Rotational spectrum - equidistance

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:

alxm

Science Advisor
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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.
 
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.
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:

alxm

Science Advisor
1,840
7
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.
 
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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