# B Electron Diffraction and the DeBroglie Wavelength

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1. Oct 24, 2016

### Guest432

I've been reading up on electron diffraction for electron microscopy, and I have been trying to understand the proof for the wavelength of an electron in a tunneling electron microscope. The proof I have been trying to emulate begins as follows:

It then says that I must account for relativistic effects so, $E^2=p^2c^2+m^2c^4$ and manages to yield this term for momentum

How did it jump from $p=\sqrt {2meV}$ to that? (Note delta $\Delta E = eV$)

Last edited: Oct 24, 2016
2. Oct 25, 2016

### vanhees71

The energy is
$$E=\sqrt{p^2 c^2+m^2 c^4}=m c^2+e V.$$
The reason is that in relativity you always include the rest energy, $mc^2$ in the kinetic energy, because then $(p^{\mu})=(E/c,\vec{p})$ is a Minkowski four-vector. Note that here (and in contemporary physics always!) $m$ is the invariant mass of the particle, i.e., in this case for an electron $mc^2 \simeq 0.511 \; \mathrm{MeV}$.