crystal structure using x – ray diffraction

1. Crystal Structure using X – ray Diffraction X-rays of the order of  1Ǻ wavelength are used to probe the structural information in solid Interatomic distances  a few Ǻ ( m,n,p) are coordinates of a point [100] represents the direction of the vector from origin to (1,0,0) Miller Indices are h,k, l if 1/h, 1/k, 1/ l are the intercepts along X,Y,Z axes Cubic : {100} = (100), (010),(001), (100), (010),(001) Tetragonal :{ 100} = (100), (010), (100), (010) (100) (001) 0 (200) X Y Z (1,0,0) Cubic unit cell

2. Bravais Lattices Triclinic Monoclinic Orthorhombic Rhombohedral ( Trigonal ) Tetragonal, Hexagonal Cubic a=b=c, α = β = γ =90 0 R H T

3. Unit cell and Crystal Classes

4. Crystal Structures of NaCl and CsCl NaCl :- lattice :FCC; 4 atoms/unit cell Basis : Na (0,0,0); Cl (1/2,1/2,1/2) FCC Coordinates: (0,0,0); (1/2,1/2,0); (1/2,0,1/2); (0,1/2,1/2) CsCl : lattice: Cubic (SC) Basis : Cs(0,0,0); Cl(1/2,1/2,1/2) SC coordinates : (0,0,0) Cubic structure : 3 variants Simple Cubic (SC) Body centered Cubic (BCC) FaceCenteredCubic (FCC)

5. Crystal Classes : 7 Bravais lattices : 14 Examples of some materials and their crystal structures

6. To produce X- rays of  1 Ǻ photons of energy :  12.4 keV need be generated because E = h  = h c /   (in m) = h c / E(Joules);  = 12398  0 / E(eV) 1  0 = 10 -8 cm= 10 -10 m = 10 -4  m (microns) Energy of n th level E n = h Z 2 R/ n 2 - in an element of At. No. Z Energy released h  ( m n ) = h Z 2 R ( ( 1/ n 2 ) – ( 1 / m 2 ) )    Z 2    1 / Z 2 Common targets used to produce X rays are Cr, Fe, Cu, and Mo Higher the Z value lower is the  of K  radiation. Production of X - rays

7. Emission Spectrum of X- rays from a Molybdenum target As Z increases,  decreases. Cr K  = 2.2909 A 0 Fe K  = 1.93597 A 0 Cu K  = 1.5418 A 0 Mo K  = 0.70926 A 0 K,L,M,N –levels L K : K  Fast e-beam knocks out inner core electrons giving rise to transitions between Inner levels. K  1 K  2 M K : K  Neutrons X-rays Also splits into two : but are too close & not resolved Characteristic X-rays & Continuum Absorption curve of Z-1

8. Monochromatic X- rays using Filters Energy levels in an atom K  L  L K  K  1 , K  2 K  L  M K  K  1 , K  2 K  L N M K If X-ray target element is of Atomic No. Z the absorption edge of the (Z-1) element overlaps the K  peak of the element Z. Hence Cu target + Ni filter gives monochromatic CuK  radiation Target Filter Cr Fe Cu Mo V Mn Ni Nb

9. Bragg’s Law Incident ray 2  2 d h k l sin  h k l = n  d h k l --- normal distance between a set of parallel planes with (hkl) as Miller Indices.  h k l --- Bragg angle for (hkl) planes n --- order of diffraction  --- Wavelength of incident radition (X-rays, here). 1.Bragg’s law selects the Bragg angle for a given set of d hkl planes 2. Scattering amplitude and hence intensity of Bragg peak is decided by the structure factor d hkl

10. Debye – Scherrer method Diffraction cones cut the Sphere of reflections (Ewald sphere) in circles. These circles cut the film in arcs.So, a pair of arcs represents one diffraction cone corresponding to one set of d hkl planes .

11. XRD pattern- Debye-Scherrer film Various diffraction cones cut the film in sets of arcs Each pair of arcs represents diffraction from one set of (hkl) planes Number of arcs between the 2 holes define the number of observed Bragg diffraction lines

12. Reciprocal Lattice (RL) picture 2 d h k l sin  h k l = n  sin  = (1 / d ) / ( 2/  ) for n = 1 For a given  , sin  goes as (1 / d ) Direct lattice vector r = m a +n b+ p c Reciprocal lattice vector G = h a * + k b * + l c * d hkl = 2  /  G hkl  A point in RL represents a set of planes in the direct lattice . Vector length  G hkl  = 2  / d hkl

13. Reciprocal Vectors

14. Orthorhombic direct and reciprocal cells Brillouin Zone

15. Ewald Sphere K K’  K = G Diffraction cone angle is 4  2 K.G = G 2 …. Bragg’s law  K  = 2  / 

16. Reflections for two different wavelengths in RL space Spheres of reflection : diameter = 2/  2 d h k l sin  h k l = n  1. (hkl) plane with smallest indices has largest d h k l & smallest  2. Maximum  or least possible d h k l observable is limited by  /2 which is Mo K  : 0.354 A 0 Cu K  : 0.771 A 0 Fe K  : 0.9679 A 0

18. Relation between d - spacings and lattice parameters

19. Analytical methods Camera diameter is chosen to be : 2r = 180/  = 57.3 mm. B A Measure S, Calculate 2  , then Sin  for each arc. Calculate d-spacings . Index the planes – i.e. identify hkl for each d-spacing Cubic : (1/d) 2 = ( h 2 + k 2 + l 2 ) / a 2 gives value of ‘a’ B – A = 180 0

21. Indexing by Graphical method – Bunn’s charts Plots of 2 log d versus c/a : for hexagonal and tetragonal structures

22. JCPDS cards - ASTM data

23. Laue method for single crystals – uses white X-radiation