tem for incommensurately modulated materials
DESCRIPTION
This presentation is a teaching lecture given on the International School on Aperiodic Crystals and explains how to index electron diffraction patterns taken from incommensurately modulated materials, with exercises, and gives some examples of HAADF-STEM and HRTEM images on incommensurately modulated materials.TRANSCRIPT
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Í Î ÂÛÅ Í ÈÇØÈÅ ÑÌ ÅØÀÍ Í Û Å ÁËÎ ×Í ÛÅ ÕÀËÜÊÎ ÃÅÍ ÈÄÛ Í ÈÊÅËß
Московский Государственный Университетим. М.В. ЛомоносоваФакультет Наук о Материалах
Исаева А.А.
Научные руководителид.х.н., проф. Б.А. Поповкинк.х.н., асс. ФНМ А.И. Баранов
КАФЕДРА НЕОРГАНИЧЕСКОЙ ХИМИИЛаборатория Направленного Неорганического Синтеза
Ë Î Ì Î Í Î ÑÎ Â -2 0 0 3
Examples of incommensurately
modulated structures and
composites studied using
transmission electron
microscopy
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Purpose of this lecture
At the end of this lecture you should be able to:
• Understand the TEM paragraph in papers about IMS and CS
• Be able to make solid comments about conclusionsclaimed from TEM by different sources (collaborators, papers,…) by knowing some possible pitfalls
• Decide whether it would be useful to do TEM on your ownIMS or CS
• Make basic interpretations of TEM data on your ownmaterials by yourself
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Outline of the lecture
The three most frequently usedtechniques in case of IMS/CS:
• Electron diffraction
• HAADF-STEM
• HRTEM
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ED: the main differences withXRD
• Shorter wavelength, almost flat sectionsthrough reciprocal space
• A nanometer size particle is enough
• Multiple diffraction
• Can be combined with direct imaging and compositional analysis
Appearance very much like single crystal XR patterns
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Incommensurately modulatedstructure example
Sb1.155Pb1.216Mn0.628O4
Related commensuratestructure
Sb2MnO4
Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134
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IMS: indexing the subcell reflections• Related commensurate structure Sb2MnO4
• P42/mbc; a=b=8.7 Å, c= 6.0 Å
• Measure R, calculate d with R.d=C (know instrument constant)
• Index each reflection according to list of d-values for each reflection (cf. XRD)
4.35 4.35
4.353.0 3.06.15
All distances in Å
6.15
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Dangers of double diffraction!• P42/mbc
• Reflection conditions: 0kl: k=2n; hhl: l=2n; 00l: l=2n; h00: h=2n
• In contradiction: 001,003,…,100,300,…,010, 030, … all caused by double diffraction
• Tilt around the row: forbidden reflections will disappear
Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134
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Subcell vs. IMS
Sb1.155Pb1.216Mn0.628O4
Sb2MnO4
Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134
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Indexing the IMS
Sb1.155Pb1.216Mn0.628O4
0200 0201 020202010202--
c02110211
-
0000 0001 000200010002--
c
021
001 00110011-
022002210222
--
002000210022
--
q≈3/4c*
02
0
022 020 022
002
000
002-
-
Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134
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Reflection conditions for m
Sb1.155Pb1.216Mn0.628O4
For thís material: no special conditions for m: if subcell reflection is seen, also hklm with m=1 and m=2 is seen.
02100200
0201 0202
0211-
0211
02210222--
0220
1100
1101 11021122-
1120-
1121
q≈3/4c*
Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134
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Reflection conditions
Sb1.155Pb1.216Mn0.628O4
0klm: k=2n
hhlm: l=2n
P42/mbc(00γ)
P42/mbc
0kl: k=2n
hhl: l=2n
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• Index the commensurate structure LaSrCuO4, this will help you with indexing the next, incommensurate one.
6.50 Å1.85 Å
6.50 Å2.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
1.85 Å
Exercise 1-1: IMS: index the non-
modulated phase
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6.50 Å1.85 Å
6.50 Å2.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
1.85 Å
001
002
200
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6.50 Å1.85 Å
6.50 Å2.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
1.85 Å
001
002
200
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0021.85 Å
0022.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
1.85 Å
010
020
110
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0021.85 Å
0022.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
1.85 Å
010
020
110
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002020
0022.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
1.85 Å
010
020
110
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002020
0022.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
1.85 Å
100
200
020
020
1.85 Å
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002020
0022.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
1.85 Å
100
200
020
020
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002020
0022.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
200
110
220
½ ½ 0
020
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002020
0022.61 Å
2.61 Å
(Simulated ED patterns)
Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å.
