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CYL729: Materials Characterization DiffractionMicroscopy Thermal Analysis A. Ramanan Department of Chemistry

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Presentation on theme: "CYL729: Materials Characterization DiffractionMicroscopy Thermal Analysis A. Ramanan Department of Chemistry"— Presentation transcript:

1 CYL729: Materials Characterization DiffractionMicroscopy Thermal Analysis A. Ramanan Department of Chemistry

2 Reference Books George M. CrankovicGeorge M. Crankovic (Editor)

3 Electro-magnetic Spectrum

4 History of X-rays n 1885-1895 Wm. Crookes sought unsuccessfully the cause of repeated fogging of photographic plates stored near his cathode ray tubes. n X-rays discovered in 1895 by Roentgen, using ~40 keV electrons (1st Nobel Prize in Physics 1901) n 1909 Barkla and Sadler discovered characteristic X-rays, in studying fluorescence spectra (though Barkla incorrectly understood origin) (Barkla got 1917 Nobel Prize) n 1909 Kaye excited pure element spectra by electron bombardment

5 History of X-rays - cont’d n 1912 von Laue, Friedrich and Knipping observe X-ray diffraction (Nobel Prize to von Laue in 1914) n 1912-13 Beatty demonstrated that electrons directly produced two radiations: (a) independent radiation, Bremsstrahlung, and (b) characteristic radiation only when the electrons had high enough energy to ionize inner electron shells. n 1913 WH + WL Bragg build X-ray spectrometer, using NaCl to resolve Pt X-rays. Braggs’ Law. (Nobel Prize 1915) n = 2d sin 

6 History of X-rays - cont’d n 1913 Moseley constructed an x- ray spectrometer covering Zn to Ca (later to Al), using an x-ray tube with changeable targets, a potassium ferrocyanide crystal, slits and photographic plates n 1914, figure at right is the first electron probe analysis of a man- made alloy T. Mulvey Fig 1.5 (in Scott & Love, 1983). Note impurity lines in Co and Ni spectra

7 History of X-rays - cont’d n Moseley found that wavelength of characteristic X- rays varied systematically (inversely) with atomic number Z Using wavelengths, Moseley developed the concept of atomic number and how elements were arranged in the periodic table. The next year, he was killed in Turkey in WWI. “In view of what he might still have accomplished (he was only 27 when he died), his death might well have been the most costly single death of the war to mankind generally,” says Isaac Asimov (Biographical Encyclopedia of Science &Technology).

8 n 1923 Manne Siegbahn published The Spectroscopy of X-rays in which he shows that the Bragg equation must be revised to take refraction into account, and he lays out the “Siegbahn notation” for X-rays n 1931 Johann developed bent crystal spectrometer (higher efficiency) Historical Summary of X-rays n 1859 Kirchhoff and Bunsen showed patterns of lines given off by incandescent solid or liquid are characteristic of that substance n 1904 Barkla showed each element could emit ≥1 characteristic groups (K,L,M) of X-rays when a specimen was bombarded with beam of x-rays n 1909 Kaye showed same happened with bombardment of cathode rays (electrons) n 1913 Moseley found systematic variation of wavelength of characteristic X-rays of different elements n 1922 Mineral analysis using X-ray spectra (Hadding) n 1923 Hf discovered by von Hevesy (gap in Moseley plot at Z=72). Proposed XRF (secondary X-ray fluorescence)

9 Summary of X-ray Properties n X-rays are considered both particles and waves, i.e., consisting of small packets of electromagnetic waves, or photons. n X-rays produced by accelerating HV electrons in a vacuum and colliding them with a target. n The resulting spectrum contains (1) continuous background (Bremsstrahlung;“white X-rays”), (2) occurrence of sharp lines (characteristic X-rays), and (3) a cutoff of continuum at a short wavelength. n X-rays have no mass, no charge (vs. electrons)

