University Physics IV · Molecular and Solid-State Physics · 12.3

Rotational Spectra & the Rigid Rotor

Point a microwave source at a gas of polar diatomic molecules and a comb of evenly spaced absorption lines comes back. That spacing is a bond length in disguise. Learn to read it out — and to see why a homonuclear molecule stays dark while rotating exactly as hard.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A rotating diatomic is a two-body problem, and the trick that tamed hydrogen tames it too: separate the centre of mass, and what remains is one fictitious particle of reduced mass μ = m₁m₂/(m₁+m₂) held at the bond length R. The entire mass distribution collapses to one number, I = μR². Freeze R and the angular equation is already solved — the spherical harmonics — so E_J = J(J+1)ħ²/2I, with each level (2J+1)-fold degenerate.

That ladder is quadratic, so its rungs spread apart; but its differences grow linearly, and since an electric-dipole photon can only change J by one, the spectrum is a comb of lines at 2B(J+1) whose every gap is the same 2B. One measured spacing therefore hands you I, and with a known μ a bond length good to three figures — which is how we know how long a chemical bond is. The costs are explicit and each shows up in the data.

The rotor is not rigid, so centrifugal stretching subtracts DJ²(J+1)² and the comb converges at high J. The molecule is not alone, so Boltzmann weighting of the (2J+1) rungs, not the transition itself, decides which line is brightest. And the whole spectrum carries the permanent dipole as a prefactor, so a homonuclear diatomic gives nothing at all — while rotating, storing rotational energy and stretching exactly as its polar neighbours do.

Simple definition: A rigid rotor is a two-atom molecule modelled as two point masses held a fixed distance apart and free to tumble; its allowed energies are E_J = J(J+1)ħ²/2I, where I = μR² is the moment of inertia about the centre of mass.
Example: ¹²C¹⁶O has μ = 6.856 u and R = 113 pm, so I = 1.456 × 10⁻⁴⁶ kg m² and B = h/8π²I = 57.64 GHz. Its J = 0 → 1 line therefore sits at 2B = 115.27 GHz — the line radio astronomers use to map molecular clouds.

Reduced mass and moment of inertiaμ = m₁m₂/(m₁ + m₂), I = μR²

¹²C¹⁶O: μ = 6.856 u with R = 113 pm gives I = 1.456 × 10⁻⁴⁶ kg m². One number carries the whole mass distribution.

μ in kg (1 u = 1.66054 × 10⁻²⁷ kg), R the bond length in m, I in kg m²

The rigid-rotor ladderE_J = J(J+1)ħ²/2I = hcB̃⋅J(J+1), J = 0, 1, 2, …

Quadratic in J, so the rungs spread: the gap above level J is 2hcB̃(J+1), widening as you climb the ladder.

B̃ = h/8π²cI in cm⁻¹; every level is (2J+1)-fold degenerate in M_J

Where the absorption lines fallν̃(J → J+1) = 2B̃(J+1), so the spacing is 2B̃

Every gap is identical, so one measured spacing gives I and then R = √(I/μ) to three significant figures.

ν̃ in cm⁻¹; as a frequency, ν = 2B(J+1) with B = h/8π²I in Hz

Electric-dipole selection ruleΔJ = ±1, ΔM_J = 0, ±1, and only if μₑₗ ≠ 0

HCl and CO absorb in the microwave; H₂, N₂ and O₂ rotate just as freely and stay completely dark.

Rotational Raman reaches the same levels with ΔJ = 0, ±2 through the anisotropic polarizability

Thermal population and the brightest lineN_J ∝ (2J+1)e^{(−hcB̃J(J+1)}/kT), Jₘₐₓ ≈ √(kT/2hcB̃) − ½

H³⁵Cl at 300 K gives Jₘₐₓ = 2.66, so level 3 is the most populated and the J = 3 → 4 line is the strongest.

hc/k = 1.4388 cm K, so Jₘₐₓ = √(T/2.8776B̃) − ½ with B̃ in cm⁻¹

Centrifugal distortionν̃(J → J+1) = 2B̃(J+1) − 4D̃(J+1)³, D̃ ≈ 4B̃³/ω̃²

Fit ν̃/(J+1) against (J+1)²: the intercept is 2B̃ and the slope is −4D̃, so one straight line gives both constants.

