Diffraction of electron through graphite film "Electrons of wavelength 434 nm  are directed into a thin film of graphite. Will they diffract?"
Ignoring relativistic effects, I used de Brogile's relation, $p = \frac{h}{\lambda}$, to determine that the velocity of the electrons is around $1680 \,\mathrm{m\,s^{-1}}$, which I believe is rather small.
I'm aware that electrons do diffract when projected through a thin metal foil (George Thomson's experiment) and through crystal (Davisson and Germer's experiment).
I would like to know if the velocity of the electron affects the diffraction of the electrons (i.e. is there a "threshold" velocity?) and if changing the metal foil from George Thomson's experiment into a graphite film would affect anything. Thanks in advance!
 A: As far as I am aware speed has no effect on diffraction only aperture size and wavelength.
I think you want to calculate the inter-atomic spacing to work out if they diffract.
EDIT: Also these electrons are very low energy I calculate <3eV. I'd be well impressed if anyone could actually make such a beam. Given this I expect their speed (lack of energy) may be an issue as iterations with the graphite atoms will be significant. Also if you have a sensible energy ignoring relativistic effects when calculating speed is a bad idea.
A: Every wave undergoes diffraction in every crystal. The question is whether the beam is monochromatic enough that the diffraction doesn't wash out, and whether the diffraction angles are large enough to be resolved by your detector.
Your electron wavelength is roughly a thousand times longer than the interatomic spacings in graphite (0.142 nm and 0.343 nm). If the 434 nm electron beam contains substantial contributions from 433.5 nm electrons and 434.5 nm electrons, it's too broad, and the diffraction peaks will destructively interfere and wash away. I'd expect that, to see diffraction in a reasonable amount of observing time, you'd need a line width somewhere around 10% of the lattice spacing you're interested in probing, in which case you'd specify $\lambda = 434.xy \pm 0.03$ nm (or better) for some final two digits $xy$.
However, this electron wavelength is ridiculous: it corresponds to 
$ E = \frac{(h/\lambda)^2}{2m} = 8\,\mathrm{\mu eV}$! The electrons in the graphite foil, with room-temperature energy 25 000 μeV, would heat up the beam, and you'd lose all of your coherence. Even if any of your "beam" escaped to the other side of the foil, you'd never see any diffraction from it.