10 X-ray Crystallography DIFFRACTION

11 What is a Unit Cell?

12 Unit cell can be chosen in different ways!

13 Unit Cells? White and black birds by the artist, M. C. Escher.

14 A unit cell chosen such that it contains minimum volume but exhibit maximum symmetry

15 {R = n 1 a 1 + n 2 a 2 + n 3 a 3 } Translational vector

16 Crystal Structure Ideal Crystal: Contain periodical array of atoms/ions Represented by a simple lattice of points A group of atoms attached to each lattice points Basis LATTICE = An infinite array of points in space, in which each point has identical surroundings to all others. CRYSTAL STRUCTURE = The periodic arrangement of atoms in the crystal. It can be described by associating with each lattice point a group of atoms called the MOTIF (BASIS)

17 Lattice parameters: a, b, c;  7 Crystal Systems

18 Bravais Lattice: an infinite array of discrete points with an arrangement and orientation that appears exactly the same from whichever of the points the array is viewed. Crystal System Bravais Lattice Essential Symmetry Conditions Cubic P, F, I 4 C 3 a=b=c  =  =  90 0  Tetragonal P, I 1 C 4 along [c-axis] a=b=c  =  =  90 0 HexagonalP 1 C 6 along [c-axis] a=b=c  =  =  90 0 RhombohedralR 1 C 3 along body diagonal a=b=c  =  =   90 0 Orthorhombic P, F, I, C 3 C 2 mutually perpedicular along the three axes a  b  c  =  =  90 0 Monoclinic P, C 1 C 2 along [b-axis] a  b  c  =  =90 0 &   90 0 TriclinicP C 2 or inversion centre a  b  c       90 0

19 14 Bravais lattices

20 Unit cell symmetries - cubic n 3 C 4 - passes through pairs of opposite face centers, parallel to cell axes n 4 C 3 - passes through cubic diagonals A cube need not have C 4 !!

21 Copper metal is face-centered cubic Identical atoms at corners and at face centers Lattice type F also Ag, Au, Al, Ni...  -Iron is body-centered cubic Identical atoms at corners and body center (nothing at face centers) Lattice type I Also Nb, Ta, Ba, Mo...

22 periodic table Hexagonal closed packed (hcp) face-centered cubic (fcc) body-centered cubic (bcc)

23 Caesium Chloride (CsCl) is primitive cubic Different atoms at corners and body center. NOT body centered, therefore. Lattice type P Also CuZn, CsBr, LiAg Sodium Chloride (NaCl) - Na is much smaller than Cs Face Centered Cubic Rocksalt structure Lattice type F Also NaF, KBr, MgO….

24 Diamond Structure: two sets of FCC Lattices Z = 8 C atoms per unit cell

25 one C 4 Why not F tetragonal? Tetragonal: P, I Yellow and green colors represents same atoms but different depths.

26 Example CaC 2 - has a rocksalt-like structure but with non-spherical carbides 2- C C Carbide ions are aligned parallel to c  c > a,b  tetragonal symmetry

27 Orthorhombic: P, I, F, C C F

28 Side centering Side centered unit cell Notation: A-centered if atom in bc plane B-centered if atom in ac plane C-centered if atom in ab plane

29 Trigonal: P : 3-fold rotation

30 Hexagonal Monoclinic Triclinic

31 Unit cell contents Counting the number of atoms within the unit cell Many atoms are shared between unit cells

32 Atoms Shared Between:Each atom counts: corner8 cells1/8 face center2 cells1/2 body center1 cell1 edge center4 cells1/4 lattice typecell contents P1 [=8 x 1/8] I2 [=(8 x 1/8) + (1 x 1)] F4 [=(8 x 1/8) + (6 x 1/2)] C 2 [=(8 x 1/8) + (2 x 1/2)]

33 e.g. NaCl Na at corners: (8  1/8) = 1 Na at face centres (6  1/2) = 3 Cl at edge centres (12  1/4) = 3 Cl at body centre = 1 Unit cell contents are 4(Na + Cl - )

