---
title: Rotational and Vibrational Spectra of Molecules
module: Molecular Spectra
moduleNumber: 2
lessonNumber: 1
order: 201
summary: >
  A diatomic molecule stores energy in three well-separated ledgers: electronic,
  vibrational, and rotational. Quantizing the rigid rotor gives levels spaced as
  ℓ(ℓ+1); quantizing the bond as a harmonic oscillator gives equally spaced
  vibrational levels. Their combination produces the P and R branches of an
  infrared absorption band, from which the bond length and force constant are
  read directly.
topics: [Molecular Spectra]
draft: false
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 9 — Molecular Structure and Spectra; §9-4 Energy Levels and Spectra of Diatomic 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](/atomic-physics/many-electron-atoms/periodic-table-atomic-spectra).
- **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 $E = \tfrac12 I\omega^2 =
L^2/2I$, 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](/atomic-physics/quantum-hydrogen-atom/schrodinger-3d-hydrogen),

$$
L^2 = \ell(\ell+1)\hbar^2, \qquad \ell = 0, 1, 2, \ldots,
$$

so the rotational energy levels are

$$
E_\ell = \frac{\ell(\ell+1)\hbar^2}{2I} = \ell(\ell+1)E_{0r}, \qquad E_{0r} \equiv \frac{\hbar^2}{2I}.
$$

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

$$
% caption: Rotational levels of a rigid rotor, with energy proportional to
% ℓ(ℓ+1); the spacing 2(ℓ+1)E_0r widens as ℓ increases, and the selection rule
% permits only steps of one unit in ℓ.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,-0.2) -- (0,5.2) node[left] {$E$};
  % levels at heights ell(ell+1)*k, k=0.24
  \foreach \l/\h in {0/0, 1/0.48, 2/1.44, 3/2.88, 4/4.8}{
    \draw[acc, thick] (0.6,\h) -- (4.4,\h);
    \node[anchor=west] at (4.5,\h) {level \l};
  }
  % transition arrows between adjacent levels
  \draw[black, ->] (1.2,0) -- (1.2,0.48);
  \draw[black, ->] (1.9,0.48) -- (1.9,1.44);
  \draw[black, ->] (2.6,1.44) -- (2.6,2.88);
  \draw[black, ->] (3.3,2.88) -- (3.3,4.8);
  \node[black, anchor=west, font=\scriptsize] at (0.62,0.24) {$2E_{0r}$};
  \node[black, anchor=west, font=\scriptsize] at (0.62,0.96) {$4E_{0r}$};
  \node[black, anchor=west, font=\scriptsize] at (0.62,2.16) {$6E_{0r}$};
\end{tikzpicture}
$$

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

$$
\Delta E_{\ell \to \ell+1} = \frac{(\ell+1)\hbar^2}{I}.
$$

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:

$$
I = \mu r_0^2, \qquad \mu = \frac{m_1 m_2}{m_1 + m_2}.
$$

> **Definition (Reduced mass).** For two masses $m_1, m_2$ bound together, the
> two-body problem separates into center-of-mass motion plus a single effective
> particle of mass $\mu = m_1 m_2/(m_1+m_2)$ orbiting at the separation $r_0$.
> Because $\mu < \min(m_1, m_2)$, the reduced mass of HCl (0.98 u) is slightly
> less than the mass of the hydrogen atom.

**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 $\Delta E =
4.77\times10^{-4}\ \text{eV}$. From $\Delta E_{0\to1} = \hbar^2/I$,

$$
I = \frac{\hbar^2}{\Delta E} = \frac{(1.055\times10^{-34}\ \text{J·s})^2}{(4.77\times10^{-4})(1.60\times10^{-19}\ \text{J})}.
$$

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](/condensed-matter/molecules-and-bonding/bonding-mechanisms) is well
approximated by a parabola, so the bond behaves as a
[harmonic oscillator](/quantum-mechanics/wave-mechanics-1d/operators-expectation-values-and-the-harmonic-oscillator).
Its energy levels are

$$
E_v = \left(v + \tfrac12\right)hf, \qquad v = 0, 1, 2, \ldots,
$$

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

$$
f = \frac{1}{2\pi}\sqrt{\frac{K}{\mu}}.
$$

$$
% caption: The molecular potential (Morse-like) is nearly parabolic near r_0; the
% low-lying vibrational levels are equally spaced by hf, and crowd together at
% high v where the true potential spreads faster than a parabola.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,-0.3) -- (0,4.4) node[left] {$U$};
  \draw[->, black] (0,0) -- (6.4,0) node[below] {$r$};
  % Morse-like well: minimum near x=1.6, y=0.4
  \draw[acc, very thick, domain=0.85:6.1, samples=120, variable=\x]
    plot ({\x},{3.6*(1 - exp(-(\x-1.6)*0.72))*(1 - exp(-(\x-1.6)*0.72))});
  % equally spaced low levels
  \foreach \v/\y in {0/0.45, 1/0.95, 2/1.45, 3/1.95}{
    \draw[black] (1.05,\y) -- (2.85,\y);
    \node[anchor=west, font=\scriptsize] at (2.9,\y) {$v=\v$};
  }
  \node[black, anchor=east, font=\scriptsize] at (1.0,0.7) {$hf$};
  \node[acc, anchor=west] at (3.8,3.2) {molecular potential};
\end{tikzpicture}
$$

