Dear Łukasz,
There must be something wrong with your calculation. A quick examination
shows that S must have a higher displacement parameter than Al and that
Ueq must be approximately 0.016.
I hope this helps.
Bob Gould
On 10/02/2014 21:50, "Łukasz Kruszewski" wrote:
Dear Rietveld friends,
I'm having some problems with calculating Uiso from the anisotropic
parameters. Here is an example - I copy first two sites of aluminite
(monoclinic):
_atom_site_aniso_label
_atom_site_aniso_U_11
_atom_site_aniso_U_22
_atom_site_aniso_U_33
_atom_site_aniso_U_12
_atom_site_aniso_U_13
_atom_site_aniso_U_23
S 0.01937 0.01562 0.01277 -0.00248 -0.00027 0.00954
Al1 0.01117 0.01169 0.01295 0.00006 -0.00012 0.00789
Using the formula for Uiso for the monoclinic system, which is:
1/3 * [U22 + 1/sin^2(beta)*(U11 + U33 + 2U13cos(beta)]
for the "S" site we obtain the value of 0,0054
for "Al": 0.01225
Meanwhile, the listed Uiso are:
_atom_site_label
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
_atom_site_U_iso_or_equiv
S 0.70076 0.37376 0.93018 0.01621
Al1 0.65581 0.45332 0.47633 0.01229
While the author-given value for Al is close to the one calculated myself,
the one for S is much different. By the way, 0.01621 * 1/3 = 0.0054.
I could find it a simple mistake, I've found such a problem in case of
many other structures. And here's my kind question: am I missing
something?
Best regards!
--
Robert Gould
Tel.UK: +44 (0)131 667 7230 or +44 (0)796 040 3872
Canada: +1 519 387 8223
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