For example: How do vary kinetic energy of atom electrons when atomic mass increases؟ For example this equation (E at Z) is liner, power, exponential or, logarithmic؟
which energy of electron ? kinetic or binding enregy ? as atomic mass increases the binding energy of electron also increases( higher for K shell than L shell,and so on.)
for more details you can read "Liquid drop model and semi-empirical mass formula" from any nuclear physics book (e.g. nuclear physics by S.B. Patel ). I hope it will help you.
As far as I understand, the question is: "How can one "pack in few lines" the estimation of E(Z) (Z - nuclear charge) dependence by simple (semiclassical) approach?"
For example, the version of answer:
1. Due to symmetry, the well-known exact (or the same in this case quasiclassical) solution of one-particle Schrodinger equation with potential U(r)=-Zeff/r (Zeff - EFFECTIVE nuclear charge, Zeff>1) the nuclear charge is well screened and Zeff~1. So, for OUTER electrons there's no (or weak) Z-dependence - E(Z)~const.
4. For most of electrons localized in (1/Z,1) area, one can use Thomas-Fermi model of atom
(only electrostatic part of electron-electron interaction is taken into account) and obtain E~Z**(4/3).
It is not so easy, however, to find the application where above mentioned simple models are more useful than well developed and implemented quantum chemistry methods...
The magnitude of the binding energy of a electron in an atomic nucleus for a given angular momentum state j having a given radial quantum numbern belonging to a principal quantum number N +0,1,2--- etc and n = N + (j+ 1/2) is equal to the kintic energy of the electron . If we denote this energy as E.(n, j) Kin.then:
E(n,j)Kin = Mod. E(n,j)Binding = mc^2 z^2 a^2/(2n^2)[ 1 + (((az)^2)/n) { 1/(j+0.5) -3/4n]]. for a better accuracy the term ibeyon (az)^4 can be added for atoms having larger z values. here in the equation mc^2 is the rest mass of electron may be taken to be 0.51 MeV. a as usual is the fine structure constant taken to be 1/137. n and j is already explained.This relation is valid for point nucleus The finite size will certainly reduce the kinentic energy some what but considering the size of the electronic orbits the effect will not be significant.theThe expression for z equal to 1 and replacing for m the reduced mass of electron and the proton will be resulted if you do not consider the additional term bracked in the above expression for a given n which will be independent of any angular momentum orbital or total.I think this gives a proper answer to your query. Harishchandra
1.(less important) Relativistic correction terms are ~ (aZ)^2*E order of magnitude (E - nonrelativistic ground state energy, a=1/137). Do you really need such accuracy ?
2.(more important) The expression E(n,j)Binding = mc^2 z^2 a^2/(2n^2)[ 1 + (((az)^2)/n) { 1/(j+0.5) -3/4n]] is proper for "hydrogen-like case"... Test it for simplest many-electron atom - helium (Z = 2, experimental value for ground state E = 2.9 a.u.). All this is well-known and described in many books...
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