determination of crystal structure (from chapter 10 of textbook 2)
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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. - PowerPoint PPT PresentationTRANSCRIPT
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.