of course in the absence ofsuch a vibration energy Ev it is well-known that the ground state energyE described in eV for two orbiting electrons could be given by the Bohr modelas
E = (-27.21) Z2.
So the combination of theenergies of the Bohr model and the vibration energies due to the opposite spinof two electrons led to my discovery of the ground state energy of two-electronatoms given by
-E = (-27.21) Z2 +(16.95 )Z - 4.1
For example the laboratorymeasurement of the ionization energy of H- yields an energy ofthe ground state -E = - 14.35 eV. In this case since Z = 1 weget
-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
Thediscovery of this simple formula based on the well-established laws ofelectromagnetism was the first fundamental equation for understanding theenergies of many-electron atoms, while various theories based on qualitativesymmetry properties lead to complications.