[3dem] RE: [Microscopy] electron diffraction limits and phase contrast

Philip Köck Philip.Koeck at ki.se
Tue Aug 27 00:10:35 PDT 2013


Hi Nestor,

I assume you are talking about the first zero of the CTF in (extended) Scherzer defocus.

In structural biology we always work at high underfocus, so we have to go far beyond the first
zero by some sort of CTF-correction and combining images with different defocus.

The resolution range I'm thinking about is between 10 Angstrom and about 2 Angstrom so I think
the size of the aperture should be important though maybe not in a completely straight forward
way as Marin pointed out.

I'm not sure if anybody has managed to get useful images of biological molecules without an aperture.
Might be worth trying.

Philip


-----Original Message-----
From: zaluzec at aaem.amc.anl.gov [mailto:zaluzec at aaem.amc.anl.gov] 
Sent: 26 August 2013 17:28
To: Philip Köck
Subject: [Microscopy] electron diffraction limits and phase contrast




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Philip

Since you have clarified your question further  I should also point out that the diffraction aperture is  not always the limiting factor in  phase contrast imaging particularly at the 1 Angstrom level that you seem to be  interested in.

When performing HR imaging experiments in this regime, one generally does not use any apertures.

In  HREM , you will need to  invoke and understand the limits created by the aberrations (Cs & Cc) of your Imaging Lenses.
Of these,  the Objective Lens is the most critical.

For high resolution work (nominally anything below about 0.3 nm) you will need to understand the Phase Contrast Transfer Function of your instrument.

I recommend that you should pickup a good  TEM textbook or review article which discusses High Resolution Phase Contrast Imaging.

The book by Carter and Williams will be a good staring point.

There are also several software programs available on the net which will calculate an approximate PCTF for you.

See for example

http://www.maxsidorov.com/ctfexplorer/webhelp/background.htm

Just to give you a ball park value, for a  200 kV FEI FEGTEM  (Cs ~ 2 mm) without aberration correctors, the spatial frequency cutoff is in the vicinity of 0.2-0.3 nm.

Cheers,

Nestor
Your Friendly Neighborhood SysOp.






On 8/26/13 7:28 AM, Philip.Koeck at ki.se wrote:
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> Thank you all very much for the answers you sent. I just picked one out to reply to.
> Thank you , Nestor, for the tables.
>
> The reason I asked is that I want to know what spatial frequencies are 
> removed by an objective aperture when recording high resolution (phase contrast) images.
> I wasn't sure whether the objective lens settings change when one switches between imaging and diffraction.
> Therefore the strange way of asking the question.
> Secondly I wasn't sure whether one can use the simple formulas given that the lens is not a thin lens.
> I don't know the position of the specimen with respect to the focal planes or the principal planes either.
>
> In any case everybody seems to point me to a simple calculation.
> For example for 200 kV and 1 Angstrom resolution cutoff I can take 
> twice the angle in Nestor's table, resulting in angle= 25 mrad.
> I can also get that from Lamda = d sin(angle), if I regard the specimen as a two dimensional grating.
>
> The next step is to use the focal length (not the distance between 
> specimen and focal plane) as camera length, giving a distance of the spot from the optical axis R = 72.5 micrometers for 2.9 mm focal length (f).
> The formula is f * tan(angle) = R.
> Most answers seem to agree about using f.
>
> Please tell me if I got anything wrong.
>
> Philip
>
>
> -----Original Message-----
> X-from: Wim Hagen [mailto:wim.hagen at embl.de]
> Sent: 24 August 2013 14:05
> To: Philip Köck
> Cc: microscopy at microscopy.com; 3dem at ncmir.ucsd.edu
> Subject: Re: [3dem] electron diffraction
>
> 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
>> _______________________________________________
>> 3dem mailing list
>> 3dem at ncmir.ucsd.edu
>> https://mail.ncmir.ucsd.edu/mailman/listinfo/3dem
>
> -------------------------------------------------------------------
>    Wim Hagen
>    Senior Engineer Electron Microscopy
>    European Molecular Biology Laboratory (EMBL)
>    Heidelberg, Germany
> --------------------------------------------------------------------
>
>
>
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