Rotational Spectroscopy
The regions of the electromagnetic spectrum and which molecular motion each one probes.
Classification of molecules by their moments of inertia: linear, spherical top, symmetric top, asymmetric top.
The gross selection rule — why a permanent dipole moment is required.
The rigid rotator: energy levels, the rotational constant \(B\), and the equally spaced spectrum.
Line intensities and the Boltzmann population maximum.
Isotope substitution and the non-rigid rotator.
7.1 Classification of Rotating Molecules
Molecules are classified by their three principal moments of inertia \(I_{a}\), \(I_{b}\), \(I_{c}\).
| Type | Moments | Examples | Microwave active? |
|---|---|---|---|
| Linear | \(I_{a}=0\), \(I_{b}=I_{c}\) | HCl, HBr, HCN, C2H2, OCS, CO2 | only if polar |
| Spherical top | \(I_{a}=I_{b}=I_{c}\) | CH4, SF6, CCl4 | no (no dipole) |
| Symmetric top — prolate | \(I_{a}| CH3Cl, CH3F, CH3CN, NH3 | yes | |
| Symmetric top — oblate | \(I_{a}>I_{b}=I_{c}\) | BF3, BCl3 | only if polar |
| Asymmetric top | \(I_{a}\ne I_{b}\ne I_{c}\) | H2O, H2CO, CH3OH, CH2CHCl | yes |
The prolate list included “CH3CH”, which is not a molecule. This is almost certainly CH3CN (acetonitrile), the standard textbook prolate symmetric top.
Most molecules are asymmetric tops. Spherical tops are the ones to remember for a different reason: their symmetry guarantees zero dipole moment, so they are microwave-inactive regardless of anything else.
Note also that CO2 is linear but non-polar (the two C=O dipoles cancel), so it too has no pure rotational spectrum — a favourite trap.
7.2 The Rigid Diatomic Rotator
Consider a rigid rotator A–B with masses \(m_{1}\) and \(m_{2}\) and fixed bond length \(r_{0}\) (no vibration).
The bond length splits about the centre of mass C: \begin{equation} r_{0} = r_{1} + r_{2} , \qquad m_{1}r_{1} = m_{2}r_{2} . \label{eq:com-condition} \end{equation} Substituting \(r_{2} = r_{0} - r_{1}\) into the second relation gives \(r_{1}(m_{1}+m_{2}) = m_{2}r_{0}\), so \begin{equation} r_{1} = \frac{m_{2}r_{0}}{m_{1}+m_{2}} , \qquad r_{2} = \frac{m_{1}r_{0}}{m_{1}+m_{2}} . \label{eq:r1r2} \end{equation}
7.2.1 Moment of inertia
\begin{align} I &= m_{1}r_{1}^{2} + m_{2}r_{2}^{2} = m_{2}r_{2}r_{1} + m_{1}r_{1}r_{2} = r_{1}r_{2}\left(m_{1}+m_{2}\right) \nonumber\\ &= \frac{m_{2}r_{0}}{m_{1}+m_{2}}\cdot\frac{m_{1}r_{0}}{m_{1}+m_{2}} \left(m_{1}+m_{2}\right) = \frac{m_{1}m_{2}}{m_{1}+m_{2}}\,r_{0}^{2} , \label{eq:moment-inertia} \end{align} that is \begin{equation} \boxed{\;I = \mred\,r_{0}^{2}\;}, \qquad \mred = \frac{m_{1}m_{2}}{m_{1}+m_{2}} \quad\text{(reduced mass)} . \label{eq:I-mu} \end{equation}
7.2.2 Energy levels
The Hamiltonian of a rigid rotator is purely rotational kinetic energy, \begin{equation} \hat H = \frac{\hat J^{2}}{2I} , \label{eq:rotor-hamiltonian} \end{equation} so the time-independent Schrödinger equation reads \(\dfrac{\hat J^{2}}{2I}\Psi = E_{J}\Psi\). The eigenfunctions are the spherical harmonics \(Y_{Jm}\), with \(\hat J^{2}Y_{Jm} = J(J+1)\hbar^{2}Y_{Jm}\). Hence \begin{equation} \boxed{\; E_{J} = \frac{\hbar^{2}J(J+1)}{2I} = \frac{h^{2}J(J+1)}{8\pi^{2}I}\;}, \qquad J = 0,1,2,3,\dots \label{eq:rotational-energy} \end{equation} where \(J\) is the rotational quantum number. Each level is \((2J+1)\)-fold degenerate, corresponding to the \(2J+1\) allowed orientations \(m = -J,\dots,+J\).
Writing \(\mathcal{E} \equiv h^{2}/8\pi^{2}I\), the first few levels are \[ E_{0} = 0, \qquad E_{1} = 2\mathcal{E}, \qquad E_{2} = 6\mathcal{E}, \qquad E_{3} = 12\mathcal{E} , \] with successive gaps \[ \Delta E_{0\to1} = 2\mathcal{E}, \qquad \Delta E_{1\to2} = 4\mathcal{E}, \qquad \Delta E_{2\to3} = 6\mathcal{E} . \]
The gaps increase linearly with \(J\), even though the levels themselves go as \(J(J+1)\). This is why the rotational spectrum consists of equally spaced lines — see \(\eqref{eq:line-spacing}\).
