Rotational spectrum - equidistance

[Petar Mali]

Difference between two rotational terms is given by

$$\tilde{\nu}=(J+1)(J+2)B-J(J+1)B=2B(J+1)$$

If we put values of $$J$$ in this expression we get that otational spectrum is equidistant.

$$T_r$$ - rotational term
$$J$$ - 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!

 

[alxm]

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.

 

[Petar Mali]

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

Look at this picture. You have rotation and vibration levels. Yes I assume that two moleculs system - rigid rotor approximation.

 

[alxm]

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.

 

[Petar Mali]

The centrifugal force is not the case in this picture

$$E_r=J(J+1)\frac{\hbar^2}{2I}$$

$$T_r=\frac{E_r}{hc}$$

For $$J=0$$ $$T_r=0$$

For $$J=1$$ $$T_r=2B$$

For $$J=2$$ $$T_r=6B$$

For $$J=3$$ $$T_r=12B$$

...

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!