Microwave spectroscopy

Shine microwave radiation through a gas of diatomic molecules and only certain sharp frequencies get absorbed — the molecules are turning like tiny dumbbells, and quantum mechanics permits only discrete rates of spin. Microwave spectroscopy reads this ladder of rotational energy levels and, remarkably, converts it into bond lengths accurate to a fraction of a picometre, without ever imaging a single molecule directly. This node builds the rigid-rotor model of a diatomic molecule from scratch: two point masses on a rigid rod, quantized rotational energy, and a selection rule that only molecules with a permanent dipole moment obey. The payoff is a beautifully simple prediction — the absorption spectrum is a ladder of equally spaced lines — and a direct route from measured line spacing back to the geometry of the molecule. The treatment stays within diatomics under the rigid-rotor approximation, exactly as CY11001 requires; centrifugal distortion and polyatomic rotors are out of scope here.

The core insight

A diatomic molecule's rotational energy is quantized as $E_J = hcBJ(J+1)$ , so its microwave absorption spectrum ( $\Delta J = \pm1$ , dipole moment required) is a series of lines spaced exactly $2B$ apart, and measuring that spacing gives the bond length via $B \to I \to r$ .

The core idea

The rigid-rotor model. Treat a diatomic molecule as two point masses $m_1$ and $m_2$ joined by a rigid, massless rod of length $r$ — the bond length — rotating freely about the centre of mass. Two combined quantities carry all the physics: the reduced mass and the moment of inertia.

\mu = \frac{m_1 m_2}{m_1 + m_2}, \qquad I = \mu r^2

where: $\mu$ is the reduced mass, $m_1, m_2$ are the atomic masses, $I$ is the moment of inertia about the centre of mass, and $r$ is the bond length. Using $\mu$ collapses the two-body rotation problem to an equivalent single mass $\mu$ orbiting at distance $r$ .

Quantized energy levels. Solving the Schrödinger equation for this rigid rotor gives discrete rotational energies labelled by the rotational quantum number $J = 0, 1, 2, \ldots$ :

E_J = \frac{h^2}{8\pi^2 I}J(J+1) = hcBJ(J+1)

where: $h$ is Planck's constant, $I$ is the moment of inertia, $J$ is the rotational quantum number, $c$ is the speed of light, and $B$ is the rotational constant (units $\text{cm}^{-1}$ when energy is expressed via $hc\tilde\nu$ ):

B = \frac{h}{8\pi^2 I c}

A larger $I$ (heavier atoms, longer bond) gives a smaller $B$ — the rotational levels bunch closer together, the same way a heavier flywheel spins up more slowly for the same push.

The selection rule and the spectrum. A transition is microwave-active only if the molecule has a permanent dipole moment — the rotating dipole is what couples to the oscillating electric field of the radiation. Homonuclear diatomics like $\text{N}_2$ have none and show no rotational spectrum at all, regardless of what $B$ would work out to. For a polar diatomic, the allowed transitions obey $\Delta J = \pm 1$ , and the absorbed wavenumber for $J \to J+1$ is:

\tilde\nu_{J\to J+1} = \frac{E_{J+1}-E_J}{hc} = 2B(J+1)

This is the central practical result: successive lines ( $J=0\to1$ , $1\to2$ , $2\to3,\ldots$ ) fall at $2B, 4B, 6B,\ldots$ — an evenly spaced ladder with spacing exactly $2B$ . A real spectrum is therefore read backward: measure the spacing, get $B$ ; from $B$ get $I$ ; from $I$ get $r$ .

Figure

Ladder diagram of rotational energy levels J equals 0 to 4 with equally spaced absorption lines below labelled 2B 4B 6B 8B

Derivation: line spacing equals 2B

Derivation ARotational line spacing from the rigid-rotor energy formula

The goal of this derivation is to show, starting only from the quantized energy $E_J = hcBJ(J+1)$ and the selection rule $\Delta J = +1$ for absorption, that the observed absorption spectrum of a polar diatomic molecule must consist of lines separated by a constant wavenumber gap of exactly $2B$ , independent of which pair of adjacent levels is involved.

Step 1. Write the energy of the lower and upper levels of an allowed transition. For absorption from level $J$ to level $J+1$ , the lower-level energy is $E_J = hcBJ(J+1)$ and the upper-level energy is $E_{J+1} = hcB(J+1)(J+2)$ , obtained simply by substituting $J+1$ for $J$ in the same formula.

Step 2. Take the energy difference and convert to wavenumber. The absorbed photon energy is $\Delta E = E_{J+1} - E_J = hcB\big[(J+1)(J+2) - J(J+1)\big]$ , and dividing by $hc$ gives the wavenumber $\tilde\nu_{J\to J+1} = B\big[(J+1)(J+2) - J(J+1)\big]$ .

Step 3. Simplify the bracket algebraically. Factor out $(J+1)$ : $(J+1)(J+2) - J(J+1) = (J+1)\big[(J+2)-J\big] = (J+1)(2) = 2(J+1)$ . So $\tilde\nu_{J\to J+1} = 2B(J+1)$ .

Step 4. Compute the spacing between two consecutive lines. The line for $J\to J+1$ sits at $2B(J+1)$ ; the next line, for $(J+1)\to(J+2)$ , sits at $2B(J+2)$ . Their difference is $2B(J+2) - 2B(J+1) = 2B$ , a constant that does not depend on $J$ at all — every gap in the ladder is the same size.