D̃ in cm⁻¹, ω̃ the vibrational wavenumber; D̃/B̃ ≈ 3 × 10⁻⁶ for CO

Two nuclei, one particle on a sphere

Born-Oppenheimer freezes the nuclei relative to the fast electrons, so the bond length R enters as a fixed parameter rather than a dynamical variable. The two-body rotation then separates the way every two-body problem does: the centre of mass drifts free, and the internal motion is one fictitious particle of reduced mass μ = m₁m₂/(m₁+m₂) moving on a sphere of radius R. Every detail of how the mass is distributed collapses into a single number, I = μR². For ¹²C¹⁶O, μ = 12 × 15.995/27.995 = 6.856 u = 1.1385 × 10⁻²⁶ kg, and with R = 113.1 pm, I = 1.456 × 10⁻⁴⁶ kg m². Watch how μ behaves when the masses are lopsided: for H³⁵Cl it is 0.9796 u, only 2% below the hydrogen mass alone, because the chlorine barely moves. The light atom does almost all the swinging and the heavy one sits close to the centre of mass — which is also why isotopic substitution at hydrogen shifts a rotational spectrum enormously and substitution at chlorine barely shifts it at all.

A quadratic ladder makes a uniformly spaced comb

Fixing R kills the radial equation and leaves the angular equation you already solved for hydrogen, so the eigenfunctions are the spherical harmonics Y_JM and the eigenvalues follow from L² → J(J+1)ħ²: E_J = J(J+1)ħ²/2I, with 2J+1 values of M_J sharing each energy. Write it as E_J = hcB̃⋅J(J+1). The rungs sit at 0, 2B̃, 6B̃, 12B̃, 20B̃ — visibly spreading apart. But the gaps between neighbouring rungs are 2B̃, 4B̃, 6B̃, 8B̃, and each one exceeds the last by the same 2B̃. Since a photon can only move you one rung, the lines land at ν̃ = 2B̃(J+1) and come out evenly spaced. For CO with B̃ = 1.9225 cm⁻¹ the levels are at 0, 3.845, 11.535 and 23.07 cm⁻¹, while the first three lines are at 3.845, 7.690 and 11.535 cm⁻¹. Levels and lines are two different lists that happen to share their first entry, and reading one as the other is the classic way to get I wrong by a factor of two.

What makes a line appear: the dipole matrix element

An electric-dipole transition rate goes as |⟨J′M′|μₑₗ⋅ε̂|JM⟩|². For a diatomic the dipole lies along the bond, so the operator is μₑₗ cos θ, and cos θ is proportional to Y₁₀. An integral over three spherical harmonics survives only when the angular momenta can couple — here ΔJ = ±1 and ΔM_J = 0, ±1, which is the photon's one unit of angular momentum being absorbed by the rotor. Two consequences follow. There are no ΔJ = 2 lines in absorption, so the comb has no extra or missing teeth. And the whole integral carries μₑₗ as a prefactor, so a molecule with no permanent dipole gives a spectrum of exactly zero intensity, not a weak one. This is why CO, present at a mole fraction near 10⁻⁴ in a molecular cloud, is the tracer radio astronomers actually use, while the H₂ that makes up the cloud is invisible in the microwave.

Degeneracy against Boltzmann sets which line is brightest

The lines are not all the same height, and two competing factors decide them. The population of level J carries the degeneracy 2J+1, rising linearly, times the Boltzmann factor exp(−hcB̃J(J+1)/kT), falling faster than any power. Their product peaks; setting its derivative to zero gives 2J+1 = √(2kT/hcB̃), so Jₘₐₓ = √(kT/2hcB̃) − ½. Since hc/k = 1.4388 cm K, that is Jₘₐₓ = √(T/2.8776B̃) − ½ with B̃ in cm⁻¹. For H³⁵Cl at 300 K it returns 2.66, so check the integers either side: level 2 has weight 5e^(−0.300) = 3.70 and level 3 has 7e^(−0.601) = 3.84, so J = 3 wins. Cool the gas to 100 K and Jₘₐₓ drops to 1.32. The band envelope is therefore a thermometer. Strictly the absorbed intensity also carries a (J+1)/(2J+1) factor from the transition moment and a stimulated-emission correction, but both drift slowly next to the exponential, so the population peak sets the envelope.

The rotor stretches, and the comb converges

Nothing holds R fixed except a bond of finite stiffness. Balance the centrifugal pull against that spring and the bond extends by x = J(J+1)ħ²/(kμR³); feeding the stretch back into the energy costs ½kx², giving E_J = hcB̃J(J+1) − hcD̃J²(J+1)², with the Kratzer estimate D̃ ≈ 4B̃³/ω̃² tying the distortion constant to the vibrational wavenumber. Lines move to ν̃ = 2B̃(J+1) − 4D̃(J+1)³. For CO, B̃ = 1.9225 cm⁻¹ and ω̃ = 2170 cm⁻¹ give D̃ = 6.04 × 10⁻⁶ cm⁻¹ against a measured 6.12 × 10⁻⁶, so the correction is six parts per million at the first line and one part in a hundred by J = 39. That cubic growth is why you must never fit a straight line to raw high-J line positions. Divide instead: ν̃/(J+1) = 2B̃ − 4D̃(J+1)², linear in (J+1)² with intercept 2B̃ and slope −4D̃.

