A hypothetical diatomic molecule has a binding length of $0.8860 nm$. When the molecule makes a rotational transition from $l = 2$ to the next lower energy level, a photon is released with $\lambda_r = 1403 \mu m$. At a vibration transition to a lower energy state, a photon is released with $\lambda_v = 4.844 \mu m$. Determine the spring constant k.

What I've done is:

1) Calculate the moment of inertia of the molecule by equating Planck's equation to the transition rotational energy:

$$E = \frac{hc}{\lambda} = \frac{2 \hbar^2}{I}$$

Thus solving for $I$:

$$I = 1.562 \times 10^{-46} kg m^2$$

2) Knowing that the moment of inertia of a diatomic molecule rotating about its CM can be expressed as $I = \rho r^2$, solve for $\rho$ (where $\rho$ is the reduced mass and $r$ is the distance from one of the molecules to the axis of rotation; thus $r$ is half the bond length. EDIT: $r$ is the bond length and NOT half of it. Curiously, if you use the half value you get a more reasonable k: around $3 N/m$):

$$\rho = \frac{I}{r^2} = 1.99 \times 10^{-28} kg$$

3) Solve for $k$ from the frequency of vibration equation:

$$f = \frac{1}{2\pi}\sqrt{\frac{k}{\rho}}$$

Knowing:

$$\omega = 2\pi f$$

We get:

$$k = \omega^2 \rho = (c/\lambda_v)^2 \rho = 0.763 N/m$$

Where $\lambda_v= 4.844 \times 10^{-6} m$

The problem I see is that the method seems to be OK but the result does not convince me. We know that the stiffness constant $k$ is is a measure of the resistance offered by a body to deformation. The unknown diatomic molecule we're dealing with seems to be much more elastic than $H_2$ (which has $k = 550 N/m$).