December 05, 2021, 08:49:02 AM
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### Topic: Difference in $Z_{eff}$ of electron of any element  (Read 304 times)

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#### Win,odd Dhamnekar

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##### Difference in $Z_{eff}$ of electron of any element
« on: November 17, 2021, 03:16:51 AM »
Using Slater's rule Zeff on  electrons in  3p orbital of Phosphorous is 4.8
http://calistry.org/calculate/slaterRuleCalculator

But If we use Eionization=-En,I=$\frac{Z_{eff}^2 \times R_H}{n^2} \Rightarrow Z_{eff}=\big( \frac{n^2 \times E_{n,I}}{R_H}\big)^{\frac12}, Z_{eff}$ on electrons in 3p orbital of Phosphorous  are different.

e.g. 1st, 2nd, and 3rd $E_{(3, I)}$ of Phosphorous are 1011.8 kJ/mol, 1907 kJ/mol, 2914.1kJ/mol repectively. If we plug in these energy values in the  above formula, we get Zeff=2.633, using 1st E(3,I), Zeff= 3.616 using 2nd E(3,I) and Zeff=4.47 using 3rd E(3,I).

Why are there such discrepancies in computations of Zeff using these two different methods?

« Last Edit: November 17, 2021, 06:33:26 AM by Win,odd Dhamnekar »
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#### mjc123

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##### Re: Difference in $Z_{eff}$ of electron of any element
« Reply #1 on: November 18, 2021, 05:36:50 PM »
Where did you get the formula for ionization energy? It looks as if it applies to a Rydberg atom - an atom with a single valence electron outside a compact core, that behaves similarly to a hydrogen-like atom with charge Zeff. It won't work for an atom with multiple valence electrons. The ionization energy is not equal to the energy level of the electron being removed, because when you remove that electron the energy levels of all the other electrons change.

#### Win,odd Dhamnekar

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##### Re: Difference in $Z_{eff}$ of electron of any element
« Reply #2 on: November 19, 2021, 09:44:37 AM »
I got the ionization formula in the following answer to the same(similar) question.

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#### mjc123

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##### Re: Difference in $Z_{eff}$ of electron of any element
« Reply #3 on: November 19, 2021, 10:23:02 AM »
If you define Zeff in that way, it is quite different from Slater's definition, and it is not surprising that you get different values. In my view it is not a very meaningful definition, because as I said P is not a Rydberg atom. When you remove an electron the interactions between the remaining electrons change, so you get different Zeff values for P+ and P2+. (You do with Slater as well.)