7.2.3 Angular momentum, angular velocity and period
\begin{align} L &= \hbar\sqrt{J(J+1)} = \frac{h\sqrt{J(J+1)}}{2\pi} , \label{eq:rotor-L}\\ \omega &= \frac{L}{I} = \frac{h\sqrt{J(J+1)}}{2\pi I} , \label{eq:rotor-omega}\\ \nu_{\text{rot}} &= \frac{\omega}{2\pi} = \frac{h\sqrt{J(J+1)}}{4\pi^{2}I} , \qquad T = \frac{2\pi}{\omega} . \label{eq:rotor-freq} \end{align}
7.3 The Rotational Constant and the Spectrum
Spectroscopists work in wavenumbers. Dividing \(\eqref{eq:rotational-energy}\) by \(hc\): \begin{equation} \varepsilon_{J} = \frac{E_{J}}{hc} = \frac{h\,J(J+1)}{8\pi^{2}Ic} = B\,J(J+1) \ \mathrm{cm}^{-1} , \label{eq:epsilon-J} \end{equation} where the rotational constant is \begin{equation} \boxed{\;B = \frac{h}{8\pi^{2}Ic} \ \mathrm{cm}^{-1}\;} \label{eq:B-constant} \end{equation} So the term values are \[ \varepsilon_{J} = 0,\ 2B,\ 6B,\ 12B,\ 20B,\dots \qquad\text{for } J = 0,1,2,3,4,\dots \]
7.3.1 Selection rule and line positions
\begin{equation} \Delta J = \pm 1 . \label{eq:rot-selection} \end{equation} For absorption, \(J \to J+1\), so \begin{equation} \Delta\varepsilon = B\left[(J+1)(J+2) - J(J+1)\right] \quad\Longrightarrow\quad \boxed{\;\Delta\varepsilon = 2B(J+1) \ \mathrm{cm}^{-1}\;} \label{eq:line-position} \end{equation} giving lines at \[ J{=}0\to1: 2B, \qquad J{=}1\to2: 4B, \qquad J{=}2\to3: 6B, \qquad \dots \]
\begin{equation} \boxed{\; \text{Separation between successive lines} = 2B \ \mathrm{cm}^{-1} \ \text{(constant)}\;} \label{eq:line-spacing} \end{equation}
Almost every rotational-spectroscopy question reduces to one of these three steps:
measure the line spacing \(\Rightarrow\) \(2B\);
\(B = h/8\pi^{2}Ic\) \(\Rightarrow\) \(I\);
\(I = \mred r_{0}^{2}\) \(\Rightarrow\) bond length \(r_{0}\).
The moment of inertia of a typical light diatomic is of order \(4.6\times 10^{-48}\,\mathrm{kg}\,\mathrm{m}^{2}\). Estimate the first two rotational transition energies.
Solution. With \(\mathcal{E} = h^{2}/8\pi^{2}I\), \[ \Delta E_{0\to1} = \frac{2h^{2}}{8\pi^{2}I} = \frac{2\times(6.6\times10^{-34})^{2}} {8\pi^{2}\times 1.6\times10^{-19}\times 4.597\times10^{-48}} = 1.5\times 10^{-2}\,\mathrm{eV} , \] \[ \Delta E_{1\to2} = 2\,\Delta E_{0\to1} \times \frac{2}{1} \ \Rightarrow \ 4.5\times 10^{-2}\,\mathrm{eV} . \]
The source labelled both of these results “energy of microwave”. They are not: \(1.5\times 10^{-2}\,\mathrm{eV}\) corresponds to \(121\,\mathrm{cm}^{-1}\), which is far infrared. Microwave photons are of order \(10^{-4}\)–\(10^{-3}\) \(\mathrm{eV}\) (\(1\)–\(10\) \(\mathrm{cm}^{-1}\)), as Table 6.1 itself states.
The value \(I = 4.597\times 10^{-48}\,\mathrm{kg}\,\mathrm{m}^{2}\) is that of H2, which is a doubly unfortunate illustration: H2 is homonuclear and therefore has no pure rotational spectrum at all, and its moment of inertia is exceptionally small, putting its rotational transitions in the far IR rather than the microwave.
Recommendation: keep the H2 arithmetic as a “smallest possible \(I\)” limiting case, but use CO (\(I = 1.46\times 10^{-46}\,\mathrm{kg}\,\mathrm{m}^{2}\), \(B = 1.93\,\mathrm{cm}^{-1}\), \(\Delta E_{0\to1} = 4.8\times 10^{-4}\,\mathrm{eV}\)) as the molecule that demonstrates microwave activity. I can rewrite this example that way on request.
The H2 molecule has reduced mass \(8.35\times 10^{-28}\,\mathrm{kg}\) and equilibrium internuclear distance \(0.742\times 10^{-10}\,\mathrm{m}\). Calculate the energy (in meV), angular momentum (SI), angular frequency (\(\mathrm{s}^{-1}\)) and period (s) of the second excited rotational level.
