This may appear to be a chemistry problem. But, after reading the Wikipedia article on copper(I) oxide, it seems to have more to do with semiconductor-physics. For example:

… light travels almost as slowly as sound in this medium.

Is that true?

What have Kramers–Kronig relations got to do with it?

To a chemist, who was never brilliant at maths, it takes a bit of understanding. I know that copper(II) oxide (Mott–Hubbard insulator [semiconductor]) is black because of intervalence charge transfer, giving rise to the generation of a highly polarising Cu(III) species. Similarly, the non-stoichiometric form of nickel(II) oxide (Mott insulator) is black because of a Ni(III) species. Again, silver(I) oxide is black … Ag(III) species.

This model does not appear to work for copper(I) oxide because the non-stoichiometry, causing the oxidation required for the balancing of charges with the oxide ions, would give Cu(II); which, by definition, is not sufficiently polarising to produce the deep, intense colour observed. Further, the reduced Cu(I) becomes Cu(0), the pure metal.

So, why is copper(I) oxide red?


2 Answers 2


No, it is not true that light moves very slowly in Cu$_2$O. Maybe some polariton extremely close to its resonant energy (I did not look into that), but not generally.

It is red for the same reason why vermillion is red. It has a bandgap of about 2 eV, so that blue and green light are absorbed. And the absorption (the imaginary part of the complex refractive index) is related by Kramers-Kronig to the real part of the refractive index, which gives high reflectivity in the red.

  • $\begingroup$ is the link between Kramers-Kronig and refractive index written-down, anywhere? As a mathematical plodder, I have to pore over things. Thank you. $\endgroup$
    – tony
    Aug 6, 2018 at 10:05
  • $\begingroup$ In that case you can start here: en.wikipedia.org/wiki/Kramers%E2%80%93Kronig_relations For numerical calculations, fast-fourier-transforms are more suitable than than the integral form. $\endgroup$
    – user137289
    Aug 6, 2018 at 10:56
  • $\begingroup$ I see you have many answers on optical phenomenon in solids. The following has remained unanswered for quite a while, do you have any thoughts? Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue? $\endgroup$
    – uhoh
    Feb 8, 2019 at 13:57

Apparently $\rm Cu_2O$ has a band gap of about $2.1\ \rm eV$ (according to the linked wikipedia page). That means it'll absorb photons with a wavelength of less than $590.4\ \rm nm$ (just do the calculation with $E=hf$). For comparison, yellow light has a wavelength in the range $570-590\ \rm nm$. Hence we detect longer wavelengths and $\rm Cu_2O$ looks reddish.

Here's an easy way to think of it: the definition of band gaps tells us that it's the energy difference between possible bands, so if a photon of a lower energy (like the ones in red light) is incident, it can't be absorbed. But yellow wavelengths and higher energy photons (green, blue, and so on) will be absorbed.


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