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Spectroscopy question

  1. Dec 6, 2007 #1
    I referred to hyperphysics to learn that for electric dipole transition,the selection rules are

    [tex]\Delta[/tex][tex]\ l =[/tex][tex]\ 1 [/tex]

    Or, [tex]\Delta[/tex][tex]\ l =[/tex][tex]\ -1 [/tex]

    And, [tex]\Delta[/tex][tex]\ m_l =[/tex][tex]\ 0 [/tex]

    Does not it include n transition?

    How is n transition different from transition among the orbitals?

    I mean n transition leads to different energy state...That I know.But does the transition among l lead also to different energy state?
     
    Last edited: Dec 6, 2007
  2. jcsd
  3. Dec 6, 2007 #2
    Selection rules in spectroscopy is not a fun area to study in my opinion, and to work out all the selection rules for a particular molecular system can be quite tiresome. Atoms are a lot easier than molecules, which is what I'm assuming you are referring to.

    n is one of the simpler quantum numbers from a spectroscopic point of view. It refers to the electronic energy level, and can go up or down by any integer amount, or even stay the same during a transition. It is the "principle quantum number", or the quantum number of an atomic shell.

    Transitions in l require a change in the orbital angular momentum of the electron in its shell, which happens every time a photon is absorbed. Higher l means more angular momentum for the electron and a higher energy in a given n shell. I can't remember much about m_l though, sorry! I think it's the projection of l on the atomic z axis.
     
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