Solution. Moment of inertia. \[ I = \mred r_{0}^{2} = 8.35\times10^{-28}\times\left(0.742\times10^{-10}\right)^{2} = 4.597\times 10^{-48}\,\mathrm{kg}\,\mathrm{m}^{2} . \]
Energy. From \(\eqref{eq:rotational-energy}\), \[ E_{J} = \frac{h^{2}J(J+1)}{8\pi^{2}I} = \frac{\left(6.62\times10^{-34}\right)^{2}J(J+1)} {8\pi^{2}\times 4.597\times10^{-48}\times 1.6\times10^{-19}}\ \eV = 7.57\,J(J+1)\ \mathrm{meV} . \] The second excited state is \(J=2\) (since \(J=0\) is the ground state), so \[ E_{2} = 7.57 \times 6 = 45.42\,\mathrm{meV} . \]
Angular momentum. From \(\eqref{eq:rotor-L}\) with \(J=2\), \[ L = \frac{h\sqrt{6}}{2\pi} = 2.58\times 10^{-34}\,\mathrm{J}\,\mathrm{s} . \]
Angular frequency and period. \[ \omega = \frac{L}{I} = \frac{6.6\times10^{-34}\sqrt6}{2\pi\times4.597\times10^{-48}} = 5.6\times 10^{13}\,\mathrm{s}^{-1} , \qquad T = \frac{2\pi}{\omega} = 1.12\times 10^{-13}\,\mathrm{s} . \]
Watch the phrase “second excited state”: the ground state is \(J=0\), so first excited is \(J=1\) and second excited is \(J=2\). Reading it as \(J=3\) is the commonest slip.
Pure rotational absorption lines of a diatomic molecule are observed at \(88.08\), \(108.78\), \(129.35\) and \(150.08\,\mathrm{cm}^{-1}\).
Calculate the moment of inertia.
Calculate the bond length, taking the molecule to be \(^{12}\mathrm{C}^{16}\mathrm{O}\).
Solution. (a) The successive separations are \[ 108.78 - 88.08 = 20.70, \quad 129.35 - 108.78 = 20.57, \quad 150.08 - 129.35 = 20.73 \ \mathrm{cm}^{-1} , \] so the mean spacing is \[ 2B = \frac{20.70 + 20.57 + 20.73}{3} = 20.66\,\mathrm{cm}^{-1} = 2066\,\mathrm{m}^{-1} . \] From \(\eqref{eq:B-constant}\), \[ 2B = \frac{2h}{8\pi^{2}Ic} \quad\Longrightarrow\quad I = \frac{2h}{8\pi^{2}c\,(2B)} = \frac{2\times 6.62\times10^{-34}} {8\pi^{2}\times 3\times10^{8}\times 2066} = 2.7\times 10^{-47}\,\mathrm{kg}\,\mathrm{m}^{2} . \]
(b) The reduced mass of \(^{12}\)C\(^{16}\)O is \[ \mred = \frac{m_{\mathrm C}m_{\mathrm O}}{m_{\mathrm C}+m_{\mathrm O}} = \frac{12\times16}{28}\times\frac{1}{6.022\times10^{23}}\ \mathrm{g} = 1.138\times 10^{-23}\,\mathrm{g} = 1.138\times 10^{-26}\,\mathrm{kg} , \] so \[ r_{0} = \sqrt{\frac{I}{\mred}} = \sqrt{\frac{2.7\times10^{-47}}{1.138\times10^{-26}}} = \sqrt{2.373\times10^{-21}} = 4.87\times 10^{-11}\,\mathrm{m} = 0.487\,\mathrm{Å} . \]
(i) Arithmetic. The source gave the answer as “\(4.8\,\mathrm{nm}\)”. The correct value is \(4.87\times 10^{-11}\,\mathrm{m} = 0.0487\,\mathrm{nm}\) — a factor of \(100\) out. A bond length of \(4.8\,\mathrm{nm}\) would be some fifty times a typical molecular bond.
(ii) The data are not CO's. A spacing of \(2B = 20.66\,\mathrm{cm}^{-1}\) implies \(B = 10.3\,\mathrm{cm}^{-1}\), whereas the literature value for CO is \(B = 1.93\,\mathrm{cm}^{-1}\). Indeed \(B \approx 10.3\,\mathrm{cm}^{-1}\) is close to HCl (\(B = 10.59\,\mathrm{cm}^{-1}\)).
The internal contradiction is visible within this chapter: a later example (page 7.4) uses the correct CO absorption frequency and recovers \(r_{0} = 1.13\,\mathrm{Å}\), the accepted CO bond length — against the \(0.487\,\mathrm{Å}\) obtained here.
Recommendation: either relabel this example as HCl (and recompute \(\mred\) with \(m_{\mathrm H}\), \(m_{\mathrm{Cl}}\)), or replace the line positions with genuine CO values (\(3.86\), \(7.72\), \(11.58\), \(15.44\,\mathrm{cm}^{-1}\)).
