determination of crystal structure (from chapter 10 of textbook 2)

Determination of Crystal Structure(From Chapter 10 of Textbook 2)

Unit cell line positionsAtom 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:

sin2;sin2 hklhkl dd

sinsin

In the analyzing step, the sin2 is used

22

2

sinsin

2.1

2

For mostradiations

filament contamination W KL radiation Diffraction by other substance

Calibration curve using known crystal

Effect of sample height displacement

R

s cos2 radians) (in 2

s: sample height displacementR: diffractometer radius

+

s

Length of cossa 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 samplewas displaced by 100 microns (0.1 mm) for (100) and(400) diffraction peaks? Assume λ = 0.154 nm and R = 225 mm.

37.7154.0sin)6.0(2 100 d87.30154.0sin)15.0(2 400 d

The displacement cause these peaks to shift by 051.01081.8225/37.7cos1.022 4

044.01063.7225/87.30cos1.022 4

(100): = 7.395o

(400): = 30.892o

5982.0154.0395.7sin2 100 ad

5999.0154.0892.30sin2 400 ad

025.0022.0

Pattern Indexing

→assign hkl values to each peak

Simplest example: indexing cubic pattern

sin2 hkld222 lkh

adhkl

2

2

222

2

222 4

sinsin2

alkhlkh

a

Constant for agiven crystal

222 lkhs Define

Values for h2 + k2 + l2 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, …

SC BCC FCC Diamond S

100 X X X 1

110 X X 2

111 X 3

200 X 4

210 X X X 5

211 X X 6

220 8

221 X X X 9

300 X X X 9

310 X X 10

311 X 11

222 X 12

320 X X X 13

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

2

2

2

22

2

1

c

l

a

kh

dhkl

hkld2/sin

2222

2

2

2222

2 )(42

sin ClkhAc

l

a

kh

dhkl

2222 4/;4/ cCaA

When l = 0 (hk0 lines), )(sin 222 khA

Possible values for l2 are 1, 4, 9, 16, …

2222 )(sin ClkhA

Possible values for h2 + k2: 1, 2, 4, 5, 8, 9, 10, …

2

2

2

22

2 3

41

c

l

a

khkh

dhkl

2222

22

2

2222 )(

43sin ClkhkhA

c

l

a

khkh

2222 4/;3/ cCaA

Possible values for h2 + hk + k2 are 1, 3, 4, 7, 9, 12, … The rest of lines are Possible values for l2 are 1, 4, 9, 16, …

2222 )(sin ClkhkhA

Indexing Hexagonal system

Example: sin2 sin2/3 sin2/4 sin2/70.097 0.0320.112 0.037 0.136 0.0450.209 0.0700.332 0.1110.390 0.1300.434 0.1450.472 0.1570.547 0.1820.668 0.2230.722 0.241

0.0240.0280.0340.0520.0830.0980.1090.1180.1370.1670.180

0.0140.0160.0190.0300.0470.0560.0620.0670.0780.0950.103

100

110

Let’s say A = 0.112

123456789

1011

sin2 sin2-A sin2-3A0.0970.1120.1360.2090.3320.3900.4340.4720.5470.6680.7220.8060.879

00.0240.0970.2200.2780.3220.3600.4350.5560.6100.6940.767

0.0540.0980.1360.2110.3320.3860.4700.543

0.097 belongs to Cl2.What is the l?There are two linesbetween 100 and 110.Probably, 10l1 and10l2 0.024 and 0.097are different ls. 0.097/0.024 ~ 40.220/0.024 ~ 90.390/0.024 ~ 16 C = 0.02441

123456789

10111213

100

110

101102

002

112

,103004

Indexing orthorhombic system:

2222

2

2

2

2

222

4sin ClBkAh

c

l

b

k

a

h

More difficult! Consider any two lines having indices hk0 and hkl Cl2 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 enoughdiffraction lines for the computer to indexing.

Effect of Cell distortion on the powder Pattern

http://www.ccp14.ac.uk/solution/indexing/index.html

Autoindexing

Determination of the number of atoms in a unit cell

M

NVn C 0

VC: unit cell volume; : densityN0: Avogodro’s number;M: molecular weight;n: number of molecules in a unit cell

66054.1)10022.6()()10( 23

03338 C

C

VN

cm

gcmVnM

in Å3

in g/cm3

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)

N

nnnnn lwkvhuifFpFI

12

22

)(2exp ; cossin

2cos1

(un vn wn): 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)

2.50:8.490009.1:16.127

4.53:

4.112

6.46Te:Cd

* Make powder diffraction and list sin2: index the pattern! Assume it is cubic

0.04620.11940.16150.1790.2340.2750.3460.3910.4610.5040.5750.6160.6880.7290.799

1234568910111213141617

0.04620.05970.053830.044750.04680.045830.043250.043440.04610.045820.047920.047380.049140.045560.047

