Absorption coefficient for liquid water

Question

Wikipedia has this plot of the absorption spectrum of liquid water:

The vertical axis seems weird to me. The Beer-Lambert Law says $T=10^{Ax}$ where $T$ is the fraction of light transmitted, $A$ is the absorption coefficient, and $x$ is the optical distance through the material. If we take the values listed on the vertical axis to be $A$ as I've described it here, the transmittance ranges from 80% transmittance though a 1 nm (!) thick sample at 80 nm wavelength ($10^{-10^8*10^{-9}}$) to 80% transmittance through a 10 m thick sample at 500 nm wavelength ($10^{-10^{-2}*10}$).

Alternatively phrased, the absorption coefficient at 2 um is 10^4. Therefore the transmittance would be $10^{-10^4*0.01}$ through a 1 cm sample. That would be one in $10^{-100}$ photons transmitted through the sample. Given that IR spectroscopy is possible, that sure doesn't seem right to me.

Does the transmission of light through water really vary by 10 orders of magnitude or am I misreading this?

Answer 1

What the plot shows is the reciprocal path length needed for a given absorbance to apply, for example slightly less than 100 m in the visible to $10^{-8}$ m in the uv, as you point out, but this is for $55.345$ molar water which is very, very concentrated compared to most solutions from which an absorption spectrum is obtained, typically $<< 1 $ molar.

As you also point out the absorbance is $A=\log_{10}(I_0/I)=\epsilon[C]l$ with $\epsilon$ as the extinction coefficient usually in units $\pu{dm^3mol^{-1}cm^{-1}}$ and which depends upon wavelength, and $[C]=55.345 \pu{ mol}$ for pure water and l the path length in cm. What seems to be plotted is $$\frac{1}{l}=\frac{\epsilon[C]}{A}$$ so is still proportional to the extinction coefficient and so still indicates regions where water absorbs greatly or otherwise. But it is not clear what value for A is used. It may be understood that $A=1$ always applies or that the value is taken at $1/e$ absorption. The values may also be based on $\log_e$ not $10$ so a factor of $2.303$ may be involved. As I'm not familiar with this much detail I can't help any more other than to suggest that you will have to check by finding a true absorption spectrum or extinction coefficient and work backwards and converting units as necessary.

In general the amount of absorption a molecule has can vary a huge amount. In water in the visible part of the spectrum absorption is low as there are only weak overtone bands from vibrational transitions and nothing else. At shorter wavelengths electronic absorption begins and these can have extinction coefficients of tens to hundreds of thousands in units of $\pu{dm^3mol^{-1}cm^{-1}}$.

A concentrated solution of a dye at, say, $1$ molar with a typical extinction coefficient of $10^4$ would have an optical density of $4$ in a $1$ cm cell or a reciprocal length of $10^6$ $\pu{m^{-1}}$ if $A=1$. Hope this helps.

Answer 2

The plot is from Wikipedia, where important/helpful links to the source information can be found, and is worth a look.

A similar and more detailed plot can be found on this page of Martin Chaplin's famous site on all things water. I can't recommend that site highly enough! The plot is shown below, and note that it's using centimeters instead of meters.

• What happened to Martin Chaplin's famous Water Structure and Science website?

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This section in Wikipedia gives us a clue. It says

Absorption coefficients for 200 nm and 900 nm are almost equal at 6.9 m−1 (attenuation length of 14.5 cm).

Those numbers are reciprocal, and reading the links shows the terms refer to $1/e$ lengths, natural units.

Both this plot and Wikipedia's are intended to give a global electromagnetic perspective rather than detailed quantitative information. I believe there really is a peak near 2 microns where a 1 cm cuvette of pure water would be nearly opaque. Remember it's using natural units, not 10, so ya, if you sit on that peak, using:

$$T = exp\left(-h(cm) \ \times \ ~100 (1/cm)\right)$$

and plugged in a 1 cm optical path, you'd get a transmission of 4E-44, or basically zero. You many find that there are different cuvettes with substantially reduced optical paths that can be used in this case. A sub-1 millimeter path length cuvette might do better.

However, a little bit above or below 2 microns, the attenuation drops to only 10 per centimeter, which gives a transmission of about 5E-05 for a standard 1 centimeter path length cuvette, and a quality instrument would have no problem measuring that with a suitable detector and integration time.

If you're interested in going farther, I'd recommend you find a plot that is nod drawn with such thick lines, or find tabulated data and plot it in detail by yourself. There are some numbers just for example in this 1975 paper but the hard short-wavelength cutoff at 2 microns may obscure the peak there. However let's look at the properties near 3 microns.

At 2.959 microns, $n, k = 1.329, 0.292$. Putting that into here

$$\frac{E}{E_0} \ = \ exp\left(j \left(n + jk\right) \frac{2\pi}{\lambda} x \right) $$

puts the electric field attenuation of 1 centimeters at $exp(-6200)$. You need to square to get transmitted intensity, so doubling the argument gives 12,400/cm or 1.24/um, which very nicely matches the "3 micron peak" in the plot, especially the inset in red(see below). This is really a mixture of states in the circa 3400 cm${}^{-1}$ water vibrational spectrum. Also, when you read, make sure you keep track of the isotopic mixture being discussed. H2O and HDO differ profoundly in their infrared optical properties.

All is well, and water continues to be amazing, profound stuff!