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Rotational spectrum - equidistance

  1. Dec 11, 2009 #1
    Difference between two rotational terms is given by

    [tex]\tilde{\nu}=(J+1)(J+2)B-J(J+1)B=2B(J+1)[/tex]

    If we put values of [tex]J[/tex] in this expression we get that otational spectrum is equidistant.

    [tex]T_r[/tex] - rotational term
    [tex]J[/tex] - rotational quantum number

    But from this picture spectrum isn't equidistant.

    http://www.mwit.ac.th/~physicslab/hbase/molecule/imgmol/rotlev.gif

    Can you tell me where is the problem? Thanks!
     
    Last edited: Dec 12, 2009
  2. jcsd
  3. Dec 12, 2009 #2

    alxm

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    I can't view the picture. But the equation you cite is valid for the rigid rotor approximation. Real systems tend not to act like that except for low rotation numbers.
     
  4. Dec 12, 2009 #3
    I put the other picture
    http://www.mwit.ac.th/~physicslab/hb...mol/rotlev.gif [Broken]

    Look at this picture. You have rotation and vibration levels. Yes I assume that two moleculs system - rigid rotor approximation.
     
    Last edited by a moderator: May 4, 2017
  5. Dec 13, 2009 #4

    alxm

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    Right well what's the question? Why the approximation fails at higher rotational states?

    Simply "centrifugal force" - the interatomic distance increases at higher rotation speeds/states, you get a different B.
     
  6. Dec 13, 2009 #5
    The centrifugal force is not the case in this picture

    [tex]E_r=J(J+1)\frac{\hbar^2}{2I}[/tex]

    [tex]T_r=\frac{E_r}{hc}[/tex]

    For [tex]J=0[/tex] [tex]T_r=0[/tex]

    For [tex]J=1[/tex] [tex]T_r=2B[/tex]

    For [tex]J=2[/tex] [tex]T_r=6B[/tex]

    For [tex]J=3[/tex] [tex]T_r=12B[/tex]

    ...

    It isn't equidistant if I calculate like this. And in picture which you see is this terms. It looks like contradiction if you look this post and my first post!
     
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