24681012141618202224263032

0.02310.029850.026920.022370.02340.022920.024710.024440.025610.02520.026140.025670.026460.02430.02497

348111216182024273235364043

0.01540.029850.020190.016270.01950.017190.019220.019550.019210.018670.017970.01760.019110.018220.01858

3811161924273235404349515659

0.01540.014930.014680.011190.012320.011460.012810.012220.013170.01260.013370.012570.013490.013020.01354

sin2 SCs

BCCs

FCCs

Diamonds

s

2sin

s

2sin

s

2sin

s

2sin

close

Veryweak

0.04620.11940.16150.2340.2750.3460.3910.4610.5040.5750.6160.6880.7290.7990.84

3811161924273235404349515659

0.01540.014930.014680.014630.014470.014420.014480.014410.01440.014370.014330.014040.014290.014270.01424

veryclose

A 486.601413.02

542.1

01413.04 2

2

a

a

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

0.04620.11980.16150.1790.2340.2750.3460.3910.4610.5040.5750.6160.6880.7290.799

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

2sin0462.0

sin2 0462.0

sin2 2 0462.0

sin3 2

1.0002.5933.4963.8745.0655.9527.4898.4639.97810.90912.44613.33314.89215.77917.294

2.0005.1896.9917.74910.13011.90514.97816.92619.95721.81824.89226.66729.78431.55834.589

3.0007.77910.48711.62315.19517.85722.46825.39029.93532.72737.33840.00044.67547.33851.883

1234568910111213141617

24681012141618202224263032

348111216182024273235364043

3811161924273235404349515659

Larger error smaller , use the first three lines to fit a more correct A. 0.01444, Use this number to divide sin2 !

0.04620.11980.16150.1790.2340.2750.3460.3910.4610.5040.5750.6160.6880.7290.799

01444.0/sin2 3.1998.29611.18412.39616.20519.04423.96127.07831.92534.90339.82042.65947.64550.48555.332

3811161924273235404349515659

LargerDeviation

New A is 0.01428

01428.0

sin2

3.2358.38911.31012.53516.38719.25824.23027.38132.28335.29440.26643.13748.17951.05055.952

.

.

.

0.1790/0.01413 = 12.66; 222 s = 12 (forbidden diffraction lines for diamond structure, OK for zinc-blend structure!) * = 5.82 g/cm3 then

67.95366054.1

82.5)486.6(

66054.1

3

CV

nM

M for CdTe is 112.4+127.6 = 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

)2

1

2

1

2

1(2

)2

1

2

10(2)

2

10

2

1(2)0

2

1

2

1(2

1

lkhi

TeCd

lkhilkhilkhi

hkl

eff

eeeF

FCC

when h, k, l all even 22)(16 TeCdhkl ffF

when h, k, l all odd 22)(16 TeCdhkl ffF

ZnS –Cd at 000 & Te at ¼ ¼ ¼ + fcc translations

)4

1

4

1

4

1(2 lkhi

TeCdFCChkl effFF

when h + k + l is odd )(16 222

TeCdhkl ffF

when h + k + l = odd multiple of 2 = even multiple of 2

22)(16 TeCdhkl ffF

FCCF

Unmixed hkl only

22)(16 TeCdhkl ffF

fCd + fTe = 100 and |fCd fTe| = 4 at sin / = 0 tofCd + fTe = 30.3 and |fCd fTe| = 1.7 at sin / ~ 1.0

sin

0.0 0.2 0.4 0.6 0.8 1.0 1.2

CdTe

48 37.7 27.5 21.8 17.6 14.3 12.052 41.3 30.3 24.0 19.5 16.0 13.3

TeCd

TeCd

TeCd Rff

ff

2

2

sin

0.0 0.2 0.4 0.6 0.8 1.0 1.2

TeCdR 625 482 426 415 381 318 379

Several hundreds

NaCl –Cd at 000 & Te at ½½½ + fcc translations

h, k, l all even: strong diffracted linesh, k, l all odd: week diffracted lines

ZnS fit better!

Order-Disorder Determination

temperaturehighlow TC

disorderorder

Example – Cu-Au system (AuCu3), TC = 390 oC

disordered ordered

Au

Cu

Cu-Au average

Substitutional solid solution A, B elements AB atoms’ arrangement

Complete Disordered structure:

the probability of each site beingoccupied: ¼ Au, ¾ Cu simple FCC with fav

)2

1

2

10(2)

2

10

2

1(2)0

2

1

2

1(2

1lkhilkhilkhi

avhkl eeefF

4/)3( CuAuav fff

For mixed h, k, l Fhkl = 0For unmixed h, k, l Fhkl = (fAu + 3fCu)

disordered

For mixed h, k, l Fhkl = (fAu fCu)For unmixed h, k, l Fhkl = (fAu + 3fCu)

Peaksshow up

Complete Ordered structure: 1 Au atom, at 000, three Cu atoms at ½ ½ 0, ½ 0 ½, 0 ½ ½.

