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B Electron Diffraction and the DeBroglie Wavelength

  1. Oct 24, 2016 #1
    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:

    upload_2016-10-25_12-22-31.png

    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. jcsd
  3. Oct 25, 2016 #2

    vanhees71

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    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}##.
     
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