[disillusioned19]

If the jth rotational energy is given by [J(J+I)ħ²]/(2μR²) of a diatomic molecule where μ is the reduced mass of both particles, I is the moment of inertia ,J is the rotational quantum number and R the bond length, what happens when we look at the lowest energy level, where J=0. Does this mean that molecules occupying the rotational ground state have zero rotational energy?

[dipesh747]

Yes. When J = 0 there is no rotational energy. Vibrational energy (if my memory serves correct) is En = J(J+1)hcb so if J=0 the rotational energy is 0. 

[juanrga]

If the jth rotational energy is given by [J(J+I)h²]/(2µR²) of a diatomic molecule where µ is the reduced mass of both particles, I is the moment of inertia ,J is the rotational quantum number and R the bond length, what happens when we look at the lowest energy level, where J=0. Does this mean that molecules occupying the rotational ground state have zero rotational energy?

The response to your question is yes, somewhat as a particle with zero momentum has zero kinetic energy (## p^2/2m ##).

But what disturbs me is your formula. Moment of inertia I has dimensions of mass times distance squared. How could I sum to J which is a number?

The expression for the rotational energy levels is
$$ E(J) = \frac{J(J+1)\hbar^2}{2I} $$
with ## I = \mu R^2 ##

[disillusioned19]

Not really disturbing, just a typographical error.

[juanrga]

Not really disturbing, just a typographical error.

Hum, if we accept that was a typographical error and that you did mean "1" instead of "I", in your formula, your original post (with the corrected formula) would be

If the jth rotational energy is given by [J(J+1)ħ²]/(2μR²) of a diatomic molecule where μ is the reduced mass of both particles, I is the moment of inertia ,J is the rotational quantum number and R the bond length, what happens when we look at the lowest energy level, where J=0. Does this mean that molecules occupying the rotational ground state have zero rotational energy?

but then in the text you write "I is the moment of inertia" alluding to a I, which would be nowhere in the formula .