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# Part 8. Diffraction Intensity

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### Part 8. Diffraction Intensity

1. 1. Part 8. Diffraction: Intensity (From Chapter 4 of Textbook 2 and Chapter 9 of Textbook 1) ♦ Scattering by an electron: An electron which has been set into oscillation by an X-ray beam is continuously accelerating and decelerating during its motion and therefore emits an EM wave → being said to scatter x-rays. The emitted X-ray has the same frequency (wavelength) as the excited X-ray beam → i.e. coherent. (the phase change on scattering from an electron is π/2) * The intensity of the scattered beam is angular dependent. J.J. Thomson: the intensity I of the beam scattered by a single electron of charge e (C) and mass m (kg), at a distance r (meters) from the electron, is given by r α P
2. 2. αα π µ 2 20 2 22 42 0 0 sinsin 4 r K I rm e II =            = µ0: 4π×10-7 mkgC-2 x y zIncident beam wavevector along x-axis, the electric field lies on the yz plane. Electron at O. Diffracted beam at P on xz plane, i.e. diffraction plane is xz. Every E can be resolved into components Ey and Ez. O P Random polarized 2θ r On average, Ey = Ez, since E is perfectly random. 222 zy EEE += 000 22 2 1 2 1 IIIEEE zyzy ==→==2 The y component accelerates the electron in the direction Oy → the intensity of the scattered beam at P is
3. 3. 20 r K II yPy = α = ∠yOP = π/2 The z component accelerates the electron in the direction Oz → the intensity of the scattered beam at P is θ2cos2 20 r K II zPz = α = ∠zOP = π/2 -2θ The total scattered intensity at P is       + =+=+= 2 2cos1 2cos 2 20 2 2020 θ θ r K I r K I r K IIII zyPzPyP Polarization factorr ↑ → I ↓; θ ↑ → I↓ If a monochromator is used with the Bragg angle θM → the polarization factor for powdered crystalline sample becomes M M θ θθ 2cos1 2cos2cos1 2 22 + + It depends on geometry and sample condition
4. 4. Another kind of scattering: Compton scattering; when X-ray encounter loosely bound or free electrons; incoherent scattering, wavelength changed, no fixed phase relation; belongs to scattering loss (attenuation). X-ray before impact hν1 (λ1), after Compton scattering hν2 (λ2). θλλλ 2 12 sin0486.0)( =−=Α∆  hν1 e e hν2 2θ ♦ Atomic scattering (or form) factor: Consider the physics of coherent scattering by atoms. The atomic scattering factor is the ratio of the scattering from an atom to that from a single bound electron. electronsingleabyscatteredamplitude atombyscatteredamplitude factorscatteringatomic =
5. 5. Assume that each electron has a spherical charge density ρ(r) around the atomic nucleus and that the X-ray wavelength λ is not too close to the absorption edge. The amplitude scattered to the plane P from a small volume of the charge density of one electron, dV, is given by ρ(r)dV. s0, s: unit vectors presenting the incident and scattered wave; The path different between the wave scattered at O and at dV is R-(x1 + x2). O s0 R r x1 dV s x2 2θ srsr ⋅−=⋅= Rxx 201 ; The differential atomic scattering factor is defines by dVer E df i e )]()[/2( 0 )( 1 ssr −⋅ = λπ ρ Ee: the magnitude of the wave from a bound electron
6. 6. Spherical integration, the integration element is dV = 2πr2 sinφ dφ dr dV S0 S S-S0 r φ S0 S S-S0 2θ Evaluate (S - S0)⋅r = | S - S0||r|cosφ (S - S0)/2 = sinθ. φθ cossin2)( 0 r=⋅−→ rSS dVer E dff i e )]()[/2( 0 )( 1 )( ssr −⋅ == ∫ λπ ρθ ∫ ∫∫ ∞= = = == r r ri e drdrer E dff 0 0 2/cossin4 sin2)( 1 )( π φ λφθπ φφπρθ Let λθπ /sin4=k 2 0 0 22 4|cos2sin2 rrdr πφπφφπ π πφ φ =−=∫ = = φcosd− ∫ ∫ ∞= = = −= r r kri e dedrrr E f 0 0 cos2 cos2)( 1 )( π φ φ φπρθ kr kr ikr ee ikr e ikrikrikr sin2 cos 0cos cos = − =−= −πφ
7. 7. ∫ ∞= = = r r e dr kr kr rr E f 0 2 sin )(4 1 )( ρπθ For θ = 0, only k = 0 → sinkr/kr = 1. For n electrons in an atom ∑ ∫ ∞= = = electronsn 0 2 sin )(4 1 )( r r e dr kr kr rr E f ρπθ Zdrrr E f r r e == ∑ ∫ ∞= = electronsn 0 2 )(4 1 )0( ρπ Number of electrons in the atom equal to 1 bound electrons Tabulated If one consider the absorption of the X-ray, the atomic form factor is expressed as a complex number. The calculation above is the real part of the factor. The imaginary part (absorption) is also tabulated for different materials and X-ray source.
