# Allowed combinations of remaining quantum numbers when n = 3

1. Dec 4, 2013

1. The problem statement, all variables and given/known data
Hey everyone

I couldnt 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?

2. Relevant equations

Dont know of any

3. The attempt at a solution

I dont have a clue...need some help T_T

2. Dec 4, 2013

### clamtrox

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?

3. Dec 4, 2013

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 dont get what all this has to do with the eigenvalues of $\hat{L}$ and $S_{z}$...

4. Dec 4, 2013

### clamtrox

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.

5. Dec 4, 2013

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...?

6. Dec 4, 2013

### clamtrox

Yap that looks correct