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crystal structure and x ray diffraction


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This is an educational ppt of Physics 2 for engineering first year students. The presentation is according to APJKTU, Lucknow syllabus

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crystal structure and x ray diffraction

  1. 1. Crystal Structure and X ray Diffraction Unit I Dr Md Kaleem Department of Applied Sciences Jahangirabad Institute of Technology (JIT), Jahangirabad, Barabanki(UP) - 225203 1/31/2017 1DR MD KALEEM/ ASSISTANT PROFESSOR
  2. 2. • Relationship between structures of engineering materials • To understand the classification of crystals • To understand mathematical description of ideal crystal • To understand Miller indices for directions and planes in lattices and crystals • To understand how to use X-Ray Diffraction for determination of crystal geometry 1/31/2017 2DR MD KALEEM/ ASSISTANT PROFESSOR
  4. 4. Solid can be divided in two categories on the basis of periodicity of constituent atoms or group of atoms • Crystalline solids consists of atoms, ions or molecules arranged in ordered repetitive array e.g: Common inorganic materials are crystalline – Metals : Cu, Zn, Fe, Cu-Zn alloys – Semiconductors: Si, Ge, GaAs – Ceramics: Alumina (Al2O3), Zirconia (Zr2O3), SiC, SrTiO3. • Non crystalline or Amorphous consists of atoms, ions or molecules arranged in random order e.g: organic things like glass, wood, paper, bone, sand; concrete walls, etc Crystalline Solids  grains  crystals 1/31/2017 4DR MD KALEEM/ ASSISTANT PROFESSOR
  5. 5. 1/31/2017 5DR MD KALEEM/ ASSISTANT PROFESSOR Crystal = Lattice + Motif Lattice : regular repeated three-dimensional arrangement of points Motif/ Basis: an entity (typically an atom or a group of atoms) associated with each lattice point
  6. 6. 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 6 Lattice  where to repeat Motif  what to repeat Lattice: Translationally periodic arrangement of points Crystal: Translationally periodic arrangement of motifs
  7. 7. Space lattice: An array of points in space such that every point has identical surroundings Unit Cell: It is basic structural unit of crystal, with an atomic arrangement which when repeated three dimensionally gives the total structure of the crystal Lattice Parameters: It defines shape and size of the unit cell Three lattice vector (a, b, c) and interfacial angle (, , ) are known as lattice parameters 1/31/2017 7DR MD KALEEM/ ASSISTANT PROFESSOR
  8. 8. Unit cell with lattice points at the corners only, called primitive cell. Unit cell may be primitive cell but all primitive cells are not essentially unit cells. 1/31/2017 8DR MD KALEEM/ ASSISTANT PROFESSOR
  9. 9. • Crystallographers classified the unit cells into seven possible distinct types of unit cells by assigning specific values to lattice vector (a, b, c) and interfacial angle (, , ) called seven crystal system. 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 9
  10. 10. Crystal System Lattice Vector Interfacial Angle Example 1 Cubic a = b = c  =  = = 90o NaCl, CaF2, Au, Ag, Cu, Fe 3 Tetragonal a = b ≠ c  =  = = 90o TiO2, NiSO4, SnO2 3 Orthorhombic a ≠ b ≠ c  =  = = 90o KNO3, BaSO4, PbCO3, Ga 4 Monoclinic a ≠ b ≠ c  =  = 90o≠  CaSO4.2H2O (Gypsum), FeSO4 5 Triclinic a ≠ b ≠ c  ≠  ≠ ≠ 90o CuSO4, K2Cr2O7 6 Trigonal a = b = c  =  = ≠ 90o As, Sb, Bi, Calcite 7 Hexagonal a = b ≠ c  =  = 90o, =120o SiO2, AgI, Ni, As, Zn, Mg 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 10
  11. 11. 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 11
  12. 12. • A. J. Bravais in 1948 shown that with the centering (face, base and body centering) added to these, 14 kinds of 3D lattices, known as Bravais lattices. 1/31/2017 12DR MD KALEEM/ ASSISTANT PROFESSOR
  13. 13. Coordination Number: It is defined as the number of nearest neighbors around any lattice point in the crystal lattice. 1/31/2017 13DR MD KALEEM/ ASSISTANT PROFESSOR
  14. 14. •Miller indices for crystallographic planes •Miller notation system (hkl) •Miller index – the reciprocals of the fractional intercepts that the plane makes with the x, y, and z axes of the three nonparallel edges of the cubic unit cell William Hallowes Miller 1/31/2017 14DR MD KALEEM/ ASSISTANT PROFESSOR
  15. 15. • Choose a plane not pass through (0, 0, 0) • Determine the intercepts of the plane with x, y, and z axes • Form the reciprocals of these intercepts • Find the smallest set of whole numbers that are in the same ratio as the intercepts 1/31/2017 15DR MD KALEEM/ ASSISTANT PROFESSOR
  16. 16. • Find the Miller Indices of the plane which cuts off intercepts in the ratio 1 a:3b:-2c along the three co-ordinate axes, where a, b and c are the primitives. • If pa, qb and rc are the intercepts of the given set of planes on X-, Y-, and Z- axes respectively then, pa: qb: rc= 1 a:3b:-2c or p:q:r=1:3:-2 so 1/p : 1/q : 1/r = 1/1 :1/3 : -1/2 LCM of 1, 3 and 2 = 6, so multiply by it 1/p : 1/q : 1/r = 6:2:-3 Thus the Miller Indices of the plane is (6 2 ) 1/31/2017 16DR MD KALEEM/ ASSISTANT PROFESSOR 3
