Vibrational-Rotational energy level of a diatomic molecule

Vibrational-Rotational energy level of a diatomic molecule is given by,

$$\ E_v = ( v + \frac{1}{2} ) \hbar \omega_0 + \frac {\hbar ^2}{2I} J( J+1 )$$

Where, \(v=0,1,2,3….\) is the vibrational quantum number.

\(\omega_0 = \sqrt \frac{k}{\mu}\) , \(\mu\) is the reduced mass of the diatomic molecule and k is the force constant.

I is the Moment of inertia of diatomoc molecule and J is the Rotational quantum number.

Vibrational energy level of a diatomic molecule

Vibrational energy level of a diatomic molecule is given by,

$$ E_v = ( v + \frac{1}{2} ) \hbar \omega_0 $$

Where, v=0,1,2,3…. is the vibrational quantum number, \(\omega_0 = \sqrt \frac{k}{\mu} \) , \( \mu \) is the reduced mass of the diatomic molecule and k is the force constant.

The selection rule for transition between vibrational states is,

$$ \Delta v= \pm 1 $$