logging in or signing up TEM for incommensurately modulated materials johader Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 94 Category: Science & Tech.. License: Some Rights Reserved Like it (3) Dislike it (0) Added: October 04, 2010 This Presentation is Public Favorites: 0 Presentation 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. Comments Posting comment... Premium member Presentation Transcript Slide 1: Examples of incommensurately modulated structures and composites studied using transmission electron microscopy Purpose of this lecture : 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 conclusions claimed from TEM by different sources (collaborators, papers,…) by knowing some possible pitfalls Decide whether it would be useful to do TEM on your own IMS or CS Make basic interpretations of TEM data on your own materials by yourself Outline of the lecture : Outline of the lecture The three most frequently used techniques in case of IMS/CS: Electron diffraction HAADF-STEM HRTEM ED: the main differences with XRD : ED: the main differences with XRD Shorter wavelength, almost flat sections through 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 Incommensurately modulated structure example : Incommensurately modulated structure example Sb1.155Pb1.216Mn0.628O4 Related commensurate structure Sb2MnO4 Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 IMS: indexing the subcell reflections : 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 4.35 3.0 3.0 6.15 All distances in Å 6.15 Dangers of double diffraction! : 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 - 021 Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 Subcell vs. IMS : Subcell vs. IMS Sb1.155Pb1.216Mn0.628O4 Sb2MnO4 Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 Indexing the IMS : Indexing the IMS Sb1.155Pb1.216Mn0.628O4 c c q≈3/4c* 020 022 020 022 002 000 002 - - Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 Reflection conditions for m : 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. 0210 0200 0201 0202 0211 - 0211 0221 0222 - - 0220 1100 1101 1102 1122 - 1120 - 1121 q≈3/4c* Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 Reflection conditions : Reflection conditions Sb1.155Pb1.216Mn0.628O4 0klm: k=2n hhlm: l=2n P42/mbc(00γ) Exercise 1-1: IMS: index the non-modulated phase : Index the commensurate structure LaSrCuO4, this will help you with indexing the next, incommensurate one. (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 Slide 13: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 001 002 200 Slide 14: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 001 002 200 Slide 15: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 010 020 110 Slide 16: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 010 020 110 Slide 17: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 010 020 110 Slide 18: (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 Slide 19: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 100 200 020 020 Slide 20: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 200 110 220 ½ ½ 0 020 Slide 21: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 200 110 220 ½ ½ 0 020 Indexed solution : Indexed solution 200 020 Zone indices : Zone indices 200 020 [010] [110] - [001] Exercise 1-2: IMS: determine the reflection condition h0l : Exercise 1-2: IMS: determine the reflection condition h0l [010] [110] - [001] 200 h0l: h+l=2n h = 2n l = 2n Exercise 1-2: IMS: determine the reflection condition h0l : [010] [110] - [001] 200 h0l: h+l=2n Exercise 1-2: IMS: determine the reflection condition h0l Exercise 1-2: IMS: determine the reflection condition hhl : Exercise 1-2: IMS: determine the reflection condition hhl [010] [110] - [001] 200 hhl: h+l=2n h = 2n l = 2n h0l: h+l=2n Exercise 1-2: IMS: determine the reflection condition hhl : Exercise 1-2: IMS: determine the reflection condition hhl [010] [110] - [001] 200 hhl: l = 2n h0l: h+l=2n Exercise 1-2: IMS: determine the reflection condition hk0 : Exercise 1-2: IMS: determine the reflection condition hk0 [010] [110] - [001] 200 hhl: l = 2n h0l: h+l=2n hk0: h+k=2n h = 2n k = 2n Exercise 1-2: IMS: determine the reflection condition hk0 : Exercise 1-2: IMS: determine the reflection condition hk0 [010] [110] - [001] 200 hhl: l = 2n h0l: h+l=2n hk0: h+k=2n Solved reflection conditions : Solved reflection conditions [010] [110] - [001] h0l:h+l=2n hhl:l=2n hk0:h+k=2n Also (from rest of the zones) hkl: h+k+l=2n. 200 Determine space group : Determine space group International Tables: I--- Most symmetrical I4/mmm Incommensurate vs. basic cell : Incommensurate vs. basic cell [010] [110] - [001] [010] [110] - [001] Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Exercise 1-3: index the IMS : 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 Slide 34: q q q 200 002 [010] Propose a modulation vector. Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 35: 200 002 [010] q=αa* α<0.5 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 36: 200 002 [010] q=αa* α<0.5 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 37: 200 002 [010] q=αa* α<0.5 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Possible solution : 200 002 [010] Possible solution q=αa* α<0.5 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the satellite indicated in green : Index the satellite indicated in green - q 001 0001 100 002 [010] 200 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the satellite indicated in green : 002 [010] 200 Index the satellite indicated in green q 0001 0001 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the next indicated satellite : Index the next indicated satellite q 0002 2002 - 2001 - 0001 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the next indicated satellite : q Index the next indicated satellite 2002 - - 0001 2002 - Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Exercise 1-4: IMS with more complete data : Exercise 1-4: IMS with more complete data Which of the indicated vectors corresponds to the modulation vector chosen on the previous slides? 020 200 110 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Exercise 1-4: IMS with more complete data : Exercise 1-4: IMS with more complete data Which of the indicated vectors corresponds to the modulation vector chosen on the previous slides? 020 200 110 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 45: Is the proposed vector still valid? yes no 020 200 110 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 46: Indicate the correct modulation vector(s) on the central white area. 020 200 110 You need q1 q2 q1 ànd q2 q1=αa*+βb* q2= -αa*+βb* α=β<0.25 hklmn Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the reflection indicated in red : 020 200 110 00010 00001 20011 20011 - 20011 - Index the reflection indicated in red Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the reflection indicated in red : 020 200 110 00010 00001 20011 - 20011 - Index the reflection indicated in red Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the others patterns consistently with this new choice. : Index the others patterns consistently with this new choice. 002 [010] 200 002 [010] 200 [110] [001] 020 200 110 002 - 00010 00001 - 20011 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the reflection indicated in green : Index the reflection indicated in green 20011 20011 20011 - - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - - 020 200 00010 00001 - 20011 Index the reflection indicated in green : Index the reflection indicated in green LaSrCuO3.52 20011 - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 20011 - 20011 - 020 200 00010 00001 - 20011 Index the reflection indicated in green : 002 [010] 200 002 [010] 200 [110] [001] 110 002 - Index the reflection indicated in green LaSrCuO3.52 10110 10111 10111 - 020 200 00010 00001 - 20011 Index the reflection indicated in green : Index the reflection indicated in green LaSrCuO3.52 10111 - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 Index the reflection indicated in green : Index the reflection indicated in green LaSrCuO3.52 00002 00002 00001 - - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - Index the reflection indicated in green : Index the reflection indicated in green LaSrCuO3.52 00002 - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - Look out for twinning! : Look out for twinning! Y0.8Sr2.2Mn2GaO8- Gillie et al., Journal of Physics and Chemistry of Solids, 65, 1 (2004) 87-93 Exercise 1-5: twinning? : Exercise 1-5: twinning? Could it be twinning and thus sufficient to use only one q-vector? yes no 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-5: twinning? : Exercise 1-5: twinning? Could it be twinning and thus sufficient to use only one q-vector? yes no 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : 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 LaSrCuO3.52 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : Exercise 1-6: derive the reflection conditions hklmn: h+k+l=2i LaSrCuO3.52 hhl:l=2n hkl:h+k+l=2n 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : 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 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : Exercise 1-6: derive the reflection conditions LaSrCuO3.52 hhl:l=2n hkl:h+k+l=2n hhlm0: l,m=2i but hhlm0:m=2i is sufficient 