Transcript Crystallography 14

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:
2d hkl sin    ;2d hkl sin    




sin  sin  
In the analyzing step, the sin2 is used

2
2
  
2
2
   sin    sin 
 

  
    1 .2
 

 filament contamination  W KL radiation
 Diffraction by other substance
 Calibration curve using known crystal
For most
radiations
Effect of sample height displacement
2 scos
 2 (in radians) 
R
s: sample height displacement
R: diffractometer radius
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


s
Length of
 scos
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 = 0.154 nm and R =
225 mm.
2d100 (  0.6) sin   0.154    7.37
2d 400 (  0.15) sin   0.154    30.87
The displacement cause these peaks to shift by

 2  2  0.1  cos 7.37 / 225  8.81  104  0.051   0.025


4




0
.
022
 2  2  0.1  cos 30.87 / 225  7.63  10  0.044
(100):  = 7.395o
(400):  = 30.892o
2d100 sin 7.395  0.154  a  0.5982
2d 400 sin 30.892  0.154  a  0.5999
Pattern Indexing
→assign hkl values to each peak
Simplest example: indexing cubic pattern
2d hkl sin   
d hkl 
a
h2  k 2  l 2
a
sin 2 
2
2
sin     2
 2
2
2
2
2
2
h k l
4a
h k l
Define
s  h2  k 2  l 2
Constant for a
given crystal
Values for h2 + k2 + l2 for cubic system
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, ..
FCC: 3, 4, 8, 11, 12, 16, 19, 20, …
Diamond: 3, 8, 11, 16, 19, …
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
1
h2  k 2 l 2

 2
2
2
d hkl
a
c
 sin    / 2d hkl
2
2
2
2
2






h

k
l
2
2
2
2
  

 sin   


A
(
h

k
)

Cl
2
2 
2
d
4
a
c


 hkl 
A  2 / 4a 2 ; C  2 / 4c 2
When l = 0 (hk0 lines),
sin 2   A( h 2  k 2 )
Possible values for h2 + k2: 1, 2, 4, 5, 8, 9, 10, …
sin 2   A( h 2  k 2 )  Cl 2
Possible values for l2 are 1, 4, 9, 16, …
Indexing Hexagonal system
1
4 h 2  hk  k 2 l 2

 2
2
2
d hkl 3
a
c
sin 2  
2  h 2  hk  k 2  2 l 2

3
a2
2
2
2
 

A
(
h

hk

k
)

Cl
2
4
c

A  2 / 3a 2 ; C  2 / 4c 2
Possible values for h2 + hk + k2 are 1, 3, 4, 7, 9, 12, …
The rest of lines are sin 2   A(h 2  hk  k 2 )  Cl 2
Possible values for l2 are 1, 4, 9, 16, …
Example:
sin2
1 0.097
2 0.112
3 0.136
4 0.209
5 0.332
6 0.390
7 0.434
8 0.472
9 0.547
10 0.668
11 0.722
sin2/3
0.032
0.037
0.045
0.070
0.111
0.130
0.145
0.157
0.182
0.223
0.241
sin2/4
0.024
0.028
0.034
0.052
0.083
0.098
0.109
0.118
0.137
0.167
0.180
Let’s say A = 0.112
sin2/7
0.014
0.016
0.019
0.030
0.047
0.056
0.062
0.067
0.078
0.095
0.103
100
110
1
2
3
4
5
6
7
8
9
10
11
12
13
sin2 sin2-A sin2-3A
0.097
0
0.112
0.136 0.024
0.209 0.097
0.332 0.220
0.054
0.390 0.278
0.098
0.434 0.322
0.136
0.472 0.360
0.211
0.547 0.435
0.332
0.668 0.556
0.386
0.722 0.610
0.470
0.806 0.694
0.543
0.879 0.767
0.097 belongs to Cl2.
002
What is the l?
There are two lines
100
between 100 and 110.
101
Probably, 10l1 and
102
10l2
110 ,103  0.024 and 0.097
004
are different ls.
0.097/0.024 ~ 4
112
0.220/0.024 ~ 9
0.390/0.024 ~ 16
 C = 0.02441
Indexing orthorhombic system:
2  h 2
2
2

k
l
2
sin    2  2  2   Ah 2  Bk 2  Cl 2
4 a
b
c 
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 enough
diffraction lines for the computer to indexing.
Effect of Cell distortion on the powder Pattern
Autoindexing
http://www.ccp14.ac.uk/solution/indexing/index.html
Determination of the number of atoms in a unit cell
Indexing the power pattern 
shape and size of unit cell (volume)
number of atoms in that unit cell.
VC N 0
n
M
VC: unit cell volume; : density
N0: Avogodro’s number;
M: molecular weight;
n: number of molecules in a unit cell
in Å3
3
in
g/cm
g
V

