Chemistry · Lesson

Rotational, vibrational and electronic spectra

Relate a photon to a state-energy gap using explicit vacuum wavelength and frequency conventions. Separate spectral positions, transition strengths and instrument-dependent observations.

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By the end, you can

Convert between photon energy, vacuum wavelength and spectroscopic wavenumber.
Calculate rigid-linear-rotor level terms and allowed absorption gaps.
Distinguish vibrational coordinates, vibronic bands and absorbance measurement conditions.

Match energy using a declared wavelength convention

For a one-photon transition, ΔE = hν = hc/λ_vac = hcṽ. Here ν is ordinary frequency, λ_vac is vacuum wavelength and ṽ = 1/λ_vac is spectroscopic wavenumber. It differs from the angular wavevector magnitude 2π/λ. Use c in units matching the length unit; 1 cm⁻¹ equals 100 m⁻¹ and 29.9792458 GHz.

The constants h and c have their SI defining values. Wavelengths reported in air instead need a refractive-index convention before applying a vacuum energy conversion. An energy gap fixes a matching photon energy but not the transition’s intensity, linewidth or experimental visibility.

Different motions have different model spectra

A rigid linear rotor has E_J = ħ²J(J+1)/(2I), with moment of inertia I and integer J = 0,1,2,… . Define its term value E_J/(hc) = B J(J+1), using consistent length units, for example c in cm s⁻¹ with B in cm⁻¹. In the ordinary electric-dipole pure-rotational model, a permanent molecular dipole and ΔJ = ±1 are required; absorption J → J+1 has wavenumber 2B(J+1).

Rotational levels still exist for a molecule whose ordinary dipole rotational spectrum is absent. A harmonic vibrational mode has equal level spacings hν; additional dipole and symmetry conditions determine which changes absorb IR light. A nonlinear isolated N-atom molecule has 3N−6 vibrational coordinates, while a linear molecule has 3N−5. These counts are modes, not guaranteed counts of distinct observable bands.

Rotational level terms and transition gaps are different sequences

Lower J	Lower level E_J/(hc) / cm⁻¹	Absorption gap / cm⁻¹	Matching ν / GHz
0	0.00	3.00	89.938
1	3.00	6.00	179.875
2	9.00	9.00	269.813
3	18.00	12.00	359.751

Original rigid linear rotor with stipulated B=1.50 cm⁻¹ and permanent dipole, using ordinary one-photon electric-dipole absorption J→J+1. Level terms are energies divided by hc; gaps are adjacent-level differences. No centrifugal distortion, line widths or line strengths are supplied. Frequencies are audit digits from the exact cm⁻¹ conversion, not experimental measurements.

An electronic transition can include nuclear-state changes

Electronic transitions can connect different vibrational and rotational states as well. Under Born–Oppenheimer and Condon approximations, a vibronic strength separates into an electronic transition-dipole factor and a squared vibrational overlap, the Franck–Condon factor. A single electronic energy difference therefore need not correspond to one narrow observed line.

Molecular environment, lifetime, temperature, unresolved states and instrument response can shape a band. A calculation of one geometry, an orbital gap and an absorption maximum are different quantities. Assigning a spectrum requires the model, selection rules and relevant sample conditions rather than only choosing the nearest-looking energy difference.

The measured signal adds an optical and chemical model

For light transmittance T_light = P/P₀, absorbance is A = −log₁₀(T_light). Here A means absorbance, distinct from Helmholtz energy in the thermodynamics chapter. In a suitable fixed-wavelength Beer–Lambert model, A = εℓc, with molar absorptivity ε, path length ℓ and concentration c. Units L mol⁻¹ cm⁻¹, cm and mol L⁻¹ cancel to give dimensionless A.

That linear relation needs appropriate bandwidth, uniform sample, stable absorbing species and adequate background correction. Concentration-dependent interactions or speciation, scattering, stray light and detector limits can cause departures. A blank corrects matched cell and solvent contributions; it does not guarantee removal of every optical artifact or establish a universal linear concentration range.

Worked examples

Convert a vacuum wavelength on particle and molar bases

Use a stipulated vacuum wavelength 500 nm, interpreted at three significant figures. Find photon energy and the energy per mole of photons.

Convert wavelength to 5.00 × 10⁻⁷ m.
E = hc/λ ≈ 3.97289 × 10⁻¹⁹ J per photon.
Divide by the exact joules per electronvolt to obtain approximately 2.47968 eV.
Multiply by N_A = 6.02214076 × 10²³ mol⁻¹ to obtain approximately 239.253 kJ per mole of photons.

Approximately 2.48 eV per photon, or 239 kJ mol⁻¹ of photons. The matching spectroscopic wavenumber is 2.00 × 10⁴ cm⁻¹.

Subtract adjacent rotor levels

For the specified rigid linear model with B=1.50 cm⁻¹, find the J=1 → 2 absorption gap and matching ordinary frequency.

The lower term is B × 1× 2 = 3.00 cm⁻¹.
The upper term is B × 2× 3 = 9.00 cm⁻¹.
Their difference is 6.00 cm⁻¹, equivalently 2B(J+1).
Multiply by 29.9792458 GHz per cm⁻¹.

The gap is 6.00 cm⁻¹ and ν ≈ 1.80 × 10² GHz. This predicts a position within the stated selection model, not its observed intensity.

Use matched Beer–Lambert units

For an invented one-species fixed-wavelength model, ε=2.00 × 10⁴ L mol⁻¹ cm⁻¹, c=2.50 × 10⁻⁵ mol L⁻¹ and ℓ=1.00 cm. Find absorbance and transmitted fraction.

A = εℓc = 0.500; the units cancel.
From A = −log₁₀T_light, T_light = 10^(−A).
10^(−0.500) ≈ 0.316228.
The transmitted fraction is approximately 31.6%, under the supplied optical and chemical assumptions.

A = 0.500 and T_light ≈ 0.316. These supplied coefficients do not identify a real substance or prove a universal linear range.

Try it yourself

How many inverse metres equal 1 cm⁻¹?
Show the answer and reasoning
100 m⁻¹.
A reciprocal centimetre must be converted as a reciprocal length.
For the stated B=1.50 cm⁻¹ model, what is the J=0 → 1 absorption gap?
Show the answer and reasoning
3.00 cm⁻¹.
Use 2B(J+1) with lower J=0.
How many vibrational coordinates does nonlinear four-atom NH₃ have?
Show the answer and reasoning
Six.
3N−6 = 6; this is not a guaranteed six-peak spectrum.
What transmitted fraction corresponds to dimensionless absorbance A=1?
Show the answer and reasoning
0.100.
T_light = 10^(−1); absorbance and transmission are not the same quantity.

What to carry forward

Keep vacuum wavelength, wavenumber and angular conventions explicit.
Subtract state energies before predicting a transition position.
Modes, allowed transitions and observed bands are different counts.
Measurement linearity requires optical and chemical conditions.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.