8. 8. ♦ Scattering by a unit cell: Next, we have to consider the diffracted beam from a group of atoms that can make up the crystal. Atoms representing unit cell → crystal Diffracted beam from the cell → diffracted beam from the crystal Find out the phase difference from a group of atoms representing a unit cell. (h00) A B C θ N M S R 1 1′3 3′ 2 2′ a The path difference between 11′ and 22′ is NCM. λθδ ===′−′ sin2 001122 hdNCM haACdh /:indicesMillernFrom 00 == The path difference between 11′ and 33′ is SBR. λλλδ a hx ha x AC AB SBR ====′−′ / 1133
9. 9. → the phase difference between 11′ and 33′ is a hx a hx π λ λ π φ 22 1133 ==′−′ If the position of atom B is specified by the fractional coordinate (normalized to the magnitude of the base vector), u = x/a. hu a hx π π φ 2 2 1133 ==′−′ Extending the argument to other two direction, the atom B has actual coordinates x y z or fractional coordinates x/a y/b z/c (= u v w), respectively. The phase difference between the wave scattered by atom B and that scattered By atoms A at the origin for the hkl reflection: )(2 lwkvhu ++= πφ If u v w is integer, no phase difference.
10. 10. Waves differing in amplitude and phase may also be added by representing them as vectors. A1 A2 A3 φ1φ3 φ2 )2sin( 111 φπν −= tAE )2sin( 222 φπν −= tAE The analytical expression for a vector representing a wave A φ iφi Ae 2 AI = real Any scattered wave can be expressed in the complex exponential form: )(2 lwkvhuii feAe ++ = πφ The resultant wave scattered by all the atoms of the unit cell is called the structure factor (F). F is obtained by simply adding together all the waves scattered by the individual atoms. Assume the unit cell contains atoms 1, 2, 3, …, N, with fractional coordinates u1v1w1, u2v2w2, …, uNvNwN. →
11. 11. )(2)(2 2 )(2 1 222111 NNN lwkvhui N lwkvhuilwkvhui efefefF ++++++ +++= πππ  ∑ ++ = N lwkvhui Nhkl NNN efF 1 )(2π F is, in general, a complex number. |F|: amplitude of the resultant wave in terms of the amplitude of the wave scattered by a single electron. electronsingleabyscatteredamplitude cellunitaofatomsallbyscatteredamplitude =F ♦ Structure factor calculations: How to choose the groups of atoms to represent a unit cell of a structure? First, determine the number of atoms in the unit cell. Second, choose the representative atoms for a cell properly (ranks of equipoints).
