I suppose that "2R" means the diameter of a powder ring, and you want to find the d-value from 2R.

If the scale of your SAED pattern is not changed by image processing,i.e. if it reflects the true camera length (CL) in mm then simply multiply CL with the wave length of the electrons.

Lambda(Angstroem) = 12.2639/sqrt(hv*(1.+.97845e-06*hv))                                              hv = high voltage[V, not kV!]

100kV -> 0.0371330 Angstrom

  200  -> 0.0250787

  300  -> 0.0196870

CL*lambda is the camera constant [CC] in Angstrom*mm, and the scale bar for 1Angstroem**(-1) has the length of CC[mm]

d(Angstrom) = CC/R

I hope my guess concerning the meaning of "2R" was correct.

With best wishes

Gerhard Miehe

I suppose that "2R" means the diameter of a powder ring, and you want to find the d-value from 2R.

If the scale of your SAED pattern is not changed by image processing,i.e. if it reflects the true camera length (CL) in mm then simply multiply CL with the wave length of the electrons.

Lambda(Angstroem) = 12.2639/sqrt(hv*(1.+.97845e-06*hv))                                              hv = high voltage[V, not kV!]

100kV -> 0.0371330 Angstrom

  200  -> 0.0250787

  300  -> 0.0196870

CL*lambda is the camera constant [CC] in Angstrom*mm, and the scale bar for 1Angstroem**(-1) has the length of CC[mm]

d(Angstrom) = CC/R

I hope my guess concerning the meaning of "2R" was correct.

With best wishes

Gerhard Miehe

If you already know what your sample is, you can try free software in link below (diff ring profiler). It uses bitmap images ans you manually calibrate the known rings... otherwise Gerhard said all the important things :D

https://code.google.com/p/diffraction-ring-profiler/

Thank you all for your answers.

Thank you Gerhard, for your detailed information.