The Beer-Lambert Law, a straight line and the units of the extinction coefficient

If you would like to test your skills working with the Beer-Lambert Law, you might like to look at the Spectrophotometry tests at Maths4Biosciences.com.

Some students struggle to understand the relationship between the Beer-Lambert Law and a straight line and work out the units of the extinction coefficient (ε).

The Beer-Lambert Law states:

A = ε . c . l

Where:

A = absorbance 

ε = extinction coefficient 

c = concentration 

l = path length (i.e. the distance the light travels through the sample)

So, what is the connection between this and a straight line, and what are the units of the extinction coefficient?

The units of the extinction coefficient

In my opinion, the extinction coefficient has some of the craziest units out there.
Absorbance (A) has no units, so the units of the extinction coefficient (ε) are determined by how the concentration (c) and path length (l) are being measured. That is, the units of the extinction coefficient must cancel out the concentration and path length units so that the absorbance can have no units!

A worked example.

A = ε . c . l

A = absorbance, units - so put in 1 

ε = extinction coefficient, units - unknown 

c = concentration, units - milli-Molar, mM 

l = path length (i.e. the distance the light travels through the sample), units - cm

So...

A = ε . c . l

And, with units:

[1] = ε . [mM] . [cm]
Rearranging...
[1] / ε = [mM] . [cm] 
ε = 1 / ([mM] . [cm]) 
or 
ε = [mM]^-1 . [cm]^-1

So, the units are mM^-1 . cm^-1

This can be checked by putting it all back together:


A = ε . c . l 
A = ([mM]^-1 . [cm]^-1) . [mM] . [cm]
This gives:


A = [mM]/[mM] . [cm]/[cm]
So, the mM and the cm cancel each other out, leaving no units for absorbance A.

A straight line

The Beer-Lambert Law:

A = ε . c . l

Where:

A = absorbance 

ε = extinction coefficient 

c = concentration, units 

l = path length

The equation for a straight line is:

y = mx + c

Where:

m = the gradient 

c = the y-intercept

If you plot concentration against absorbance, then x = concentration and y = absorbance. Plus, from the Beer-Lambert Law, we know that if the concentration is zero, then absorbance must be zero.

A = ε . c . l A = ε . 0 . l A = 0  So...

y = mx + c absorbance = m . concentration + c

From above, if concentration = 0, then absorbance = 0, hence c must be zero
y = mx + c absorbance = m . concentration + c* 0 = m . 0 + c  c = 0
(* note, this c is the y-intercept and not the concentration)

Therefore...
y = mx + c absorbance = m . concentration + 0 or y = mx


Comparing:

y = mx absorbance = m . concentration
With (and rearranging):

A = ε . c . l 
A = (ε . l) . c 
y = m . x If y = absorbance, and x = concentration, then m (the gradient) must equal extinction coefficient (ε) multiplied by the path length, l, or ε . l. As l is typically 1 cm, the gradient, m, must equal the extinction coefficient (ε).

You may also find the following two videos helpful.

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