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

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

5. Pass through a monochromator first (Bragg angle  M )  the polarization factor is ? (Homework) x y z O P polarization is not complete random anymore 22 r x y z O P Random polarized 2M2M r P

6. M

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

8. Differential atomic scattering factor (df) : E e : the magnitude of the wave from a bound electron O s0s0 R r x1x1 dV s x2x2 22 path different (O and dV): R-(x 1 + x 2 ). Electron density Phase difference

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

10. = 2

11. S0S0 S S-S 0 22 Evaluate (S - S 0 )  r = | S - S 0 ||r|cos  |(S - S 0 )|/2 = sin . Let

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

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

15. 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 = x 0 e i  t Real part and imaginary part

16. f +  f + i  f  realimaginary  E E     if   0 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)

17. Examples: Si, 400 diffraction peak, with Cu K  (0.1542 nm) 0.30.4 8.227.20 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

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

19. 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.

20. F (in general) a complex number. N atoms in a unit cell; f n : atomic form factor of atom n 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).

21.  Example 1: Simple cubic 1 atoms/unit cell; 000 and 100, 010, 001, 110, 101, 011, 111: equipoints of 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: equipoints of 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

22.  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

23.  Example 4: Diamond Cubic 8 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; 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

24. 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

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

26. h + 2k l 3m3m3m13m13m3m3m13m1 even odd even odd 1 0 0.25 0.75 4f 2 0 f 2 3f 2

27. 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

28. Multiplicity p is the one counted in the point group stereogram. In cubic (h  k  l) p = 483x2x2 3 = 48 p = 243x2 3 = 24 p = 243x2 3 = 24 p = 123x2 2 = 12 p = 82 3 = 8 p = 63x2 = 6

29. 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

30. 2B2B Intensity Diffraction Angle 2  I max I max /2  22 Integrated Intensity B 2 12 AB a 22 C D Na 1 N BB 22 11 22 11 path difference for 11-22 = AD – CB = acos  2 - acos  1 = a[cos(  B -  ) - cos (  B +  )] = 2asin(  )sin  B ~ 2a  sin  B. 2Na  sin  B =  completely cancellation (1- N/2, 2- (N/2+1) …) 1

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

33. diffracted energy: equally distributed (2  Rsin2  B )  the relative intensity per unit length  1/sin2  B. 2B2B 3 Lorentz–polarization factor: (omitting constant) Lorentz factor:

36. Absorption factor: X-ray absorbed during its in and out of the sample. Hull/Debye-Scherrer Camera: A(  ); A(  )  as . l dx  I0I0 22 A B C dI D  x Incident beam: I 0 ; 1cm 2 incident angle . Beam incident on the plate:  : linear absorption coefficient a: volume fraction of the specimen that are at the right angle for diffraction b: diffracted intensity/unit volume 1cm Diffractometer:  volume = l  dx  1cm = ldx. actual diffracted volume = aldx Diffracted intensity: Diffracted beam escaping from the sample:

37. If  =  =   Infinite thickness ~ dI D (x = 0)/dI D (x = t) = 1000 and  =  =  ).

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

39. 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

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

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

42. Example Debye-Scherrer powder pattern of Cu made with Cu radiation 12345678 linehklh2+k2+l2h2+k2+l2 sin 2  sin  (o) (o)sin  / (Å -1 ) f Cu 1234567812345678 111 200 220 311 222 400 331 420 3 4 8 11 12 16 19 20 0.1365 0.1820 0.364 0.500 0.546 0.728 0.865 0.910 0.369 0.427 0.603 0.707 0.739 0.853 0.930 0.954 21.7 25.3 37.1 45.0 47.6 58.5 68.4 72.6 0.24 0.27 0.39 0.46 0.48 0.55 0.60 0.62 22.1 20.9 16.8 14.8 14.2 12.5 11.5 11.1 Cu: Fm-3m, a = 3.615 Å

43. 191011121314 line|F|2|F|2 P Relative integrated intensity Calc.( x10 5 ) Calc. Obs. 1234567812345678 7810 6990 4520 3500 3230 2500 2120 1970 8 6 12 24 8 6 24 12.03 8.50 3.70 2.83 2.74 3.18 4.81 6.15 7.52 3.56 2.01 2.38 0.71 0.48 2.45 2.91 10.0 4.7 2.7 3.2 0.9 0.6 3.3 3.9 Vs S s m w s If h, k, l are unmixed If h, k, l are mixed 111 200 220 311 222 400 331 420 Structure Factor 

44. 0 0.1 0.2 0.3 0.4 29 27.19 23.63 19.90 16.48

45. p = 8 (2 3 = 8)

46. Dynamic Theory for Single crystal Kinematical theory Dynamical theory K0K0 K0K0 K1K1 S0S0 S   K1K1 K2K2 K2K2 K1K1 K2K2 (hkl) Refraction PRIMARY EXTINCTION K 0 & K 1 :  /2; K 1 & K 2 :  /2 K 0 & K 2 :  ; destructive interference I  |F| not |F| 2 ! Negligible absorption e: electron charge; m: electron mass; N: # of unit cell/unit volume.

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