Calculating Sodium Atom Ground State Degeneracy

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Homework Help Overview

The discussion revolves around calculating the degeneracy of the ground state of the sodium atom using Hartree's theory. Participants are exploring the electronic configuration and its implications for degeneracy.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the electronic configuration of sodium and its relation to ground state degeneracy, with some questioning whether the ground state corresponds to the highest occupied n level state.

Discussion Status

There is an ongoing exploration of the concept of degeneracy in the context of Hartree's theory. Some participants have provided clarifications regarding the nature of the ground state and its degeneracy, while others express uncertainty about the question's context.

Contextual Notes

Participants note that the question is a standalone exam question, which may contribute to the confusion regarding its requirements.

hanson1011
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Hi guys,
Just got a question I'm a little stuck on and would love a push in the right direction

Q) Using Hartree's theory calculate the degeneracy of the ground state of the Sodium atom.

Its a previous exam question and I'm struggling to find much descriptive information about the topic so any help is great

Many thanks

H
 
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I'm not sure about this, but using the fact that the electronic configuration for the sodium atom is 1s2 2s2 2p6 3s1, I'd say the ground state is twice degenerate (two states can be the ground state).

(electronic configurations in chemistry are based on the hartree approximation).
 
Cheers for the reply Amok,

So does the ground state correspond to the highest occupied n level state?
 
No, the ground state is the ground electronic configuration. 1s2 2s2 2p6 3s1 is the ground state. I said it is twice degenrate (actually that was a mistake, it is once-degenerate), because the electron in the 3s orbital can have spin up or spin down.

Hartree suggested that the SE for a many-electron atom could be approximately solved by using a "mean-field" approximation (like a Hartree-Fock approximation without exchange). This results in a wave function that is a product of wave functions each depending on a single-electron coordinate (the positions of each electron are uncorrelated):

[tex]\psi(x_1, x_2, x_3, ... x_n)= \phi(x_1) \phi(x_2) \phi(x_3) ... \phi(x_n)[/tex]

Where the phis are spin-orbitals (single electron wavefunctions) and x is a position and spin coordinate. They are the product of a "position wavefunction" with a "spin wavefunction".

In the case of the sodium atom this would give:

[tex]\psi(x_1, x_2, x_3, ... x_{11})= \phi(x_1) \phi(x_2) \phi(x_3) ... \phi(x_{11})[/tex]

The last orbital can be written as:

[tex]\phi(x_{11}) = \varphi(r_{11}) \alpha[/tex]

or as:

[tex]\phi(x_{11}) = \varphi(r_{11}) \beta[/tex]

Where alpha and beta are spin wavefunctions. The ground state is therefore once-degenerate.

Now knowing what you have learned, I'm not sure this is the answer they want. The question seems kinda weird to me. In which context does this question appear?
 
Last edited:
Thanks very much, that makes a lot more sense! Its a stand alone question with no prior sections.

Many thanks
 

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