**Worked example — force constant of CO.** The vibrational frequency of CO is
measured as $f = 6.42\times10^{13}\ \text{Hz}$. Solving $f =
(1/2\pi)\sqrt{K/\mu}$ for $K$ with $\mu = 6.86\ \text{u}$,

$$
K = (2\pi f)^2\mu = (2\pi \cdot 6.42\times10^{13})^2 (6.86)(1.66\times10^{-27}\ \text{kg}) = 1.86\times10^3\ \text{N/m}.
$$

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.

$$
% caption: Electronic, vibrational (v), and rotational (ℓ) levels of a diatomic
% molecule; each vibrational level carries its own rotational ladder, and the
% ladders themselves sit inside the electronic states.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % ground electronic, v=0 rotational ladder
  \foreach \l/\y in {0/0.0, 1/0.22, 2/0.52, 3/0.9}{
    \draw[acc, thick] (0.5,\y) -- (3.2,\y);
  }
  \node[acc, anchor=west, font=\scriptsize] at (3.25,0.45) {$v=0$};
  % v=1 rotational ladder
  \foreach \l/\y in {0/1.7, 1/1.92, 2/2.22, 3/2.6}{
    \draw[acc, thick] (0.5,\y) -- (3.2,\y);
  }
  \node[acc, anchor=west, font=\scriptsize] at (3.25,2.15) {$v=1$};
  \node[black, anchor=east] at (0.45,1.3) {ground electronic};
  % excited electronic level (higher)
  \draw[black, thick] (0.5,4.2) -- (3.2,4.2);
  \draw[black, thick] (0.5,4.55) -- (3.2,4.55);
  \node[black, anchor=east] at (0.45,4.35) {excited electronic};
  % annotate the rotational ladder
  \draw[black, ->] (5.3,1.05) -- (4.4,0.6);
  \node[black, anchor=west, font=\scriptsize] at (5.35,1.05) {rotational levels};
\end{tikzpicture}
$$

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

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.

$$
% caption: The vibration-rotation absorption band of a diatomic molecule; equally
% spaced lines flank a central gap at the vibration frequency f, the P branch
% below and the R branch above, with a spacing that measures E_0r.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (10.6,0) node[below] {frequency};
  % P branch lines (left of gap)
  \foreach \x in {1.0,1.9,2.8,3.7}
    \draw[acc, very thick] (\x,0) -- (\x,1.4);
  % gap centered at 5.3
  \draw[black, dashed] (5.3,0) -- (5.3,1.7) node[above] {$f$};
  % R branch lines (right of gap)
  \foreach \x in {6.9,7.8,8.7,9.6}
    \draw[acc, very thick] (\x,0) -- (\x,1.4);
  \node[acc, anchor=north] at (2.35,-0.1) {P branch};
  \node[acc, anchor=north] at (8.25,-0.1) {R branch};
  \draw[black, <->] (1.0,1.55) -- (1.9,1.55) node[midway, above, font=\scriptsize] {$2E_{0r}$};
\end{tikzpicture}
$$

## 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](/statistical-mechanics/foundations/classical-statistics-and-equipartition).
A rotational level $\ell$ has degeneracy $2\ell + 1$ (the number of $m_\ell$
values), so

$$
n(E_\ell) \propto (2\ell+1)\,e^{-\ell(\ell+1)E_{0r}/kT}.
$$

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 $\d n/\d\ell = 0$,

$$
\ell_{\max} = \frac12\left(\sqrt{\frac{2kT}{E_{0r}}} - 1\right).
$$

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 $\ell \approx 3$.

$$
% caption: Thermal population of rotational levels: the degeneracy 2ℓ+1 raises
% low levels while the Boltzmann factor suppresses high ones, so the population
% peaks near ℓ ≈ 3 at room temperature rather than at the ground level.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.4,0);
  \draw[->, black] (0,0) -- (0,3.6) node[left] {population};
  \node[anchor=north, font=\scriptsize] at (3.3,-0.5) {rotational level number};
  % bars peaking around ell=3
  \foreach \l/\h in {0/0.9,1/1.9,2/2.7,3/3.0,4/2.8,5/2.3,6/1.6,7/1.0,8/0.55,9/0.28}{
    \fill[acc!18, draw=acc] (\l*0.7+0.25,0) rectangle (\l*0.7+0.65,\h);
  }
  \foreach \l in {0,3,6,9}
    \node[anchor=north, font=\scriptsize] at (\l*0.7+0.45,0) {\l};
  \draw[black, dashed] (2.35,0) -- (2.35,3.35) node[above, font=\scriptsize] {peak};
\end{tikzpicture}
$$

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](/condensed-matter/molecular-spectra/anharmonicity-and-rovibrational-structure)
sharpens the rigid-rotor and harmonic-oscillator pictures into the anharmonic,
centrifugally distorted structure of a real band.