C
nM  VC (1083 cm 3 )  ( 3 ) N 0 (6.022  1023 ) 
cm
1.66054
Determination of Atom positions
Relative intensities determine atomic positions
(a trial and error process)
N
 1  cos2 2 
 ; F   f n exp2i ( hun  kvn  lwn )
I  F p 2
n 1
 sin  cos  
2
(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)
46.6 53.4
Cd : Te 
:
112.4 127.6
 1 : 1.0009  49.8 : 50.2
* Make powder
diffraction and list
sin2: index the
pattern! Assume it is
cubic
sin2
0.0462
0.1194
Very 0.1615
weak 0.179
0.234
0.275
0.346
0.391
0.461
0.504
0.575
0.616
0.688
0.729
0.799
SC
s
1
2
3
4
5
6
8
9
10
11
12
13
14
16
17
0.0462
0.0597
0.05383
0.04475
0.0468
0.04583
0.04325
0.04344
0.0461
0.04582
0.04792
0.04738
0.04914
0.04556
0.047
sin 2 
s
BCC
s
FCC
s
2
4
6
8
10
12
14
16
18
20
22
24
26
30
32
3
4
8
11
12
16
18
20
24
27
32
35
36
40
43
0.0231
0.02985
0.02692
0.02237
0.0234
0.02292
0.02471
0.02444
0.02561
0.0252
0.02614
0.02567
0.02646
0.0243
0.02497
sin 2 
s
Diamond
s
close
0.0154
0.02985
0.02019
0.01627
0.0195
0.01719
0.01922
0.01955
0.01921
0.01867
0.01797
0.0176
0.01911
0.01822
0.01858
sin 2 
s
3
8
11
16
19
24
27
32
35
40
43
49
51
56
59
0.0154
0.01493
0.01468
0.01119
0.01232
0.01146
0.01281
0.01222
0.01317
0.0126
0.01337
0.01257
0.01349
0.01302
0.01354
sin 2 
s
Diamond structure and Zinc blend structure:
forbidden peaks! Which one has more peaks?
 removing line 4 first
0.0462
0.1194
0.1615
0.234
0.275
0.346
0.391
0.461
0.504
0.575
0.616
0.688
0.729
0.799
0.84
3
8
11
16
19
24
27
32
35
40
43
49
51
56
59
0.0154
0.01493
0.01468
0.01463
0.01447
0.01442
0.01448
0.01441
0.0144
0.01437
0.01433
0.01404
0.01429
0.01427
0.01424
very
close
2
4a
2
 0.01413 

1.542
a
 6.486 A
2  0.01413
Alternative:
Assume the first line is from s = 1, s = 2 and s = 3, …
sin 2 
sin 2 
0.0462
0.0462
0.1198
0.1615
0.179
0.234
0.275
0.346
0.391
0.461
0.504
0.575
0.616
0.688
0.729
0.799
1.000
2.593
3.496
3.874
5.065
5.952
7.489
8.463
9.978
10.909
12.446
13.333
14.892
15.779
17.294
2  sin 2 
0.0462
1
2
3
4
5
6
8
9
10
11
12
13
14
16
17
2.000
5.189
6.991
7.749
10.130
11.905
14.978
16.926
19.957
21.818
24.892
26.667
29.784
31.558
34.589
3  sin 2 
0.0462
2
4
6
8
10
12
14
16
18
20
22
24
26
30
32
3.000
7.779
10.487
11.623
15.195
17.857
22.468
25.390
29.935
32.727
37.338
40.000
44.675
47.338
51.883
3
4
8
11
12
16
18
20
24
27
32
35
36
40
43
3
8
11
16
19
24
27
32
35
40
43
49
51
56
59
Larger error smaller , use the first three lines to
fit a more correct A.
 0.01444, Use this number
to divide sin2 !
sin 2 
sin 2  / 0.01444
0.0462
0.1198
0.1615
0.179
0.234
0.275
0.346
0.391
0.461
0.504
0.575
Larger 0.616
Deviation0.688
0.729
0.799
3.199
8.296
11.184
12.396
16.205
19.044
23.961
27.078
31.925
34.903
39.820
42.659
47.645
50.485
55.332
3
8
11
16
19
24
27
32
35
40
43
49
51
56
59
0.01428
.
.
.
3.235
8.389
11.310
12.535
16.387
19.258
24.230
27.381
32.283
35.294
40.266
43.137
48.179
51.050
55.952
New A is 0.01428
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
VC 
(6.486)3  5.82
nM 
1.66054

