[3dem] electron diffraction

[Wim Hagen]

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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