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : 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 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Determining the superspace group : Determining the superspace group Online tables Yamamoto (Acta Cryst.A 1996, A52, 509) I4/mmm(α α0, -αα0)00mg LaSrCuO3.52 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Use of doing ED : Use of doing ED Deriving approximate cell parameters, modulation vector and superspace group Not useful for: precise parameters, precise modulation vector Excellent if multiphase sample or satellites very weak in XRD/ND Be aware of: Twinning Multiple diffraction (for forbidden refls., for using intensities,…) The possibility of missing important sections of reciprocal space Relation composition – modulation vector : (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 a* a* a* b* b* b* Ba4In6-xMgxO13-x/2 x = 0 x = 0.2 x = 0.4 ED of a composite structure (CS) : ED of a composite structure (CS) a1= a2, b1=b2, с1 с2, q = gc1* = c2* hk00 average cell hkl0 first subsystem hk0m second subsystem hklm 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 ED of a composite structure (CS) : ED of a composite structure (CS) Sr1.3Co0.8Mn0.2O3 2022 - Mandal et al., Chem. Mater., 19, 25 (2007) 6158 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern [SrF0.82(OH)0.18]2.5[Mn6O12] 0001 0010 Which reflection is 0011? - Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 Which reflection is 0011? - [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 Which reflection is 0012? - [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 Which reflection is 0012? - [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 Which reflection does the blue arrow point at? 0022 0021 0012 - - - [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 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 Exercise 2: composite structure: index the pattern - solution : Exercise 2: composite structure: index the pattern - solution [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Streaks : Streaks order in the plane, disorder between these planes: streak on ED pattern disorder in two directions: diffuse intensity plane: streak or background intensity depending on the intersection [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 HRTEM and HAADFSTEM-images : HRTEM and HAADFSTEM-images contrast related to Zn direct information about the heavy atom positions brighter with thickness also correct orientation very important! scanned image: positions imprecise HRTEM complicated phase transfer function no direct interpretation possible contrast changes with thickness and defocus correct orientation very important! one-shot image: position of the contrast objects reliable HAADF-STEM Slide 78: Heavy atoms from HAADF-STEM “Pb2Fe2O5” Abakumov et al., Ang.Chemie-Int.Ed., 45, 40 (2006) 6697-6700 High resolution transmission electron microscopy (HREM) : High resolution transmission electron microscopy (HREM) Bi2Sr1.2La0.8CuO6.46-xF2x Hadermann et al., JSSC, 156, 2 (2001) 455-451 Relation structure – image in HRTEM : image influenced by contrast transfer function of the microsope no direct correspondence visual comparison with simulations needed Pb Fe O Defocus Thickness -700 Å -500 Å -300 Å -100 Å 10 Å 30 Å 50 Å 70 Å 90 Å Relation structure – image in HRTEM simple perovskite Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Exercise 3: predict what the ED pattern will roughly look like : 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” Exercise 3: predict what the ED pattern will roughly look like : 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” Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Slide 88: Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Domains with varying q : Domains with varying q Sr1.3Co0.8Mn0.2O3 Mandal et al., Chem. Mater., 19, 25 (2007) 6158 Comments to the applicability of conclusions from TEM : Comments to the applicability of 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 compositional information also. Purpose of this lecture : 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 conclusions claimed from TEM by different sources (collaborators, papers,…) by knowing some possible pitfalls Decide whether it would be useful to do TEM on your own IMS or CS Make basic interpretations of TEM data on your own materials by yourself You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.