  21. 21. • It is infinite periodic three dimensional array of reciprocal lattice points whose spacing varies inversely as the distances between the planes in the direct lattice of the crystal. 1/31/2017 21DR MD KALEEM/ ASSISTANT PROFESSOR
  22. 22. Take some point as an origin From this origin, lay out the normal to every family of parallel planes in the direct lattice; Set the length of each normal equal to 2p times the reciprocal of the interplanar spacing for its particular set of planes; Place a point at the end of each normal. 1/31/2017 22DR MD KALEEM/ ASSISTANT PROFESSOR
  23. 23. • Any diffraction pattern of a crystal is a map of the reciprocal lattice of the crystal whereas the microscopic image is a map of the direct lattice. • While the primitive vectors of a direct lattice have the dimensions of length those of the reciprocal lattice have the dimensions of (length)− 1. • Direct lattice or crystal lattice is a lattice in ordinary space or real space. Reciprocal lattice is in reciprocal space or k-space or Fourier space. • The direct lattice is the reciprocal of its own reciprocal lattice. • The reciprocal lattice of a simple cubic lattice is also a simple cubic lattice. • The reciprocal lattice of a face centered cubic lattice is a body centered cubic lattice. • The reciprocal lattice of a body centered cubic lattice is a face centered cubic lattice, and 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 23
  24. 24. NaCl has a cubic unit cell. It is best thought of as a face- centered cubic array of anions with an interpenetrating fcc cation lattice (or vice-versa) The cell looks the same whether we start with anions or cations on the corners. Each ion is 6-coordinate and has a local octahedral geometry. 1/31/2017 24DR MD KALEEM/ ASSISTANT PROFESSOR
  25. 25. • The Bravais space lattice of NaCl is truly fcc with a basis of one Na+ ion one Cl- ion separated by one half the body diagonal (a√3/2) of a unit cube. • There are four pair of Na+ and Cl- ions present per unit cell. • The position of ions in unit cell are • Na+ : (½, ½, ½), (0,0, ½), (0, ½,0), (½,0,0) • Cl- : (0,0,0), (½, ½,0), (½,0, ½), (0, ½, ½) 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 25
  26. 26.  For electromagnetic radiation to be diffracted the spacing in the grating should be of the same order as the wavelength  In crystals the typical inter-atomic spacing ~ 2-3 Å so the suitable radiation is X-rays  Hence, X-rays can be used for the study of crystal structures 1/31/2017 26DR MD KALEEM/ ASSISTANT PROFESSOR
  27. 27. The path difference between rays = 2d Sin  For constructive interference: n = 2d Sin 1/31/2017 27DR MD KALEEM/ ASSISTANT PROFESSOR
  28. 28. • Q. A beam of X-rays of wavelength 0.071 nm is diffracted by (110) plane of rock salt with lattice constant of 0.28 nm. Find the glancing angle for the second-order diffraction. • Given data are: • Wavelength (λ) of X-rays = 0.071 nm, Lattice constant (a) = 0.28 nm Plane (hkl) = (110), Order of diffraction = 2 Glancing angle θ = ? Bragg’s law is 2d sin θ = nλ 1/31/2017 28DR MD KALEEM/ ASSISTANT PROFESSOR
  29. 29. Substitute in Bragg’s equation 1/31/2017 29DR MD KALEEM/ ASSISTANT PROFESSOR
  30. 30. Bragg’s spectrometer method is one of the important method for studying crystals using X-rays. The apparatus consists of a X-ray tube from which a narrow beam of X- rays is allowed to fall on the crystal mounted on a rotating table. The rotating table is provided with scale and vernier, from which the angle of incidence, θ can be measured. 1/31/2017 30DR MD KALEEM/ ASSISTANT PROFESSOR
  31. 31. • Bragg’s spectrometer is used to determine the structure of crystal. • The ratio of lattice spacing for various groups of planes are obtained by using Bragg’s Law. • The ratio would be different for different crystals • By comparing those known standard ratios with experimentally determined ratios, crystal structure can be obtained. 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 31
  32. 32. • If for a particular crystal having interplaner spacing d1, d2, d3 strong Bragg’s reflection occur at glancing angle θ1, θ2, θ3 then from Bragg’s law • 2d1sin θ1=λ, 2d2sin θ2=λ, 2d3sin θ3=λ • So, d1: d2: d3 = 1/sin θ1= 1/sin θ2=1/sin θ3 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 32
  33. 33. • For KCl Crystal, Bragg’s obtained strong Bragg’s reflection at θ1= 5o23’, θ2=7o37’, θ3=9o25’’for planes (100), (110) and (111) • So, d100: d110: d111= 1/sin 5o23’= 1/sin 7o37’=1/sin 9o25’ = 1:1/√2:1/√3 • This corresponds to theoretical result for simple cubic lattice . Therefore it is concluded that KCl crystal has simple cubic structure. 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 33
  34. 34. • When light encounters charged particles, the particle interact with light and cause some of the light to be scattered. This is called Compton Scattering. 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 34
  35. 35. • Arthur H. Compton in 1923 observed that when electromagnetic wave of short wavelength (X ray) strikes an electron, an increase in wavelength of X-rays or gamma rays occurs when they are scattered. 1/31/2017 DR MD KALEEM/ ASSISTANT PROFESSOR 35   cos1 cm h e if