GR0177 #84



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grephy 20121106 09:39:13  why s is not taken into account?
In transition C, there IS a change in s, right? Then why this transition can happen?!
jonrichens 20121108 02:39:37 
I believe that the spin selection rule only comes into play for multielectron atoms (where the directional spin breaks the symmetry, or something). This question
is about hydrogen and the orientation of spin is not conserved in the same way.
Hydrogen selection rules
(delta)j = 0, + 1 not 0 > 0
(delta)l = + 1
(delta)mj = 0, + 1
Helium selection rule (spin orientation important)
(delta)j = 0, + 1 not 0 > 0
(delta)l = 0, + 1
(delta)mj = 0, + 1
(delta)s = 0 not 0 > 0
hope that helps

  antithesis 20071001 17:36:51  Can you clarify?
You state the selection rules for l and m, but when you go through the ABC options, you only mentions the l and j rules (I know they are related).
Thanks
Richard 20071030 11:12:31 
The condition on the total angular momentum quantum number comes from the fact that
I think it may be best to remember that for an electricdipole transition the following selection rules apply:
with the exception of a to transition.
Of course, these rules are different when you have a quadrupole or octupole transition. I believe the only difference is an additional or possibility on and for quadrupole and octupole respectively.

 

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