X-ray Diffraction

Interaction of Waves Reciprocal Lattice and Diffraction X-ray Scattering by Atoms The Integrated Intensity Basic Principles of Interaction of Waves

Periodic waves characteristic:

 Frequency : number of waves (cycles) per unit time – =cycles/time. [] = 1/sec = Hz.

 Period T: time required for one complete cycle – T=1/=time/cycle. [T] = sec.

 Amplitude A: maximum value of the wave during cycle.

 Wavelength : the length of one complete cycle. [] = m, nm, Å.

A simple wave completes cycle in 360 degrees E(t, x)  Aexpkxt  x   x  2  Aexp 2  t   Aexp 2 t    2 , k   c      Basic Principles of Interaction of Waves

Consider two waves with the same wavelength and amplitude but displaced a

distance x0. The phase shift:  x    2 0    

E1(t)  Aexpt

o E1(t)  Aexpt  Waves #1 and #2 are 90 out of phase

x0

Waves #1 and #2 are 180o out of phase

When similar waves combine, the outcome can be constructive or destructive interference Superposition of Waves

Resulting wave is algebraic sum of the amplitudes at each point

Small difference in phase Large difference in phase Superposition of Waves

Thomas Young's diagram of double slit interference (1803)

퐿휆 Δ푥 = 푑 d

L 휆 The angular spacing of the fringes is given by 휃 ≈ , where 휃푓 ≪ 1 푓 푑 The discovery of X-ray diffraction and its use as a probe of the structure of matter

• The reasoning: x-rays have a wavelength similar to the interatomic distances in crystals, and as a result, the crystal should act as a diffraction grating.

• 1911, von Laue suggested to one of his research assistants, Walter Friedrich, and a doctoral student, Paul Knipping, that they try out x-rays on crystals. Max von Laue

• April 1912, von Laue, Friedrich and Knipping had performed their pioneering experiment on copper sulfate.

First diffraction pattern from NaCl crystal The discovery of X-ray diffraction and its use as a probe of the structure of matter

• They found that if the interatomic distances in the crystal are known, then the wavelength of the X-rays can be measured, and alternatively, if the wavelength is known, then X-ray diffraction experiments can be used to determine the interplanar spacings of a crystal.

• The three were awarded Nobel Prizes in Physics for their discoveries.

ZnS Laue photographs along Friedrich & Knipping's improved set-up four-fold and three-fold axes The Laue Equations

If an X-ray beam impinges on a row of atoms, each atom can serve as a source of scattered X-rays. The scattered X-rays will reinforce in certain directions to produce zero-, first-, and higher-order diffracted beams. The Laue Equations

Consider 1D array of scatterers spaced a apart. Let x-ray be incident with wavelength .

acos cos0   h

In 2D and 3D:

acos  cos0   h 2D

bcos   cos 0   k 3D

ccos  cos 0   l

The equations must be satisfied simultaneously, it is in general difficult to produce a diffracted beam with a fixed wavelength and a fixed crystal. Lattice Planes Bragg’s Law

AB  BC  2d sin Bragg’s Law

If the path AB + CD is a multiple of the x-ray wavelength λ, then two waves will give a constructive interference:

n  AB CD  2d sin n

2d sin  n

The diffracted waves will interfere destructively if equation is not satisfied.

Equation is called the Bragg equation and the angle θ is the Bragg angle. Bragg’s Law

The incident beam and diffracted beam are always coplanar. The angle between the diffracted beam and the transmitted beam is always 2.

n Since sin  1:  sin 1 For n = 1:   2d 2d UV radiation   500 Å For most crystals d ~ 3 Å   6 Å Cu K1  = 1.5406 Å Rewrite Bragg’s law: d   2 sin    2dsin   2d sin n

1 h2  k 2  l 2  d 2 a2 a d100   a 12  0  0 a 1 d200   a 22  0  0 2 Reciprocal lattice and Diffraction

* dhkl s s 0  d*  hb  kb  lb  hkl 1 2 3

s  s0  s s 0   It is equivalent to the Bragg law since s s  2sin   0

s  s0 2sin * 1   dhkl    dhkl

  2dhkl sin Reciprocal lattice and Diffraction

s  s0 2sin * 1   dhkl    dhkl Sphere of Reflection – Ewald Sphere

1 1 * 1 OC  sin  dhkl     2dhkl sin  2 2dhkl

* OB  dhkl Sphere of Reflection – Ewald Sphere Kinematical x-ray diffraction

Silicon lattice constant: For cubic crystal: aSi = 5.43 Å Bragg’s law: 1 h2  k 2  l 2 X-ray wavelength:  2d sin  n  = 1.5406 Å d 2 a2 Two perovskites: SrTiO3 and CaTiO3

Differences:

 Peak position – d-spacing.

