Molecular spectroscopy — rotational (rigid rotor), vibrational (harmonic oscillator, Morse potential), rotational-vibrat

1. Rotational spectroscopy — the rigid rotor

WHAT we need: the quantised rotational energy levels.

Formula Deriving rotational energy from scratch

Classical rotational energy: $E = \tfrac12 I\omega^2 = \dfrac{L^2}{2I}$ , where angular momentum $L=I\omega$ . Why this form? It mirrors $\tfrac12 mv^2 = p^2/2m$ with $I\leftrightarrow m$ , $L\leftrightarrow p$ .

Quantum mechanics quantises angular momentum: $L^2 = J(J+1)\hbar^2$ , $J=0,1,2,\dots$ Why $J(J+1)$ ? It is the eigenvalue of the $\hat L^2$ operator for spherical harmonics — the same maths as the H-atom angular part.

Substituting: $E_J = \frac{J(J+1)\hbar^2}{2I}, \qquad I=\mu r^2,\quad \mu=\frac{m_1 m_2}{m_1+m_2}$ Spectroscopists use wavenumbers ( $\tilde\nu = E/hc$ , cm $^{-1}$ ): $F(J)=\frac{E_J}{hc}=B\,J(J+1),\qquad \boxed{B=\frac{\hbar^2}{2Ihc}=\frac{h}{8\pi^2 c I}}$ $B$ is the rotational constant (cm $^{-1}$ ).

HOW the spectrum looks. Selection rule $\Delta J=\pm 1$ (molecule must have a permanent dipole). Absorption $J\to J+1$ : $\tilde\nu = F(J+1)-F(J) = B[(J+1)(J+2)-J(J+1)] = 2B(J+1)$ So lines appear at $2B,4B,6B,\dots$ — equally spaced by $2B$ . Measure the spacing → get $B$ → get $I$ → get $r$ . That is the whole point.

Worked example Find the bond length of CO

$B_{\text{CO}}=1.93$ cm $^{-1}$ . Find $r$ .

$I=\dfrac{h}{8\pi^2 cB}$ . Why? Just invert $B=h/8\pi^2cI$ .
$\mu=\dfrac{12\cdot16}{28}\,\text{u}=6.86\,\text{u}=1.14\times10^{-26}$ kg. Why $\mu$ ? Two-body rotation reduces to one effective mass.
$I = \dfrac{6.626\times10^{-34}}{8\pi^2(3\times10^{10})(1.93)} \approx 1.45\times10^{-46}$ kg m $^2$ . (use $c$ in cm/s since $B$ in cm $^{-1}$ )
$r=\sqrt{I/\mu}=\sqrt{1.45\times10^{-46}/1.14\times10^{-26}}\approx 1.13\times10^{-10}$ m $=113$ pm. ✓ matches CO.

2. Vibrational spectroscopy

2a. Harmonic oscillator

Selection rule $\Delta v=\pm1$ (and dipole must change). All transitions absorb at the same $\tilde\nu_e$ → harmonic model predicts one line. Reality shows overtones and convergence ⇒ we need the Morse potential.

2b. Morse potential (anharmonicity)

Figure — Molecular spectroscopy — rotational (rigid rotor), vibrational (harmonic oscillator, Morse potential), rotational-vibrat

3. Rotational–vibrational coupling (ro-vibrational spectrum)

Selection rules: $\Delta v=+1$ , $\Delta J=\pm1$ .

$\Delta J=+1$ → R branch (higher $\tilde\nu$ )
$\Delta J=-1$ → P branch (lower $\tilde\nu$ )
$\Delta J=0$ (forbidden for most diatomics) → would be the Q branch (a gap at band centre).

4. Electronic spectroscopy — the Franck–Condon principle

If the upper-state well is displaced ( $r_e' > r_e''$ ), the vertical line from $v''=0$ lands on a high $v'$ → intensity peaks at $v'>0$ , giving the characteristic Franck–Condon intensity envelope.

Common mistake Steel-manning the classic errors

(1) "Use mass of one atom for $I$ ." Feels right because rotation looks like one ball swinging. Fix: both atoms move about the centre of mass → use reduced mass $\mu$ . $I=\mu r^2$ . (2) "Harmonic oscillator can dissociate." Feels right since real bonds break. Fix: a parabola has infinite walls — it never dissociates. You need the Morse flattening for $D_e$ . (3) "Ground vibrational energy is zero." Feels right (no excitation). Fix: zero-point energy $=\tfrac12\tilde\nu_e\ne0$ from Heisenberg. (4) "Rotational lines are perfectly evenly spaced." True only for a rigid rotor; centrifugal distortion + $B_v$ change makes P/R branches asymmetric. (5) "Brightest electronic line is always $0\to0$ ." Only if wells are aligned. If displaced, the vertical Franck–Condon overlap peaks at higher $v'$ .

Recall Feynman: explain to a 12-year-old

Imagine a molecule as two balls on a spring. It can spin (rotation — needs tiny energy, like microwaves), wobble in and out on the spring (vibration — needs more energy, like heat/IR), and its electrons can jump to a new arrangement (needs lots of energy, like a flash of UV light). Because spinning, wobbling, and electron-jumping each cost very different amounts of energy, we can tell them apart by which colour of light the molecule drinks. The gaps between the spinning lines tell us how long the spring is; how the wobble lines crowd together tells us how strong the spring is and when it snaps. When electrons jump, they jump so fast the balls don't have time to move — like a photo with no blur — and that "frozen snapshot" rule (Franck–Condon) tells us which wobble it lands in.

