# Stiffness constant $k$ for a diatomic molecule

1 vote
176 views

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 $k = 550 N/m$).

edited Aug 22, 2019

1 vote

The energy levels of a quantum rigid rotator are given by $$E=(\ell+1)\ell \frac{\hbar^2}{2I}$$ where $I=\rho r^2$ is moment of inertia, $\rho$ is reduced mass, $r$ is bond length and $\ell$ is the rotational quantum number.

The difference in energy between the states $\ell=2$ and $\ell=1$ is $$\Delta E=(6-2)\frac{\hbar^2}{2I}=\frac{h^2}{2\pi^2 \rho r^2}\equiv \frac{hc}{\lambda_r}$$ therefore $$\rho=\frac{h\lambda_r}{2\pi^2cr^2}$$

The energy levels of a quantum harmonic oscillator (mass on a spring) are given by $$E=(n+\frac12)\hbar \omega$$ where $\omega^2=k/\rho$ and $k$ is spring constant. The difference in energy between any two adjacent values of $n$ is $$\Delta E=\hbar\omega =\frac{h}{2\pi}\omega\equiv \frac{hc}{\lambda_v}$$ therefore $$\omega^2=\frac{k}{\rho}=(\frac{2\pi c}{\lambda_v})^2$$

Substitute for $\rho$ from above : $$k=\frac{h\lambda_r}{2\pi^2cr^2} \frac{4\pi^2c^2}{\lambda_v ^2}=\frac{2hc\lambda_r}{r^2\lambda_v^2}\approx 30.2 N/m$$

Comment :

This value for $k$ is not typical for real molecules. HCl has one of the smallest values at $480 N/m$. For H-H it is $573N/m$. See table here.

The values given for $\lambda_r$ and $\lambda_v$ are typical of real molecules.

The value $r=884pm$ is exceptionally large. Bond lengths for real molecules typically range from $100pm$ to $300pm$. That for H-H is $74pm$. See table here or graph at bottom of this page.

However the largest bond length for a diatomic molecule is $5200pm$ for He-He (helium dimer). The bond is 5000 times weaker than in H-H. It flies apart at low rotational speeds.

So I think the answer is that a non-typical (but not impossible) value of $r$ has been used.

answered Feb 17 by (28,876 points)
selected Mar 3