## 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,

$$\Delta v= \pm 1$$

## What is the formula of rotational energy of diatomic molecule?

Rotational energy of diatomic molecule is given by,
$$E_J = \frac {\hbar ^2}{2I} J( J+1 )$$
Where,
I = Moment of inertia of diatomoc molecule.
J = Rotational quantum number.