200
110
220
½ ½ 0
020
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Indexed solution
002200
002110 200
110
020
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Zone indices
002200
002110 200
110
020
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Exercise 1-2: IMS: determine the reflection condition h0l
002200
002110 200
110
h0l: h+l=2n
h = 2n
l = 2n
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002200
002110 200
110
h0l: h+l=2n
Exercise 1-2: IMS: determinethe reflection condition h0l
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Exercise 1-2: IMS: determinethe reflection condition hhl
002200
002110 200
110
hhl: h+l=2n
h = 2n
l = 2n
h0l: h+l=2n
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Exercise 1-2: IMS: determinethe reflection condition hhl
002200
002110 200
110
hhl:
l = 2n
h0l: h+l=2n
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Exercise 1-2: IMS: determinethe reflection condition hk0
002200
002110 200
110
hhl: l = 2nh0l: h+l=2n hk0: h+k=2n
h = 2n
k = 2n
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Exercise 1-2: IMS: determinethe reflection condition hk0
002200
002110 200
110
hhl: l = 2nh0l: h+l=2n hk0: h+k=2n
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Solved reflection conditions
002200
002110
h0l:h+l=2n hhl:l=2n hk0:h+k=2n
Also (from rest of the zones) hkl: h+k+l=2n.
110-
200
110
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Determine space group
• International Tables: I---
• Most symmetrical I4/mmm
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Incommensurate vs. basiccell
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Exercise 1-3: index the IMS
Identify and index the subcell reflections.
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Propose a modulation vector.
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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• q=αa*
• α<0.5
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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• q=αa*
• α<0.5
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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• q=αa*
• α<0.5
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Possible solution
• q=αa*
• α<0.5
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Index the satellite indicated in green
-001
0001
100
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Index the satellite indicated in green
0001
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Index the next indicatedsatellite
0002
2002-
2001-
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Index the next indicatedsatellite
2002-
-
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Exercise 1-4: IMS with more complete data
• Which of the indicatedvectors corresponds to the modulation vector chosen on the previousslides?
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Exercise 1-4: IMS with more complete data
• Which of the indicatedvectors corresponds to the modulation vector chosen on the previousslides?
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Is the proposedvector still valid?
yes
no
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Indicate the correct modulationvector(s) on the central white area.
You need
q1
q2
q1 ànd q2
q1=αa*+βb*
q2= -αa*+βb*
α=β<0.25
hklmnHadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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20011
20011-
20011-
Index the reflection indicatedin red
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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20011-
Index the reflection indicatedin red
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Index the others patternsconsistently with this new choice.
Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350
LaSrCuO3.52
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Index the reflection indicatedin green
20011
20011
20011-
-
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Index the reflection indicatedin green
LaSrCuO3.52
20011-
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Index the reflection indicatedin green
LaSrCuO3.52
10110
10111
10111-
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Index the reflection indicatedin green
LaSrCuO3.52
10111-
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Index the reflection indicatedin green
LaSrCuO3.52
00002
00002
00001-
-
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Index the reflection indicatedin green
LaSrCuO3.52
00002-
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Look out for twinning!
Y0.8Sr2.2Mn2GaO8-
Gillie et al., Journal of Physics and Chemistry of Solids, 65, 1 (2004) 87-93
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Exercise 1-5: twinning?
• Could it be twinning and thus sufficient to useonly one q-vector?
yes
no
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Exercise 1-5: twinning?
• Could it be twinning and thus sufficient to useonly one q-vector?
yes
no
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Exercise 1-6: derive the reflection conditions
hklmn:
h+k+l+m+n=2i
h+k+l=2i
m+n=2i
hhl:l=2n
hkl:h+k+l=2n
LaSrCuO3.52
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Exercise 1-6: derive the reflection conditions
hklmn:
h+k+l=2i
LaSrCuO3.52
hhl:l=2n
hkl:h+k+l=2n
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Exercise 1-6: derive the reflection conditions
hhlm0: l=2i
h=2i
l,m=2i
LaSrCuO3.52
hhl:l=2n
hkl:h+k+l=2n
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Exercise 1-6: derive the reflection conditions
LaSrCuO3.52
hhl:l=2n
hkl:h+k+l=2n
hhlm0:
l,m=2i
buthhlm0:m=2iis sufficient
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Exercise 1-6: derive the reflection conditions
• hklmn:h+k+l=2i
• hhlm0:m=2i
(-hhl0n: n=2i)
LaSrCuO3.52
hhl:l=2n
hkl:h+k+l=2n
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Determining the superspace group
• Online tables Yamamoto (ActaCryst.A 1996, A52, 509)
• I4/mmm(α α0, -αα0)00mg
LaSrCuO3.52
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Use of doing ED
• Deriving approximate cell parameters, modulation vector and superspace group
• Not useful for: precise parameters, precisemodulation vector
• Excellent if multiphase sample or satellitesvery weak in XRD/ND
• Be aware of:
– Twinning
– Multiple diffraction (for forbidden refls., for usingintensities,…)
– The possibility of missing important sections of reciprocal space
