the properties of the most important semiconductors depend on the crystalline structure of the host I'm an undergraduate physics student currently studying for the Solid State Physics Exam.
This is the introductory paragraph of Chapter 1 from "Introduction to Solid State Physics" by Charles Kittel:
"The serious study of solid state physics began with the discovery of x-ray
diffraction by crystals and the publication of a series of simple calculations of
the properties of crystals and of electrons in crystals. Why crystalline solids
rather than nonclystalline solids? The important electronic properties of solids
are best expressed in crystals. Thus the properties of the most important semiconductors
depend on the crystalline structure of the host, essentially because
electrons have short wavelength components that respond dramatically to the
regular periodic atomic order of the specimen. Noncrystalline materials, notably
glasses, are important for optical propagation because light waves have a
longer wavelength than electrons and see an average over the order, and not
the less regular local order itself."
Well, I have problems understanding two sentences:
1)"because electrons have short wavelength components that respond dramatically to the
regular periodic atomic order of the specimen." What does it mean "that respond dramatically to the regular periodic atomic order of the specimen"
2)What does it mean "light waves have a longer wavelength than electrons and see an average over the order, and not the less regular local order itself."
 A: Your two questions are basically asking the same thing. In both cases the answer boils down to the scattering of waves.
Think back to the Young's slits measurement that we all did in school:

The experiment requires that the spacing between the slits be of around the same size as the wavelength of the light. The reason for this is that there must be a significant change in the phase of the light moving from one slit to the other otherwise the light from the two slits won't interfere and we won't see any scattering.
To make this quantitative, the phase shift along a distance $d$ is given by:
$$ \Delta \phi = 2\pi\frac{d}{\lambda} $$
and to see light scattering and an intereference pattern we require that $\Delta\phi \gg 0$.
Now consider your second question, why light isn't scattered by the lattice in a semiconductor. The wavelength of visible light is in the range $\lambda = 400 - 700$nm, and the lattice spacing in a semiconductor is in the range $d = 0.1 - 1$nm, so for visible light the ratio of $d/\lambda$ is around $10^{-3}$ and the phase change of the light over such a small distance is negligable. It's like trying to do a Young's slits experiment with the slit spacing only $0.001$ of the light wavelength. You won't see an interference pattern bacause the phase change between the slits is too small.
To get the light to scatter you have to reduce the wavelength until it's comparable to $d$. If you do this you get X-rays, and X-rays do indeed scatter from crystal lattices. This is the basis of using X-ray diffraction to study crystal structure.
Now back to your first question: if you treat the electrons in the bands of a semiconductor as plane waves then they will have a wavelength given by the de Broglie equation. This wavelength is comparable to the lattice spacing (for some of the electrons - not all), and therefore the electrons are strongly scattered by the semiconductor lattice.
So the difference between the ebehaviour of electrons and light is just down to the ratio of their wavelength to the semiconductor lattice spacing.
