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

  1. Dec 10, 2016 #1
    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?
     
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  3. Dec 10, 2016 #2

    Charles Link

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    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 ## ?
     
  4. Dec 10, 2016 #3
    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?
     
  5. Dec 10, 2016 #4

    Charles Link

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    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
  6. Dec 10, 2016 #5
    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?
     
  7. Dec 10, 2016 #6

    Charles Link

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    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
  8. Dec 10, 2016 #7
    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?
     
  9. Dec 10, 2016 #8

    Charles Link

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    I am no expert on this, but also on a learning curve. A google (hyperphysics) shows the S=1 state is symmetric so the only allowed L's for S=1 are are L=1 and L=3 since these states are antisymmetric. (The total electron state must be antisymmetric). With the maximum L=3 and maximum S=1, the available J's are 4,3, and 2 and the minimum J=2 would be the ground state by Hund's 3rd rule. There is also, I think, the possibility of S=0 and L=4. Exactly how the d shells fill in is beyond my expertise in this area. Even the case of the p orbitals with two p electrons would be something I would need to research further.
     
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