From: javadi_hossein@hotmail.com (Hossein Javadi)
Newsgroups:  sci.physics.plasma
Subject:  A Mathematical Mode for Effective Nuclear Charge
Date:  31 Mar 2004 18:06:15 -0800
Organization: http://groups.google.com


A Mathematical Mode for Effective Nuclear Charge

Introduction;
The effective nuclear charge (Z*) is the "pull" that the specific
electron "feels" from the nucleus. Of the first time Slater did give a
simple rule for calculate the effective nuclear charge on any electron
in any atoms in 1930. Clementi and Raimondi did their work on
effective nuclear charges in the early 1960s. The results of
Clementi's method is diffenrence of Staler's rule. For example
Clementi calculated as Sc from a 4s perspective Z*=4.632 , but
Staler's Z* is equal 3..
Staler's rule and Clementi's method based on expriments. There is no
any analystic method for why and how the strongly of nuclear charge
does lose? By according CPH theory force and energy are equivalence.
Force converts to energy and energy changes to force. I will give a
Mathematical Mode for Effective Nuclear Charge by CPH Theory.

Mathematical Mode;
By according Theory of CPH, when a force works on an object/particle,
when W (work) is positive, force converts to energy and when W<o,
energy changes to force.
Suppose an atom with Z proton is at stationary state, its Nuclear
Charge Fz and Effective Nuclear Charge on an electron is Fz*. In
during Fz reaches to electron, its works on other electrons is W, so
Fw converts to energy E=W and electron feels the effective nuclear
charge equal Fz* that given by;

Fz*=Fz - Fw

Group each electron like this:
(1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p)(5d)(5f)...
Electrons to the right of the electron you have chosen do not
contribute.

Examples;
Suppose two objects A and B absorb each other. By according CPH Theory
a force particle leaves A and pulls it toward B, when force particle
reaches to B, another leaves B and pulls it toward A and so on. In the
following examples please attend that electrons are moving in their
orbits, but Fz (nuclear charge) moves faster than electrons.

Hydrogen;
Hydrogen atom contains one proton and one electron in 1s, so Fw=o and
Fz=Fz*. Because there is no any other electron in hydrogen atom and
Fw=0. Clementi supposed Fz*=1

Hellium;
Hellium contains two protons and two electrons in 1s. Fz=2 from two
protons moves toward electron1. Electron2 has electric charge and
magnetic field. So, Fz acts on electron2. But direction of Fz is
toward the electron1. So, electron2 does change direction of Fz. It
deppends to distance between electrons in this orbit. Suppose this
effect is nothing.
But, Fz  works on electron2, energy of electron2 increses and Fz
loses a part of its strong. So, the effective nuclear charge Fz* on
electron1 given by;
Fz*=Fz - Fw
Energy of electron2 increases equal E=W. It leaves its orbit. But
electric force leaves it toward nuclear and pulls electron2 toward
nuclear. Also, electric force of electron1 acts on it. Then electron2
comes back to its orbit and loses energy E, and E converts to electric
force equal Fw. Then Fw does add to Fz* that is coming back of
electron1 and Fz=Fz*+Fw reaches to nuclear. So, nuclear feels that
effective force of electron1 is equal Fz.
The effective nuclear charge Fz* on electron2 is same as electron1. By
according Clementi calculate Fz*=1.688

Lithium;
Lithium has 3 protons and 3 electrons, two electrons are in 1s and one
electron is in 2s.

For 1s;
Fz=3 from 3 protons moves toward electron1 in 1s orbit . This case is
same as Helliu, but radius of 1s orbits is smaller than in Hellium and
distance between electrons is less than Hellium's orbit. So deviation
direction of Fz is less than in Hellium. It shows the effect of
deviation direction for Fz is less than Hellium. By according
Clementi's calculate Fz*=2.691. Do compare with Hellium that
Fz*=1.688.

For 2s;
There is one eletron in orbit 2s in Lithium. So, this electron feels
Fz* that is coming of over the orbit 1s. Fz=3 leaves nuclear toward
it. Fz works on two electrons in orbit 1s.
Fz loses Fw1 for act on electron1, and Fw2 for act on electron2. So,
when Fz reaches to orbit 1s, It comes up to F'z=Fz - (Fw1+Fw2).
In during F'z is passing of orbit 1s, it works on the sum of electron1
and electron2. Suppose this work is equal Fw3.
So, Fw=Fw1+Fw2+Fw3 and Fz*=Fz-Fw reaches to electron in orbit 2s. By
according Clementi's calculation Fz*=1.279.
When Fz* reaches to electron electron, then an other electric force
particle equal Fz* leaves it toward nuclear. When it reaches to orbit
1s works on it. In during Fz* is passing of orbit 1s, energy E=W
converts to force Fw and Fz=Fz*+Fw reaches to nuclear.

Summary;
For calculation the effective nuclear charge Fz* on any electron in
any atoms, we must calculate Fw. W is the sum of works that Fz acts on
electrons and orbits (or suborbits) befor Fz reaches to it.
For example; For Na from a 3s,
Na has 11 protons and 11 electrons, so;
A. 2 electrons in 1s, Fw1
B. an orbit n1, Fw2
C. 6 electrons in 2p, Fw3
D. a suborbit 2p, Fw4
E. 2 electrons in 2s, Fw5
F. an orbit n2, Fw6
So, Fw=Fw1+Fw2+Fw3+Fw4+Fw5+Fw6,
and Fz*=Fz-Fw

Sincerely
Hossein Javadi

For more explain see;
http://groups.yahoo.com/group/cph_theory/files/PDFofCPH.pdf