TEM for incommensurately modulated materials johader Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 94 Category: Science & Tech.. License: Some Rights Reserved Like it (3) Dislike it (0) Added: October 04, 2010 This Presentation is Public Favorites: 0 Presentation 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. Comments Posting comment... Premium member Presentation Transcript Slide 1: Examples of incommensurately modulated structures and composites studied using transmission electron microscopy Purpose of this lecture : 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 conclusions claimed from TEM by different sources (collaborators, papers,…) by knowing some possible pitfalls Decide whether it would be useful to do TEM on your own IMS or CS Make basic interpretations of TEM data on your own materials by yourself Outline of the lecture : Outline of the lecture The three most frequently used techniques in case of IMS/CS: Electron diffraction HAADF-STEM HRTEM ED: the main differences with XRD : ED: the main differences with XRD Shorter wavelength, almost flat sections through 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 Incommensurately modulated structure example : Incommensurately modulated structure example Sb1.155Pb1.216Mn0.628O4 Related commensurate structure Sb2MnO4 Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 IMS: indexing the subcell reflections : 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 4.35 3.0 3.0 6.15 All distances in Å 6.15 Dangers of double diffraction! : 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 - 021 Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 Subcell vs. IMS : Subcell vs. IMS Sb1.155Pb1.216Mn0.628O4 Sb2MnO4 Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 Indexing the IMS : Indexing the IMS Sb1.155Pb1.216Mn0.628O4 c c q≈3/4c* 020 022 020 022 002 000 002 - - Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 Reflection conditions for m : 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. 0210 0200 0201 0202 0211 - 0211 0221 0222 - - 0220 1100 1101 1102 1122 - 1120 - 1121 q≈3/4c* Abakumov et al., Chemistry of Materials, 17, 5, 2005, 1123-1134 Reflection conditions : Reflection conditions Sb1.155Pb1.216Mn0.628O4 0klm: k=2n hhlm: l=2n P42/mbc(00γ) Exercise 1-1: IMS: index the non-modulated phase : Index the commensurate structure LaSrCuO4, this will help you with indexing the next, incommensurate one. (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 Slide 13: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 001 002 200 Slide 14: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 001 002 200 Slide 15: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 010 020 110 Slide 16: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 010 020 110 Slide 17: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 010 020 110 Slide 18: (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 Slide 19: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 1.85 Å 100 200 020 020 Slide 20: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 200 110 220 ½ ½ 0 020 Slide 21: (Simulated ED patterns) Given data: cell parameters of LaSrCuO4: a=b=3.7 Å, c=13 Å. 200 110 220 ½ ½ 0 020 Indexed solution : Indexed solution 200 020 Zone indices : Zone indices 200 020 [010] [110] - [001] Exercise 1-2: IMS: determine the reflection condition h0l : Exercise 1-2: IMS: determine the reflection condition h0l [010] [110] - [001] 200 h0l: h+l=2n h = 2n l = 2n Exercise 1-2: IMS: determine the reflection condition h0l : [010] [110] - [001] 200 h0l: h+l=2n Exercise 1-2: IMS: determine the reflection condition h0l Exercise 1-2: IMS: determine the reflection condition hhl : Exercise 1-2: IMS: determine the reflection condition hhl [010] [110] - [001] 200 hhl: h+l=2n h = 2n l = 2n h0l: h+l=2n Exercise 1-2: IMS: determine the reflection condition hhl : Exercise 1-2: IMS: determine the reflection condition hhl [010] [110] - [001] 200 hhl: l = 2n h0l: h+l=2n Exercise 1-2: IMS: determine the reflection condition hk0 : Exercise 1-2: IMS: determine the reflection condition hk0 [010] [110] - [001] 200 hhl: l = 2n h0l: h+l=2n hk0: h+k=2n h = 2n k = 2n Exercise 1-2: IMS: determine the reflection condition hk0 : Exercise 1-2: IMS: determine the reflection condition hk0 [010] [110] - [001] 200 hhl: l = 2n h0l: h+l=2n hk0: h+k=2n Solved reflection conditions : Solved reflection conditions [010] [110] - [001] h0l:h+l=2n hhl:l=2n hk0:h+k=2n Also (from rest of the zones) hkl: h+k+l=2n. 200 Determine space group : Determine space group International Tables: I--- Most symmetrical I4/mmm Incommensurate vs. basic cell : Incommensurate vs. basic cell [010] [110] - [001] [010] [110] - [001] Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Exercise 1-3: index the IMS : 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 Slide 34: q q q 200 002 [010] Propose a modulation vector. Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 35: 200 002 [010] q=αa* α<0.5 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 36: 200 002 [010] q=αa* α<0.5 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 37: 200 002 [010] q=αa* α<0.5 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Possible solution : 200 002 [010] Possible solution q=αa* α<0.5 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the satellite indicated in green : Index the satellite indicated in green - q 001 0001 100 002 [010] 200 q Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the satellite indicated in green : 002 [010] 200 Index the satellite indicated in green q 0001 0001 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the next indicated satellite : Index the next indicated satellite q 0002 2002 - 2001 - 0001 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the next indicated satellite : q Index the next indicated satellite 2002 - - 0001 2002 - Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Exercise 1-4: IMS with more complete data : Exercise 1-4: IMS with more complete data Which of the indicated vectors corresponds to the modulation vector chosen on the previous slides? 020 200 110 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Exercise 1-4: IMS with more complete data : Exercise 1-4: IMS with more complete data Which of the indicated vectors corresponds to the modulation vector chosen on the previous slides? 020 200 110 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 45: Is the proposed vector still valid? yes no 020 200 110 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Slide 46: Indicate the correct modulation vector(s) on the central white area. 020 200 110 You need q1 q2 q1 ànd q2 q1=αa*+βb* q2= -αa*+βb* α=β<0.25 hklmn Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the reflection indicated in red : 020 200 110 00010 00001 20011 20011 - 20011 - Index the reflection indicated in red Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the reflection indicated in red : 020 200 110 00010 00001 20011 - 20011 - Index the reflection indicated in red Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the others patterns consistently with this new choice. : Index the others patterns consistently with this new choice. 002 [010] 200 002 [010] 200 [110] [001] 020 200 110 002 - 00010 00001 - 20011 Hadermann et al., Journal of Materials Chemistry, 17, 22, 2007, 2344-2350 LaSrCuO3.52 Index the reflection indicated in green : Index the reflection indicated in green 20011 20011 20011 - - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - - 020 200 00010 00001 - 20011 Index the reflection indicated in green : Index the reflection indicated in green LaSrCuO3.52 20011 - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 20011 - 20011 - 020 200 00010 00001 - 20011 Index the reflection indicated in green : 002 [010] 200 002 [010] 200 [110] [001] 110 002 - Index the reflection indicated in green LaSrCuO3.52 10110 10111 10111 - 020 200 00010 00001 - 20011 Index the reflection indicated in green : Index the reflection indicated in green LaSrCuO3.52 10111 - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 Index the reflection indicated in green : Index the reflection indicated in green LaSrCuO3.52 00002 00002 00001 - - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - Index the reflection indicated in green : Index the reflection indicated in green LaSrCuO3.52 00002 - 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - Look out for twinning! : Look out for twinning! Y0.8Sr2.2Mn2GaO8- Gillie et al., Journal of Physics and Chemistry of Solids, 65, 1 (2004) 87-93 Exercise 1-5: twinning? : Exercise 1-5: twinning? Could it be twinning and thus sufficient to use only one q-vector? yes no 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-5: twinning? : Exercise 1-5: twinning? Could it be twinning and thus sufficient to use only one q-vector? yes no 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : 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 LaSrCuO3.52 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : Exercise 1-6: derive the reflection conditions hklmn: h+k+l=2i LaSrCuO3.52 hhl:l=2n hkl:h+k+l=2n 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : 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 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : Exercise 1-6: derive the reflection conditions LaSrCuO3.52 hhl:l=2n hkl:h+k+l=2n hhlm0: l,m=2i but hhlm0:m=2i is sufficient 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Exercise 1-6: derive the reflection conditions : 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 