[Ifeffit] Nearest neighbor statistics in a random alloy

[Felix E. Feiten]

Dear Qingying and Anatoly,

thank you very much for your quick help!

It is very much appreciated.

With kind regards

Felix


On 23/06/2017 22:57, Anatoly Frenkel wrote:
> ... and if the random alloy is a nanoparticle, the equations are 
> modified (they will be the same as in Qingying's email in the bulk 
> alloy limit and if the NPs are sufficiently large).
> See here, Eq. (11), for more details:
>
> Chem. Soc. Reviews *41*, 8163-8178 (2012) 
> <https://you.stonybrook.edu/frenkel/files/2016/10/Bimetallics-CSR-19p1ya2.pdf>
>
> Anatoly
> -------------------------------------------------------------------
>
> On Fri, Jun 23, 2017 at 9:28 AM, Qingying Jia <qjia at hawk.iit.edu 
> <mailto:qjia at hawk.iit.edu>> wrote:
>
>     Hi Felix,
>
>
>     Your assumptions are not correct because the model you use is not
>     a representative unit cluster model. Basically the following three
>     equations are always correct for foil:
>
>      NPt-Pt+NPt-Co=12
>
>     NCo-Co+NCo-Pt=12
>
>     NCo-Pt/NPt-Co=9
>
>     And the specific values for the coordination number can be
>     determined if you know the orderness of the structure. As for the
>     random distribution of Pt and Co in foil,
>
>     NPt-Pt/NPt-Co=9
>
>     or equivalently,
>
>     NCo-Pt/NCo-Co=9,
>
>     Then you can get
>
>     NPt-Pt=10.8, and NPt-Co=1.2
>
>     The results will be different if the structure is ordered.
>
>
>       Regards,
>
>
>     On Fri, Jun 23, 2017 at 9:02 AM, Felix E. Feiten
>     <feiten at cat.hokudai.ac.jp <mailto:feiten at cat.hokudai.ac.jp>> wrote:
>
>         Dear all,
>
>         I am fitting a Pt9Co1 foil sample.
>         I assume the sample to be basically fcc platinum with every
>         10th atom randomly replaced by cobalt.
>
>         Here are the assumptions I make:
>
>         1. every atom has 12 nearest neighbors
>
>         2. every Pt atom has (on average) 10.7 Pt nearest neighbors
>         and 1.3 Co nearest neighbors
>
>         3. every Co atom has (on average) 11.7 Pt nearest neighbors
>         and 0.3 Co nearest neighbors
>
>         I have been told that assumptions 2 and 3 are not precise.
>         Here is what I'm confused about:
>
>         In order to determine the number of Pt or Co nearest neighbors
>         I looked at an ensemble of 13 atoms:
>         1 absorber and 12 scatterers/nearest neighbors.
>         I assume that the Pt/Co ratio in this ensemble is 9/1 and
>         therefor, statistically, 1.3 out of those 13 atoms should be
>         Co. That means that if Co is the central atom only 0.3 Co
>         atoms are remaining for the 12 nearest neighbors. On the other
>         hand, if Pt is the central atom, (all) 1.3 Co atoms are
>         remaining for the 12 nearest neighbors.
>
>         I have been told that the ratio of Pt/Co should be 9/1 in the
>         12 nearest neighbors, not in the 12 atoms including the
>         central atom (quote: "because we have a bulk sample").
>
>         But I do not understand why.
>
>         Can someone please explain the correct answer to me?
>
>         With kind regards,
>
>         Felix
>
>
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