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Daniel M. T�bbens Tel.: +49 30 8062-2793
Hahn-Meitner-Institut Fax : +49 30 8062-2999
Glienicker Stra�e 100
D-14109 Berlin, Deutschland http://www.hmi.de/people/toebbens/
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On Fri, 12 Feb 1999, Dr. Jaap Vente wrote:
> On Fri, 12 Feb 1999, Armel Le Bail wrote:
>
> > >Sure! And you can allways fit a peak with consists out of two components,
> > >but at times you can be really struggeling to find out whether you see
> > >peak splitting due to alp1/alp2 or due to a slight reduction in symmetry
> > >or stcking faults. If you can rule out the machine straight away, than
> > >life becomes a bit easier.
> >
> > You cannot always fit any peak which consists out of two
> > components when the position, shape and intensities of both
> > components are tied. You can fit if and only if your model is
> > consistent with those constrained data.
>
> I am NOT talking about a model at all, this is not Rietveld! The only
> assumption I make is that alp1 and alp2 behave in the same way, and thus
> have the the FWHM, a given intensity ratio, and a related position. If you
> are looking at one reflection you can easily fit that to the proper peak
> shape. If you have two reflections than it is more complex, and that was
> exactly what I was trying to say above.
>
But do you even know that the FWHM behave the same? I remember a lecture
at a DGK meeting a while ago where an examination of various diffraction
patterns showed that they don't. Therefore one should use different
relation functions for alp1 and alp2 (I didn't do this up to now, since my
laboratory data were not THAT good).
Daniel Toebbens
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Daniel M. T�bbens Tel.: +49 30 8062-2793
Hahn-Meitner-Institut Fax : +49 30 8062-2999
Glienicker Stra�e 100
D-14109 Berlin, Deutschland http://www.hmi.de/people/toebbens/
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