Are These Spectroscopic Notations Physically Possible?

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SUMMARY

The discussion centers on the validity of quantum states represented by the spectroscopic notations 2S3/2, 3D2, and 5P3. The quantum numbers derived are: for 2S3/2, l = 0, s = 1/2, j = 3/2; for 3D2, l = 2, s = 1, j = 2; and for 5P3, l = 1, s = 2, j = 3. The state 2S3/2 is confirmed impossible due to the rule that when l = 0, j must equal s, which cannot be 3/2. The states 3D2 and 5P3 are deemed possible as their quantum numbers adhere to the angular momentum addition rules.

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njdevils45
Moved from a technical forum, so homework template missing
Question

(a) Write down the quantum numbers for the states described in spectroscopic notation as 2S3/2, 3D2 and 5P3.
(b) Determine if any of these states are impossible, and if so, explain why. (Please note that these could describe states with more than one electron.)

My Attempt

A)

I came up with the quantum numbers for everything that I'm pretty sure is correct. Was hoping I could get a double check on this.
2S3/2 l = 0, s = 1/2, j = 3/2
3D2 l = 2, s = 1, j = 2
5P3 l = 1, s = 2, j = 3

B)
I'm almost 100% confident that 2S3/2 is impossible because when l = 0, j must = 1/2.

I'm not sure about the other two because I don't know all the rules for what s can be compared to l and j.

Any help would be greatly appreciated. Thank you!
 
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njdevils45 said:
I'm almost 100% confident that 2S3/2 is impossible because when l = 0, j must = 1/2.
Not exactly. When ##L=0##, ##J=S##.

njdevils45 said:
I'm not sure about the other two because I don't know all the rules for what s can be compared to l and j.
What are the rules of addition of angular momenta? In other words, given ##L## and ##S##, what are the possible values of ##J=L+S##?
 
(A) is correct
You are correct that 2S3/2 is impossible.

The rule is that the maximum value of J is L+S, the minimum value of J is |L-S| and J takes all values in between in steps of 1.
 
DrClaude said:
Not exactly. When ##L=0##, ##J=S##.

So to say that when l = 0, j must = 1/2 for the first state is correct right?

DrClaude said:
What are the rules of addition of angular momenta? In other words, given ##L## and ##S##, what are the possible values of ##J=L+S##?

Does this mean that j could only equal 1,2, or 3 for the second and third one? Therefore the value for j is possible for those two?
Is the value for s possible for the other two though? Is s allowed to be integer? I know for a fact that the values for l is possible for all three of them, and assuming that s is correct, the value for j is possible for the 2nd and third state, but are the values of s possible for the 2nd and third state?
 
kuruman said:
(A) is correct
You are correct that 2S3/2 is impossible.

The rule is that the maximum value of J is L+S, the minimum value of J is |L-S| and J takes all values in between in steps of 1.

So for the second and third states, j is correct assuming the value for s is correct? In that case is the value for s possible for those two states?
 
njdevils45 said:
So to say that when l = 0, j must = 1/2 for the first state is correct right?
Correct.

njdevils45 said:
Does this mean that j could only equal 1,2, or 3 for the second and third one? Therefore the value for j is possible for those two?
Yes on both.

njdevils45 said:
Is the value for s possible for the other two though? Is s allowed to be integer? I know for a fact that the values for l is possible for all three of them, and assuming that s is correct, the value for j is possible for the 2nd and third state, but are the values of s possible for the 2nd and third state?
The spectroscopic notation is based on ##S##, the total spin of the electrons. It can have a half-integer or an integer value.
 
DrClaude said:
The spectroscopic notation is based on ##S##, the total spin of the electrons. It can have a half-integer or an integer value.

So in that case the 2nd and third states for the problem are both definitely possible?
 
njdevils45 said:
So in that case the 2nd and third states for the problem are both definitely possible?
Yes.
 
DrClaude said:
Yes.
Thank you!
 

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