Carbon monoxide absorbs at \(1.153\times 10^{11}\,\mathrm{Hz}\) due to the \(J=0 \to J=1\) transition.
What is the wavelength, and in which part of the spectrum does it lie?
What is the energy in eV?
Calculate the reduced mass.
Given \(E = \dfrac{\hbar^{2}}{2\mred r^{2}}J(J+1)\), find the interatomic distance.
Solution. (i) \[ \lambda = \frac{c}{\nu} = \frac{3\times10^{8}}{1.153\times10^{11}} = 2.6\times 10^{-3}\,\mathrm{m} = 2.6\,\mathrm{mm} , \] which lies in the microwave region.
(ii) \[ E = \frac{1240\,\mathrm{eV}\,\mathrm{nm}}{\lambda\,[\mathrm{nm}]} = \frac{1240}{2.6\times10^{6}} = 4.77\times 10^{-4}\,\mathrm{eV} . \]
(iii) \[ \mred = \frac{m_{\mathrm C}m_{\mathrm O}}{m_{\mathrm C}+m_{\mathrm O}} = \frac{12\times16}{12+16}\times 1.67\times10^{-27} = 1.145\times 10^{-26}\,\mathrm{kg} . \]
(iv) With \(E_{0}=0\) and \(E_{1} = \dfrac{2\hbar^{2}}{2\mred r^{2}} = \dfrac{\hbar^{2}}{\mred r^{2}}\), \[ \Delta E = E_{1}-E_{0} = \frac{\hbar^{2}}{\mred r^{2}} \quad\Longrightarrow\quad r = \left(\frac{\hbar^{2}c^{2}}{\mred c^{2}\,\Delta E}\right)^{1/2} . \] Using natural units (\(\hbar c = 197.3\,\mathrm{MeV}\,\mathrm{fm}\), \(\mred c^{2} = 6.857 \times 931.5\,\mathrm{MeV}\), \(\Delta E = 477\times 10^{-12}\,\mathrm{MeV}\)): \[ r = \left(\frac{197.3^{2}}{6.857\times931.5\times477\times10^{-12}}\right)^{1/2} = 1.13\,\mathrm{Å} , \] in excellent agreement with the accepted CO bond length of \(1.128\,\mathrm{Å}\).
Note how much faster the natural-units route is: expressing everything in \(\mathrm{MeV}\) and \(\mathrm{fm}\) via \(\hbar c = 197.3\,\mathrm{MeV}\,\mathrm{fm}\) avoids a stack of powers of ten. Worth practising — it also appears in nuclear physics questions.
At what temperature would the average kinetic energy of the molecules in a hydrogen sample equal the H2 bond dissociation energy of \(4.5\,\mathrm{eV}\)?
Solution. Equating \(\tfrac32 kT\) to the dissociation energy, \[ \frac{3}{2}kT = 4.5\,\mathrm{eV} \quad\Longrightarrow\quad T = \frac{2}{3}\cdot\frac{4.5}{8.62\times10^{-5}} = 3.5\times 10^{4}\,\mathrm{K} . \]
The \(4.5\,\mathrm{eV}\) here is the H2 molecular bond dissociation energy, not the \(13.6\,\mathrm{eV}\) ionization energy of the hydrogen atom from Chapter 1. The source called it simply “the binding energy of hydrogen”, which invites confusion between the two.
7.4 Intensity of Spectral Lines
The intensity of a rotational line is governed by the population of the originating level, which follows the Boltzmann distribution weighted by the \((2J+1)\) degeneracy: \begin{equation} \boxed{\; \frac{N_{J}}{N_{0}} = (2J+1)\, \exp\!\left[-\frac{E_{J}-E_{0}}{k_{B}T}\right]\;} \label{eq:boltzmann-rot} \end{equation}
Two competing factors act here: the degeneracy \((2J+1)\) grows linearly with \(J\), while the exponential falls. Their product peaks at an intermediate \(J\). Differentiating and setting the result to zero: \begin{equation} \boxed{\;J_{\max} = \sqrt{\frac{k_{B}T}{2hcB}} - \frac{1}{2}\;} \label{eq:Jmax} \end{equation}
An earlier version of this figure plotted normalised populations computed with \(Bhc/k_{B}T = 0.09\), which corresponds to \(B \approx 18.8\,\mathrm{cm}^{-1}\) and so did not match either of the labelled values. The curves above are computed directly from (6.12) with \(k_{B}T/hc = 208.5\,\mathrm{cm}^{-1}\), and their maxima now agree with \(J_{\max}\) from \(\eqref{eq:Jmax}\): \(\sqrt{208.5/10} - \tfrac12 = 4.07\) for \(B = 5\,\mathrm{cm}^{-1}\), and \(\sqrt{208.5/20} - \tfrac12 = 2.73\) for \(B = 10\,\mathrm{cm}^{-1}\).
