# Using Hund's rules to find ground state L, S, J

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1. Dec 10, 2016

### astrocytosis

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

Using Hund's rules, find the ground state L, S and J of the following atoms: (a) fluorine, (b) magnesium, and (c) titanium.

2. Relevant equations

J = L + S

3. The attempt at a solution

I'm having trouble understanding what L, S and J mean on a basic level. I read the textbook chapter multiple times but I'm still not grasping their physical significance and how they are related to s, l and j.

I think I need to look at the outer shell, so for Mg for example the outer shell is an s orbital which I know means L = 0. And J = L + S. But I'm unsure about how to find S. I think it might be 1/2 + -1/2 = 0 since there are 2 electrons and one must be spin up and the other spin down? This makes J = 0. So the answer would be 0, 0, 0. Is that correct? I didn't really make use of Hund's rules here.

Using this reasoning I would get S = 1/2, L = 1, J =1+1/2 for F and S = 1, L = 2, J = 3 for Ti. Is this the correct approach? Is there a good way to make sense of S, L, and J so I don't get so confused about what they represent?

2. Dec 10, 2016

Titanium has $3d^2 4s^2$. (I googled it.) The $S=1$ looks correct, but what is the maximum $L$ that you can make? Then, to apply the 3rd rule, with the shell less than or equal to be half-filled, what is the minimum $J$ ?

3. Dec 10, 2016

### astrocytosis

I'm having trouble understanding how L can have more than one value. I thought that L = 2 for d shells and L = 0 for S shells always. Maybe it can be 3 since n = 4 for 4s2?

4. Dec 10, 2016

In the d=2 orbital, $l=2$ for each electron. You are adding angular momentum: $\vec{L}=\vec{l_1}+\vec{l_2}$. With 2 electrons each of l=2, you can generate an $L=4$ state. The possible $L$ states (for $l_1=2$ and $l_2=2$) are $L=4,3,2,1, \, and \, 0$. The topic that covers it in more detail is found under the category Clebsch-Gordon coefficients, and I think Racah coefficients, but if you read anything that gives a good summary of the addition of angular momentum, that should be sufficient for the purpose at hand. (Similarly you are adding $\vec{S}=\vec{s_1}+\vec{s_2}$. Finally, you take the $L$ and $S$ and get the $\vec{J}=\vec{L}+\vec{S}$. In all 3 cases, you are adding angular momentum. The process is a little bit odd, but good explanations of it can be found if you google it and read the wikipedia summary, etc.)

Last edited: Dec 10, 2016
5. Dec 10, 2016

### astrocytosis

Ok, that makes sense; somehow I missed that L is just the sum of each electron's l. So for Ti S = 1, L = 4, and J = 1 since L = l + l = 0 + 0 = 0 for minimum J?

6. Dec 10, 2016

You want the minimum $J$ (since the shell is half-filled or less) with the maximum $L$ and maximum $S$. Since $\vec{J}=\vec{L}+\vec{S}$, with the way it works, when $L=4$ (which is the correct maximum L) and $S=1$ (which is the correct maximum S), the available J's are 5,4, and 3. The minimum of these is 3. (With this state comes an $m_J$, but they are not asking for that.) $\\$ Additional note: Here (for Titanium) we are basically working with the 2 $3d$ electrons since the $4s^2$ forms a closed shell. The $\vec{L}=\vec{l_1}+\vec{l_2}$ is first computed, and then the $\vec{S}=\vec{s_1}+\vec{s_2}$ and then finally the $\vec{J}=\vec{L}+\vec{S}$

Last edited: Dec 10, 2016
7. Dec 10, 2016

### astrocytosis

How do you obtain the other available J's? Also wouldn't the maximum L be 3 because of the way the d subshells fills up; there are 5 subshells and each subshell has a different l (from -2 to 2) so you would get 2(1) + 1(1) = 3 if there are 2 electrons in two different subshells?

8. Dec 10, 2016