Making a pellet with correct density and thickness: As we know the mean free path of a pellet is: lambda=4*R/3/rou, where R is the particle radius, rou is volume density. We want to make a pellet because our sing crystal sample is too thick to transmit light. Then we want to grind sample to small particles that can transmit light and put them into pellets. The keys are: 1) pellet thickness d should be slightly larger than the mean free path: d > lambda =4*R/3/rou 2) effective thickness (deff) should be small enough so that light can penetrate: deff =d*rou < 1/alpha, where alpha is the absorption coefficient. Thus the correct thickness is: 4R/3/rou < d < 1/alpha/rou is the real condition that we want to meet. There is a implicit condition here: R < 3/alpha/4, which is very important. In conclusion, to make a good pellet: 1) particle size has to be R < 3/alpha/4 2) sample thickness 4R/3/rou < d < 1/alpha/rou, which can be tuned by the volume concentration rou. Example: If we have alpha=1e7 m^{-1}, then the condition of a good pellet is : 1) R < 75 nm. 2) d < 1e7/rou This condition can not be realized because the particle size needs to be very small.

where is the most color of the world coming from?

All the colors of insulators are decided by transmittance instead of reflectance. This has been a huge confusion that I just make it clear myself not so long ago. In most insulators, the refractive index n is between 1 and 3, therefore the reflectance R=(1-n)²/(1+n)² is between 10% and 20%. This is not a small amount of light, however, this part of light is normally not so colorful either. That's because n is not rapidly frequency dependent. We can see this from our everyday experience. Any flat surface will reflect white light, no matter what color the surface is.

If everything reflects white light, how come we see a colorful world? The reason is that surface roughness disperses reflection, making this part of light only important when you have a flat surface and you are looking at a appropriate angle. So, the colorful light we see on normal stuff is actually from double transmittance, because it does not care about the surface roughness. Remember, this part of light can be as much as 80-90% of the total light illuminated on the material. It acts like reflectance because it is transmitted twice. It is colorful because the color is filtered by the media. (As we know, absorption is truly a strong frequency dependent quantity.)

The question immediately follows is that if what we see is transmittance, why most of the stuff is not transparent at all. This is another confusing phenomenon that took me a long time to understand. Actually this has to do with scattering again. If you have a big piece of NaCl single crystal, it should be transparent. But if you grind it into powder, it will keep the color, but become not transparent due to too much scattering at the powder surface. In everyday life, the stuff we see all contain so called pigments, which are colorful transparent small particles. They act like powders, which keeps the color, but not the transparency. If you look everything under optical microscopy, you can see that. Another easy way to verify this is that there is no opaque and colorful (not black) single crystal.

Back to the transition metal oxides. Their band gap is often large enough to leave all the visible range in the gap. In that case, in principle, all the visible lights should go through, making these oxides white or colorless. However this is only true for Sc₂O₃ and TiO₂ when there is not 3d electrons left on the metal ion. Other oxides are normally black and their powder shows colors. The reason is there is the d-d transition. For example, NiO, the electronic configuration is 3d_⁸, (5t_2g,3e_g), which can be excited to 3d⁸* (4t_2g, 3e_g). The energy is in the visible range. The oscillator strength (the absolute value of the absorption) is however much smaller (normally 2 to 3 order of magnitude) than the O-p to Ni-d charge transfer excitation, simply because these kind of excitations are symmetry forbidden.

If it is so small, should it be so important? The answer is yes, because the absorption does not need to be large to make a 1 mm thick sample colorful or even black. We can easily estimate this: to make a 1mm thick sample black, the absorption coeffcient α> 1/1mm = 1000 1/m. (definition of alpha is that transmitted light intensity I=I₀*exp(-α*d), where d is the thickness). If we take one step further, we can estimate extinction coefficient: k=α*c/(ω*2), where c is the speed of light, ω is the angular frequency. For a light with wave length 500 nm, α = 100 /m corresponds to k= 4e-3. This is so small that you can not it in a figure that is scaled to a charge transfer excitations. In fact, you have to use logarithmic scale to bring this detail up. This also explains why the powder is colorful. Because the transmittance depends on the absorption coefficient alpha and sample thickness. As long as the sample is small enough, it is not black anymore. Instead it will show the intrinsic color which corresponds to the small excitation in the visible range. After all, green color for NiO, red color for Fe₂O₃ (the rust) are their intrinsic properties.

bandgap_and _colors

Color and band gap of 3d transition metal oxides

mateials	Band gap (eV)	IBulk	Film (100 nm)	Powder (10 micron)
alpha Fe2O3 (Hametite)	2.34	black
Fe3O4	0.05	black
BiFeO3	2.7	black	yellow
NiO	3.4	black
TiO2	3.03	colorless