Rotational and Vibrational Spectra of Molecules

A molecule emits or absorbs radiation when it changes energy state, so its spectrum maps its energy levels. The energy of a diatomic molecule separates into three parts whose magnitudes are so different they can be treated independently:

• Electronic, from exciting the electrons, of order $1\text{eV}$ — the same scale as atomic transitions.
• Vibrational, from the atoms oscillating along the bond, of order $0.1\text{eV}$.
• Rotational, from the molecule turning about its center of mass, of order $10^{-3}\text{eV}$.

The vibrational and rotational energies are $10^{-2}$ to $10^{-3}$ of the electronic energy. This lesson quantizes the two mechanical motions and combines them.

Rotational energy levels

Classically the kinetic energy of a rigid rotor is $E=\frac{1}{2}I{\omega }^2=L^2/2I$, where $I$ is the moment of inertia and $L=I\omega$ the angular momentum. Quantum mechanically the angular momentum is quantized by the same condition that governs orbital motion in an atom,

so the rotational energy levels are

The constant $E_{0r}$ is inversely proportional to the moment of inertia. The levels spread apart as $\ell$ grows: the gaps go as $2,6,12,20,\dots$ in units of $E_{0r}$.

Only molecules with a permanent electric dipole moment have a pure rotational spectrum; symmetric molecules such as ${\text{H}}_2$, ${\text{O}}_2$, or ${\text{CO}}_2$ do not radiate by rotating alone. For polar molecules the selection rule is $\Delta \ell =\pm 1$, giving an energy separation between adjacent states

The moment of inertia of a diatomic molecule about its center of mass reduces to that of a single reduced mass $\mu$ at the bond length:

Worked example — bond length of CO. The $\ell =0\to 1$ transition of CO is observed at wavelength $\lambda =2.6\text{mm}$, corresponding to $\Delta E=4.77\times 10^{-4}\text{eV}$. From $\Delta E_{0\to 1}={\hslash }^2/I$,

The reduced mass of CO is $\mu =(12)(16)/(12+16)=6.86\text{u}$. Then $r_0=\sqrt{I/\mu }=0.113\text{nm}$. Measuring a rotational line yields the bond length directly.

Vibrational energy levels

Near the equilibrium separation, the molecular potential energy of the previous lesson is well approximated by a parabola, so the bond behaves as a harmonic oscillator. Its energy levels are

equally spaced by $\Delta E=hf$, where $f$ is the classical vibration frequency. For two masses on a spring of force constant $K$,

Worked example — force constant of CO. The vibrational frequency of CO is measured as $f=6.42\times 10^{13}\text{Hz}$. Solving $f=(1/2\pi )\sqrt{K/\mu }$ for $K$ with $\mu =6.86\text{u}$,

A typical vibrational quantum is $E\sim hf\approx 0.2\text{eV}$, roughly 1000 times a rotational quantum and about 8 times the thermal energy $kT\approx 0.026\text{eV}$ at room temperature. So at ordinary temperatures collisions readily excite the low rotational levels but leave nearly all molecules in the vibrational ground state $v=0$.

The vibration-rotation band

A molecule carries rotational and vibrational energy at once. Stacking the rotational ladder on each vibrational level gives the full level scheme.

Infrared absorption studies the ground electronic state, exciting only vibrational and rotational levels. At ordinary temperatures nearly all molecules start in $v=0$, and the dominant absorption is $v=0\to 1$. But the molecules are spread over many rotational levels, so the vibrational jump is accompanied by a rotational change $\Delta \ell =\pm 1$. This splits the band into two branches:

• R branch ($\ell \to \ell +1$): energies $hf+2E_{0r},hf+4E_{0r},hf+6E_{0r},\dots$
• P branch ($\ell \to \ell -1$): energies $hf-2E_{0r},hf-4E_{0r},hf-6E_{0r},\dots$

The lines are equally spaced by $2E_{0r}$, with a gap of $4E_{0r}$ centered on the pure vibrational frequency $f$ — no line falls exactly at $f$ because a vibrational transition must be accompanied by a rotational one.

Line intensities and level populations

The lines are not equally intense. The intensity of an absorption line is set by how many molecules occupy the starting level, which is the level's degeneracy times the Boltzmann factor. A rotational level $\ell$ has degeneracy $2\ell +1$ (the number of $m_{\ell }$ values), so

The degeneracy $2\ell +1$ pushes the population up at small $\ell$; the Boltzmann factor pulls it down at large $\ell$. The two compete, and the population peaks at an intermediate $\ell$. Setting $\mathbf{d}n/\mathbf{d}\ell =0$,

For a typical molecule at room temperature, ${\ell }_{\max }\approx 3$: the ground rotational level is not the most populated, and the brightest absorption lines come from $\ell \approx 3$.

Two further complications appear in real spectra. The line spacing is not exactly constant: at high $\ell$ the molecule stretches, its moment of inertia grows, and $E_{0r}$ shrinks, so the high-$\ell$ lines crowd together. And a molecule such as HCl shows each line doubled, because natural chlorine is a mixture of ${}^{35}$Cl (75.5%) and ${}^{37}$Cl (24.5%) with slightly different reduced masses. Both effects turn the spectrum into a precise probe of molecular structure. Absorption and emission are two of several ways light interacts with molecules; the next lesson sharpens the rigid-rotor and harmonic-oscillator pictures into the anharmonic, centrifugally distorted structure of a real band.