X-ray Diraction Interaction o Waves Reciprocal Lattice and Diraction X-ray Scattering by Atoms The Integrated Intensity
Basic Principles o Interaction o Waves Periodic waves characteristic: Frequency : number o waves (cycles) per unit time =cycles/time. [] = 1/sec = Hz. Period T: time required or one complete cycle T=1/=time/cycle. [T] = sec. Amplitude A: maximum value o the wave during cycle. Wavelength : the length o one complete cycle. [] = m, nm, Å. E( t, x) Aexp kxt Aexp x c t x Aexp t A simple wave completes cycle in 360 degrees, k
Basic Principles o Interaction o Waves Consider two waves with the same wavelength and amplitude but displaced a distance x 0. The phase shit: x 0 E ( t) 1 Aexp t E ( t) Aexp t x 0 1 Waves #1 and # are 90 o out o phase Waves #1 and # are 180 o out o phase When similar waves combine, the outcome can be constructive or destructive intererence
Superposition o Waves Resulting wave is algebraic sum o the amplitudes at each point Small dierence in phase Large dierence in phase
Superposition o Waves Thomas Young's diagram o double slit intererence (1803) Δx = Lλ d d L θ λ The angular spacing o the ringes is given by, where θ 1 d
The discovery o X-ray diraction and its use as a probe o the structure o 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 diraction grating. 1911, von Laue suggested to one o his research assistants, Walter Friedrich, and a doctoral student, Paul Knipping, that they try out x-rays on crystals. Max von Laue April 191, von Laue, Friedrich and Knipping had perormed their pioneering experiment on copper sulate. First diraction pattern rom NaCl crystal
The discovery o X-ray diraction and its use as a probe o the structure o matter They ound that i the interatomic distances in the crystal are known, then the wavelength o the X-rays can be measured, and alternatively, i the wavelength is known, then X-ray diraction experiments can be used to determine the interplanar spacings o a crystal. The three were awarded Nobel Prizes in Physics or their discoveries. Friedrich & Knipping's improved set-up ZnS Laue photographs along our-old and three-old axes
The Laue Equations I an X-ray beam impinges on a row o atoms, each atom can serve as a source o scattered X-rays. The scattered X-rays will reinorce in certain directions to produce zero-, irst-, and higher-order diracted beams.
The Laue Equations Consider 1D array o scatterers spaced a apart. Let x-ray be incident with wavelength. a cos cos h 0 In D and 3D: D a cos cos b cos cos 0 0 h k 3D c cos cos l 0 The equations must be satisied simultaneously, it is in general diicult to produce a diracted beam with a ixed wavelength and a ixed crystal.
Lattice Planes
Bragg s Law AB BC d sin
Bragg s Law I the path AB + CD is a multiple o the x-ray wavelength λ, then two waves will give a constructive intererence: n AB CD d sin n d sin n The diracted waves will interere destructively i equation is not satisied. Equation is called the Bragg equation and the angle θ is the Bragg angle.
Bragg s Law The incident beam and diracted beam are always coplanar. The angle between the diracted beam and the transmitted beam is always. n Since sin 1: sin 1 For n = 1: d d For most crystals d ~ 3 Å 6 Å UV radiation 500 Å Cu K 1 = 1.5406 Å Rewrite Bragg s law: d sin dsin n d sin 1 d d d 100 00 h k a a 1 l 0 0 a 0 0 a 1 a
Reciprocal lattice and Diraction * d hkl s s 0 * d b hkl hb1 kb l 3 s 0 s s 0 s It is equivalent to the Bragg law since s s 0 sin s s sin * 1 0 d hkl d hkl d sin hkl
Reciprocal lattice and Diraction s s sin * 1 0 d hkl d hkl
Sphere o Relection Ewald Sphere 1 OC sin d 1 * hkl 1 d hkl d hkl sin * OB d hkl
Sphere o Relection Ewald Sphere
Kinematical x-ray diraction Silicon lattice constant: a Si = 5.43 Å 1 d For cubic crystal: k a X-ray wavelength: = 1.5406 Å h l Bragg s law: d sin n
Two perovskites: SrTiO 3 and CaTiO 3 Dierences: Peak position d-spacing. Peak intensity atom type: Ca vs Sr. O Ti Sr/Ca a STO = 3.905 Å a CTO = 3.795 Å
Scattering by an Electron Elementary scattering unit in an atom is electron Classical scattering by a single ree electron Thomson scattering equation: I I 0 e 4 4 m c R 1 cos The polarization actor o an unpolarized primary beam I R = ew cm: e 4 4 m c 7.9410 0 10 6 6 1 mg o matter has ~10 0 electrons I I cm Another way or electron to scatter is maniested in Compton eect. Cullity p.17
Scattering by an Atom Atomic Scattering Factor amplitude o thewavescatteredbyan atom amplitude o thewavescatteredbyoneelectron
Scattering by an Atom Scattering by a group o electrons at positions r n : Scattering actor per electron: e exp i / s s 0 r dv Assuming spherical symmetry or the charge distribution = (r ) and taking origin at the center o the atom: e sin kr 4r 0 r dr kr For an atom containing several electrons: n en n 0 sin kr 4r nr dr kr k 4 sin atomic scattering actor sin Calling Z the number o electrons per atom we get: 0 n 4r n r dr Z
Scattering by an Atom amplitudeo thewave scattered byan atom amplitudeo thewave scattered by one electron The atomic scattering actor = Z or any atom in the orward direction ( = 0): I(=0) =Z As increases decreases unctional dependence o the decrease depends on the details o the distribution o electrons around an atom (sometimes called the orm actor) is calculated using quantum mechanics
Electron vs nuclear density Powder diraction patterns collected using Mo K radiation and neutron diraction
Scattering by an Atom
