# Electron Configuration of Excited State

1. Feb 24, 2016

### samjohnny

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

An exercise examining the tin atom (Sn). Tin has a ground state electron configuration of $[Kr]4d^{10}5s^25p^2$.

a) Write down the electronic configuration of the first excited state.

b) Illustrate with a vector diagram the allowed total angular momentum $J$ values for this excited state.

2. Relevant equations

N/a

3. The attempt at a solution

Hi all,

For a) As I understand it, excited states for atoms are achieved when an atom in the outermost shell makes a transition to an atomic orbital with greater energy, according to the selection rule of $ΔL=±1$. However I'm unsure as to what would constitute the first excited state.

I think it would be one of the two following:

(1) $[Kr]4d^{10}5s^25p^16s^1$

(2) $[Kr]4d^{10}5s^25p^15d^1$

The $5d$ level has a higher energy than that of the $6s$, so perhaps (1) is the first excited state?

For b) I'm not entirely sure about it, but I presume that the correct answer for a) must be determined first.

Last edited: Feb 24, 2016
2. Feb 25, 2016

### Staff: Mentor

Careful here. When considering the excited states by themselves, do not refer to selection rules. The first excited state of an atom may not be reachable by a dipole transition from the ground state. The best example is hydrogen, where the first excited state is 2s.

The rule that works most of the time is to take a valence electron and put it in the next highest energy orbital.

Yes.

Yes.

3. Feb 28, 2016

### samjohnny

I see. Then the electron configuration of the first excited state would be: $[Kr]4d^{10}5s^25p^16s^1$.

Then for part b), since the total angular momentum of an atom is given by that of its outer electron, the total $J$ will be defined by the single outer electron in the $6s$ orbital. Therefore, we have $l=5$ and $s=\frac{1}{2}$. And therefore the total angular momentum values we have are $j=s±l=\frac{11}{2}$ and $\frac{9}{2}$. Therefore $J=\sqrt{j(j+1)}ħ=\frac{143}{4}ħ$ and $\frac{121}{4}ħ$. I have a feeling this isn't right..

4. Feb 28, 2016

### Staff: Mentor

This is not correct. It is only filled (sub)shells that have an overall contribution of 0 to the total orbital angular momentum and spin. Here, there are two subshells that are not filled.

???

5. Feb 28, 2016

### samjohnny

Ah OK, so both the $5p$ and $6s$ shells have contribution to the total angular momentum. And I went completely off track with $l$. For the $6s$ orbital $l=0$ and $s=\frac{1}{2}$. And so would that mean that $j=s+l=\frac{1}{2}$, and so $J=\sqrt{j(j+1)}=\frac{3}{4}ħ$?

And for $5p$, $l=1$ therefore $j=\frac{3}{2}$ and $\frac{1}{2}$. So $J=\sqrt{j(j+1)}=\frac{15}{4}ħ$ and $\frac{3}{4}ħ$. Am I on the right track?

6. Feb 28, 2016

### Staff: Mentor

For Sn, LS-coupling should apply, so this is not the correct way to go about it.

7. Feb 28, 2016

### samjohnny

Ok, then I'll implement LS coupling as found on this page. And so let $L_1$ be the orbital angular momentum of the $5p$ electron, and $L_2$ that of the $6s$ electron. Then $L=L_1+L_2=1+0=1$. And similarly for the spins: $S=\frac{1}{2}±\frac{1}{2}=1$ or $0$. Therefore, $J=L+S=1$ or $2$.

I wonder if the atomic nucleus might be perhaps too large for this type of coupling?

8. Feb 28, 2016

### Staff: Mentor

That is correct, but
is not complete. You have to find all possible J's for each combination of L and S.

Tin is at the limit where LS coupling applies, and it could be argued that jj-coupling should be used. I've checked the NIST website, and they use LS term symbols for Sn.

9. Feb 28, 2016

### samjohnny

So including all possible $Js$ gives $J=0,1,2$?

10. Feb 28, 2016

### Staff: Mentor

You found that L = 1 and S = 0 or 1. That gives you two terms, 1P and 3P. Each term has its own possible set of J values.