1.66054
 953.67
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
Fhkl
1 1
1
1
1 1
2 i ( h  k  l 0 )
2 i ( h  k 0  l )
2 i ( h 0  k  l ) 

2
2 2
 1  e 2 2  e 2
e



1 1 1
2 i ( h  k  l ) 

2 2 2
f

f
e
Te
 Cd



FCC  FFCC
when h, k, l all even  Fhkl  16( f Cd  fTe )2
2
2

F

16
(
f

f
)
when h, k, l all odd
hkl
Cd
Te
ZnS –Cd at 000 & Te at ¼ ¼ ¼ + fcc translations
2
1 1 1
2 i ( h  k  l ) 

Unmixed hkl only
Fhkl  FFCC   f Cd  fTee 4 4 4 


2
2
2

F

16
(
f

f
when h + k + l is odd
hkl
Cd
Te )
when h + k + l
2
2

F

16
(
f

f
)
= odd multiple of 2
hkl
Cd
Te
2
2

F

16
(
f

f
)
= even multiple of 2
hkl
Cd
Te
sin 

0.0
0.2
0.4
0.6
0.8
1.0
1.2
Cd 48 37.7 27.5 21.8 17.6 14.3 12.0
Te 52 41.3 30.3 24.0 19.5 16.0 13.3
fCd + fTe = 100 and |fCd  fTe| = 4 at sin / = 0 to
fCd + 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
RCd Te 625 482 426 415 381 318 379
 fCd  fTe 2  R
Several hundreds
Cd Te
2
 fCd  fTe 
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
Substitutional solid solution  A, B elements
 AB atoms’ arrangement
order
low
disorder
TC
temperature
high
Example – Cu-Au system (AuCu3), TC = 390 oC
Cu
Au
Cu-Au average
disordered
ordered
Complete Disordered structure:
the probability of each site being
disordered
occupied: ¼ Au, ¾ Cu
 simple FCC with fav f av  ( f Au  3 f Cu ) / 4
Fhkl
1 1
1
1
1 1
2 i ( h  k  l 0 )
2 i ( h  k 0  l )
2 i ( h 0  k  l ) 

2
2 2
 f av 1  e 2 2  e 2
e



For mixed h, k, l  Fhkl = 0
For unmixed h, k, l  Fhkl = (fAu + 3fCu)
Complete Ordered structure:
1 Au atom, at 000,
three Cu atoms at ½ ½ 0, ½ 0 ½, 0 ½ ½.
Fhkl
ordered
1 1
1
1
1 1
2 i ( h  k  l 0 )
2 i ( h  k 0  l )
2 i ( h 0  k  l ) 

2 i ( h 0  k 0  l 0 )
2
2 2
  f Au e
 f Cu e 2 2  f Cu e 2
 f Cu e




Fhkl  f Au  f Cu ei ( h  k )  ei ( h l )  ei ( k l )

For mixed h, k, l  Fhkl = (fAu  fCu)
For unmixed h, k, l  Fhkl = (fAu + 3fCu)
Peaks
show up
Define a long range order parameter S:
rA  FA
S
1  FA
rA: fraction of A sites occupied by the right
atoms; 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
Average atomic factor for A-site
rAu  S (1  FAu )  FAu  0.75S  0.25
A-site is the 000 equipoint
rAu :fraction of Au atoms in 000 site
 the average atomic form factor
f avA  rAu f Au  (1  rAu ) f Cu 
0.75Sf Au  0.25 f Au  0.75 f Cu  0.75Sf Cu
f avA  0.75S ( f Au  f Cu )  0.25 f Au  0.75 f Cu
= fav
Average atomic factor for B-site
(1  rA): is the fraction of Au occupying the B site
 in B site
(1  rA)/3 Au and 1 (1rA)/3 Cu = (2 + rA)/3 Cu

f avB
(1  rAu ) f Au ( 2  rAu ) f Cu


3
3
f avB  0.25 f Au  0.25Sf Au  0.25Sf Cu  0.75 f Cu
f avB  0.25S ( f Cu  f Au )  0.25 f Au  0.75 f Cu
The structure factor is
Fhkl
1 1
1
1
1 1

