Determination of Crystal Structure (From Chapter 10 of Textbook 2) Unit cell  line positions Atom position  line intensity (known chemistry) Three steps to determine an unknown structure: (1) Angular position of diffracted lines  shape and size of the unit cell. (2) sizes of unit cell, chemical composition, density  # atoms/unit cell (3) Relative intensities of the peaks  positions of the atoms within the unit cell

Preliminary treatment of data:  Ensure true random orientation of the particles of the sample  Remove extraneous lines from (1) K  or other wavelength: In the analyzing step, the sin 2  is used  For most radiations  filament contamination  W K L radiation  Diffraction by other substance  Calibration curve using known crystal

Effect of sample height displacement s: sample height displacement R: diffractometer radius   +  s  Length of a larger error for low angle peaks  for the most accurate unit cell parameters it is generally better to use the high angle peaks for this calculation. R

Example for sample height displacement Assume a crystal; cubic structure; a = 0.6 nm. Consider the error that can be introduced if the sample was displaced by 100 microns (0.1 mm) for (100) and (400) diffraction peaks? Assume λ = nm and R = 225 mm. The displacement cause these peaks to shift by (100):  = o (400):  = o

Pattern Indexing →assign hkl values to each peak Simplest example: indexing cubic pattern Constant for a given crystal Define

Values for h 2 + k 2 + l 2 for cubic system SC: 1, 2, 3, 4, 5, 6, 8, … BCC: 2, 4, 6, 8, 10, 12, 14,.. FCC: 3, 4, 8, 11, 12, 16, 19, 20, … Diamond: 3, 8, 11, 16, 19, … SCBCCFCCDiamondS 100  XXX1 110  XX2 111  X   X4 210  XXX5 211  XX6 220   XXX9 300  XXX9 310  XX  X   X  XXX13

SC: 1, 2, 3, 4, 5, 6, 8, … BCC: 2, 4, 6, 8, 10, 12, 14,.. s is doubled in BCC, No s = 7 in SC

Indexing Tetragonal system When l = 0 (hk0 lines), Possible values for l 2 are 1, 4, 9, 16, … Possible values for h 2 + k 2 : 1, 2, 4, 5, 8, 9, 10, …

Possible values for h 2 + hk + k 2 are 1, 3, 4, 7, 9, 12, … The rest of lines are Possible values for l 2 are 1, 4, 9, 16, … Indexing Hexagonal system

Example: sin 2  sin 2  /3 sin 2  /4 sin 2  / Let’s say A =

sin 2  sin 2  -A sin 2  -3A belongs to Cl 2. What is the l? There are two lines between 100 and 110. Probably, 10l 1 and 10l 2  and are different ls /0.024 ~ /0.024 ~ /0.024 ~ 16  C = ,

Indexing orthorhombic system: More difficult! Consider any two lines having indices hk0 and hkl  Cl 2  put it back  get A and B.  guess right (consistent)  not right, try another guesses C Indexing Monoclinic and Triclinic system Even more complex, 6 variables must have enough diffraction lines for the computer to indexing.

Effect of Cell distortion on the powder Pattern

Autoindexing

Determination of the number of atoms in a unit cell V C : unit cell volume;  : density N 0 : Avogodro’s number; M: molecular weight; n: number of molecules in a unit cell in Å 3 in g/cm 3 Indexing the power pattern  shape and size of unit cell (volume)  number of atoms in that unit cell.

Determination of Atom positions Relative intensities determine atomic positions (a trial and error process) (u n v n w n ): position of nth atom in a unit cell. Trial and error: known composition, known number of molecules, known structure  eliminate some trial structure. Space groups and Patterson Function (selection of trial structures)

Example: CdTe Chemical analysis which revealed: 49.8 atomic percent as Cd (46.6 weight percent) 50.2 atomic percent as Te (53.4 weight percent) * Make powder diffraction and list sin2  : index the pattern! Assume it is cubic

sin 2  SC s BCC s FCC s Diamond s close Very weak

very close Diamond structure and Zinc blend structure: forbidden peaks! Which one has more peaks?  removing line 4 first

Alternative: Assume the first line is from s = 1, s = 2 and s = 3, …

Larger error smaller , use the first three lines to fit a more correct A.  , Use this number to divide sin 2  ! Larger Deviation New A is

