Electron Configuration of Excited State

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Homework Statement



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.

Homework Equations



N/a

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.
 
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Answers and Replies

  • #2
DrClaude
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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##.
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.

However I'm unsure as to what would constitute the first excited state.
The rule that works most of the time is to take a valence electron and put it in the next highest energy orbital.

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?
Yes.

For b) I'm not entirely sure about it, but I presume that the correct answer for a) must be determined first.
Yes.
 
  • #3
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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.
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
DrClaude
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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.
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.

Therefore, we have ##l=5##
???
 
  • #5
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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.


???
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
DrClaude
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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?
For Sn, LS-coupling should apply, so this is not the correct way to go about it.
 
  • #7
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For Sn, LS-coupling should apply, so this is not the correct way to go about it.
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
DrClaude
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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##.
That is correct, but
Therefore, ##J=L+S=1## or ##2##.
is not complete. You have to find all possible J's for each combination of L and S.

I wonder if the atomic nucleus might be perhaps too large for this type of coupling?
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
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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.
So including all possible ##Js## gives ##J=0,1,2##?
 
  • #10
DrClaude
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So including all possible ##Js## gives ##J=0,1,2##?
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.
 

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