12. 12. • Example 1: Simple cubic There are 1 atoms per unit cell, 000 and 100, 010, 001, 110, 101, 011, 111 are all equipoints of rank 1. Choose any one will have the same result. ffeF lkhi hkl == ++ )000(2π 22 fFhkl =→ for all hkl • Example 2: Body centered cubic There are 2 atoms per unit cell, 000 and 100, 010, 001, 110, 101, 011, 111 are all equipoints of rank 1. Another equipoints of rank 1 is ½ ½ ½. Two points to choose are 000 and ½ ½ ½. )1( )( ) 2 1 2 1 2 1 (2 )000(2 lkhi lkhi lkhi hkl effefeF ++ ++ ++ +=+= π π π fFhkl 2= when h+k+l is even 22 4 fFhkl =→ 0=hklF when h+k+l is odd 0 2 =→ hklF
13. 13. • Example 3: Face centered cubic There are 4 atoms per unit cell, 000 and 100, 010, 001, 110, 101, 011, 111 are all equipoints of rank 1. Another equipoints of rank 3 is ½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½. Four atoms chosen are 000, ½ ½ 0, ½ 0 ½, 0 ½ ½. ]1[ )()()( ) 2 1 2 1 0(2) 2 1 0 2 1 (2)0 2 1 2 1 (2 )000(2 lhilkikhi lkhilkhilkhi lkhi hkl eeef fefefefeF +++ ++++++ ++ +++= +++= πππ πππ π fFhkl 4= when h, k, l is unmixed (all evens or all odds) 22 16 fFhkl =→ 0=hklF when h, k, l is mixed 0 2 =→ hklF • Example 4: Diamond Cubic There are 8 atoms per unit cell, 000 and 100, 010, 001, 110, 101, 011, 111 are all equipoints of rank 1.
14. 14. Another equipoints of rank 3 is ½ ½ 0, ½ 0 ½, 0 ½ ½, ½ ½ 1, ½ 1 ½, 1 ½ ½. The other equipoints of rank 4 is ¼ ¼ ¼, ¾ ¾ ¼, ¾ ¼ ¾, ¼ ¾ ¾. Eight atoms chosen are 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. ) 4 3 4 3 4 1 (2) 4 3 4 1 4 3 (2) 4 1 4 3 4 3 (2) 4 1 4 1 4 1 (2 ) 2 1 2 1 0(2) 2 1 0 2 1 (2)0 2 1 2 1 (2 )000(2 lkhilkhilkhilkhi lkhilkhilkhi lkhi hkl fefefefe fefefefeF ++++++++ ++++++ ++ ++++ +++= ππππ πππ π       + ×      +++= ++ ++++++ ) 4 1 4 1 4 1 (2 ) 2 1 2 1 0(2) 2 1 0 2 1 (2)0 2 1 2 1 (2 1 1 lkhi lkhilkhilkhi hkl e eeefF π πππ FCC structure factor Two FCC shifted by ¼ ¼ ¼
15. 15. )1(4 ifFhkl += when h, k, l are all odd 22 32 fFhkl =→ fFhkl 8= when h, k, l are all even and h + k + l = 4n 22 64 fFhkl =→ 0)11(4 =−= fFhkl when h, k, l are all even and 0 2 =→ hklFh + k + l ≠ 4n 0=hklF when h, k, l are mixed 0 2 =→ hklF • Example 5: Close packed hexagonal cell There are two atoms in a unit cell The eight atoms in the corner are equipoints of rank 1. Choose 000 to represent it. The other point is 1/3 2/3 1/2. (000) (001) (100) (010) (110) ( 1/3 2/3 1/2) equipoints ) 2 1 3 2 3 1 (2 )000(2 lkhi lkhi hkl fefeF ++ ++ += π π Set [h + 2k]/3+ l/2 = g
16. 16. )1( 2 ig hkl efF π += gfgfeefF igig hkl ππππ 2222222 cos4)2cos22()1)(1( =+=++= −       + + == ) 23 2 (cos4cos4 22222 lkh fgfFhkl ππ h + 2k l 3m 3m 3m±1 3m±1 even odd even odd       + + ) 23 2 (cos2 lkh π 2 hklF 1 0 0.25 0.75 4f 2 0 f 2 3f 2 • Example 6: NaCl (two kinds of atoms), can be considered as 4 Na at 000 FCC structure + 4 Cl at ½ ½ ½ FCC structure
17. 17. The structure factor:       + ×      +++= ++ ++++++ ) 2 1 2 1 2 1 (2 ) 2 1 2 1 0(2) 2 1 0 2 1 (2)0 2 1 2 1 (2 1 lkhi ClNa lkhilkhilkhi hkl eff eeeF π πππ FCC structure factor )(0 )( lkhi clNahkl effF ++ +×= π when h, k, l mixed 0 2 =→ hklF )(4 clNahkl ffF +×= when h, k, l all even 22 )(16 ClNahkl ffF +=→ )(4 clNahkl ffF −×= when h, k, l all odd 22 )(16 ClNahkl ffF −=→ You can also think of the structure as simple FCC and the atom is replaced by a pair of atoms (1 Na + 1 Cl) with the new structure factor       + ++ ) 2 1 2 1 2 1 (2 lkhi ClNa eff π
