[3dem] electron diffraction
Wim Hagen
wim.hagen at embl.de
Sat Aug 24 05:05:15 PDT 2013
Hi Philip,
A simple estimation:
Angular distance in back focal plane [rad] = 2 * Bragg angle [rad] = Lambda (relativistic wavelength) [m] / d (sample spacing) [m] = R (distance) [m] / F (focal length) [m]
All CM/Tecnai Twin systems have a focal length of ~2.9 mm. For 200KV, a 0.1 nm spacing gives ~25 mrad so ~72.5 micrometers distance from optical axis in back focal plane.
Put in the radius of your objective apertures for R and you can calculate where your objective apertures cut off resolution for each KV.
Best,
Wim
On Aug 23, 2013, at 4:18 PM, Philip Köck wrote:
> Hi everybody,
>
> I'm looking for a simple estimation that I can't find in any textbook I've looked in, nor on the internet.
> I wonder if anybody can help me.
>
> If I put a diffraction grating with spacing 1 Angstrom into the specimen holder and record an image of it.
> At what distance from the optical axis will the corresponding diffraction spot (for 1 Angstrom spacing)
> appear in the back focal plane of the objective lens.
>
> Assume a TEM with 200 kV and a twin lens, one that's typically used for protein structure determination.
>
> Thanks,
>
> Philip
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Wim Hagen
Senior Engineer Electron Microscopy
European Molecular Biology Laboratory (EMBL)
Heidelberg, Germany
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