## A Mathematical Mode for Effective Nuclear Charge

Expand Messages
• A Mathematical Mode for Effective Nuclear Charge Introduction; The effective nuclear charge (Z*) is the pull that the specific electron feels from the
Message 1 of 1 , Mar 30, 2004

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