CHEM 305

Absorption of light: Beer-Lambert Law

Up to this point, we have learned how electromagnetic waves are generated,how molecules can scatter light (and how we can determine molecularweight from the amount of scattering, using a Zimm plot) and how helical molecules interact with circularly polarized light.

In this section, we will learn how compounds absorb ultraviolet (UV) or visiblelight:

I0I

monochromatic

P0 is the incident radiant power or intensityP is the radient intensity that remainsb is the path length

b

CHEM 305

Physical principle

Recall the potential energy function that we had for a diatomic molecule.

CHEM 305

Transmittance T = I / I0% Transmittance %T = 100 T

Absorbance A = log10 I0 / IA = log10 1 / TA = log10 100 / %TA = 2 - log10 %T

We can describe how the light is absorbed by either the transmittance orthe absorbance:

Both the transmittance and absorbance are related:

CHEM 305

Beer-Lambert Law

The Beer-Lambert law relates the absorbance to the concentration:

A=bcwhere A is absorbance (no units, since A = log10 P0 / P ), is the molar absorbtivity or extinction coefficient with units of L mol-1 cm-1, b is the path length of the sample i.e. the path length of the cuvette in which the sample is contained (in cm) and finally, c is the concentration of the compound in solution, expressed in mol L-1.

Of course, we can also express transmittance in terms of concentration, butthe Beer-Lambert law is more useful because the relationship between absorbance and concentration is linear.

CHEM 305

Derivation of the Beer-Lambert law

In order to derive the law, we need to approximate the absorbing molecules inthe cuvette as opaque disks, of cross-sectional area,

http://www.chem.ufl.edu/~itl/3417_s98/spectroscopy/beerslaw.htm

I0 = intensity of light entering the sample

Iz = intensity at point zin the sample cell

dI = intensity of lightabsorbed by the slab

I = intensity of the lightleaving the sample

CHEM 305

Total opaque area in the slab = N dz A

i.e. fraction of opaque areas in the slab times the total area. This fraction ofopaque areas is also a measure of the fraction of light absorbed:

dI = - N dzIz

We can integrate this equation from z=0 to z=b, to give us

ln(I) ln(I0) = - N bor

ln (I0/I) = N b.N, the number of molecules per cm3, can be related to concentration by

N * (1000 cm3/L) = c (M)NA

and 2.303 * log(x) = ln(x)

CHEM 305

Therefore, we have

log (I0/I) = NA c b2.303*1000

or A = b c with

= NA2.303*1000

This shows how the extinction coefficient is related to the cross-sectional areas ofabsorption. E.g.

Absorption atoms 10-12 cm2 3*108 M-1cm-1molecules 10-16 cm2 3*104 M-1cm- 19 kDa protein 10-17 cm2 7*103 M-1cm- 1

CHEM 305

Non-linearity of Beer-Lamberts law

Beers law is linear in most cases, except:

at high concentrations if there is scattering of light due to particulates in the sample if the sample fluoresces or phosphoresces if the radiation is not monochromatic if there is stray light

Taking a closer look at concentrationeffects high concentration results innon-linearity because:

at high concentration, we have strongelectrostatic interactions betweenmolecules at high concentrations, we may get changes in refractive index if we have a system in chemical equilibrium,

equilibrium may shift at highconcentrations.

CHEM 305

Application of absorbance measurements

1 Estimate the concentration of a protein which has 1 or more Trp residues

You can simulate your own sequences athttp://www.proteinchemist.com/Multbot.html

if c is toohigh, dilute!

for all 20 amino acids together

CHEM 305

2 HPLC Trace

set absorbance wavelength = 280 nm

time

CHEM 305

3 Changes in configuration