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.