Scattering by a Unit Cell For atoms A & C F hkl amplitude scattered by atoms in unit cell amplitude scatteredby single electron 1 MCN dh00sin a dh00 AC h For atoms A & B AB 31 RBS MCN AC phase 31 hx 31 a AB AC x a / h I atom B position: u x / a hx 31 hu a For 3D: hu kvlw
Scattering by a Unit Cell We can write: lw kv hu i i e Ae N lw kv hu i n N i n i i i n n n n e e F e e e F 1 1 3 1... 3 1 hkl hkl hkl F F F I
Scattering by a Unit Cell Examples Unit cell has one atom at the origin F e i 0 In this case the structure actor is independent o h, k and l ; it will decrease with as sin/ increases (higher-order relections)
Scattering by a Unit Cell Examples Unit cell is base-centered F e i 0 e i h/ k / ihk 1 e F h and k unmixed F 0 h and k mixed (00), (400), (0) (100), (11), (300) F hkl 4 F hkl 0 orbidden relections
Body-Centered Unit Cell Examples F e i For body-centered cell h0k0l0 i h/ k / l / ihk l e 1 e F when (h + k + l ) is even F 0 when (h + k + l ) is odd (00), (400), (0) (100), (111), (300) F hkl 4 F hkl 0 orbidden relections
Body-Centered Unit Cell Examples For body centered cell with dierent atoms: F Cl Cl e h0k 0l0 ih / k / l / i Cs Cs e i hkl e F Cl Cs when (h + k + l) is even Cs + Cl + F Cl Cs when (h + k + l) is odd (00), (400), (0) (100), (111), (300) F hkl F hkl Cs Cl Cs Cl
Face Centered Unit cell The cc crystal structure has atoms at 000, ½½0, ½0½ and 0½½: F hkl N 1 n e i hu kv lw ihk ihl ik l n n n 1 e I h, k and l are all even or all odd numbers ( unmixed ), then the exponential terms all equal to +1 F = 4 I h, k and l are mixed even and odd, then two o the exponential terms will equal -1 while one will equal +1 F = 0 e e F 16, h, k and l unmixed even and odd hkl 0, h, k and l mixed even and odd
The Structure Factor F hkl N 1 n e i hu kv n n lw n F hkl amplitude scattered by all atoms in a unit cell amplitude scattered by a single electron The structure actor contains the inormation regarding the types ( ) and locations (u, v, w ) o atoms within a unit cell A comparison o the observed and calculated structure actors is a common goal o X-ray structural analyses The observed intensities must be corrected or experimental and geometric eects beore these analyses can be perormed
Integrated Intensity Peak intensity depends on Structural actors: determined by crystal structure Specimen actors: shape, size, grain size and distribution, microstructure Instrumental actors: radiation properties, ocusing geometry, type o detector We can say that: I q F q
Integrated Intensity I hkl ( q) K p L P AT E F( q) hkl hkl K scale actor, required to normalize calculated and measured intensities. p hkl multiplicity actor. Accounts or the presence o symmetrically equivalent points in reciprocal lattice. L Lorentz multiplier, deined by diraction geometry. P polarization actor. Account or partial polarization o electromagnetic wave. A absorption multiplier. Accounts or incident and diracted beam absorption. T hkl preerred orientation actor. Accounts or deviation rom complete random grain distribution. E hkl extinction multiplier. Accounts or deviation rom kinematical diraction model. F hkl the structure actor. Deined by crystal structure o the material
The Multiplicity Factor The multiplicity actor arises rom the act that in general there will be several sets o hkl -planes having dierent orientations in a crystal but with the same d and F values Evaluated by inding the number o variations in position and sign in h, k and l and have planes with the same d and F The value depends on hkl and crystal symmetry For the highest cubic symmetry we have: 100, 100, 010, 010, 001, 001 p 100 = 6 110, 110,110, 1 10,101,101, 101, 101, 011, 011, 011, 01 1 p 110 = 1 111,111,111, 111,11 1, 111, 1 11, 1 1 1 p 111 =8
The Polarization Factor The polarization actor p arises rom the act that an electron does not scatter along its direction o vibration In other directions electrons radiate with an intensity proportional to (sin ) : The polarization actor (assuming that the incident beam is unpolarized): 1 cos P
The Lorentz-Polarization Factor The Lorenz actor L depends on the measurement technique used and, or the diractometer data obtained by the usual θ-θ or ω-θ scans, it can be written as L 1 sin sin 1 cos sin The combination o geometric corrections are lumped together into a single Lorentz-polarization (LP ) actor: 1 cos LP cos sin The eect o the LP actor is to decrease the intensity at intermediate angles and increase the intensity in the orward and backwards directions
The Absorption Factor Angle-dependent absorption within the sample itsel will modiy the observed intensity Absorption actor or ininite thickness specimen is: A Absorption actor or thin specimens is given by: t A 1 exp sin where μ is the absorption coeicient, t is the total thickness o the ilm
The Extinction Factor Extinction lowers the observed intensity o very strong relections rom perect crystals primary extinction secondary extinction In powder diraction usually this actor is smaller than experimental errors and thereore neglected
Scattering by C atom expressed in electrons 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 actor to decrease with sin/ (or S = 4sin/ )more rapidly than it would normally: The temperature actor is given by: sin e B where the thermal actor B is related to the mean square displacement o the atomic vibration: B 8 u This is incorporated into the atomic scattering actor: e e M M 0 ~
Diracted Beam Intensity I F hkl F hkl F hkl I ( q) KAp( LP) F( q) C I b where K is the scaling actor, I b is the background intensity, q = 4sinθ/λ is the scattering vector or x-rays o wavelength λ For thin ilms: I t sin 1 cos cos sin C ( q) K 1 exp F( q) I b