Two phase structural refinement using FullProf

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correct phase fraction. Normalized Occ:- Occupation number. Occupation ... 1*16/192= 0.083333. O. 100% =1. 32e (1/2,0,z 32. 192. 1*32/192= 0.166667.
Two phase structural refinement using FullProf Here I will explain how to perform two phase structural refinement using FullProf

Aga Shahee [email protected]

What we need for Rietveld Refinement? 1) We need (X-ray or Neutron) Powder diffraction data of Sample For that we have to do Powder diffraction experiment of sample 2) We need structural Information in order to develop PCR file For that we will look for Crystallographic Information File (CIF): CIF is a standard text file format for representing crystallographic information. One can find many sources for CIF, Here I will use freely available Crystallographic data base sources Namely Crystallography Open Database http://www.crystallography.net/cod/ 3) We need a Rietveld refinement softwear to perform structural refinement Here I will use freely available FullProf Software https://www.ill.eu/sites/fullprof/

Crystallography Open Database site

FullProf Software site

For Two Phase Rietveld Refinement For Rietveld using Fullprof,  The data quality should be good i.e. signal to noise ratio high enough  A good number of distinct peaks for each phase  A step by step refinement is needed, this means either refine the data with each phase separately or refine the data with one phase and once the first phase refinement is done add the second phase  Also keep in mind some parameters are much more important than the others.

Important point to be noted The occupation of atomic site should be normalized. It is very essential to have normalized site occupancy in order to get the correct phase fraction.

Normalized Occ:- Occupation number Occupation number = chemical occupancy × site multiplicity site multiplicity = multiplicity of special position/general multiplicity

Here I will perform structural refinement of Two Phase Pattern  We have room temperature XRD data BaTiO3 and NiFe2O4 composite multiferroic material  The data contains tetragonal BaTiO3 with space group P4mm and cubic NiFe2O4 with space group Fd-3m Data taken using Lab Source Copper k-Alpha

How to Normalized Occupancy (Occupation number) BaTiO3 contains one Ba atom, one Ti atom and 3 Oxygen, In Tetragonal P4mm lattice, Ba fully occupies single 1a site, Ti occupies fully single 1b site and O occupies two sites (1b and 2c) single 1a site. The 1b site of Ti and Oxygen is different.

Normalized Occ:- Occupation number Occupation number =     

chemical occupancy × multiplicity of special position /general multiplicity

General multiplicity of space group P4mm is 8 Multiplicity of special position for Ba at 1a site is 1 Multiplicity of special position for Ti at 1b site is 1 Multiplicity of special position for O at 2c site is 2 Multiplicity of special position for O at 1b site is 1

BaTiO3: Tetragonal Lattice, Space group: P4mm and General multiplicity: 8 Atom

Chemical Occ.

Site

Site Multi. General Multi.

Occupancy

Ba

100% =1

1a(0,0,z)

1

8

1*1/8= 0.125

Ti

100% =1

1b(0.5,0.5,Z)

1

8

1*1/8= 0.125

O1

100% =1

2c (1/2,0,z

2

8

1*2/8= 0.250

O2

100% =1

1b(0.5,0.5,Z)

1

8

1*1/8= 0.125

How to Normalized Occupancy (Occupation number) NiFe2O4 contains one Ba atom, one Ti atom and 3 Oxygen, In Tetragonal P4mm lattice, Ba fully occupies single 1a site, Ti occupies fully single 1b site and O occupies two sites (1b and 2c) single 1a site. The 1b site of Ti and Oxygen is different.

Normalized Occ:- Occupation number Occupation number =    

chemical occupancy × multiplicity of special position /general multiplicity

General multiplicity of space group Fd-3m is 192 Multiplicity of special position for Ni at 8a site is 8 Multiplicity of special position for Fe at 16d site is 16 Multiplicity of special position for O at 32e site is 32

BaTiO3: Tetragonal Lattice, Space group: P4mm and General multiplicity: 8 Atom Chemical Occ.

Site

Site Multi. General Multi.

Occupancy

Ni

100% =1

8a (0,0,0)

8

192

1*8/192= 0.041667

Fe

100% =1

16d (5/8,5/8,5/8)

16

192

1*16/192= 0.083333

O

100% =1

32e (1/2,0,z

32

192

1*32/192= 0.166667

Refinement Strategy: Two Phase Rietveld Refinement Using FullProf through 5-steps  1st step of data format

 2nd step of background selection  3rd step of PCR develop and for that we will download a CIF file  4th step One Phase Structure refinement  Finally 5th step Two Phase Structure refinement