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

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

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

M

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

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

Spherical integration dV = dr(rd  ) (rsin  d  ) continuity-equation-in.html r: 0 -   : 0 -   :    rsin  d  rsin(  +d  )d  r dr dd dd

= 2

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

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

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

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

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)

Examples: Si, 400 diffraction peak, with Cu K  ( nm) Anomalous Scattering correction Atomic scattering factor in this case: i = i  f and  f  : International Table for X-ray Crystallography V.III

Structure factor atoms  unit cell plane (h00) A B C  NM SR 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)

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.

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).

 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

 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

 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

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

 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

h + 2k l 3m3m3m13m13m3m3m13m1 even odd even odd f 2 0 f 2 3f 2

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

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

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

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 = 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

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-  

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:

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:

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

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 .

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

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

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

Example Debye-Scherrer powder pattern of Cu made with Cu radiation linehklh2+k2+l2h2+k2+l2 sin 2  sin  (o) (o)sin  / (Å -1 ) f Cu Cu: Fm-3m, a = Å

line|F|2|F|2 P Relative integrated intensity Calc.( x10 5 ) Calc. Obs Vs S s m w s If h, k, l are unmixed If h, k, l are mixed Structure Factor 

p = 8 (2 3 = 8)

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.

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