# Homework Help: Calculating J,L,S

1. Jul 29, 2013

### LagrangeEuler

1. The problem statement, all variables and given/known data
Structure of natrium
$1s^2 2s^2 2p^6 3s^1$
How to calculate $L$ and $S$?

2. Relevant equations

3. The attempt at a solution
From that $S=\frac{1}{2}$. I know that I need to get $L=0$ and $J=\frac{1}{2}$, but I'm not sure why?

2. Jul 29, 2013

### Mandelbroth

I appreciate your devotion to Latin, but it is clearer to everybody if you just say sodium. :tongue:

You've given the valence shell structure of a sodium atom. I don't know what $L$ and $S$ are. Could you clarify?

3. Jul 29, 2013

### tman12321

Learn how to read spectroscopic notation. That will give you L and S. Since you're already given the configuration, use Hund's third rule to find J.

4. Jul 31, 2013

### LagrangeEuler

I don't understand why $L=0$?

5. Jul 31, 2013

### DrDu

Full (sub-)shells don't contribute to L. What's the l of the only open subshell?

6. Aug 12, 2013

### LagrangeEuler

For only open subshell l=0. But for example for $4f^3$, $L=6$ and for $4f^9$ $L=5$.

Last edited: Aug 12, 2013
7. Aug 13, 2013

### jwelch

Keep in mind what you are trying to do, which is describe the lowest energy level of the atom in terms of the outer shell electrons. Hund's rules are heuristics of more general rules that apply in most normal situations to atoms. The ordering of the levels implies a hierarchy, a listing from greatest to least of the order of magnitudes of these effects. By that logic, spin-spin interactions >> orbit-orbit interactions >> spin-orbit interactions.

S is determined by the spin-spin interaction, which has a term in the Hamiltonian proportional to -S1*S2 when describing two electrons. You start with the highest S values, therefore, because they have the lowest energy. You see this effect play out in the way orbitals are filled: first with spin up (+1/2) electrons in each orbital, then spin down (-1/2) in each orbital.

Sooooooooo for the first half of the level (in this case, 1 orbital = 2 electrons) with S = +1/2. If it was the p-orbital rather than the s-orbital, you would fill the first three with +1/2. Then add up all the S values you have. That's step one.

On the next level of magnitude, we see orbit-orbit interactions, which has to do with the angular momentum of each electron. A single particle has its own angular momentum in an atom. If the next particle comes in and orbits in the same direction, then the L is greater. However, if the next particle runs in the opposite direction, then there is an increase in the repulsive force. The more they run into each other, the move the repulsion force increase, which takes more energy to overcome. Keep in mind this idea while you add more particles, where atom momentum will reach a maximum, level off, then fall down (i.e. angular momentum peaks in the middle, while E increases with each step). This number is given from the primary quantum numbers. Your particle is in the s-orbital, which has L total = 0. p-orbital has value -1 to 1 for L, etc.

J is the lowest available option. This term represents spin-orbit coupling. I always think of spin-orbit coupling as an extension of the previous two items. The addition of all the momenta on top of each other will lead to a maximum, followed by a gradual drop (building up to maximum achievable, then falling back off). There is a coupling term with delta E = j*(j+1) - l*(l+1) - s*(s+1). The switch of the sign/the decreasing of the term occurs because of the eventual accumulation of too many couplings to effectively neutralized.

In summary:

S = sum of s values in outer shell; assign following the convention of +1/2 for the first half of available spots, -1/2 for the rest
L = sum of l values in outer shell; assign following the shell convention for l (l = 0 for s, l = 1, 0, -1 for p, etc.)
J = sum of L and S; start from |L-S| then work up from lowest.

8. Aug 17, 2013

### LagrangeEuler

Thanks for the answer a lot. I think I understand it now in better level.
So for $4f^9$ $l=3$ and $m_l=-3,-2,-1,0,1,2,3$. All of them have the same energy because energy is defined by principal quantum number $n$.
$+3 \uparrow$
$+2 \uparrow$
$+1 \uparrow$
$+0 \uparrow$
$-1 \uparrow$
$-2 \uparrow \downarrow$
$-3 \uparrow \downarrow$
So $S=0+0+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}$
and $L=-1+0+1+2+3=5$
Ok. Thanks. Just to ask is it the same as
$+3 \uparrow \downarrow$
$+2 \uparrow \downarrow$
$+1 \uparrow$
$+0 \uparrow$
$-1 \uparrow$
$-2 \uparrow$
$-3 \uparrow$
or even maybe
$+3 \uparrow \downarrow$
$+2 \uparrow$
$+1 \uparrow \downarrow$
$+0 \uparrow$
$-1 \uparrow$
$-2 \uparrow$
$-3 \uparrow$
?