The ratio of two line intensities is \begin{equation} \frac{I_{J}}{I_{J'}} = \frac{2J+1}{2J'+1}\, \exp\!\left[-\frac{E_{J}-E_{J'}}{k_{B}T}\right] . \label{eq:intensity-ratio} \end{equation} At high temperature \(\left(E_{J}-E_{J'}\right)/k_{B}T \ll 1\), the exponential tends to unity and \begin{equation} \frac{I_{J}}{I_{J'}} \approx \frac{2J+1}{2J'+1} . \label{eq:intensity-ratio-hot} \end{equation}
Note from \(\eqref{eq:Jmax}\) that \(J_{\max}\) grows as \(\sqrt T\) and falls as \(1/\sqrt B\). Heavier molecules (large \(I\), small \(B\)) have their intensity maximum at much higher \(J\) — which is why their spectra show many more observable lines.
For a transition \(J=4 \to J=3\), calculate the ratio of the population of the final state to that of the initial state at \(5000\,\mathrm{K}\), given that the transition energy corresponds to a wavelength \(\lambda\).
Solution. The ratio, including degeneracies, is \[ \frac{N_{F}}{N_{I}} = \frac{2J_{F}+1}{2J_{I}+1}\, \exp\!\left[-\frac{E_{F}-E_{I}}{k_{B}T}\right] = \frac{7}{9}\,\exp\!\left[+\frac{hc}{\lambda k_{B}T}\right] , \] since the final state \(J=3\) lies below the initial state \(J=4\), making \(E_{F}-E_{I} = -hc/\lambda\).
The source's solution wrote \(\frac{N_F}{N_I} = \frac79 e^{-hc/\lambda k_B T}\), i.e.\ with the exponent negative, and obtained \(8.2\times 10^{-5}\).
That sign is wrong. The lower level is always the more populated one at thermal equilibrium, so the ratio must exceed the degeneracy factor \(7/9\), not fall five orders of magnitude below it.
Separately, the wavelength given in the source was \(3122\,\mathrm{Å}\) (\(312\,\mathrm{nm}\), ultraviolet), corresponding to \(3.97\,\mathrm{eV}\). A pure rotational transition is of order \(10^{-3}\)–\(10^{-4}\) \(\mathrm{eV}\); \(3.97\,\mathrm{eV}\) is an electronic transition energy. With realistic rotational data the exponent is tiny and the answer reduces to the degeneracy ratio \(7/9 = 0.78\), consistent with \(\eqref{eq:intensity-ratio-hot}\).
Recommendation: either supply a physically sensible rotational wavelength (a few mm) and keep the correct sign, or relabel the exercise as a generic Boltzmann-ratio problem not tied to rotational levels.
7.5 The Effect of Isotope Substitution
Substituting a heavier isotope changes the reduced mass but leaves the bond length essentially unaltered (the electronic structure is unchanged). Writing primed quantities for the heavier isotopologue, \(\mred' > \mred\), and since \begin{equation} B \propto \frac{1}{I} \propto \frac{1}{\mred} \quad\Longrightarrow\quad \frac{B}{B'} = \frac{\mred'}{\mred} > 1 \quad\Longrightarrow\quad B > B' . \label{eq:isotope-B} \end{equation}
The heavier isotopologue therefore has the smaller rotational constant, and its whole spectrum is compressed towards lower wavenumber — shifted to the left, with the shift growing linearly along the series \(2B\), \(4B\), \(6B\), …
Isotope shifts in rotational spectra are how the masses of isotopes — and their relative abundances, from line intensities — are measured. It is also how interstellar \(^{13}\)CO is distinguished from \(^{12}\)CO in radio astronomy.
What is the change in the rotational constant \(B\) of H2 when (i) one hydrogen atom, (ii) both hydrogen atoms are replaced by deuterium? Find also the shift in the \(J=0\to1\) line in each case.
Solution. For H2, \(\mred_{\mathrm{HH}} = \dfrac{m_{H}m_{H}}{2m_{H}} = \dfrac{m_{H}}{2}\), and \(\Delta\varepsilon(J{=}0\to1) = 2B\).
(i) HD. With \(m_{D} = 2m_{H}\), \[ \mred_{\mathrm{HD}} = \frac{m_{H}\cdot 2m_{H}}{m_{H}+2m_{H}} = \frac{2m_{H}}{3} , \qquad \frac{B'}{B} = \frac{\mred_{\mathrm{HH}}}{\mred_{\mathrm{HD}}} = \frac{m_{H}/2}{2m_{H}/3} = \frac{3}{4} . \] So \(B' = \tfrac34 B\), a change of \[ B - B' = \tfrac14 B . \] The line moves from \(2B\) to \(2B' = \tfrac32 B\), a shift of \[ 2B - 1.5B = 0.5B . \]
(ii) D2. \[ \mred_{\mathrm{DD}} = \frac{2m_{H}\cdot 2m_{H}}{4m_{H}} = m_{H} , \qquad \frac{B'}{B} = \frac{m_{H}/2}{m_{H}} = \frac{1}{2} , \] so \(B' = \tfrac12 B\), a change of \(B - B' = \tfrac12 B\). The line moves from \(2B\) to \(2B' = B\), a shift of \(B\).
