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UV-Vis spectroscopy measures how much ultraviolet or visible light a substance absorbs at different wavelengths. Students use it to connect molecular structure, electronic transitions, and solution concentration. This cheat sheet helps organize the equations and graph skills needed to interpret spectra and solve quantitative lab problems.

It is especially useful for calibration curve analysis and Beer-Lambert law calculations.

The most important relationships connect wavelength, frequency, photon energy, absorbance, transmittance, path length, and concentration. Photon energy is found with E=hνE = h\nu or E=hcλE = \frac{hc}{\lambda}, while solution concentration is often found using A=εbcA = \varepsilon bc. A calibration curve relates absorbance to concentration with a line such as A=mc+bA = mc + b.

Higher absorbance usually means less transmitted light and, within the linear range, a greater concentration of the absorbing species.

Key Facts

  • Frequency and wavelength are related by c=λνc = \lambda \nu, where c=3.00×108 m/sc = 3.00 \times 10^8\ \text{m/s} in a vacuum.
  • Photon energy is calculated with E=hνE = h\nu or E=hcλE = \frac{hc}{\lambda}, so shorter wavelength light has higher energy.
  • Absorbance is defined by A=log10(T)A = -\log_{10}(T), where transmittance is T=II0T = \frac{I}{I_0}.
  • Percent transmittance is %T=100T\%T = 100T, so absorbance can also be written as A=log10(%T100)A = -\log_{10}\left(\frac{\%T}{100}\right).
  • The Beer-Lambert law is A=εbcA = \varepsilon bc, where ε\varepsilon is molar absorptivity, bb is path length, and cc is concentration.
  • For a Beer-Lambert calibration curve, the slope is often m=εbm = \varepsilon b when the graph is absorbance versus concentration.
  • The wavelength of maximum absorbance is written as λmax\lambda_{\max} and is usually chosen for concentration measurements.
  • Beer-Lambert law is most reliable for dilute solutions that give absorbance values in the instrument's linear range, often about 0.10.1 to 1.01.0.

Vocabulary

Absorbance
Absorbance is a logarithmic measure of how much light a sample absorbs, calculated by A=log10(T)A = -\log_{10}(T).
Transmittance
Transmittance is the fraction of incoming light that passes through a sample, given by T=II0T = \frac{I}{I_0}.
Wavelength
Wavelength is the distance between matching points on a wave and is represented by λ\lambda.
Molar absorptivity
Molar absorptivity, ε\varepsilon, measures how strongly a substance absorbs light at a specific wavelength.
Path length
Path length, bb, is the distance light travels through the sample, commonly 1.00 cm1.00\ \text{cm} in a standard cuvette.
Calibration curve
A calibration curve is a graph made from standards that relates absorbance to known concentration so an unknown concentration can be found.

Common Mistakes to Avoid

  • Using percent transmittance directly in A=log10(T)A = -\log_{10}(T) is wrong because TT must be a decimal fraction, so 45%45\% must become 0.450.45.
  • Forgetting to convert nanometers to meters in E=hcλE = \frac{hc}{\lambda} gives energy values that are too small or too large by powers of 1010.
  • Treating absorbance and transmittance as directly proportional is wrong because absorbance depends logarithmically on transmittance through A=log10(T)A = -\log_{10}(T).
  • Using a calibration curve outside its measured concentration range can give unreliable results because Beer-Lambert behavior may no longer be linear.
  • Measuring at a random wavelength reduces accuracy because concentration analysis is usually best at λmax\lambda_{\max}, where absorbance is strongest and sensitivity is highest.

Practice Questions

  1. 1 A solution has %T=25.0%\%T = 25.0\%. Calculate its absorbance using A=log10(%T100)A = -\log_{10}\left(\frac{\%T}{100}\right).
  2. 2 A compound has λmax=520 nm\lambda_{\max} = 520\ \text{nm}. Calculate the photon energy in joules using E=hcλE = \frac{hc}{\lambda}, h=6.626×1034 Jsh = 6.626 \times 10^{-34}\ \text{J}\cdot\text{s}, and c=3.00×108 m/sc = 3.00 \times 10^8\ \text{m/s}.
  3. 3 A sample has A=0.640A = 0.640, ε=1.60×104 Lmol1cm1\varepsilon = 1.60 \times 10^4\ \text{L}\cdot\text{mol}^{-1}\cdot\text{cm}^{-1}, and b=1.00 cmb = 1.00\ \text{cm}. Find the concentration using A=εbcA = \varepsilon bc.
  4. 4 Explain why a chemist would choose λmax\lambda_{\max} instead of a weakly absorbed wavelength when building a calibration curve.