)2

1

2

10(2)

2

10

2

1(2)0

2

1

2

1(2)000(2 lkhi

Cu

lkhi

Cu

lkhi

Culkhi

Auhkl efefefefF

)()()( lkilhikhiCuAuhkl eeeffF

ordered

Define a long range order parameter S:

A

AA

F

FrS

1

rA: fraction of A sites occupied by the rightatoms; FA: fraction of A atoms in the alloy

complete order: rA = 1 S = 1;complete disorder: rA = FA S = 0

0 S 1

AuCu3 : order parameter S

25.075.0)1( SFFSr AuAuAu

A-site is the 000 equipointrAu :fraction of Au atoms in 000 site the average atomic form factor

Average atomic factor for A-site

CuCuAuAu

CuAuAuAuavA

SfffSf

frfrf

75.075.025.075.0

)1(

CuAuCuAuavA ffffSf 75.025.0)(75.0

= fav

(1 rA): is the fraction of Au occupying the B site in B site (1 rA)/3 Au and 1 (1rA)/3 Cu = (2 + rA)/3 Cu

3

)2(

3

)1( CuAuAuAuavB

frfrf

CuCuAuAuavB fSfSfff 75.025.025.025.0

CuAuAuCuavB ffffSf 75.025.0)(25.0

Average atomic factor for B-site

The structure factor is

)2

1

2

10(2)

2

10

2

1(2)0

2

1

2

1(2)000(2 lkhilkhilkhi

avBlkhi

avAhkl eeefefF

For mixed h, k, l avBavAhkl ffF

)()(25.0)(75.0 CuAuavAuCuavCuAuhkl ffSfffSfffSF

For unmixed h, k, l avBavAhkl ffF 3

avavAuCuavCuAuhkl ffffSfffSF 43)(25.03)(75.0

CuAuCuAuavhkl fffffF 3)75.025.04( 4

Intensity |F|2 superlattice lines S2

B

BB

F

FrS

1

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

2)( BAD ffkI k: a constant for any one composition

f decreases as sin/ increases ID as sin/

Weak signal, very difficult to measure

fav

fZn

fCu

Example – Cu-Zn system (CuZn), TC = 460 oC

disordered ordered

)1( )()2

1

2

1

2

1(2)000(2 lkhi

av

lkhi

avlkhi

avhkl efefefF

* Completely random: a BCC structure

ZnCuav fff 5.05.0

* Completely order: a CsCl

For h + k + l even Fhkl = fCu + fZn

For h + k + l odd Fhkl = 0

)()2

1

2

1

2

1(2)000(2 lkhi

ZnCu

lkhi

Znlkhi

Cuhkl effefefF

For h + k + l even Fhkl = fCu + fZn

For h + k + l odd Fhkl = fCu fZn

A

AA

F

FrS

1

Define a long range order parameter S:

For h + k + l even Fhkl = fCu + fZn

For h + k + l odd Fhkl = S(fCu fZn)

(practice yourself)

1

0

TCT

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.

2

2

|3|

||

line) al(FundamentIntensity

line) ice(superlattIntensity

CuAu

CuAu

ff

ff

Assume sin/ = 0 f = z

09.0~|29379|

|2979|

line) al(FundamentIntensity

line) ice(superlattIntensity 2

2

About 10%, can be measured without difficulty.

Assume sin/ = 0.2

11.0~|6.21365|

|6.2165|

line) al(FundamentIntensity

line) ice(superlattIntensity 2

2

* Case CuZn: the atomic number Z of Cu and Zn is close atomic form factor is close!!

2

2

||

||

line) al(FundamentIntensity

line) ice(superlattIntensity

ZnCu

ZnCu

F

S

ff

ff

I

I

Assume sin/ = 0 f = z

Assume sin/ = 0.2

0003.0|3029|

|3029|

||

||2

2

2

2

ZnCu

ZnCu

F

S

ff

ff

I

I

0003.0|4.226.21|

|4.226.21|

||

||2

2

2

2

ZnCu

ZnCu

F

S

ff

ff

I

I

About 0.03%, very difficult to measure

choose a proper wavelength to resolve the case!

Resonance between the radiation and theK shell electrons larger absorption f

Produce extradifference in

fCu fZn

0013.0|)7.230()6.329(|

|)7.230()6.329(|

||

||2

2

2

2

ZnCu

ZnCu

F

S

ff

ff

I

I

Using Zn K radiation. f for Cu is -3.6 and for Zn is -2.7

About 0.13%, possible to be detected.