# Absorption coefficient units

## Main Question or Discussion Point

Why is the absorption coefficient ($\alpha$) of water is always given using the units $\text{length}^{-1}$, while for other materials it is often given by $\text{length}^{2}.\text{mass}^{-1}$?

For instance, this paper uses $cm^{-1}$ for water and yet $cm^2 / \mu g$ for other substances (such as chlorophyll, cellulose and lignin). I have seen the same in other literature.

For a given wavelength, I am trying to calculate the extinction length $L=4.6/\alpha$ (the thickness of material required to absorb 99% of incident light). For water this calculation is straightforward. For instance, at a wavelength of 10 micron, the absorption coefficient of water is $\alpha \approx 1000\ cm^{-1}$, corresponding to $L \approx 40 \ \mu m$.

But how do we handle this calculation when the absorption coefficient of the material is instead given by $cm^2 / \mu g$?

Any explanation is greatly appreciated.

mjc123
Homework Helper
What is the relation between length-1 and length2.mass-1? Does that suggest anything to you?
What is the difference between water and these other substances, in terms of the form in which they are presented for spectroscopy?
What extra information do you need to calculate the extinction length?

roam
What is the relation between length-1 and length2.mass-1? Does that suggest anything to you?
What is the difference between water and these other substances, in terms of the form in which they are presented for spectroscopy?
What extra information do you need to calculate the extinction length?
Did they multiply it by a volume and divide it by a mass? Since density is equal to mass divided by volume ($\rho=m/V$), it appears like they are dividing by density of something.

For water, I presume they pour it into a cuvet and then place it in a spectrophotometer. I am not sure how those other substances were measured. Sometimes substances are dissolved in a liquid, but I think the paper that I linked to uses some kind of remote sensing to estimate $\alpha$.

What other information do we need to calculate the extinction length?

mjc123
Homework Helper
Density, or mass concentration.

I am assuming that water is measured as a pure substance (or a solvent in great excess over solute), then you only need an absorption coefficient in length-1 as concentration is not a variable. I assume the other things are measured in dilute solution in water, so you need to know the concentration (in µg/cm3 to judge from your units). Then the absorbance is given by
A = log(I0/It) = αcL
where α is in cm2/µg, c is in µg/cm3 and L in cm.
(The units I am familiar with are L/mol/cm, where concentration is expressed in moles per litre.)

[Actually, considering your definition of extinction length, I think that should be ln rather than log, to give the factor of 4.6. The extinction length of a solution of solute is then 4.6/(αc), and is concentration dependent.]

roam
Hi mjc123,

So, if I have the coefficient given in cm2/µg, I simply need to multiply it by the desired concentraion, and then use $L=4.6/\alpha c$ to find the extinction length?

And I think the quantity given in cm2/µg is not exactly the $\alpha$ which is seen in the following Beer-Lambert law (which should be a decadic logarithm relationship):

$$I(z)=I_{0}e^{-\alpha z}.$$

Some references explicitly say that the units for $\alpha$ has to be length-1. In a few textbooks, the quantity that you referred to is called the "molar extinction coefficient" (denoted by $\epsilon$), so that:

$$A = \log_{10} \left( \frac{I_0}{I_t} \right) = \epsilon \times c \times l,$$

where they define $l$ as the length of the cell that contains the solution (not the extinction length $L$). I think maybe some authors are conflating $\epsilon$ and $\alpha$ by calling both of them the "absorption coefficient".

As a side-note, the paper that I linked to uses µg/cm2 as the unit for concentration $c$. This has made me very confused since this is mass.length-2, rather than mass.length-3...

mjc123