Flashcards

What model is the "rigid rotor" and its key assumption?

Diatomic as two masses on a fixed-length massless rod; bond length constant during rotation.

Rotational energy in wavenumbers

F(J)=BJ(J+1)

with

B=h/8\pi^2 cI

I=\mu r^2

Spacing between adjacent rotational absorption lines

2B

(lines at

2B,4B,6B,\dots

), so

\Delta J=+1

gives

\tilde\nu=2B(J+1)

Why use reduced mass in

I

Both atoms orbit the centre of mass; the two-body problem reduces to one effective mass

\mu=m_1m_2/(m_1+m_2)

Harmonic oscillator levels and

\tilde\nu_e

G(v)=\tilde\nu_e(v+\tfrac12)

\tilde\nu_e=\frac{1}{2\pi c}\sqrt{k/\mu}

What is zero-point energy and why nonzero?

\tfrac12\tilde\nu_e

; Heisenberg forbids resting exactly at the potential minimum.

Morse potential expression

V(r)=D_e[1-e^{-a(r-r_e)}]^2

, flattening to

D_e

r\to\infty

Morse (anharmonic) energy levels

G(v)=\tilde\nu_e(v+\tfrac12)-\tilde\nu_e x_e(v+\tfrac12)^2

; levels converge.

Why does harmonic model fail for dissociation?

Parabola has infinite walls and equal spacing; can't break or converge — Morse fixes this.

P vs R branch

\Delta J=+1

, higher

\tilde\nu

; P:

\Delta J=-1

, lower

\tilde\nu

; Q (

\Delta J=0

) usually forbidden, leaving a gap at band origin.

R-branch line positions

\tilde\nu_R=\tilde\nu_0+2B(J+1)

P-branch line positions

\tilde\nu_P=\tilde\nu_0-2BJ

What causes P/R branch asymmetry?

Centrifugal distortion and

B_v=B_e-\alpha_e(v+\tfrac12)

(rotation–vibration coupling).

State the Franck–Condon principle

Electronic transitions are vertical (nuclei frozen); intensity is greatest for max vibrational-wavefunction overlap.

Franck–Condon factor

|\langle\psi'_{v'}|\psi''_{v''}\rangle|^2

, the squared vibrational overlap integral.

Order of energy scales

Electronic

\gg

Vibrational

\gg

Rotational (eV

\gg

0.1 eV

\gg

10⁻³ eV).

Selection rule for pure rotational absorption

\Delta J=\pm1

and molecule must have a permanent dipole.

Connections

Born-Oppenheimer Approximation — justifies separating electronic/nuclear motion.
Reduced Mass and Two-Body Problem — basis of $I=\mu r^2$ and $\tilde\nu_e$ .
Quantum Harmonic Oscillator — vibrational ladder & zero-point energy.
Angular Momentum in QM — $J(J+1)\hbar^2$ eigenvalues.
Heisenberg Uncertainty Principle — why zero-point energy exists.
Dissociation Energy and Birge-Sponer Plot — extracting $D_0$ from Morse spacings.
Boltzmann Distribution — populations $\Rightarrow$ rotational line intensities.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek molecule energy ko teen alag-alag "buckets" mein store karta hai, aur teeno ka size bahut different hai: rotational (ghoomna — sabse kam energy, microwave), vibrational (spring ki tarah andar-bahar wobble — IR), aur electronic (electron jump — sabse zyada energy, UV-Vis). Kyunki teeno bahut alag size ke hain, hum inhe alag-alag treat kar sakte hain. Spectroscopy ka matlab — light maaro, dekho molecule kaun si energy "pee" raha hai, aur ulta hisaab lagao ki bond ki length kitni hai, spring kitna strong hai, bond kab tootega.

Rotational (rigid rotor): molecule ko do balls ek stiff rod par socho. Energy $F(J)=BJ(J+1)$ , aur lines $2B$ , $4B$ , $6B$ ... yaani har line ke beech $2B$ ka gap. Yeh gap naapo to $B$ mil jaata hai, $B$ se $I$ , aur $I=\mu r^2$ se bond length $r$ . Yaad rakho — single atom ka mass mat lo, reduced mass $\mu$ lo, kyunki dono atoms centre of mass ke aas-paas ghoomte hain.

Vibrational: chhote vibrations ke liye parabola (harmonic) theek hai, levels barabar spacing ke. Lekin parabola kabhi tootta nahi — isliye Morse potential use karte hain jo door jaake flatten ho jaata hai $D_e$ par. Morse mein levels upar jaate-jaate paas-paas aate jaate hain (converge), aur isi se dissociation energy nikalti hai. Aur ek important cheez — zero-point energy $\tfrac12\tilde\nu_e$ kabhi zero nahi hoti (Heisenberg ki wajah se).

Ro-vibrational + electronic: jab IR vibration excite karta hai, $J$ bhi badalta hai — isse P branch (neeche) aur R branch (upar) bante hain, beech mein ek gap (band origin). Franck–Condon principle bolta hai electron itni tezi se jump karta hai ki nuclei freeze ho jaate hain — transition seedha vertical hota hai, aur jis vibrational level par wavefunction ka overlap maximum hota hai wahi line sabse bright hoti hai. Bas yeh saari ideas yaad rakho to poora spectroscopy clear ho jaayega.

Test yourself — Physical Chemistry (Advanced)

Connections