Electronic structure of the atoms


by Sean Torrebadel
Tags: atoms, electronic, structure
Sean Torrebadel
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#1
Dec1-07, 05:18 AM
P: 92
I'm a little confused about something very simple. From a chemistry perspective I am told about how many electrons fit into each type of orbital. For instance, 1s and 2s can only have two each, the 2p orbitals total 6. It is composed of the Px, Py, and Pz and only two can be in each set, while the exclusion principle applies.

When I look at the spectrum of He I in the Nist lines database

http://physics.nist.gov/PhysRefData/ASD/lines_form.html

where you enter He I and then 10 -100000 I get a series of wavelengths at the beginning of the data beginning with 584 Angstroms. This uninterupted series represents that transition of electrons from the 2s2p orbitals to the 1s. The problem that I am having is understanding why there are 10 p orbitals listed. Shouldn't there be only six.

To further complicate the problem I have an equation that reproduces this series mathematically, and it includes 5 more lines in the same sequence, which are not even classified by the Nist, but recorded there.

Am I to suppose that there are 15p? What is going on here?
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malawi_glenn
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#2
Dec1-07, 06:20 AM
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Have you tried the Aufbau-priciple?
Sean Torrebadel
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#3
Dec1-07, 04:54 PM
P: 92
Okay, I think I get it now. The configuration is not from say the six different 2p orbitals which are essentially degenerate, but from the 2p, 3p, 4p, 5p, 6p, 7p etc to the 1s.

Sean Torrebadel
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#4
Dec1-07, 08:07 PM
P: 92

Electronic structure of the atoms


Would it be true to say that the 2p, 3p, 4p, 5p, and so forth are all quantum relative to one another. I mean are they suppose to be an integral amount of some energy, E, so that the 2p is 2E, the 3p is 3E, the 4p is 4E?
ZapperZ
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#5
Dec1-07, 08:12 PM
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Are you sure? Look at the simple Rydberg type atom. Look at the transition of n=3 to 1 and compare that with n=2 to 1. Is one a multiple of the other?

Zz.
Sean Torrebadel
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#6
Dec1-07, 09:15 PM
P: 92
No, your right, the energies of the n=1, n=2, n=3, n=4, n=5, etc are all E/n^2. Where E is the ionization energy. Would it be true then that the 2p, 3p, 4p, 5p etc should follow suit? Thanks Zap


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