The Beer-Lambert Law - Chemistry LibreTexts

The Absorbance of a Solution

For each wavelength of light passing through the spectrometer, the intensity of the light passing through the reference cell is measured. This is usually referred to as \(I_o\) - that's \(I\) for Intensity.

The intensity of the light passing through the sample cell is also measured for that wavelength - given the symbol, \(I\). If \(I\) is less than \(I_o\), then the sample has absorbed some of the light (neglecting reflection of light off the cuvette surface). A simple bit of math is then done in the computer to convert this into something called the absorbance of the sample - given the symbol, \(A\). The absorbance of a transition depends on two external assumptions.

1. The absorbance is directly proportional to the concentration (\(c\)) of the solution of the sample used in the experiment.
2. The absorbance is directly proportional to the length of the light path (\(l\)), which is equal to the width of the cuvette.

Assumption one relates the absorbance to concentration and can be expressed as

\[A \propto c \label{1}\]

The absorbance (\(A\)) is defined via the incident intensity \(I_o\) and transmitted intensity \(I\) by

\[ A=\log_{10} \left( \dfrac{I_o}{I} \right) \label{2}\]

Assumption two can be expressed as

\[A \propto l \label{3}\]

Combining Equations \(\ref{1}\) and \(\ref{3}\):

\[A \propto cl \label{4}\]

This proportionality can be converted into an equality by including a proportionality constant (\(\epsilon\)).

\[A = \epsilon c l \label{5}\]

This formula is the common form of the Beer-Lambert Law, although it can be also written in terms of intensities:

\[ A=\log_{10} \left( \dfrac{I_o}{I} \right) = \epsilon l c \label{6} \]

The constant \(\epsilon\) is called molar absorptivity or molar extinction coefficient and is a measure of the probability of the electronic transition. On most of the diagrams you will come across, the absorbance ranges from 0 to 1, but it can go higher than that. An absorbance of 0 at some wavelength means that no light of that particular wavelength has been absorbed. The intensities of the sample and reference beam are both the same, so the ratio \(I_o/I\) is 1 and the \(\log_{10}\) of 1 is zero.

Example \(\PageIndex{1}\)

In a sample with an absorbance of 1 at a specific wavelength, what is the relative amount of light that was absorbed by the sample?

Solution

This question does not need Beer-Lambert Law (Equation \(\ref{5}\)) to solve, but only the definition of absorbance (Equation \(\ref{2}\))

\[ A=\log_{10} \left( \dfrac{I_o}{I} \right)\nonumber\]

The relative loss of intensity is

\[\dfrac{I-I_o}{I_o} = 1- \dfrac{I}{I_o}\nonumber\]

Equation \(\ref{2}\) can be rearranged using the properties of logarithms to solved for the relative loss of intensity:

\[ 10^A= \dfrac{I_o}{I}\nonumber\]

\[ 10^{-A}= \dfrac{I}{I_o}\nonumber\]

\[ 1-10^{-A}= 1- \dfrac{I}{I_o}\nonumber \]

Substituting in \(A=1\)

\[ 1- \dfrac{I}{I_o}= 1-10^{-1} = 1- \dfrac{1}{10} = 0.9\nonumber\]

Hence 90% of the light at that wavelength has been absorbed and that the transmitted intensity is 10% of the incident intensity. To confirm, substituting these values into Equation \(\ref{2}\) to get the absorbance back:

\[\dfrac{I_o}{I} = \dfrac{100}{10} =10 \label{7a}\]

and

\[\log_{10} 10 = 1 \label{7b}\]