What are the possible L and S states for the 2s2p3p configuration?

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SUMMARY

The discussion focuses on determining the possible total angular momentum (L) and total spin (S) states for the electron configurations 2s2p3p and 2p^23p. For the 2s2p3p configuration, three distinct electrons are identified, each occupying different quantum states, which simplifies the enumeration process. In contrast, the 2p^23p configuration requires careful consideration of the Pauli exclusion principle, as it restricts the occupancy of identical quantum states by electrons. Participants emphasize the importance of constructing a table to systematically list the possible states for clarity.

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Homework Statement


a) Enumerate the possible L and S for the [itex]2s2p3p[/itex] configuration.

b) Find the possible L and S for the [itex]2p^23p[/itex] configuration.

Homework Equations



[tex]^{2S+1}L_J[/tex]

[tex]S=0 (Singlet)[/tex]

[tex]S=1 (Triplet)[/tex]

The Attempt at a Solution



First of all, doesn't "enumerate" and "find the possible states" mean the same thing? In my previous courses, we've been asked to find all the possible states for a particular atom, like Boron (for ex., Griffiths Problem 5.12 b) which is simple enough, all I do is write down a table and start listing off the possible total angular momenta.

But for part a), we are asked to enumerate [itex]2s2p3p[/itex]. To me this seems like there are three different electrons: the first has n=2,l=0 and the second n=2,l=1, and the third n=3,l=1. How would I go about doing this? I'm just confused that all. Can someone give me a push in the right direction?

Thanks yall

IHateMayonnaise
 
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Well I too haven't seen a problem like that. But after reading it, I believe you are correct about it being 3 electrons each in a different n,l state. Just make the table as usual, and you don't have to worry about the Pauli exclusion principle since no two electrons can ever be in the same state for (a).

Just be careful about (b) since the Pauli exclusion principle applies there.
 

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