Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue?

Question

Most optics texts will mention that alkali metals can become transparent in the near ultraviolet in the sections on reflections from metals, plasma frequency, and electron density. I remembered this again just now when I saw Table XXVII in Born and Wolf's Principles of Optics (6th ed.) and saw that the critical wavelength for cesium has an observed value of 4400Å, which is plain old blue light that we see all the time. The table caption says:

The critical wavelength $\lambda_c$ below which the alkali metals become transparent, and above which they are opaque and highly reflective.

...alkali metals become transparent... This sounds like it would be quite amazing to actually see - metal that becomes transparent in blue or even near-UV light! I would like to see this - even an image published somewhere - anything! Or if it is actually not really true as stated in Born & Wolf, what else is there to consider?

I think I've seen pieces of cesium under mineral oil, and plenty of pictures on the internet and it just looks like metal. Definitely not transparent for blue light in bulk from what I've seen.

Is the 4400Å value wrong, or am I misunderstanding something, or would the cesium still need to be relatively thin or perhaps very cold to be detectably transparent in blue visible light? Are there any examples of this surprising effect due to the low plasma frequency (large critical wavelength) that can be linked to or shown here?

Is there some data I can see that's not behind a paywall? A photo of blue light passing through bulk cesium metal?

Answer 1

Simple metals are transparent at short wavelengths. It is maybe a problem that physicists do not usually write what their samples look like (chemists often note color and smell). I found this old paper from 1931 that starts:

It is a matter of common knowledge that an ordinary gold leaf appears green by transmitted light while silver appears blue.

I did not know this "common knowledge" about silver, such thin foils are not common. I have seen it for gold, but I cannot find photos. It would be very illustrative to have a photo of transparent films of the alkali metals but those are a bit difficult to prepare. I found a movie about alkali metal dissolved in ammonia: first blue (solvated electrons are blue), then "bronze".

There should be some gold leaf around, I will look for it and try to make a photo.

Answer 2

We used cesium based atomic line filters in an air to air laser communications system, but the wavelength was near 850 nm (near infrared).

But a quick search finding this suggests Cs is also useful in the blue spectrum, so possibly UV.

Atomic line filters heat the Cs to create a vapor state in the Cs, trapped between two optical windows that admit light.

Answer 3

Yes, this is true under certain circumstances, of course. I will keep things simple and say that the best way to answer your question is to look at the refractive index and transmittance spectrum of caesium. For that, I invite you to visit this link

https://refractiveindex.info/?shelf=main&book=Cs&page=Smith

Where you can find these data for many materials. But, concentrating caesium, if we look at the refractive index (figure below) We see that $ n=k $ at around $ \lambda = 410 \text{nm} $, which means that the plasma wavelength is 410 nm. Just to be clear on how I made this conclusion, the plasma wavelength is defined when the real part of the relative permettivity $ \epsilon_r $ is equal to 0. How does that relate to the refractive index? Well, the relative permettivity is nothing but the square of the complex refractive index, meaning $$ \begin{align} \epsilon_r & = {\widetilde n}^2 \\ & = {(n +ik)}^2 \\ & = (n^2 - k^2) +i2nk \end{align} $$ We can clearly see that $\operatorname{Re}(\epsilon_r) =0$ when $ n = k $.

Now, concerning the transparency, knowing the refractive index, one can calculate the transmittance of thin films using Transfer Matrix method, which relies on Fresnel coefficients (out of the scope of this discussion). The website offers a small tool that calculates the transmittance (figure below). I did it for a 50 nm thick caesium layer. You can see that the transmittance drastically increases for wavelengths below 500 nm. Of course, if you take thicker layers, the transmittance decreases. You need to remember the absorption coefficient is directly related to $ k $ and at the plasma frequency $ k = n $ but is not 0. Therefore, the electromagnetic wave will still be attenuated when passing through the caesium. So taking a thinner layer will insure that a decent amount of light will still go through. Now I invite you to check the website and check the refractive index and transmittance of other Alkali metals.

Answer 4

It is absolutely accurate that two continuous functions (the magnitude and imaginary component of the optical index, in this case) will cross if they trend in opposite directions. Therefore, there must be a very specific wavelength at which the the real and imaginary components are exactly equal.

There are many reasons you may not see the transparency for practical reasons. Metal purity being one. Phonon scattering being another...