I wonder the reason for TiO2 thin films to be transparent for visible light but not for UV. I made a quick search and I found that it is due to the band gap of TiO2. It absorbs UV light but not visible light. I imagine this occurs because of the different wavelengths of these two types of radiation. But what is the relation between the wavelength of a certain type of radiation and the width of the band gap a semiconducting material? And how does this effect its optical properties?

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    $\begingroup$ I once visited a lab where they were showing a little experiment. They gave us a silicon wafer (monocrystalline) and a piece of glass. They had set a UV camera connected to a screen. When I put the glass between me and the camera my face didn't appear on the screen (even though I was able to see it with my eyes). When I tried the same with the silicon wafer I didn't see with my eyes but my face appeared on the screen. Silicon is also transparent for UVs and not for visible light. $\endgroup$
    – ChocoPouce
    Mar 27, 2014 at 10:03

1 Answer 1


The energy per photon of light with wavelegth $\lambda$ is given by:

$$ E = \frac{hc}{\lambda} $$

If the energy per photon is smaller than the band gap the light cannot excite electrons from the valence to conduction band so it will pass through the material without being absorbed. If the energy is larger than the band gap the light will excite electrons and will be (partially) absorbed. The cutoff wavelength is given by simply rearranging the above formula to get:

$$ \lambda \approx \frac{hc}{\Delta E} $$

where $\Delta E$ is the band gap. I've used the approximately equal sign because band gaps are rarely sharp and the light absorbtion will increase over a wavelength range of around the cutoff wavelength. If you want to establish the band gap accurately you'd use a Tauc plot.


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