Bond vibrations

Chemical bond, if stretched too far, will break. A typical potential energy curve for a chemical bond as a function of $R$, the separation between the two nuclei in the bond is given in the figure below:

Figure: Potential energy curve of a chemical bond as a function of $R$. The small blue curve is an approximate harmonic oscillator curve fit to the true potential energy curve at low energies.

If the energy of the bond is not too high, then the potential energy curve is well approximated by a harmonic oscillator curve (shown in blue in the figure). The true curve is given by a function of the form

$$V(R) = D_e \left(1-e^{-a(R-R_e)}\right)^2$$

where $D_e$ is the dissociation energy. This curve is well approximated by a simpler harmonic oscillator function at low energy

$$V(R) = {1 \over 2}k\left(R-R_e\right)^2$$

Thus, as long as the energy is not too high, the energy levels are those of a harmonic oscillator

$$E_n = (n+1)h\nu_0\;\;\;\;\;\;\;\;\;\;n=0,1,2,....$$

where $\nu_0$ is the intrinsic frequency

$$\nu_0 = {1 \over 2\pi}\sqrt{{k \over \mu}}$$

where $\mu$ is the reduced mass.

The energy change in a transition from energy level $n$ to level $n+1$ is

$$\Delta E_{(n+1),n} = (n+1+1)h\nu_0 - (n+1)h\nu_0$$

$$= (n+2-n-1)h\nu_0$$

$$= h\nu_0$$

Hence, the frequency at which transitions occur is

$$h\nu = \Delta E_{(n+1),n} = h\nu_0$$

$$\nu = \nu_0$$

Example: In NaH, a photon of wavelength 8.53$\times$10$^{-6}$ m can induce a vibrational transition from the $n=0$ to the $n=1$ level. what is the force constant $k$ of the NaH bond?

First, find the frequency $\nu$ of the photon:

$$\nu = {c \over \lambda} = {3\times 10^{8} m/s \over 8.53\times 10^{-6}\;{\rm m}} = 3.515\times 10^{13}\;{\rm Hz}$$

Now, since $\nu = \nu_0$, the intrinsic frequency is also 3.515$\times$10$^{13}$Hz. Since

$$\nu_0 = {1 \over 2\pi}\sqrt{{k \over \mu}}$$

$$k = 4\pi\nu_0^2 \mu$$

Plugging in

$$k = 4\pi(3.515\times 10^{13}\;{\rm Hz})^2(1.603\times 10^{-27}\;{\rm kg}) = 78.2 N/m$$

We noted in the last lecture that the frequency plotted on the $x$ axis of a spectrum is almost always in units known as wavenumbers (cm$^{-1}$), which is the inverse of the wavelength. The conversion from Hz to wavenumbers proceeds via the relation

$$\nu = {c \over \lambda} \;\;\;\;\;\;\;\;\;\;{\nu \over c} = {1 \over \lambda}$$

This means that the conversion from Hz to cm$^{-1}$ requires that we divide the frequency by 2.998$\times$10$^{10}$ cm/s:

$$\nu\;\left({\rm in\ cm}^{-1}\right) = {\nu\;({\rm in\ Hz}) \over 2.998\times 10^{10}\;{\rm cm/s}}$$

Hence, the frequency in the example above $3.515\times 10^{13}$ Hz is

$${3.515\times10^{13}\;{\rm Hz} \over 2.998\times10^{10}\;{\rm cm/s}} = 1170\;{\rm cm}^{-1}$$