 Peak intensity – atom type: Ca vs Sr.

O Ti

Sr/Ca

aSTO = 3.905 Å

aCTO = 3.795 Å Scattering by an Electron

Elementary scattering unit in an atom is electron Classical scattering by a single free electron – Thomson scattering equation:   2

e4 1 cos2 2  I  I   0 2 4 2   m c R  2  The polarization factor of an unpolarized primary beam e4  7.941026 cm2 m2c4

I 26 If R = few cm: 10 I0 Another way for electron to scatter is manifested in Compton effect. 1 mg of matter has ~1020 electrons Cullity p.127 Scattering by an Atom

Atomic amplitudeof thewavescatteredbyan atom Scattering f  Factor amplitudeof thewavescatteredbyoneelectron Scattering by an Atom

Scattering by a group of electrons at positions rn:

Scattering factor per electron:

f  exp 2i /  s s r  dV e    0   Assuming spherical symmetry for the charge distribution  = (r ) and taking origin at the center of the atom:

 2 sin kr fe  4r r dr 0 kr For an atom containing several electrons:

 2 sin kr 4 sin f  fen  4r n r dr k   0 n n kr 

sin f – atomic scattering factor f   Calling Z the number of electrons per atom we get:

 2 4r n rdr  Z 0 n Scattering by an Atom

amplitudeof thewave scattered byan atom f  amplitudeof thewave scattered byoneelectron

The atomic scattering factor f = Z for any atom in the forward direction (2 = 0): I(2=0) =Z2

As  increases f decreases  functional dependence of the decrease depends on the details of the distribution of electrons around an atom (sometimes called the form factor)

f is calculated using quantum mechanics Electron vs nuclear density

Powder diffraction patterns collected using Mo K radiation and neutron diffraction Scattering by an Atom Scattering by a Unit Cell

amplitude scattered by atoms in unit cell Fhkl  For atoms A & C amplitude scatteredby single electron

 21  MCN  2dh00 sin   a d  AC  h00 h For atoms A & B

AB AB x   RBS  MCN     31 AC AC a / h  phase   2   2hx   2 31  31  a If atom B position: u  x / a 2hx    2hu 31 a

For 3D:   2hu  kvlw Scattering by a Unit Cell

We can write: Ae i  fe 2ihukvlw

i1 i2 i3 F  f1e  f2e  f3e ...

N N in 2ihun kvn lwn  F   fne   fne 1 1

 2 I  FhklFhkl  Fhkl Scattering by a Unit Cell

Examples

Unit cell has one atom at the origin

F  fe 2i0  f

In this case the structure factor is independent of h, k and l ; it will decrease with f as sin/ increases (higher-order reflections) Scattering by a Unit Cell

Examples

Unit cell is base-centered F  fe 2i0  fe 2ih/ 2k /2  f 1 eihk  

F  2 f h and k unmixed

F  0 h and k mixed

2 2 (200), (400), (220)   Fhkl  4 f (100), (121), (300)   F 2  0 “forbidden” hkl reflections Body-Centered Unit Cell

Examples

For body-centered cell

F  fe 2ih0k0l0  fe 2ih/ 2k /2l / 2  f 1 eihkl  

F  2 f when (h + k + l ) is even

F  0 when (h + k + l ) is odd

2 2 (200), (400), (220)   Fhkl  4 f (100), (111), (300)   F 2  0 “forbidden” hkl reflections Body-Centered Unit Cell

Examples

For body centered cell with different atoms:

2ih0k0l0 2ih/ 2k / 2l / 2 F  fCle  fCse ihkl   fCl  fCse

F  fCl  fCs when (h + k + l) is even Cs+ Cl+

F  fCl  fCs when (h + k + l) is odd

2 2 (200), (400), (220)   Fhkl   fCs  fCl  (100), (111), (300)   2 2 Fhkl   fCs  fCl  Face Centered Unit cell

The fcc crystal structure has atoms at 000, ½½0, ½0½ and 0½½:

