# Activation energy of vibrational modes in diatomic gases

My question is related to the calculation of the heat capacity of diatomic molecules. It is easy two see that such atoms have a Hamiltonian $$H=\frac{P_x^2+P_y^2+P_z^2}{2M}+\frac{L^2}{2I}+\frac{p_r^2}{2\mu}+\frac{1}{2}\mu\omega^2r^2$$ which means three square terms for center of mass (translational degrees of freedom), two square terms for angular momentum (rotational degrees of freedom) and two square terms for vibrational modes.

One proceeds and solves the quantuma mechanical Hamiltonians individually leading to energies:

• Translational: $E=\hbar^2n^2/2mL^2$ where $L$ is the size of the container. As a result the activation energy of the translational modes are zero ($\propto 1/L^2$).
• Rotational degrees of freedom lead to energies $E=\hbar^2\ell(\ell+1)/2I$, so the activation energy is of the order $\propto \hbar^2/ma_B^2:=k_B\Theta_r$.
• Finally, vibrational modes have energies $E=\hbar\omega(n+1/2)$, leading to activation energies $\propto \hbar\omega:=k_B\Theta_v$.

The problem is I have no idea how does one estimate the value of $\omega$ based on the parameters in the problem. It should be $\omega^2=\sqrt{K/\mu}$ with $K=d^2U/d r^2$ with $U$ the binding potential between the two atoms. But what is this potential? Actually is there a smart way of estimating what $\Theta_v$ should be.

The goal is to be able to tell whether $\Theta_v<\Theta_r$ or $\Theta_v>\Theta_r$ to finish the calculation of heat capacity of diatomic gas.