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Re: [sdpd] size broadening and diffracting volume
At 10:29 10/01/03 +0100, you wrote:
>
>>Being not an expert on size broadening effects I would like to know why an
>>'increasing diffracting volume leads to an increased symmetric broadening'.
>><http://www.cristal.org/opinions/>.
>
> Explanation without any formula :
> If you consider the Bragg Brentano reflexion geometry, the X-ray
> penetrating depth is of the order of microns for highly absorbing
> compounds and can easily attain one millimeter for very low-absorbing
> compounds. This means that you can really put highly absorbing
> compounds in the diffracting plane. But for very-low absorbing compounds
> a large sample volume is quite under of the diffracting plane. Now, just
> think to what happens if your sample is placed under or above the
> diffracting plane : your peaks are displaced either at lower or larger
>"""" in this geometry in case
> of low-absorbing coumpounds due to the sample thickness. That
> broadening reflects the distribution of sample outside of the diffracting
> plane.
>
> This is why Al2O3, in such a reflexion geometry, will never show peaks
> is
> certainly very well crystallized and the more compact as possible).
>
Not yet clear. What you mean by 'diffracting plane' is not the plane that
contains the
incoming wave vector k_in and the reflected wave vector k_out (as is very
often stated), but rather the plane that is tangentiel
to the focussing circle and perpendicular to the momentum transfer vector
k_out-k_in. Right?
If this is so, where do you define the physical surface of the sampleholder
to be, right
within the plane that is exactly tangentiel to the focussing circle? If
this is the case, then all the diffracting volume, except the surface
layer, is below the diffracting plane. For a highly absorbing
specimen there is no shift, since the effective diffracting volume is in
focussing condition.
For a transparant specimen the effective diffracting plane is below the
plane that is tangentiel to the focussing circle, and so there is a shift.
I agree that there are in fact many diffracting planes,
starting from the surface and then going inside the sample. So if you
define the 'diffracting plane'
as a hypothetical effective plane that is already displaced with respect to
the tangential plane, then
there is indeed a distribution of 'diffracting planes' around this 'effective
diffracting plane' that leads, apart from a shift, to a broadening of the
peak. It is however hard to see why this broadening is symmetric, since the
absorption itself is not linear. There is much less
diffraction from planes underneath the effective diffracting plane than
from the planes above the
effective diffracting plane, except, maybe, if your sample is infinitely
thick. So maybe the (a)symmetry of the broadening depends on the absorption
coefficient and at the same time on the thickness of the sample.
Is there anybody who has rigorously derived this, maybe Klug and Alexander
(I have not, unfortunately, access to this book)?
Having said this, I can add to what Lutterotti said: 'Should we have
different standards for different absorption?' the following: Could we
just, what I did yesterday, use a very low absorbing good crystalline
organic compound, spray a very thin layer on a non-absorbive Silicon sample
holder and use the resulting peak widths as 'the' instrumental widths. The
widths I got, FWHM(2theta)=0.04 at 5 degr. 2theta,
FWHM(2theta)=0.065 at 22 degr. 2theta were amazingly low for me in the
present set-up. Non-sintered NIST LaB6 gives FWHM(2theta)=0.010 for the
first peak. Is there much difference between the FWHM's of sintered and
non-sintered material?
regards, Arie
*******************************************
A. van der Lee
Institut Européen des Membranes (UMR 5635)
Université de Montpellier II - cc 047
Place E. Bataillon
34095 Montpellier Cedex 5
FRANCE
visiting address: 300 Av. Prof. E. Jeanbrau
tél.: 00-33-(0)4.67.14.91.35
FAX.: 00-33-(0)4.67.14.91.19
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