As an exercise in the \(\mred\) algebra this is exact. Physically, though, both H2 and D2 are homonuclear and so have no pure rotational spectrum. Only HD — which has a small but non-zero dipole moment, because the centre of mass no longer coincides with the centre of charge — is genuinely microwave active. Worth stating explicitly so students do not draw the wrong general conclusion.
The \(J=0\to1\) transition occurs at \(1.153\times 10^{11}\,\mathrm{s}^{-1}\) in \(^{12}\)C\(^{16}\)O and at \(1.102\times 10^{11}\,\mathrm{s}^{-1}\) in \(^{?}\)C\(^{16}\)O. Estimate the mass number and reduced mass of the unknown isotope.
Solution. Since \(\nu = 2Bc\) for the \(J=0\to1\) line and \(B \propto 1/\mred\), \[ \frac{B_{1}}{B_{2}} = \frac{1.153\times10^{11}}{1.102\times10^{11}} = 1.046 \quad\Longrightarrow\quad \frac{\mred_{2}}{\mred_{1}} = 1.046 . \] For \(^{12}\)C\(^{16}\)O, \[ \mred_{1} = \frac{12\times16}{28\times6.022\times10^{23}} = 1.138\times 10^{-23}\,\mathrm{g} , \] and for the unknown, of mass number \(A\), \[ \mred_{2} = \frac{16A}{(16+A)\times6.022\times10^{23}}\ \mathrm{g} . \] Setting \(\mred_{2} = 1.046\,\mred_{1}\): \[ \frac{16A}{16+A} = 1.046 \times 1.138\times10^{-23}\times6.022\times10^{23} = 7.17 \quad\Longrightarrow\quad A = 12.9 . \] Hence \[ \boxed{A \approx 13} \quad\text{--- the isotope is } ^{13}\mathrm{C} , \] and \[ \mred_{2} = \frac{16\times13}{29\times6.022\times10^{23}} = 1.19\times 10^{-23}\,\mathrm{g} . \]
The source stopped at \(A = 12.85\) without identifying the isotope; since mass numbers are integers the physical answer is \(^{13}\)C. The source also printed the final reduced mass with “\(6.022\times10^{-23}\)” — the Avogadro constant is \(6.022\times10^{+23}\).
7.6 The Non-Rigid Rotator
In reality the separation between successive lines is not exactly constant: it decreases slightly as \(J\) increases. The reason is centrifugal distortion. A rapidly rotating molecule experiences a centrifugal force that stretches the bond, increasing \(r_{0}\) and hence \(I\), and therefore decreasing \(B\). The energy of a non-rigid rotator is consequently a little lower than the rigid-rotator prediction, increasingly so at high \(J\).
The corrected term values are \begin{equation} \boxed{\; \varepsilon_{J} = B\,J(J+1) - D\,J^{2}(J+1)^{2} \ \mathrm{cm}^{-1}\;} \label{eq:non-rigid} \end{equation} where \(D\) is the centrifugal distortion constant, with \begin{equation} D \ll B , \qquad D \approx 10^{-3}B . \label{eq:D-magnitude} \end{equation}
The transition wavenumbers for \(J \to J+1\) become \begin{equation} \boxed{\; \Delta\varepsilon = 2B(J+1) - 4D(J+1)^{3} \ \mathrm{cm}^{-1}\;} \label{eq:non-rigid-lines} \end{equation}
The distortion constant is not independent: it is related to the vibrational wavenumber \(\bar\omega\) of the bond by \[ D = \frac{4B^{3}}{\bar\omega^{2}} . \] A stiffer bond (large \(\bar\omega\)) stretches less and gives a smaller \(D\). This links Chapters 7 and 7 and is occasionally asked directly.
7.7 Applications of Rotational Spectra
Rotational spectra are used to determine molecular parameters with very high precision:
the rotational constant \(B\), directly from the line spacing;
the moment of inertia \(I\), from \(B = h/8\pi^{2}Ic\);
the bond length \(r_{0}\), from \(I = \mred r_{0}^{2}\);
the reduced mass, and hence isotopic mass numbers and abundances;
the bond stiffness, via the distortion constant \(D = 4B^{3}/\bar\omega^{2}\).
Microwave spectroscopy gives the most accurate bond lengths of any technique — frequencies can be measured to one part in \(10^{10}\). This is why it is the reference method for small-molecule geometry, and why radio astronomers can identify individual molecular species in interstellar clouds from their rotational lines alone.
