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 A Derivation of Beer's Law for Optical Absorption

Beer's Law is expressed as differential equation

#### (1/I)dI/dx = -αρ

where I is the intensity of radiation, x is the distance traveled in a medium, ρ is the molecular volume density of absorbers in the medium, and α is a parameter, called the absorption coefficient, characteristic of the medium. The law can be expressed in different terms such as the density being the linear density of absorbers. Then the absorption coefficient is correspondingly changed. For the purposes here it is most convenient to have the density being the number of absorbing molecules per unit volume.

The solution to the differential equation is

#### I(x) = I(0)e-αρx

This means that the proportion of the radiation transmitted is

#### I(x)/I(0) = e-αρx

Thus the proportion absorbed is

## Derivation

Consider a beam of light of cross section area A passing through a medium of depth L. Suppose that for a quantum of light to pass through the medium unabsorbed there has to be cylindrical tunnel of cross section area a where there are no absorbing molecules.

In the diagram the area that must be free of the centers of the absorbing moleculars, shown in white and designated as a, is the same as the cross section area of an absorbing molecule, shown in red.

The probability p that the center of one molecule is not located in that area a is

#### p = (A-a)/A = 1-a/A

The probability P that all of N molecules are located outside of the area a is then

Since p=1-a/A

#### ln(p) = ln(1-a/A) and for small values of -a/A ln(p) is approximately equal to -a/A

The number of molecules in the beam is the volume times the molecular density; i.e.,

#### N = (A*L)ρ

where ρ is the molecular volume density of the molecules.

Thus

#### P = e−(a/A)(A*L)ρwhich reduces to P = e−aρL

The light intensity at a distance L in the medium is then

#### I(L) = I(0)e−aρLand thus dI/dL = −aρ

Thus the absorption coefficient is equal to the cross section area that a quantum needs for clear passage. This cross section area is the effective cross section area of an absorbing molecule.

The proportion not absorbed is the same as the probability of a quantum not being absorbed, P. Thus,

#### I(L)/I(0) = P = e−aρLand hence I(L) =I(0)e−aρL

The proportion absorbed is then 1-P; i.e.,

If L varies then

#### dI/dL = −aρI(0)e−aρL = −aρI(L) and hence (1/I(L))dI/dL = −aρ

This is Beer's Law with α=a, the cross section area of an absorbing molecule.

## The Dimensions of the Parameters

In the above derivation the absorption coefficient had the dimensions of area, m² and the concentration ρ was in molecules per unit volume, 1/m³. The path length is in meters. This means the product αρ has the dimension inverse length, 1/m. This product, αρ is sometimes called the absorption coefficient. This quantity is shown below plotted versus the wave length of the radiation.

The molecular density is related to the temperature and pressure through the gas law. One form of an absorption coefficient is per unit of pressure.

(To be continued.)