34 (0,0,0) (0, ½, ½) (½, ½, 0) (½, 0, ½) Fractional Coordinates

35 Cs (0,0,0) Cl (½, ½, ½)

36 Density Calculation n: number of atoms/unit cell A: atomic mass V C : volume of the unit cell N A : Avogadro’s number (6.023x10 23 atoms/mole) Calculate the density of copper. R Cu =0.128nm, Crystal structure: FCC, A Cu = 63.5 g/mole n = 4 atoms/cell, 8.94 g/cm 3 in the literature


38 Miller Indices describe which plane of atom is interacting with the x-rays

39 How to Identify Miller indices (hkl)? direction: [hkl] family of directions: planes: (hkl) family of planes: {hkl} [001] [010] [001] a b c to identify planes: Step 1 : Identify the intercepts on the x-, y- and z- axes. Step 2 : Specify the intercepts in fractional coordinates Step 3 : Take the reciprocals of the fractional intercepts

40 Miller indices (hkl) e.g.: cubic system: to identify planes: Step 1 : Identify the intercepts on the x-, y- and z- axes (a/2, ∞, ∞) Step 2 : Specify the intercepts in fractional co-ordinates (a/2a, ∞, ∞) = (1/2,0,0) Step 3 : Take the reciprocals of the fractional intercepts (2, 0, 0) (210) (110) (111) (100)

41 Miller Indices


43 Crystallographic Directions And Planes Lattice Directions [uvw] Individual directions: [uvw] Symmetry-related directions: Miller Indices: 1. Find the intercepts on the axes in terms of the lattice constant a, b, c 2. Take the reciprocals of these numbers, reduce to the three integers having the same ratio(hkl) {hkl} Set of symmetry-related planes: {hkl}

44 (100)(111) (200) (110)


46 In cubic system, [hkl] direction perpendicular to (hkl) plane

47 Wilhelm Conrad Röntgen Wilhelm Conrad Röntgen discovered 1895 the X-rays. 1901 he was honoured by the Noble prize for physics. In 1995 the German Post edited a stamp, dedicated to W.C. Röntgen.

48 The Principles of an X-ray Tube Anode focus Fast electrons Cathode X-Ray

49 The Principle of Generation of X-ray X-ray Fast incident electron nucleus Atom of the anodematerial electrons Ejected electron (slowed down and changed direction)

50 The Principle of Generation the Characteristic Radiation KK LL KK K L M Emission Photoelectron Electron

51 The Generating of X-rays Bohr`s model

52 The Generating of X-rays M K L K  K  K  K  energy levels (schematic) of the electrons Intensity ratios K  K  K 

53 The Generating of X-rays Anode Mo Cu Co Fe (kV) 20.0 9.0 7.7 7.1 K  1 : 0,70926 K  2 :0,71354 K  1 :0,63225 Filter K  1 : 1,5405 K  2 :1,54434 K  1 :1,39217 K  1 : 1,78890 K  2 :1,79279 K  1 :1,62073 K  1 : 1,93597 K  2 :1,93991 K  1 :1,75654 Zr 0,08mm Mn 0,011mm Fe 0,012mm Ni 0,015mm Wavelength (Angström)

54 The Generating of X-rays Emission Spectrum of a Molybdenum X-Ray Tube Bremsstrahlung = continuous spectra characteristic radiation = line spectra

55 Interaction between X-ray and Matter d wavelength Pr intensity Io incoherent scattering Co (Compton-Scattering) coherent scattering Pr (Bragg´s-scattering) absorbtion Beer´s law I = I 0 *e-µd fluorescense > Pr photoelectrons

56 C. Gordon Darwin C. Gordon Darwin, grandson of C. Robert Darwin developed 1912 dynamic theory of scattering of X-rays at crystal lattice

57 P. P. Ewald P. P. Ewald 1916 published a simple and more elegant theory of X-ray diffraction by introducing the reciprocal lattice concept. Compare Bragg’s law (left), modified Bragg’s law (middle) and Ewald’s law (right).