18. 18. ♦ Multiplicity Factor: The following factors discussed are related to the diffraction from polycrystalline sample. The multiplicity factor, p, for hkl planes may be defined as the number of permutations of position and sign of ±h, ±k, ±l. It represents the number of equivalent planes to yield the same diffraction cone. E.g. d100, d010, d001, d-100, d0-10, d00-1 form part of the same diffraction cone for cubic crystal → p = 6. If the crystal is tetragonal, d100, d010, d-100, d0-10, form part of the same diffraction cone → p = 4; d001, d00-1 → p = 2 ♦ Lorentz factor: Certain trigonometrical factors, including: Angular distribution of the diffraction (finite spreading of the intensity peak, fraction of crystal contributing to a
19. 19. 2θB Intensity Diffraction Angle 2θ Imax Imax/2 θ 2θ Integrated Intensity B diffraction peak, and the intensity spreading in a cone. • Integrated intensity: (finite spreading) from constructive interference at Bragg angle to complete destructive interference → transition region θB 2θ θ1 θ2 θθθ ∆+= B1 θθθ ∆−= B2 1 2 1′ 2′ A B a θ1 θ2 path difference for 11′-22′ = AD – CB = acosθ2 - acosθ1 = a[cos(θB-∆θ) - cos (θB+∆θ)] = 2asin(∆θ)sinθB ~ 2a∆θ sinθB. C D Na 1′ N′ 2Na∆θ sinθB = λ → completely cancellation (1- N/2, 2- (N/2+1) …)
20. 20. BNa θ λ θ sin2 =∆ Maximum angular range of the peak Imax ∝ ∆θ ∝ 1/sinθB, Half maximum B ∝ 1/cosθB (will be shown later) → integrated intensity of the reflection ∝ ImaxB ∝ (1/sinθB)(1/cosθB) ∝ 1/sin2θB. • The second geometrical factor arises from the fact that the integrated intensity depends on the number of crystals orientated at or near the Bragg angle. → estimated the fraction of the crystal around the angle. See Fig. 4-16. ∆θ crystal plane θθπ ∆⋅−=∆ rrN B )90sin(2  2 cos 4 )90sin(2 2 BB r rr N N θθ π θθπ ∆ = ∆⋅− = ∆  Fraction of crystal:
21. 21. In accessing relative intensity, one should compare the total diffracted energies in different cone. See, Fig. 4-17. For different cones, the radius of the diffracted energy intersecting the film (Hull/Debye- Scherrer film) is different. Assume the total diffracted energy is equally distributed in the cone (2πRsin2θB) → the relative intensity per unit length ∝ 1/sin2θB. • Lorentz factor θθ θ θ θ θ θ cossin4 1 2sin cos 2sin 1 cos 2sin 1 factorLorentz 2 2 = =            = BB B B • Lorentz–polarization factor: (omitting constant) θθ θ cossin 2cos1 factoronpolarizati-Lorentz 2 2 + =
22. 22. ♦ Absorption factor: The factor accounts for the X-ray being absorbed during its in and out of the sample. Two geometries are considered. • Geometry for Hull/Debye-Scherrer Camera: difficult to calculate, different path different absorption, see Fig. 4-19. In general, the absorption factor is written as A(θ). A(θ) varies with θ, qualitatively, A(θ) ↑ as θ↑. • Geometry for Diffractometer: l dx γ I0 2θ A B C dID β x Incident I0, is 1cm2 in cross section, and is incident on the powder plate at an angle γ. The beam incident on the plate is . )( 0 AB eI µ− µ: linear absorption coefficient Let a: volume fraction of the specimen containing particles having 1cm