2 i ( h  k  l 0 )
2 i ( h  k 0  l )
2 i ( h 0  k  l )  

2
2 2 
  f avAe 2i ( h 0 k 0l 0 )  f avB  e 2 2  e 2
e




For mixed h, k, l  Fhkl   f avA  f avB 
Fhkl  0.75S ( f Au  f Cu )  f av  0.25S ( f Cu  f Au )  f av  S ( f Au  f Cu )
For unmixed h, k, l 
Fhkl   f avA  3 f avB 
Fhkl  0.75S ( f Au  f Cu )  f av  3  0.25S ( f Cu  f Au )  3 f av  4 f av
Fhkl  4 f av  4(0.25 f Au  0.75 f Cu )  f Au  3 f Cu
Intensity  |F|2  superlattice lines  S2
Using different S definition
rB  FB
S
1  FB
What would you get?
Homework!
Intensity  weak diffuse background 
If atoms A and B completely random in a solid solution
 diffuse scattering
I D  k ( f A  f B )2
k: a constant for any one composition
f decreases as sin/ increases  ID  as sin/ 
Weak signal, very difficult to measure
Example – Cu-Zn system (CuZn), TC = 460 oC
fav
fZn
fCu
disordered
ordered
* Completely random: a BCC
structure
1 1 1
Fhkl  f av e
2 i ( h 0  k 0  l 0 )
 f av e
2 i ( h  k  l )
2 2 2
 f av (1  ei ( h  k l ) )
f av  0.5 f Cu  0.5 f Zn
For h + k + l even  Fhkl = fCu + fZn
For h + k + l odd  Fhkl = 0
* Completely order: a CsCl
Fhkl  f Cu e 2i ( h 0 k 0l 0 )  f Zn e
1 1 1
2 i ( h  k  l )
2 2 2
 f Cu  f Zn ei ( h  k l )
For h + k + l even  Fhkl = fCu + fZn
For h + k + l odd  Fhkl = fCu  fZn
Define a long range order parameter S:
S
rA  FA
1  FA
(practice yourself)
For h + k + l even  Fhkl = fCu + fZn
For h + k + l odd  Fhkl = S(fCu  fZn)
Different system
1
0
T
TC
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.
Intensity (superlattice line)
| f Au  f Cu |2

Intensity (Fundamental line) | f Au  3 f Cu |2
Assume sin/ = 0  f = z
Intensity (superlattice line)
| 79  29 |2

~ 0.09
2
Intensity (Fundamental line) | 79  3  29 |
Assume sin/ = 0.2
Intensity (superlattice line)
| 65  21.6 |2

~ 0.11
2
Intensity (Fundamental line) | 65  3  21.6 |
About 10%, can be measured without difficulty.
* Case CuZn: the atomic number Z of Cu and Zn is close
 atomic form factor is close!!
Intensity (superlattice line)
I S | f Cu  f Zn |2


Intensity (Fundamental line) I F | f Cu  f Zn |2
Assume sin/ = 0  f = z
I S | f Cu  f Zn |2 | 29  30 |2


 0.0003
2
2
I F | f Cu  f Zn |
| 29  30 |
Assume sin/ = 0.2
I S | f Cu  f Zn |2 | 21.6  22.4 |2


 0.0003
2
2
I F | f Cu  f Zn |
| 21.6  22.4 |
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
fCu  fZn
Using Zn K radiation. f for Cu is -3.6 and for Zn is
-2.7 
I S | f Cu  f Zn |2 | ( 29  3.6)  (30  2.7) |2


 0.0013
2
2
I F | f Cu  f Zn |
| ( 29  3.6)  (30  2.7) |
About 0.13%, possible to be detected.