Rotational energy of a molecule in the angular momentum state j is given by \(E_j=\frac{\hbar^2}{2 I} j(j+1)\), where I is the moment of inertia of the molecule. The probability that the molecule will be in its ground state at temperature T (such that \(k_B T \gg \frac{\hbar^2}{2 I}\)), is
1
\(\frac{3}{2} \frac{\hbar^2}{I k_B T}\)
2
\(\frac{2}{3} \frac{\hbar^2}{I k_B T}\)
3
\(\frac{1}{2} \frac{\hbar^2}{I k_B T}\)
4
\(\frac{\hbar^2}{I k_B T}\)