Reading a bond length out, and what it actually is

Invert the chain: a measured spacing gives B̃ = spacing/2, then I = h/8π²cB̃, then R = √(I/μ). For H³⁵Cl, a 20.88 cm⁻¹ spacing gives B̃ = 10.44 cm⁻¹, I = 2.681 × 10⁻⁴⁷ kg m² and R = 128.4 pm. Two checks are worth running. First, isotopes: Born-Oppenheimer says R is set by the electrons and so is identical for H³⁵Cl and H³⁷Cl, while μ differs by 0.15%, so B̃ must scale as 1/μ — predicting B̃(H³⁷Cl) = 10.424 cm⁻¹, a comb 0.032 cm⁻¹ tighter, with the doublet intensities in chlorine's 3:1 abundance ratio. Second, honesty about what R is. A rotational spectrum of the vibrational ground state measures B̃₀, an average of ħ²/2μR² over the v = 0 wavefunction rather than the value at the potential minimum, with B̃ᵥ = B̃ₑ − α̃ₑ(v + ½). So HCl's spectroscopic r₀ = 128.4 pm sits about 1 pm above its equilibrium rₑ = 127.5 pm.

Change one variable at a time

Make the relationship visible.

Interactive model

Rotational constant B1.9 cm⁻¹

Temperature T300 K

Distortion D/B0 ×10⁻⁴

Raise B and the whole comb stretches while every gap stays equal — the bond-length readout falls as 1/√B. Drop T to 20 K and the heights collapse onto J = 1 with the spacing untouched. Then push D/B (exaggerated here) to 15 × 10⁻⁴ and watch the far lines crowd left of the dashed rigid mark.

LINE SPACING 2B3.80 cm⁻¹

BOND LENGTH IF μ = 6.86 u113.8 pm

BRIGHTEST J6.9

J = 9→10 LINE SHIFTED BY0.0 %

Live interpretationLINE SPACING 2B: 3.80 cm⁻¹. BOND LENGTH IF μ = 6.86 u: 113.8 pm. BRIGHTEST J: 6.9. J = 9→10 LINE SHIFTED BY: 0.0 %

Catch the common trap

Explain before calculating.

Misconception check

Students hear that N₂ has no rotational spectrum and conclude that N₂ does not rotate, or that ΔJ = ±1 somehow forbids the transition. Neither is true. A homonuclear diatomic has exactly the same ladder, E_J = hcB̃J(J+1) — for N₂, B̃ ≈ 2.0 cm⁻¹ — it is thermally populated up to J ≈ 7 at room temperature, it contributes R to the molar heat capacity, and it stretches under centrifugal force like anything else. What vanishes is the transition moment, not the rotation: the electric-dipole matrix element carries the permanent dipole as a prefactor, and that is zero by symmetry, so the integral is zero however perfectly ΔJ = ±1 is satisfied. Rotational Raman spectroscopy reaches those same levels through the anisotropic polarizability, with ΔJ = 0, ±2, and returns a bond length of 110 pm for N₂.

The pure rotational absorption spectrum of ¹²C¹⁶O has its first line at 3.845 cm⁻¹, and every later line lies 3.845 cm⁻¹ above the one before. What does that uniform spacing tell you about the molecule's rotational energy levels?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe lowest pure-rotational absorption line of ¹²C¹⁶O lies at 115.271 GHz. Treating the molecule as a rigid rotor, find B, predict the next two lines, and obtain the moment of inertia.

The lowest line is J = 0 → 1, and ν(J → J+1) = 2B(J+1), so this line sits at 2B. Hence B = 115.271/2 = 57.636 GHz.
The next two are J = 1 → 2 at 4B = 230.54 GHz and J = 2 → 3 at 6B = 345.81 GHz. Every gap is the same 2B = 115.27 GHz, even though the levels behind them are not evenly spaced.
Invert B = h/8π²I: I = h/(8π²B) = 6.626 × 10⁻³⁴ / (78.957 × 5.7636 × 10¹⁰) = 6.626 × 10⁻³⁴ / 4.551 × 10¹² = 1.456 × 10⁻⁴⁶ kg m².
Check against the real sky. The measured lines are 230.538 and 345.796 GHz, so the rigid rotor over-predicts by 4 MHz and then 17 MHz — small, one-signed, and growing about four times faster each step. That is centrifugal distortion arriving exactly on cue.

