# Basic spectroscopy help needed

Hello everyone,

hope you are all well.

I would like to ask a simple spectroscopy question please.

"An electron has been excited from the ground state to the 3p state. Find the orbital angular momentum of the electron"

This is my interpritation of the question,

3p means n=3
thus, l[angular momentum quantum number] = 0,1,2

thus, to calculate the orbital angular momentum.... we have 3 calculations

using
L=[sqrt(l(l+1))] X h[bar]

so one will be 0

the other root2 h[bar], and

the last root6 h[bar]?

Hope im right.

QM is easier to work than spectroscopy.

Thanks for reading

very much appreciated.

Rob

## Answers and Replies

jtbell
Mentor
The fact that the atom is in a p state (3p) tells you something about the orbital angular momentum. Check your textbook for the significance of s, p, d, f...

The fact that the atom is in a p state (3p) tells you something about the orbital angular momentum. Check your textbook for the significance of s, p, d, f...

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/orbdep.html#c1

does l=1?

s p d f ...
0 1 2 3....[l]?

EDIT: so 3p means n=3 l=1? what about the orbital magnetic number m?, and spin, and total quantum number j?
ie, is 3p not a total description of the electron, would you need to say....3,p,0,1/2,1/2?

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I have been looking at this for a while, and think I might have a handle of it but would like some opinions...

So, if the electron in the 3p state 'relaxes' and drops to say, the n=1,2 level, am I right in thinking that the electron configuration of the electron is...

1)
If it relaxes to the 1s state, n=1, l=0, m=0

2)
If it relaxes to the 2p state, n=2, l=0, m=0
3)
If it relaxes to the 2p state,
n=2, l=1, m=-1
n=2, l=1, m=0
n=2, l=1, m=+1

secondly, because I am dealing with hydrogen am I right in saying that the energy of the electron only depends upon n?

I hope this is right, i am getting rather confused by it all.

Thank you for reading.

R

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