Formula Summary
Gross selection rule permanent electric dipole moment required. Active: HCl, CO, H2O, NH3. \ Inactive: H2, N2, O2, CO2, CH4
Specific selection rule \(\Delta J = \pm1\)
Rotor classification
linear \(I_a=0,I_b=I_c\); \ spherical \(I_a=I_b=I_c\); \
prolate \(I_a
Rigid rotator \[ I = \mred r_0^{2}, \qquad \mred = \frac{m_1m_2}{m_1+m_2}, \qquad E_J = \frac{\hbar^{2}J(J+1)}{2I} = \frac{h^{2}J(J+1)}{8\pi^{2}I} \] degeneracy \(= 2J+1\); \(L = \hbar\sqrt{J(J+1)}\), \ \(\omega = L/I\), \ \(T = 2\pi/\omega\)
Wavenumbers \[ \varepsilon_J = BJ(J+1)\ \mathrm{cm}^{-1}, \qquad B = \frac{h}{8\pi^{2}Ic}\ \mathrm{cm}^{-1} \] \[ \varepsilon_J = 0,\ 2B,\ 6B,\ 12B,\ 20B,\dots \] Line positions \(\Delta\varepsilon = 2B(J+1)\): lines at \(2B, 4B, 6B,\dots\); \ spacing \(= 2B\), constant
Intensities \[ \frac{N_J}{N_0} = (2J+1)e^{-(E_J-E_0)/k_BT}, \qquad J_{\max} = \sqrt{\frac{k_BT}{2hcB}} - \frac12 \] at high \(T\): \ \(I_J/I_{J'} \approx (2J+1)/(2J'+1)\)
Isotope substitution \(B \propto 1/\mred\); heavier isotopologue \(\Rightarrow\) smaller \(B\) \(\Rightarrow\) spectrum shifts to lower wavenumber
Non-rigid rotator (centrifugal distortion) \[ \varepsilon_J = BJ(J+1) - DJ^{2}(J+1)^{2}, \qquad \Delta\varepsilon = 2B(J+1) - 4D(J+1)^{3} \] \[ D \approx 10^{-3}B, \qquad D = \frac{4B^{3}}{\bar\omega^{2}} \] spacing decreases slowly with \(J\)
Working method line spacing \(\to 2B\) \ \(\to\) \ \(I = h/8\pi^{2}Bc\) \ \(\to\) \ \(r_0 = \sqrt{I/\mred}\)
Practice Problems
These problems are original to this book. They cover the standard examinable manipulations of rotational spectroscopy — dipole selection rules, extracting \(B\) and \(r_{0}\) from a line spacing, isotope shifts, centrifugal distortion and the thermal intensity envelope. Answers with the key steps follow; full solutions are deliberately withheld, because the value of a problem set lies in committing to a method before checking it.
Constants: \(h = 6.626\times 10^{-34}\,\mathrm{J}\,\mathrm{s}\); \(c = 2.998\times 10^{10}\,\mathrm{cm}\,\mathrm{s}^{-1}\); \(k_{B} = 1.381\times 10^{-23}\,\mathrm{J}\,\mathrm{K}^{-1}\); \(1\,\mathrm{u} = 1.6605\times 10^{-27}\,\mathrm{kg}\); \(k_{B}T/hc = 208.5\,\mathrm{cm}^{-1}\) at \(300\,\mathrm{K}\).
Which of the following will show a pure rotational (microwave) absorption spectrum, and why? \[ \text{N}_{2},\quad \text{CO},\quad \text{CH}_{4},\quad \text{CHCl}_{3}, \quad \text{CS}_{2},\quad \text{H}_{2}\text{S},\quad \text{BF}_{3}. \]
The pure rotational spectrum of a diatomic molecule consists of equally spaced lines \(3.845\,\mathrm{cm}^{-1}\) apart. Its reduced mass is \(1.139\times 10^{-26}\,\mathrm{kg}\). Calculate (a) the rotational constant \(B\); (b) the moment of inertia \(I\); (c) the bond length \(r_{0}\); (d) the transition giving the most intense line at \(300\,\mathrm{K}\); (e) the number of revolutions per second the molecule makes in the \(J=1\) and \(J=10\) states.
The rotational constant of H35Cl is \(10.5909\,\mathrm{cm}^{-1}\). Assuming the bond length is unchanged by isotopic substitution, calculate \(B\) for H37Cl and for D35Cl. Atomic masses (u): \(^{1}\)H \(=1.0078\), \(^{2}\)D \(=2.0141\), \(^{35}\)Cl \(=34.9689\), \(^{37}\)Cl \(=36.9659\).
Carbon monoxide has \(B = 1.9225\,\mathrm{cm}^{-1}\) and a centrifugal distortion constant \(D = 6.12\times 10^{-6}\,\mathrm{cm}^{-1}\). Estimate the vibrational wavenumber and the force constant of the bond. The equilibrium vibrational wavenumber of CO is measured to be \(2170\,\mathrm{cm}^{-1}\); comment on the agreement, and state what the relation you used assumes.
A radio telescope observes the \(J = 1 \to 0\) line of \(^{12}\)C\(^{16}\)O in a molecular cloud. (a) At what wavenumber does the corresponding line of \(^{13}\)C\(^{16}\)O appear? Take \(^{12}\)C \(=12.0000\), \(^{13}\)C \(=13.0034\), \(^{16}\)O \(=15.9949\) u, and assume the bond length is the same for both. (b) What resolving power \(\tilde\nu/\Delta\tilde\nu\) is needed to separate the two lines? (c) Suggest how observing several successive rotational lines of the same species would let you estimate the temperature of the cloud.
