Spectroscopy & Smart Materials Science (8 hrs) · Engineering Chemistry

Spectroscopy: fundamentals of rotational, vibrational & electronic spectra; Beer-Lambert law

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Spectroscopy Fundamentals & Beer–Lambert Law

1. What spectroscopy is

Molecules absorb/emit electromagnetic radiation at discrete energies matching gaps between quantized levels:

ΔE=Eupper−Elower=hu=hcλ=hcuˉ\Delta E = E_{upper} - E_{lower} = h u = \frac{hc}{\lambda} = hc\bar{ u}

where $\bar{
u} = 1/\lambda$ is the wavenumber (cm⁻¹, the unit of IR spectra).

Energy ladder: electronic (largest gap) > vibrational > rotational (smallest). So UV-Vis photons drive electronic jumps, IR drives vibrations, microwaves drive rotations.

2. The three regions compared

Type Region Transition Energy order Selection rule / condition
Rotational (microwave) λ∼\lambda \sim mm–cm J→J+1J \to J+1 rotation levels smallest molecule must have permanent dipole (HCl yes, N₂/O₂ no)
Vibrational (IR) 4000–400 cm⁻¹ v=0→1v = 0 \to 1 stretching/bending medium dipole moment must change during vibration (CO₂ asymmetric stretch active, symmetric stretch inactive)
Electronic (UV-Vis) 200–800 nm HOMO → LUMO (σ→σ\*\sigma\to\sigma^\*, n→π\*n\to\pi^\*, π→π\*\pi\to\pi^\*) largest spin-allowed (ΔS=0\Delta S = 0); conjugation shifts λmax\lambda_{max} up

Rotational lines are equally spaced: separation =2B= 2B, giving bond length. Vibrational frequency follows Hooke's law: $\bar{
u} = \frac{1}{2\pi c}\sqrt{k/\mu}—strongerbond(bigger — stronger bond (bigger k)andlighteratoms(smallerreducedmass) and lighter atoms (smaller reduced mass \mu$) absorb at higher wavenumber.

3. Beer–Lambert law (quantitative analysis)

Absorbance is proportional to concentration and path length:

A=log⁡10I0I=εclA = \log_{10}\frac{I_0}{I} = \varepsilon c l
  • AA = absorbance (unitless), I0,II_0, I = incident/transmitted intensity.
  • ε\varepsilon = molar absorptivity (L·mol⁻¹·cm⁻¹), cc = concentration (mol/L), ll = path length (cm, usually 1).
  • Transmittance T=I/I0T = I/I_0; A=−log⁡10TA = -\log_{10}T.

Numericals:

  • ε=5000\varepsilon = 5000, c=2×10−5c = 2 \times 10^{-5} M, l=1l = 1 cm → A=0.10A = 0.10.
  • A=0.699A = 0.699 means T=10−0.699≈0.20T = 10^{-0.699} \approx 0.20 (20% transmitted).
  • Calibration curve: plot AA vs known cc, read unknown from line — standard exam numerical.

Deviations: law fails at high concentration (> 0.01 M, molecular interactions), if light isn't monochromatic, or if the analyte dissociates/associates.

Exam Points

  • Energy-order + region + selection rule table (3–5 marks, frequently asked as "distinguish rotational, vibrational, electronic spectra").
  • Beer–Lambert statement, formula, one numerical + calibration-curve method (5 marks favourite); remember AA uses log base 10, and units of ε\varepsilon.
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