Allowed combinations of remaining quantum numbers when n = 3

Dixanadu
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Homework Statement


Hey everyone

I couldn't really fit a good description into the title. Basically here's what it is.

Consider an atom in which all the states with principal quantum number n = 3 and angular momentum quantum number l = 2 are occupied by electrons (this is called a closed subshell). List the allowed combinations of the remaining quantum numbers m,m_{s} which are related to the eigenvalues of \hat{L}_{z},S_{z}. Hence, how many electrons are in the n = 3, l = 2 closed subshell?


Homework Equations



Dont know of any

The Attempt at a Solution



I don't have a clue...need some help T_T
 
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What's the spin of an electron? That determines the allowed values for ms. What's the thing that tells you what values m can take?
 
Spin of an electron is either +1/2 or -1/2...that's m_{s} right? as far as m goes, is that the one that ranges from -l to +l...so it has 2l+1 values right? I just don't get what all this has to do with the eigenvalues of \hat{L} and S_{z}...
 
So m is the eigenvalue corresponding to operator \hat{L}_z, and ms corresponds to \hat{S}_z.

You basically got it, now you just need to figure out which values can m and ms take together, and how many combinations there are.
 
for m = 1/2:
m = -2, -1, 0, 1, 2

for m = -1/2
m= -2, -1, 0, 1, 2

So total is 10 electrons...?
 
Yap that looks correct
 

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