Answers
P6.1 A pure rotational spectrum requires a permanent electric dipole moment. Active: CO (heteronuclear diatomic); CHCl3 (\(C_{3v}\), dipole along the C–H axis); H2S (bent, like water). Inactive: N2 (homonuclear); CH4 (\(T_{d}\)); CS2 (linear and symmetric, S=C=S); BF3 (trigonal planar, \(D_{3h}\)). Note that CS2 and BF3 contain polar bonds but have no net dipole — the bond moments cancel by symmetry.
P6.2 (a) The spacing is \(2B\), so \(B = 1.9225\,\mathrm{cm}^{-1}\). (b) \(I = h/8\pi^{2}cB = 1.455\times 10^{-46}\,\mathrm{kg}\,\mathrm{m}^{2}\). (c) \(r_{0} = \sqrt{I/\mred} = 1.130\times 10^{-10}\,\mathrm{m} = 1.130\,\mathrm{Å}\) (the molecule is CO). (d) \(J_{\max} = \sqrt{k_{B}T/2hcB} - \tfrac12 = \sqrt{208.5/3.845} - 0.5 = 6.9\), so the most intense line is \(J = 7 \to 8\). (e) From \(\tfrac12 I\omega^{2} = hcBJ(J+1)\), \(f = \omega/2\pi\): \(f(J{=}1) = 1.6\times 10^{11}\,\mathrm{s}^{-1}\) and \(f(J{=}10) = 1.2\times 10^{12}\,\mathrm{s}^{-1}\).
P6.3 \(B \propto 1/\mred\). Reduced masses: \(\mred(\text{H}^{35}\text{Cl}) = 0.97207\), \(\mred(\text{H}^{37}\text{Cl}) = 0.97366\), \(\mred(\text{D}^{35}\text{Cl}) = 1.90437\) u. Hence \(B(\text{H}^{37}\text{Cl}) = 10.574\,\mathrm{cm}^{-1}\) and \(B(\text{D}^{35}\text{Cl}) = 5.406\,\mathrm{cm}^{-1}\). The chlorine substitution moves \(B\) by only \(0.16\,\%\) because the light hydrogen carries almost all of the reduced mass; deuteration nearly halves it.
P6.4 From \(D = 4B^{3}/\tilde\omega^{2}\), \(\tilde\omega = \sqrt{4B^{3}/D} = \sqrt{4\times7.106/6.12\times10^{-6}} = 2155\,\mathrm{cm}^{-1}\), and \(k = 4\pi^{2}c^{2}\tilde\omega^{2}\mred = 1.88\times 10^{3}\,\mathrm{N}\,\mathrm{m}^{-1}\). This is about \(0.7\,\%\) below the measured \(2170\,\mathrm{cm}^{-1}\). The relation assumes a harmonic bond — it comes from balancing the centrifugal force against a Hooke's-law restoring force — so it returns the harmonic wavenumber, while a real bond softens as it stretches.
P6.5 (a) \(\mred(^{12}\text{CO}) = 6.8562\), \(\mred(^{13}\text{CO}) = 7.1724\) u, so \(B(^{13}\text{CO}) = 1.9225\times(6.8562/7.1724) = 1.8377\,\mathrm{cm}^{-1}\). The \(J=1\to0\) lines lie at \(2B\): \(3.845\,\mathrm{cm}^{-1}\) and \(3.675\,\mathrm{cm}^{-1}\), a separation of \(0.170\,\mathrm{cm}^{-1}\). (b) \(\tilde\nu/\Delta\tilde\nu = 3.845/0.170 \approx 23\) — undemanding, and the higher-\(J\) lines separate proportionally further, since the shift grows as \((J+1)\). Resolution is never the difficulty here; sensitivity is, because \(^{13}\)C is only about \(1\,\%\) as abundant. (c) The intensity envelope over \(J\) peaks at \(J_{\max} \approx \sqrt{k_{B}T/2hcB} - \tfrac12\). Identifying the strongest line, or better, fitting the Boltzmann envelope through several lines, returns \(T\) directly.
Previous Year Questions
The source carries a single combined previous-year bank headed “Chapter 6 to Chapter 9”, covering rotational, vibrational, Raman and electronic spectroscopy together. Rather than split it, it is typeset in full — GATE, CSIR-NET/JRF and JEST, with worked solutions — at the end of Chapter 10. A set of original practice problems on this chapter's material appears immediately above.
The questions bearing most directly on this chapter are, in that bank: GATE Q1 (symmetric top), Q3 (rigid-rotator levels), Q4 (equal rotational energies), Q5 (line spacing to Raman shift), Q6 (methanol eigenvalues), Q11 (spectral ranges), Q20 (energy scaling) and Q23 (rotational levels of HD); NET Q1 (isotopic moments of inertia), Q2 (HD level spacing), Q3 (HCl moment of inertia), Q5 (largest allowed transition) and Q10 (rotational dissociation of OH); and JEST Q1 (rotational constant of H2).