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66(Abakumov et al., Chem. Mater. 20
(13), pp. 4457-4467)
Relation composition –modulation vector• Most TEM have EDX available for composition
determination
• Determination of direct links between composition and modulation vector, also in multiphased samples
q q q
2q 2q 2q
a* a* a*
b* b* b*
Ba4In6-xMgxO13-x/2
x = 0 x = 0.2 x = 0.4
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ED of a composite structure (CS)
a1= a2, b1=b2, с1 с2, q = c1* = c2*
hk00 average cellhkl0 first subsystemhk0m second subsystemhklm satellites
Sr1.3Co0.8Mn0.2O3
A1+xA’xB1-xO3 first subsystem[(B, A’)O3]second subsystem [A]
Mandal et al., Chem. Mater., 19, 25 (2007) 6158
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ED of a composite structure (CS)
Sr1.3Co0.8Mn0.2O3
2022-
Mandal et al., Chem. Mater., 19, 25 (2007) 6158
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Exercise 2: compositestructure: index the pattern
[SrF0.82(OH)0.18]2.5[Mn6O12]
Which reflection is 0011?-
Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189
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Exercise 2: compositestructure: index the pattern
Which reflection is 0011?-
[SrF0.82(OH)0.18]2.5[Mn6O12]
Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189
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Exercise 2: compositestructure: index the pattern
Which reflection is 0012?-
[SrF0.82(OH)0.18]2.5[Mn6O12]
Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189
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Exercise 2: compositestructure: index the pattern
Which reflection is 0012?-
[SrF0.82(OH)0.18]2.5[Mn6O12]
Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189
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Exercise 2: compositestructure: index the pattern
Which reflection does the blue arrow point at? 002200210012
---
[SrF0.82(OH)0.18]2.5[Mn6O12]
Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189
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Exercise 2: compositestructure: index the pattern
Which reflection does the blue arrow point at? 0022
-
[SrF0.82(OH)0.18]2.5[Mn6O12]
Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189
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Exercise 2: compositestructure: index the pattern -solution
[SrF0.82(OH)0.18]2.5[Mn6O12]
Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189
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Streaks
• order in the plane, disorder betweenthese planes: streak on ED pattern
• disorder in twodirections: diffuse intensity plane: streak or background intensity depending onthe intersection
[SrF0.82(OH)0.18]2.5[Mn6O12]
Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189
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HRTEM and HAADFSTEM-images
– contrast related to Zn
– direct information about the heavy atom positions
– brighter with thickness also
– correct orientation veryimportant!
– scanned image: positionsimprecise
HRTEM– complicated phase transfer
function
– no direct interpretationpossible
– contrast changes withthickness and defocus
– correct orientation veryimportant!
– one-shot image: position of the contrast objects reliable
HAADF-STEM
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78
cm
am
Heavy atoms from HAADF-STEM
“Pb2Fe2O5” Abakumov et al., Ang.Chemie-Int.Ed., 45, 40 (2006) 6697-6700
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High resolution transmissionelectron microscopy (HREM)
Bi2Sr1.2La0.8CuO6.46-xF2x Hadermann et al., JSSC, 156, 2 (2001) 455-451
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image influenced by contrast transfer function of the
microsope
no direct correspondence
visual comparison with simulations needed
PbFe
O
Defocus
Th
ick
nes
s
-700 Å -500 Å -300 Å -100 Å
10
Å3
0 Å
50
Å7
0 Å
90
Å
Relation structure – image in HRTEM
simple perovskite
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Exercise 3: predict what the ED pattern will roughly look like
Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389
“Pb2Fe2O5”
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Exercise 3: predict what the ED pattern will roughly look like
What will be the subcell reflections?
Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389
“Pb2Fe2O5”
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Exercise 3: predict what the ED pattern will roughly look like
What will be the subcell reflections?
Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389
“Pb2Fe2O5”
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Exercise 3: predict what the ED pattern will roughly look like
Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389
“Pb2Fe2O5”
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Exercise 3: predict what the ED pattern will roughly look like
Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389
“Pb2Fe2O5”
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Exercise 3: predict what the ED pattern will roughly look like
Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389
“Pb2Fe2O5”
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Exercise 3: predict what the ED pattern will roughly look like
Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389
“Pb2Fe2O5”
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88Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389
“Pb2Fe2O5”
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Domains with varying q
Sr1.3Co0.8Mn0.2O3Mandal et al., Chem. Mater., 19, 25 (2007) 6158
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Comments to the applicabilityof conclusions from TEM• If the satellites cannot be seen on the XRPD-NPD patterns
it is no use to try to refine the structure using the ED cell.
• Precision of parameters determined from ED is not high. ED is good for determining parameters (cell parameters and direction and length of q) qualitatively.
• Good for determining symmetry.
• Can give direct space information and compositionalinformation also.
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Purpose of this lecture
At the end of this lecture you should be able to:
• Understand the TEM paragraph in papers about IMS and CS
• Be able to make solid comments about conclusionsclaimed from TEM by different sources (collaborators, papers,…) by knowing some possible pitfalls
• Decide whether it would be useful to do TEM on your ownIMS or CS
• Make basic interpretations of TEM data on your ownmaterials by yourself