0.1790/ = 12.66; 222  s = 12 (forbidden diffraction lines for diamond structure, OK for zinc-blend structure!) *  = 5.82 g/cm 3 then M for CdTe is = 240  n = 3.97 ~ 4 There are 4 Cd and 4 Te atoms in a unit cell. * FCC based structure with 4 molecules in a unit cell – NaCl and ZnS; CdTe: NaCl or ZnS

NaCl –Cd at 000 & Te at ½½½ + fcc translations FCC when h, k, l all even when h, k, l all odd ZnS –Cd at 000 & Te at ¼ ¼ ¼ + fcc translations when h + k + l is odd when h + k + l = odd multiple of 2 = even multiple of 2 Unmixed hkl only

f Cd + f Te = 100 and |f Cd  f Te | = 4 at sin  / = 0 to f Cd + f Te = 30.3 and |f Cd  f Te | = 1.7 at sin  / ~ Cd Te Several hundreds NaCl –Cd at 000 & Te at ½½½ + fcc translations h, k, l all even: strong diffracted lines h, k, l all odd: week diffracted lines

ZnS fit better!

Order-Disorder Determination temperature highlow TCTC disorderorder Example – Cu-Au system (AuCu 3 ), T C = 390 o C disordered ordered Au Cu Cu-Au average Substitutional solid solution  A, B elements  AB atoms’ arrangement

Complete Disordered structure: the probability of each site being occupied: ¼ Au, ¾ Cu  simple FCC with f av For mixed h, k, l  F hkl = 0 For unmixed h, k, l  F hkl = (f Au + 3f Cu ) disordered

For mixed h, k, l  F hkl = (f Au  f Cu ) For unmixed h, k, l  F hkl = (f Au + 3f Cu ) Peaks show up Complete Ordered structure: 1 Au atom, at 000, three Cu atoms at ½ ½ 0, ½ 0 ½, 0 ½ ½. ordered

Define a long range order parameter S: r A : fraction of A sites occupied by the right atoms; F A : fraction of A atoms in the alloy complete order: r A = 1  S = 1; complete disorder: r A = F A  S = 0 0  S  1

AuCu 3 : order parameter S A-site is the 000 equipoint r Au :fraction of Au atoms in 000 site  the average atomic form factor Average atomic factor for A-site = f av

(1  r A ): is the fraction of Au occupying the B site  in B site (1  r A )/3 Au and 1  (1  r A )/3 Cu = (2 + r A )/3 Cu  Average atomic factor for B-site

The structure factor is For mixed h, k, l  For unmixed h, k, l  Intensity  |F| 2  superlattice lines  S 2

Using different S definition What would you get? Homework!

Intensity  weak diffuse background  If atoms A and B completely random in a solid solution  diffuse scattering k: a constant for any one composition f decreases as sin  / increases  I D  as sin  /  Weak signal, very difficult to measure

f av f Zn f Cu Example – Cu-Zn system (CuZn), T C = 460 o C disorderedordered

* Completely random: a BCC structure * Completely order: a CsCl For h + k + l even  F hkl = f Cu + f Zn For h + k + l odd  F hkl = 0 For h + k + l even  F hkl = f Cu + f Zn For h + k + l odd  F hkl = f Cu  f Zn

Define a long range order parameter S: For h + k + l even  F hkl = f Cu + f Zn For h + k + l odd  F hkl = S(f Cu  f Zn ) (practice yourself)

1 0 TCTC T T  Different system

Relative intensity from the superlattice line and the fundamental line: * Case AuCu3: ignoring the multiplication factor and Lorentz-polarization factor, just look at the |F| 2. Assume sin  / = 0  f = z About 10%, can be measured without difficulty. Assume sin  / = 0.2

* Case CuZn: the atomic number Z of Cu and Zn is close  atomic form factor is close!! Assume sin  / = 0  f = z Assume sin  / = 0.2 About 0.03%, very difficult to measure choose a proper wavelength to resolve the case!

Resonance between the radiation and the K shell electrons  larger absorption   f Produce extra difference in f Cu  f Zn Using Zn K  radiation.  f for Cu is -3.6 and for Zn is -2.7  About 0.13%, possible to be detected.