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Determining the superspace group : Determining the superspace group Online tables Yamamoto (Acta Cryst.A 1996, A52, 509) I4/mmm(α α0, -αα0)00mg LaSrCuO3.52 002 [010] 200 002 [010] 200 [110] [001] 110 002 - 10111 - 10111 - 020 200 00010 00001 - 20011 20011 - 20011 - 00002 - Use of doing ED : Use of doing ED Deriving approximate cell parameters, modulation vector and superspace group Not useful for: precise parameters, precise modulation vector Excellent if multiphase sample or satellites very weak in XRD/ND Be aware of: Twinning Multiple diffraction (for forbidden refls., for using intensities,…) The possibility of missing important sections of reciprocal space Relation composition – modulation vector : (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 a* a* a* b* b* b* Ba4In6-xMgxO13-x/2 x = 0 x = 0.2 x = 0.4 ED of a composite structure (CS) : ED of a composite structure (CS) a1= a2, b1=b2, с1 с2, q = gc1* = c2* hk00 average cell hkl0 first subsystem hk0m second subsystem hklm 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 ED of a composite structure (CS) : ED of a composite structure (CS) Sr1.3Co0.8Mn0.2O3 2022 - Mandal et al., Chem. Mater., 19, 25 (2007) 6158 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern [SrF0.82(OH)0.18]2.5[Mn6O12] 0001 0010 Which reflection is 0011? - Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 Which reflection is 0011? - [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 Which reflection is 0012? - [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 Which reflection is 0012? - [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 Which reflection does the blue arrow point at? 0022 0021 0012 - - - [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Exercise 2: composite structure: index the pattern : Exercise 2: composite structure: index the pattern 0001 0010 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 Exercise 2: composite structure: index the pattern - solution : Exercise 2: composite structure: index the pattern - solution [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 Streaks : Streaks order in the plane, disorder between these planes: streak on ED pattern disorder in two directions: diffuse intensity plane: streak or background intensity depending on the intersection [SrF0.82(OH)0.18]2.5[Mn6O12] Abakumov et al., Chem.Mater. 19, 5 (2007) 1181-1189 HRTEM and HAADFSTEM-images : HRTEM and HAADFSTEM-images contrast related to Zn direct information about the heavy atom positions brighter with thickness also correct orientation very important! scanned image: positions imprecise HRTEM complicated phase transfer function no direct interpretation possible contrast changes with thickness and defocus correct orientation very important! one-shot image: position of the contrast objects reliable HAADF-STEM Slide 78: Heavy atoms from HAADF-STEM “Pb2Fe2O5” Abakumov et al., Ang.Chemie-Int.Ed., 45, 40 (2006) 6697-6700 High resolution transmission electron microscopy (HREM) : High resolution transmission electron microscopy (HREM) Bi2Sr1.2La0.8CuO6.46-xF2x Hadermann et al., JSSC, 156, 2 (2001) 455-451 Relation structure – image in HRTEM : image influenced by contrast transfer function of the microsope no direct correspondence visual comparison with simulations needed Pb Fe O Defocus Thickness -700 Å -500 Å -300 Å -100 Å 10 Å 30 Å 50 Å 70 Å 90 Å Relation structure – image in HRTEM simple perovskite Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Exercise 3: predict what the ED pattern will roughly look like : 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” Exercise 3: predict what the ED pattern will roughly look like : 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” Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Exercise 3: predict what the ED pattern will roughly look like : Exercise 3: predict what the ED pattern will roughly look like Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Slide 88: Hadermann et al., Solid State Sciences, 10; 4 (2008) 382-389 “Pb2Fe2O5” Domains with varying q : Domains with varying q Sr1.3Co0.8Mn0.2O3 Mandal et al., Chem. Mater., 19, 25 (2007) 6158 Comments to the applicability of conclusions from TEM : Comments to the applicability of 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 compositional information also. Purpose of this lecture : 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 conclusions claimed from TEM by different sources (collaborators, papers,…) by knowing some possible pitfalls Decide whether it would be useful to do TEM on your own IMS or CS Make basic interpretations of TEM data on your own materials by yourself