58 Bragg’s Description The incident beam will be scattered at all scattering centres, which lay on lattice planes. The beam scattered at different lattice planes must be scattered coherent, to give an maximum in intensity. The angle between incident beam and the lattice planes is called . The angle between incident and scattered beam is 2 . The angle 2  of maximum intensity is called the Bragg angle. W.H. Bragg (father) and William Lawrence.Bragg (son) developed a simple relation for scattering angles, now call Bragg’s law.

59 Bragg’s Law A powder sample results in cones with high intensity of scattered beam. Above conditions result in the Bragg equation or

60 35KeV ~ 0.1-1.4A Cu K 1.54 A Mo: X-Ray Diffraction

61 Structure Determination

62 Light Interference fringes Constructive Destructive Diffraction

63 Diffraction Conditions



66 For cubic system Lattice spacing


68 Bragg’s Law For cubic system: But not all planes have the diffraction !!!

69 (200) (211) Powder diffraction X-Ray

70 Powder X-ray Diffraction Tube Powder Film

71 The Elementary Cell a b c a = b = c == = 90 o

72 Relationship between d-value and the Lattice Constants Bragg´s law The wavelength is known Theta is the half value of the peak position d will be calculated Equation for the determination of the d-value of a tetragonal elementary cell h,k and l are the Miller indices of the peaks a and c are lattice parameter of the elementary cell if a and c are known it is possible to calculate the peak position if the peak position is known it is possible to calculate the lattice parameter

73 D8 ADVANCE Bragg-Brentano Diffractometer A scintillation counter may be used as detector instead of film to yield exact intensity data. Using automated goniometers step by step scattered intensity may be measured and stored digitally. The digitised intensity may be very detailed discussed by programs. More powerful methods may be used to determine lots of information about the specimen.

74 The Bragg-Brentano Geometry Tube measurement circle focusing- circle   2 Detector Sample

75 The Bragg-Brentano Geometry Divergence slit Detector- slit Tube Antiscatter- slit Sample Mono- chromator

76 Powder Diffraction Pattern

77 What is a Powder Diffraction Pattern? A powder diffractogram is the result of a convolution of a) the diffraction capability of the sample (F hkl ) and b) a complex system function. The observed intensity y oi at the data point i is the result of y oi =  of intensity of "neighbouring" Bragg peaks + background The calculated intensity y ci at the data point i is the result of y ci = structure model + sample model + diffractometer model + background model

78 Which Information does a Powder Pattern offer? peak position - dimension of the elementary cell peak intensity - content of the elementary cell peak broadening - strain/crystallite size/nanostr.

79 Powder Pattern and Structure The d-spacings of lattice planes depend on the size of the elementary cell and determine the position of the peaks. The intensity of each peak is caused by the crystallographic structure, the position of the atoms within the elementary cell and their thermal vibration. The line width and shape of the peaks may be derived from conditions of measuring and properties - like particle size - of the sample material.

80 (110) (200) (211) Powder diffraction X-Ray

81 Example: layered silicates mica 2*theta d 7.212.1 14.46.1 224.0 growth oriented along c-axis (hkl) (001) (002) (003) C~12.2 A (00l)

82 What we will see in XRD of simple cubic, BCC, FCC?


84 Observable diffraction peaks Ratio Simple cubic SC: 1,2,3,4,5,6,8,9,10,11,12.. BCC: 2,4,6,8,10, 12…. FCC: 3,4,8,11,12,16,24….

85 Ex: An element, BCC or FCC, shows diffraction peaks at 2  : 40, 58, 73, 86.8,100.4 and 114.7. Determine:(a) Crystal structure?(b) Lattice constant? (c) What is the element? 2thetatheta(hkl) 40.0200.1171(110) 58.0290.2352(200) 86.843.40.47214(220) 100.450.20.59035(310) 114.757.350.70906(222) a =3.18Å, BCC,  W

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