Electron Configuration and energy levels

AI Thread Summary
The discussion focuses on determining the possible quantum states for an excited magnesium atom with the electron configuration 1s² 2s² 2p⁶ 3p 3d. Participants clarify that the ground state configuration of magnesium is 1s² 2s² 2p⁶ 3s², and upon excitation, one electron from the 3s orbital is promoted to the 3p and 3d orbitals. The conversation emphasizes the need to consider quantum numbers n, l, m (or ml), and ms for the outermost electrons to describe the atom's quantum states accurately. It also touches on the application of L-S coupling for calculating total spin and orbital angular momentum values. The final goal is to express the results in the notation 2S+1{L}_J for the excited states.
Nusc
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1. Homework Statement
The electrons of a nelectronically-excited neutral magnesium atom have the coniguration 1s^2 2s^2 2p^6 3p 3d. Provide spec notation for the possible quant states of the atom as a whole


2. Homework Equations



3. The Attempt at a Solution

We neglect 1s^2 2s^2 2p^6 and only focus on 3p and 3d. I don't know what to do because we have two high energy levels to consider. If there was just one the then problem would be trivial.
 
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Okay so n = 3 therefore l = n-1 = 2.

Thus, when l = 2 we have d.

So d is our designator of choice.
 
Was the above correct? Because if I apply the same reasoning to this following problem:

an excited e-state of aluminimum atom is labelled by the config: 1s2 2s2 2p6 3s 3p2. What are the possible e-states for a atom with this config?

Here we have a choice between 3s and 3p2. n = 3 in this case therefore l = n-1 = 2 which would correspond with a d.

What the heck am I doing?
 
i don't know if i understood your problem correctly;

Mg is 1s2 2s2 2p6 3s2

when it is excited, one electron from 3s will be promoted to an empty 3p orbital.

hence:

*Mg 1s2 2s2 2p6 3s1 3p1
 
This is a good discussion of electron configuration
http://en.wikipedia.org/wiki/Electron_configuration

http://www.chemicalelements.com/show/electronconfig.html - chart

http://www.webelements.com/


Anyway, I was wondering if one has an example of what is meant by "Provide spec notation for the possible quant states of the atom as a whole"

Mg has the ground state electron configuration of 1s2 2s2 2p6 3s2, and Al has the g.s. e-config of 1s2 2s2 2p6 3s2 3p1 as indicated by Kushal.

The problem statement for an excited Mg atom states 1s2 2s2 2p6 3p 3d, so it appears one 3s electron is excited to 3p and one to 3d.


Is the problem asking for the possible quantum states described by n, l, m (or ml), ms? Does one need to provide descriptions for the n=1 and n=2 electrons, or just the outermost (valence electrons)?

See - http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c3

For n, what are the possible values for l, and then for l, what are the possible values for m. Then ms has two possibilities.

l is constrained such that l = 0, 1, . . . n-1,
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydcol.html#c2

then m = -l, -l+1, . . . , 0, 1, . . . , l-1, l (there are 2l+1 values)
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydazi.html#c2
http://en.wikipedia.org/wiki/Magnetic_quantum_number
 
The problem wants the configuration based on the 2S+1{L}_J labelling.
 
Here's an example: What are the poss gs of Cl+.

1s2 2s2 2p6 3s2 3p4

We need only consider the 3p4.

S = 1/2 + 1/2 + 1/2 + 1/2 : + indicates direct sum notation

So S = {0,1,2}

Since we only consider the four p electron, l =1

Therefore ml = -1,0,1

=> At least one of the spin particles must have opposite spin to the others so at least 2o f hte elctrons pair cancellin their psins. THus S= 1/2 + 1/2 = {0,1}
So the spin degeneracies are 2S+1 = {1,3}

Then L = 1 + 1 + 1 + 1 = {0,1,2,3,4}


|L-S| <= J <= L+S

Gives us a whole number of values using the 2S+1 {L}_J

Does that help?
 
I think the problem is asking for the possible quantum states described by n, l, m (or ml), ms
 
  • #10
nusc said:
I think the problem is asking for the possible quantum states described by n, l, m (or ml), ms
That is what I'm thinking.

OK, so in the example of the excited Mg, there is 1 p and 1 d electron. Then, what are the possible states for the p and for the d electron?
 
  • #11
You mean with a given p state and d state you want the orbital ang mom states?


FOr p, l =1
so ml = -1 , 0, 1

and for d, l = 2.
so ml = -2,-1 , 0, 1,2
N =3 in this case
IS that what you want?
 
  • #12
Focus on p, l =1 and d, l = 2.

J runs from |L+S| to |L-S|.

See this discussion of optical spectra.
 
  • #13
Nusc said:
The problem wants the configuration based on the 2S+1{L}_J labelling.

From that statement I will assume that we are doing L-S coupling (as opposed to j-j coupling).

Each electron has s=1/2. So what are the possible values of the total spin S?

One electron has l=1 (a p state) and the other electron has l=2 (a d state). What are the possible values of the total orbital angular momentum L?

Now, for each combination (S,L) you have (there will be a total of 6 combinations), figure out the possible values of J. Then, in each case, give the result in the notation \,^{2S+1} L_J. For example, if the total spin is 1, L is 2 and J is 2, you have a\,^3 D_2 state.

I hope this helps
 
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