This question already has an answer here:
Why do azimuthal quantum numbers have a value lesser than Principal Quantum number. example- if n=1 it will have subshell "s" as 0
Asked
Viewed
48 times
1
$\begingroup$
$\endgroup$
asked May 8 '15 at 4:36
marked as duplicate by Geoff Hutchison, ron, Ben Norris, Martin - マーチン♦, Klaus-Dieter Warzecha May 29 '15 at 4:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
-
2$\begingroup$ In order to understand the exact reason for this you will have to study quantum mechanics. It is simply something that evolves naturally from the Schroedinger equation of the hydrogen atom. You should be able to find the proof for that in basically every theoretical physics textbook on the subject. $\endgroup$ – Philipp May 8 '15 at 8:49
-
$\begingroup$ See chemistry.stackexchange.com/q/31639/5017 for more $\endgroup$ – Geoff Hutchison May 28 '15 at 20:38
-
$\begingroup$ @Philipp, I might guess the OP would like an explanation a little bit friendlier to the novice than your average theoretical physics textbook $\endgroup$ – Ben Norris May 28 '15 at 22:53
-
$\begingroup$ @BenNorris You are certainly right, but I couldn't think of an intuitive way to show why the azimuthal quantum numbers are bounded. From your answers on this site I gather that you are pretty proficient in quantum mechanics so I guess you know about the derivation of the angular momentum operator's eigenvalue spectrum for the hydrogen atom. I feel this topic lacks approachablilty via non-mathematical means and simply cannot be explained without delving into the math but if you can think of a nice layman-explanation I would be very interested to see it. $\endgroup$ – Philipp May 31 '15 at 21:00