23. 23. the correct orientation for diffraction of the incident beam; b: the fraction of the incident energy which is diffracted by one unit volume. → volume = l × dx × 1cm = ldx. → actual diffracted volume = abldx → diffracted beam → diffracted beam absorbed before escaping from the sample dxeablI AB)( 0 µ− dxeeablI BCAB )()( 0 µµ −− βγγ sin ; sin ; sin 1 x BC x ABl === dxe abI dI x D       +− = βγ µ γ sin 1 sin 1 0 sin If γ = β = θ → dxe abI dI x D θµ θ sin/20 sin − = µθ µ µ θ µ 2sin 2 2 0 0 sin 2 0 0 abIx de abI dII x x x x x DD ∫∫ ∞= = −∞= = =      ==
24. 24. a varies with θ, being accounted for in the Lorentz factor already. The case discussed is for infinite thickness (defined in the textbook, dID(x = 0)/dID(x = t) = 1000 and γ = β = θ). ♦ Temperature factor: Also known as Debye Waller factor. Atoms in lattice vibrate (Debye model). Temperature ↑ → (1) lattice constants ↑ 2θ ↓; (2) Intensity of diffracted lines ↓; (3) Intensity of the background scattering ↑. d u d u high θB low θB Lattice vibration is more important at high θB (u/d). High θB decrease more! Introducing the temperature factor e-2M . e-2M sinθ / λ 1
25. 25. • Formally, the factor is included in f as Because F = |f 2 | → factor e-2M shows up • What is M? M eff − = 0 22 22 2 2 2 sinsin2 22       =      =         = λ θ λ θ ππ BB Bu d u M u 0=u 02 ≠u : Mean square displacement in the direction normal to the diffraction plane, difficult to calculate, Debye has given the following expression: 2 u 2 2 2 sin 4 )( 6             + Θ = λ θ φ Bx x mk Th M h: Plank’s constant; T: absolute temperature; m: mass of vibrating atom; k: Boltzmann’s constant; Θ: Debye temperature of the substance; x = Θ/T; φ(x): tabulated function (appendix 15)
26. 26. 2 4 2 2 1015.16 Θ × = Θ A T mk Th Put all numbers in and change m to atomic weight (A): • Temperature diffuse scattering: general coherent scattering in all directions by the displacement (u) → contributing to the general background of the pattern; ↑ as θ ↑ • Temperature effect on the diffraction intensity, see Fig. 4-22. The diffraction peak width B ↑ slightly as T ↑
27. 27. ♦ Summary of the intensities of diffraction peaks from polycrystalline samples: • Hull/Debye-Scherrer Camera: M eApFI 2 2 2 2 )( cossin 2cos1 −       + = θ θθ θ • Diffractometer: M epFI 2 2 2 2 cossin 2cos1 −       + = θθ θ A(θ) independent of θ. • Effects that make the above intensity equation invalid: (1) Preferred orientation: like films, wire, not true random orientation of the crystal → Lorentz factor depends on how the crystals orientated (2) Extinction: for crystals with small sizes (10-4 -10-5 cm)→ kinematical theory: intensity is the sum of all diffraction planes
28. 28. For large perfect crystal → dynamical theory is required; takes into account the multiple reflections between diffracted planes (absorption also increases) Actually to match the intensity calculation with the measured intensity is difficult. But, at least the magnitude (very strong, strong, weak) is qualitatively matched.