UV-visible, infrared and Raman measurements

An optical signal needs an interaction and a reference

Optical measurements observe how a sample affects radiation. Ultraviolet and visible absorption commonly probe electronic transitions. Molecular infrared measurements often probe vibrational transitions. Raman measurements use inelastic scattering, in which the scattered photon has a different energy from the incident photon. These observations constrain different features of a sample.

Transmission compares a measured radiant power P with an appropriate reference P₀ under the specified measurement convention: T=P/P₀. Decadic absorbance is A=−log₁₀ T=log₁₀(P₀/P). Both are dimensionless, but they are different quantities. Reference correction, scattering, reflection and stray light affect what a measured ratio can mean.

A strong-looking band is not automatically a high concentration, and an absent band is not automatically an absent constituent. The interaction, optical path, chemical forms, matrix, detection and calibration all belong to the interpretation.

Absorbance follows a path–concentration product under stated conditions

The Beer–Lambert model A=εlc relates absorbance to molar absorption coefficient ε, path length l and absorber concentration c. With ε in L mol⁻¹ cm⁻¹, l in cm and c in mol L⁻¹, the product is dimensionless. SI units can instead use ε in m² mol⁻¹, l in m and c in mol m⁻³.

An original teaching model assigns ε=15000 L mol⁻¹ cm⁻¹, l=0.500 cm and c=20.0 μmol L⁻¹. Absorbance is 0.150 and transmitted fraction T=10⁻⁰·¹⁵=0.7079458. The same calculation in SI uses ε=1500 m² mol⁻¹, l=0.00500 m and c=0.0200 mol m⁻³.

The plotted linear relation is hypothetical. It assumes a monochromatic, homogeneous, isotropic, non-scattering system with unchanged absorber speciation and valid linear response. Chemical equilibria, broad bandwidth, stray light or detector effects can invalidate a simple concentration–absorbance relation. A high-quality straight line over one range does not prove suitability in another.

Independent responses can distinguish a defined two-component mixture

For an additive two-absorber model at a 1.00 cm path, assign coefficient rows (10000, 2000) and (3000, 8000) L mol⁻¹ cm⁻¹ for A and B at two selected wavelengths. Given c_A=20.0 μmol L⁻¹ and c_B=30.0 μmol L⁻¹, predicted absorbances are 0.260 and 0.300. These are invented coefficients and responses, not measured spectra.

Writing concentrations numerically in μmol L⁻¹ gives equations A₁=0.0100c_A+0.00200c_B and A₂=0.00300c_A+0.00800c_B, with the coefficients in the corresponding inverse-concentration units. Subtracting 0.300 times the first response equation from the second gives 0.222=0.00740c_B in that numerical convention, recovering c_B=30.0 and then c_A=20.0.

The coefficient rows supply different dependencies on the two constituents. If the second row were exactly twice the first, there would be only one independent relation and no unique two-concentration inversion. Extra constituents, changing speciation, background and uncertain or nearly dependent coefficients also limit inference. Two wavelengths alone are not universal mixture identification.

A wavenumber is a reciprocal length, not a frequency in hertz

For a photon in vacuum, frequency ν=c₀/λ and energy E_ph=hν=hc₀/λ, where c₀ is the speed of light and h the Planck constant. Spectroscopic wavenumber ṽ=1/λ is often given in cm⁻¹. Multiplying its numerical value by 100 converts it to m⁻¹. The symbol c₀ here is a speed, distinct from absorber concentration c above.

Assign ṽ=1600 cm⁻¹. Then ṽ=160000 m⁻¹, λ=6.25 μm and ν=4.796679328×10¹³ Hz. Using exact SI h=6.62607015×10⁻³⁴ J s and c₀=299792458 m s⁻¹ gives photon energy 3.17831337×10⁻²⁰ J, or 0.198374717 eV with 1 eV=1.602176634×10⁻¹⁹ J.

This calculation converts a given photon or transition energy; it does not establish an observed band, an active vibrational mode or a molecular identity. In the usual vibrational absorption model, infrared activity requires an appropriate changing dipole response along the vibration. Frequency coincidence alone cannot identify a functional group uniquely.

Raman shift and scattered wavelength are different coordinates

In normal vibrational Raman scattering, a Stokes photon loses energy to the sample and has a lower wavenumber and longer wavelength than the incident photon. An anti-Stokes photon gains energy and has a higher wavenumber and shorter wavelength. The usual nonresonant activity criterion concerns changing molecular polarizability along the vibration. It is different from the infrared dipole criterion.

Use invented laser wavelength λ₀=532 nm and vibrational shift Δṽ=1000 cm⁻¹. Incident wavenumber is 10⁷/532=18796.9925 cm⁻¹. Stokes and anti-Stokes scattered wavenumbers are respectively 17796.9925 and 19796.9925 cm⁻¹, giving wavelengths 561.892691 and 505.127231 nm. The photon energy difference is hc₀Δṽ=1.98644586×10⁻²⁰ J after converting the shift to m⁻¹.

Keeping the same shift while changing the incident wavelength changes the scattered wavelengths. Positions alone do not give a Stokes/anti-Stokes intensity ratio: populations, response and mechanism matter. The normal-scattering model should not be described universally as absorption into a real excited electronic state followed by emission; resonance and other mechanisms require their own scope.

Different optical sensitivities can constrain alternatives

Infrared and Raman observations can be complementary because their response criteria differ. Some vibrations can contribute to both. A simple “symmetric means Raman, asymmetric means IR” slogan is insufficient as a universal selection rule; the actual molecular symmetry and mode responses govern.

Use references and calibrated responses suited to the analyte, chemical form and matrix. A spectral comparison may support a candidate, while quantitative concentration requires an appropriate response model and uncertainty. These original examples contain no measured spectra, library matches or instrument operating recommendations.

Worked examples

Try it yourself

1. Within the same linear model, halve path length while doubling concentration. What happens to A and T?

   Show the answer and reasoning

   They remain 0.150 and 0.7079458 for the stated original values.

   The path–concentration product is unchanged; this arithmetic does not demonstrate a real new working range.

2. Replace the second optical coefficient row by twice the first. Can the two unknown concentrations now be uniquely recovered from those responses?

   Show the answer and reasoning

   No; the rows supply only one independent relation.

   The determinant is zero. A change proportional to (1, −5) in the two concentration numbers leaves both responses unchanged.

3. Double the assigned wavenumber to 3200 cm⁻¹. Find wavelength and photon energy relative to the original case.

   Show the answer and reasoning

   Wavelength 3.125 μm; energy doubles to 6.35662674×10⁻²⁰ J.

   Wavelength is reciprocal to wavenumber while frequency and photon energy are proportional to it.

4. Change the incident wavelength to 785 nm while keeping the assigned 1000 cm⁻¹ shift. Find the Stokes wavelength and identify what remains fixed.

   Show the answer and reasoning

   Stokes wavelength 851.871948 nm; the shift and photon energy difference remain fixed.

   Subtract 1000 cm⁻¹ from 10⁷/785, then take the reciprocal wavelength conversion. Equal shift does not mean equal scattered wavelength.

What to carry forward

• Specify the interaction and reference behind an optical response.
• Absorbance and transmission are different dimensionless quantities.
• A concentration model requires unchanged chemical and optical conditions.
• Wavenumber conversion needs consistent reciprocal-length units.
• Raman positions do not establish intensity ratios or unique identity.