AnswerB = 57.636 GHz; the next lines fall at 230.54 and 345.81 GHz; I = 1.456 × 10⁻⁴⁶ kg m². The 4 MHz and 17 MHz over-predictions are centrifugal distortion, not experimental error.

MediumThe far-infrared absorption spectrum of H³⁵Cl shows lines evenly spaced by 20.88 cm⁻¹. Taking μ = 0.9796 u, find the bond length. Then identify which line is strongest at 300 K and say where it lies.

Uniform spacing is 2B̃, so B̃ = 20.88/2 = 10.44 cm⁻¹.
I = h/(8π²cB̃). With 8π²c = 78.957 × 2.998 × 10¹⁰ cm s⁻¹ = 2.367 × 10¹², I = 6.626 × 10⁻³⁴/(2.367 × 10¹² × 10.44) = 2.681 × 10⁻⁴⁷ kg m².
μ = 0.9796 × 1.66054 × 10⁻²⁷ = 1.6267 × 10⁻²⁷ kg, so R = √(I/μ) = √(2.681 × 10⁻⁴⁷ / 1.6267 × 10⁻²⁷) = √(1.6485 × 10⁻²⁰) = 1.284 × 10⁻¹⁰ m = 128.4 pm.
Brightest line: Jₘₐₓ = √(T/2.8776B̃) − ½ = √(300/30.04) − 0.5 = 3.160 − 0.5 = 2.66. That is not an integer, so test the neighbours.
Weights are (2J+1)e^{(−1.4388B̃J(J+1)}/T), and 1.4388 × 10.44/300 = 0.05007. J = 2: 5e^(−0.3004) = 3.70. J = 3: 7e^(−0.6008) = 3.84. J = 4: 9e^(−1.0014) = 3.31. Level 3 carries the largest population.
The strongest absorption is therefore J = 3 → 4, at 2B̃ × 4 = 83.52 cm⁻¹.

AnswerB̃ = 10.44 cm⁻¹, I = 2.681 × 10⁻⁴⁷ kg m² and R = 128.4 pm. At 300 K the strongest line is J = 3 → 4 at 83.52 cm⁻¹.

Hard¹²C¹⁶O has B̃₀ = 1.9225 cm⁻¹, vibrational wavenumber ω̃ = 2170 cm⁻¹ and R₀ = 113.1 pm. Estimate its centrifugal distortion constant, find the first line pulled down by a full 1% of its rigid-rotor position, and say how far the bond has stretched by then.

Kratzer: D̃ ≈ 4B̃³/ω̃² = 4(1.9225)³/(2170)² = 28.42/4.7089 × 10⁶ = 6.04 × 10⁻⁶ cm⁻¹. The measured value is 6.12 × 10⁻⁶ cm⁻¹, so a two-parameter estimate lands within 1.4%.
The line sits at ν̃ = 2B̃(J+1) − 4D̃(J+1)³, so the fractional pull-down is 4D̃(J+1)³ / 2B̃(J+1) = 2D̃(J+1)²/B̃ — it grows as the square of the line number, not linearly.
Set that to 0.010: (J+1)² = 0.010 × 1.9225/(2 × 6.04 × 10⁻⁶) = 0.019225/1.207 × 10⁻⁵ = 1593, so J + 1 = 39.9. The J = 39 → 40 line is the first to reach 1%.
Its position: rigid 2B̃ × 40 = 153.80 cm⁻¹, distortion −4 × 6.04 × 10⁻⁶ × 40³ = −1.55 cm⁻¹, so the line lands at 152.25 cm⁻¹ rather than 153.80 cm⁻¹.
Bond stretch. Balancing centrifugal force against the bond spring and using D̃ = 2B̃²/kR², the extension satisfies x/R = D̃J(J+1)/B̃ = 6.04 × 10⁻⁶ × 39 × 40 / 1.9225 = 4.90 × 10⁻³, so x = 0.0049 × 113.1 pm = 0.55 pm.
Consistency check: B̃ ∝ R⁻², so a 0.49% stretch lowers B̃ by 0.98% — the same 1% shift, reached by a completely independent route.

AnswerD̃ ≈ 6.0 × 10⁻⁶ cm⁻¹. The J = 39 → 40 line is the first pulled down 1%, landing at 152.25 cm⁻¹ instead of 153.80 cm⁻¹, by which point the bond has stretched 0.55 pm — about half a percent of 113 pm.