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Diffraction : Intensity (From Chapter 4 of Textbook 2 and Chapter 9 of Textbook 1)

Diffraction : Intensity (From Chapter 4 of Textbook 2 and Chapter 9 of Textbook 1). Electron  atoms  group of atoms or structure  Crystal (poly or single). Scattering by an electron :. .   =  /2. . P. r.  0 : 410 -7 mkgC -2. by J.J . Thomson. .

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Diffraction : Intensity (From Chapter 4 of Textbook 2 and Chapter 9 of Textbook 1)

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  1. Diffraction: Intensity (From Chapter 4 of Textbook 2 and Chapter 9 of Textbook 1) Electron  atoms  group of atoms or structure  Crystal (poly or single)

  2. Scattering by an electron:  = /2  P r 0: 410-7 mkgC-2 by J.J. Thomson  a single electron charge e (C), mass m (kg), distance r (meters)

  3. z    P r 2 y O Random polarized x  = yOP = /2 y component  = zOP = /2 -2 z component Polarization factor

  4. Pass through a monochromatorfirst (Bragg angle M)  the polarization factor is ? polarization is not complete random anymore z z Random polarized P P P r r y 2 y 2M O O x x (Homework)

  5. M

  6. Atomic scattering (or form) factor a single free electron atoms

  7. x2 path different (Oand dV): R-(x1 + x2). x1 dV r O s s0 2 R Differential atomic scattering factor (df) : Ee: the magnitude of the wave from a bound electron Phase difference Electron density

  8. Spherical integration dV = dr(rd)(rsind) rsind r: 0 -   : 0 -   : 0 - 2 dr  rsin(+d)d d r d  http://pleasemakeanote.blogspot.tw/2010/02/9-derivation-of-continuity-equation-in.html

  9. = 2

  10. Evaluate (S - S0)r = | S - S0||r|cos |(S - S0)|/2 = sin. S S-S0 S0 2 Let

  11. For n electrons in an atom Tabulated For  = 0, only k = 0  sinkr/kr = 1. Number of electrons in the atom equal to 1 bound electrons

  12. Anomalous Scattering: Previous derivation: free electrons! Electrons around an atom: free? k free electron harmonic oscillator m Assume Resonance frequency Forced oscillator Assume Assume

  13. Same frequency as F(t), amplitude(, 0)  = 0  C is ; in reality friction term exist  no  Oscillator with damping (friction  v) assume c = m Assume x = x0eit Real part and imaginary part

  14.  if   0    E Resonance : X-ray frequency; 0: bounded electrons around atoms  0  electron escape  # of electrons around an atom   f  (f correction term) imaginary part correction: f   (linear absorption coefficient) f +f + if real imaginary

  15. Examples: Si, 400 diffraction peak, with Cu K (0.1542 nm) 0.3 0.4 8.22 7.20 Anomalous Scattering correction Atomic scattering factor in this case: 7.526-0.2+0.4i = 7.326+0.4i f and f: International Table for X-ray Crystallography V.III

  16. Structure factor atoms  unit cell How is the diffraction peaks (hkl) of a structure named? Unit cell How is an atom located in a unit cell affect the h00 diffraction peak? Miller indices (h00): 1 1 3 3 A  2 2 S R B path difference:11 and 22 (NCM) a plane (h00) C N M why:? Meaningful! path difference: 11 and 33 (SBR)

  17. phase difference (11 and 33) position of atom B: fractional coordinate of a: ux/a. the same argument  B: x,y,z x/a,y/b,z/cu,v,w Diffraction from (hkl) plane  F: amplitude of the resultant wave in terms of the amplitude of the wave scattered by a single electron.

  18. N atoms in a unit cell; fn: atomic form factor of atom n F(in general) a complex number. How to choose the groups of atoms to represent a unit cell of a structure? 1. number of atoms in the unit cell 2. choose the representative atoms for a cell properly (ranks of equipoints).

  19.  Example 1: Simple cubic 1 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipointsof rank 1; Choose any one will have the same result! for all hkl  Example 2: Body centered cubic 2 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipointsof rank 1; ½ ½ ½: equipoints of rank 1; Two points to choose: 000 and ½ ½ ½. when h+k+l is even when h+k+l is odd

  20.  Example 3: Face centered cubic 4 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of rank 1; ½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½: : equipoints of rank 3; Four atoms chosen: 000, ½ ½ 0, ½ 0 ½, 0 ½ ½. when h, k, l is unmixed (all evens or all odds) when h, k, l is mixed

  21.  Example 4: Diamond Cubic 8 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipointsof rank 1; ½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½: equipoints of rank 3; ¼ ¼ ¼, ¾ ¾ ¼, ¾ ¼ ¾, ¼ ¾ ¾: equipoints of rank 4; Eight atoms chosen: 000, ½ ½ 0, ½ 0 ½, 0 ½ ½ (the same as FCC), ¼ ¼ ¼, ¾ ¾ ¼, ¾ ¼ ¾, ¼ ¾ ¾! FCC structure factor

  22. when h, k, l are all odd when h, k, l are all even and h + k + l = 4n when h, k, l are all even and h + k + l 4n when h, k, l are mixed