N 2ihun kvn lwn  ihk  ihl  ikl  Fhkl   fne  f 1 e  e  e  1 If h, k and l are all even or all odd numbers (“unmixed”), then the exponential terms all equal to +1  F = 4f If h, k and l are mixed even and odd, then two of the exponential terms will equal -1 while one will equal +1  F = 0

2 16f 2, h, k and l unmixed even and odd Fhkl  0, h, k and l mixed even and odd The Structure Factor

N 2ihun kvn lwn  Fhkl   fne 1

amplitude scattered by all atoms in a unit cell F  hkl amplitude scattered by a single electron

The structure factor contains the information regarding the types (f ) and locations (u, v, w ) of atoms within a unit cell A comparison of the observed and calculated structure factors is a common goal of X-ray structural analyses The observed intensities must be corrected for experimental and geometric effects before these analyses can be performed Integrated Intensity

Peak intensity depends on Structural factors: determined by crystal structure Specimen factors: shape, size, grain size and distribution, microstructure Instrumental factors: radiation properties, focusing geometry, type of detector

We can say that: 퐼 푞 ∝ 퐹 푞 2 Integrated Intensity

2 Ihkl (q)  K  phkl  L P AT  Ehkl  F(q)

K – scale factor, required to normalize calculated and measured intensities. phkl – multiplicity factor. Accounts for the presence of symmetrically equivalent points in reciprocal lattice. L – Lorentz multiplier, defined by diffraction geometry. P – polarization factor. Account for partial polarization of electromagnetic wave. A – absorption multiplier. Accounts for incident and diffracted beam absorption.

Thkl – preferred orientation factor. Accounts for deviation from complete random grain distribution.

Ehkl – extinction multiplier. Accounts for deviation from kinematical diffraction model.

Fhkl – the structure factor. Defined by crystal structure of the material The Multiplicity Factor

The multiplicity factor arises from the fact that in general there will be several sets of hkl -planes having different orientations in a crystal but with the same d and F 2 values Evaluated by finding the number of variations in position and sign in h, k and l and have planes with the same d and F 2 The value depends on hkl and crystal symmetry For the highest cubic symmetry we have:

100, 100, 010, 010, 001, 001 p100 = 6

110, 110,110, 1 10,101,101, 101, 101, 011, 011, 011, 01 1 p110 = 12

111,111,111, 111,11 1, 111, 1 11, 1 1 1 p111=8 The Polarization Factor

The polarization factor p arises from the fact that an electron does not scatter along its direction of vibration In other directions electrons radiate with an intensity proportional to (sin )2:

The polarization factor (assuming that the incident beam is unpolarized):

1 cos2 2 P  2 The Lorentz-Polarization Factor

The Lorenz factor L depends on the measurement technique used and, for the diffractometer data obtained by the usual θ-2θ or ω-2θ scans, it can be written as 1 1 L   sin sin 2 cos sin 2  The combination of geometric corrections are lumped together into a single Lorentz-polarization (LP ) factor:

1 cos2 2 LP  cos sin 2 

The effect of the LP factor is to decrease the intensity at intermediate angles and increase the intensity in the forward and backwards directions The Absorption Factor

Angle-dependent absorption within the sample itself will modify the observed intensity

• Absorption factor for infinite thickness specimen is:

 A  2 • Absorption factor for thin specimens is given by:

 2t  A 1 exp   sin 

where μ is the absorption coefficient, t is the total thickness of the film The Extinction Factor

Extinction lowers the observed intensity of very strong reflections from perfect crystals

primary extinction secondary extinction

In powder diffraction usually this factor is smaller than experimental errors and therefore neglected The Temperature Factor

As atoms vibrate about their equilibrium positions in a crystal, the electron density is spread out over a larger volume This causes the atomic scattering factor to decrease with sin/ (or |S| = 4sin/ )more rapidly than it would normally:

The temperature factor is given by: sin2  B e 2 where the thermal factor B is related to the mean square displacement of the atomic vibration: B  82 u2 This is incorporated into the

atomic scattering factor: Scattering by C atom by atom expressed C Scattering in electrons M 2 2M f  f0e  f ~ e Diffracted Beam Intensity

 2 I  FhklFhkl  Fhkl

2 IC (q)  KAp(LP) F(q)  Ib

where K is the scaling factor, Ib is the background intensity, q = 4sinθ/λ is the scattering vector for x-rays of wavelength λ

For thin films:

2   2t  1 cos 2 2 IC (q)  K1 exp  2 F(q)  Ib   sin  cos sin 