Re: [sdpd] A QUESTION

[Jonathan WRIGHT <wright...@esrf.fr>]

At 11:42 18/10/2002 +0800, you wrote:
>DEAR ALL,
>        There is a questions on GSAS, when doing the exercise of ZrO2 in 
> the zip file attached in GSAS,first I extracted Fos with one H atom and 
> get satisfied refinement Rp=13,and then by the Fourier option I 
> calculated the PTSN map and got the result by one atom to be the 0,0,0 
> coordinate. but in the next step

You calculated a Patterson map and found a large peak at 0,0,0 and then 
assumed this was an atom? Patterson maps always have big peaks at the 
origin, so you probably want to find out how to assign peaks in a Patterson 
map and then use them to determine co-ordinates. I think there's a chapter 
in most crystallography textbooks explaining what a Patterson map is and 
how to solve a structure from it. The key thing to understanding is print 
the map out onto pieces of paper and use a pencil to write down the 
co-ordinates, before trying to get the computer to do any more for you.

>of refinement i found that the result i got was not agree with the 
>coordinats offered by the class.zip file, in which no atom was on the 
>0,0,0 point.
>my question is if this method is reasonable in solving the ZrO2 structure 
>by powder data, and if it is resonable then how to determine the correct 
>atom coordinates from the patterson result in this example?    Thank you 
>very much.

Since Zr is a heavy atom then Patterson methods would be a very sensible 
way to start, provided there aren't too many atoms to find. The nice thing 
about a Patterson map is that each point depends on *all* of the extracted 
structure factors - so if you get a few wrong due to peak overlaps they 
don't destroy the overall picture.

The non origin peaks in the map correspond to interatomic vectors in the 
structure. (ie: a peak at 0.1,0.2,0.4 corresponds to a pair of atoms with 
that offset between them, the pair could still be anywhere in the cell). 
The heights are proportional to the scattering powers of the atoms, so you 
will ideally see peaks of height 40*40=1600 (Zr-Zr peaks), 40*8=320 (Zr-O) 
and 8*8=64 (O-O). At the origin you get the self vectors, so the height 
would be 40*40+2*8*8=1728 for ZrO2 in the asymmetric unit. In practice the 
O-O vectors will almost certainly be obscured, and it might be hard to find 
the Zr-O peaks. Once you find the Zr atoms then the O's will probably stick 
up in a difference fourier.

If you have access to the physical review online archive then you can read 
all about it: "A Fourier Series Method for the Determination of the 
Components of Interatomic Distances in Crystals", A. L. Patterson, Phys. 
Rev. 46, 372-376 (1934)

The difficulty with Patterson maps is that if you have n atoms you get n*n 
peaks. In principle the mess can be sorted out, and sometimes in practice 
too. Try "A new approach to crystal structure analysis", M. J. Buerger, 
Acta Cryst 4, 531, or better (if you can get hold of it) the book by the 
same author "Vector space and its applications in crystal structure 
investigations" (1959).


Cheers,

Jon




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