Step 5. State the result. Because the spacing is the same $2B$ everywhere in the spectrum, a single measured gap between any two adjacent absorption lines determines $B$ directly, without needing to know which $J$ values produced those particular lines.

Worked example

Problem. The pure rotational spectrum of $^1\text{H}^{35}\text{Cl}$ shows absorption lines spaced $20.68~\text{cm}^{-1}$ apart. Find (a) the rotational constant $B$ , (b) the moment of inertia $I$ , and (c) the bond length $r$ . (Atomic masses: $m_H = 1.008~\text{u}$ , $m_{Cl} = 34.97~\text{u}$ ; $1~\text{u} = 1.6605\times10^{-27}~\text{kg}$ .)

Step 1. Get $B$ from the spacing. Since consecutive lines are separated by $2B$ , $B = 20.68/2 = 10.34~\text{cm}^{-1}$ .

Step 2. Get $I$ from $B$ . Rearranging $B = \dfrac{h}{8\pi^2 I c}$ gives $I = \dfrac{h}{8\pi^2 c B}$ . Using $c = 3.00\times10^{10}~\text{cm s}^{-1}$ so units match $B$ in $\text{cm}^{-1}$ :

I = \frac{6.626\times10^{-34}}{8\pi^2(3.00\times10^{10})(10.34)} = \frac{6.626\times10^{-34}}{2.449\times10^{13}} = 2.706\times10^{-47}~\text{kg m}^2

Step 3. Get the reduced mass. $\mu = \dfrac{m_H m_{Cl}}{m_H + m_{Cl}} = \dfrac{35.25}{35.98}~\text{u} = 0.9797~\text{u} = 1.6267\times10^{-27}~\text{kg}$ .

Step 4. Solve $I = \mu r^2$ for $r$ . $r = \sqrt{I/\mu} = \sqrt{2.706\times10^{-47}/1.6267\times10^{-27}} = \sqrt{1.663\times10^{-20}} = 1.29\times10^{-10}~\text{m}$ .

Final answer: $B = 10.34~\text{cm}^{-1}$ , $I = 2.706\times10^{-47}~\text{kg m}^2$ , and $r = 129~\text{pm}$ — matching the known H–Cl bond length ( $\approx 127$ – $130~\text{pm}$ ) closely.

Competitive stretch

For $^{12}\text{C}^{16}\text{O}$ , $B = 1.931~\text{cm}^{-1}$ . CO is heteronuclear with a small permanent dipole ( $\approx 0.11~\text{D}$ ), so it is microwave-active. Using $\tilde\nu_{J\to J+1} = 2B(J+1)$ : the $J=0\to1$ line sits at $2(1.931)(1) = 3.862~\text{cm}^{-1}$ , the $J=1\to2$ line at $2(1.931)(2) = 7.724~\text{cm}^{-1}$ , and the $J=2\to3$ line at $2(1.931)(3) = 11.586~\text{cm}^{-1}$ — each pair spaced by exactly $2B = 3.862~\text{cm}^{-1}$ , confirming the ladder prediction. Plotting line position against the integer index $(J+1)$ gives a straight line of slope $2B$ , which is how real spectra (with dozens of lines) fit $B$ by regression rather than reading a single gap.

Common traps

Using the wrong mass in $I = \mu r^2$ . It is the reduced mass $\mu$ , never a single atomic mass and never the total mass $m_1 + m_2$ — the most common numerical slip in this topic.
Skipping the dipole-moment check. A homonuclear diatomic such as $\text{N}_2$ or $\text{Cl}_2$ shows no microwave spectrum at all, no matter what $B$ computes to on paper. Always confirm microwave activity before crunching numbers.
Misreading which transition a line belongs to. Adjacent-line spacing is always $2B$ regardless of $J$ , but the absolute position of a specific line is $2B(J+1)$ , not $2BJ$ — mixing these gives an off-by-one error.
Unit inconsistency for $B$ . $B$ in $\text{cm}^{-1}$ uses $B = h/8\pi^2Ic$ ; $B$ in Hz uses $B = h/8\pi^2I$ with no $c$ . Mixing the two mid-calculation produces an error of order $10^{10}$ .

Ask the tutor: if you want more bond-length-from-spectrum problems, help telling which molecules are microwave-active, or a walkthrough of how isotopic substitution shifts the rotational constant, just ask.

Why does a molecule need a permanent dipole moment to show a microwave spectrum?

The rotating molecule interacts with the oscillating electric field of the microwave radiation through its dipole moment — a spinning dipole creates a fluctuating electric field that the radiation's field can couple to and drive transitions between rotational levels. A molecule with no permanent dipole, like $\text{N}_2$ or $\text{Cl}_2$ , has nothing for the field to grab onto, so however precisely you compute $B$ for it, no absorption occurs. This is a strict gate: check dipole moment first, before any arithmetic.

If I only measure one line spacing, do I need to know its J value?

No. Because every adjacent gap in the ladder equals $2B$ regardless of $J$ , a single measured spacing between any two neighbouring lines gives $B$ directly. You only need to identify the actual $J$ value if you want the absolute wavenumber position of one specific line, since that position is $2B(J+1)$ , which does depend on $J$ .

Does a larger bond length mean a larger or smaller rotational constant B?

Larger $r$ gives a larger moment of inertia $I = \mu r^2$ , and since $B = h/8\pi^2Ic$ is inversely proportional to $I$ , a longer bond means a smaller $B$ — the lines sit closer together. This is why light, tightly bound molecules like HCl have relatively large $B$ and widely spaced lines, while heavier or longer-bonded molecules crowd their rotational lines closer together.