  23.  Example 5: HCP 2 atoms/unit cell 8 corner atoms: equipointsof rank 1; 1/3 2/3 ½: equipoints of rank 1; Choose 000, 1/3 2/3 1/2. ( 1/3 2/3 1/2) (001) (000) (010) Set [h + 2k]/3+ l/2=g (100) (110) equipoints

  24. h + 2k l 3m 3m 3m1 3m1 even odd even odd 1 0 0.25 0.75 4f 2 0 f 2 3f 2

  25. Multiplicity Factor Equal d-spacings equal B E.g.: Cubic (100), (010), (001), (-100), (0-10), (00-1): Equivalent  Multiplicity Factor = 6 (110), (-110), (1-10), (-1-10), (101), (-101), (10-1),(-10-1), (011), (0-11), (01-1), (0-1-1): Equivalent  Multiplicity Factor = 12 lower symmetry systems  multiplicities . E.g.: tetragonal (100) equivalent: (010), (-100), and (0-10) not with the (001) and the (00-1). {100}  Multiplicity Factor = 4 {001}  Multiplicity Factor = 2

  26. Multiplicity p is the one counted in the point group stereogram. In cubic (h  k  l) p = 48 3x2x23 = 48 p = 24 3x23 = 24 p = 24 3x23 = 24 p = 12 3x22 = 12 p = 8 23 = 8 p = 6 3x2 = 6

  27. Lorentz factor:  dependence of the integrated peak intensities 1. finite spreading of the intensity peak 2. fraction of crystal contributing to a diffraction peak 3. intensity spreading in a cone

  28. 1 2 Imax  Imax/2 Intensity 2 B Integrated Intensity 2 B 2B 1 Diffraction Angle 2 1 2 2 path difference for 11-22 = AD – CB = acos2 - acos1 = a[cos(B-) - cos (B+)] = 2asin()sinB~ 2a sinB. D C 1 2 a A B 1 N 2Na sinB =   completely cancellation (1- N/2, 2- (N/2+1) …) Na

  29. Maximum angular range of the peak Imax   1/sinB, Half maximum B  1/cosB (will be shown later)  integrated intensity  ImaxB  (1/sinB)(1/cosB)  1/sin2B. 2 number of crystals orientated at or near the Bragg angle  /2- Fraction of crystal: r   crystal plane

  30. 3 diffracted energy: equally distributed (2Rsin2B)  the relative intensity per unit length  1/sin2B. 2B Lorentz factor: Lorentz–polarization factor: (omitting constant)

  31. Absorption factor: X-ray absorbed during its in and out of the sample. Hull/Debye-ScherrerCamera: A(); A()  as . Diffractometer: dID 1cm I0   C Incident beam: I0; 1cm2 incident angle . Beam incident on the plate: A x dx B l 2 a: volume fraction of the specimen that are at the right angle for diffraction b: diffracted intensity/unit volume : linear absorption coefficient  volume = l  dx 1cm = ldx. actual diffracted volume = aldx Diffracted intensity: Diffracted beam escaping from the sample:

  32. If  =  =   Infinite thickness ~ dID(x = 0)/dID(x = t) = 1000 and  =  = ).

  33. Temperature factor (Debye Waller factor): Atoms in lattice vibrate (Debye model)  (1) lattice constants  2 ;  (2) Intensity of diffracted lines ;  (3) Intensity of the background scattering . Temperature  u u d d low B high B Lattice vibration is more significant at high B (u/d)  as B 

  34. Formally, the factor is included in f as Because F = |f 2|  factor e-2M shows up What is M? : Mean square displacement Debye: h: Plank’s constant; T: absolute temperature; m: mass of vibrating atom; : Debye temperature of the substance; x = /T; (x): tabulated function

  35. m atomic weight (A): 1 e-2M 0 sin /  I TDS 2 or sin/ Temperature (Thermal) diffuse scattering (TDS)  as   I as  peak width B  slightly as T 

  36. Summary Intensities of diffraction peaks from polycrystalline samples: Diffractometer: Other diffraction methods: Match calculation? Exactly: difficult; qualitatively matched. Perturbation: preferred orientation; Extinction (large crystal)

  37. Example Debye-Scherrer powder pattern of Cu made with Cu radiation Cu: Fm-3m, a = 3.615 Å

  38. Structure Factor  If h, k, l are unmixed If h, k, l are mixed 111 200 220 311 222 400 331 420

  39. 0 0.1 0.2 0.3 0.4 29 27.19 23.63 19.90 16.48

  40. (23 = 8) p = 8

  41. Dynamic Theory for Single crystal Kinematical theory Dynamical theory S0 S K0 K0 K1 Refraction K1  PRIMARY EXTINCTION  K2 K2 K0 & K1 : /2; K1 & K2 : /2 K0 & K2 :  ; destructive interference (hkl) K1 K2 Negligible absorption I |F| not |F|2! e: electron charge; m: electron mass; N: # of unit cell/unit volume.

  42. Width of the diffraction peak (~ 2s) FWHM for Darwin curve = 2.12s 5 arcs < < 20 arcs

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