Is the 52D5/2 Level of Na Atom Degenerate?

PatF
Messages
17
Reaction score
0

Homework Statement



This is a problem that was on a quiz. The quiz has already been graded and handed back and the faculty member graded very gently. (This means I passed when I think should not have.) I have thought about the answer I gave at the time and it was pretty bad. My current answer is below. Tell me if it right/wrong/stupefyingly boneheaded.

Q: "The valence electron of the Na atom is at the 52D5/2 level. Is this level degenerate? If so, how many folds? Give the values of all the appropriate quantum numbers for each degenerate member. Exhibit these values in tabular form with one line for each degenerate member."


Homework Equations



No equations.


The Attempt at a Solution



The above notation is spectroscopic notation telling me that the principal quantum number is 5 and the orbital angular momentum number is 2. Therefore, spin must be 1/2 since we have the subscript 5/2. In this configuration, we could have mj equal to -5/2, -3/2,-1/2,1/2,3/2,5/2.

So, my answer should be:

A) The level is degenerate.
B) There are 6 folds.
C) The table would look as follows:

n l mj s
5 2 -5/2 1/2
5 2 -3/2 1/2
5 2 -1/2 1/2
5 2 1/2 1/2
5 2 3/2 1/2
5 2 5/2 1/2

Is the above right or wrong?

Does anyone have a good reference for translating between quantum numbers - which I think I understand - and spectroscopic notation which is a lot more confusing?

Thanks.
 
Physics news on Phys.org
PatF said:

Homework Statement



This is a problem that was on a quiz. The quiz has already been graded and handed back and the faculty member graded very gently. (This means I passed when I think should not have.) I have thought about the answer I gave at the time and it was pretty bad. My current answer is below. Tell me if it right/wrong/stupefyingly boneheaded.

Q: "The valence electron of the Na atom is at the 52D5/2 level. Is this level degenerate? If so, how many folds? Give the values of all the appropriate quantum numbers for each degenerate member. Exhibit these values in tabular form with one line for each degenerate member."


Homework Equations



No equations.


The Attempt at a Solution



The above notation is spectroscopic notation telling me that the principal quantum number is 5 and the orbital angular momentum number is 2. Therefore, spin must be 1/2 since we have the subscript 5/2.

You don't have to deduce the spin since
the superscript "2" of the spectroscopic notation {}^2D_{5/2} tells you that the spin is
<br /> 2S+1=2\;,<br />
I.e.,
<br /> S=1/2<br />

In this configuration, we could have mj equal to -5/2, -3/2,-1/2,1/2,3/2,5/2.

So, my answer should be:

A) The level is degenerate.
B) There are 6 folds.
C) The table would look as follows:

n l mj s
5 2 -5/2 1/2
5 2 -3/2 1/2
5 2 -1/2 1/2
5 2 1/2 1/2
5 2 3/2 1/2
5 2 5/2 1/2

Is the above right or wrong?

Does anyone have a good reference for translating between quantum numbers - which I think I understand - and spectroscopic notation which is a lot more confusing?

The spectroscopic notation and the quantum numbers are related by
<br /> {}^{(2S+1)}{\mathcal{L}}_J\;,<br />
where
\mathcal{L} is S or P or D or F or G etc
depending on whether the angular momentum quantum number
L is 1 or 2 or 3 or 4 or 5 etc, respectively
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top