X-ray Powder Diffraction II Peak Intensities Chemistry 754 Solid State Chemistry Lecture #10 Outline • Elastic scattering of X-rays by atoms • Diffraction peak intensities – – – – Structure Factors Multiplicity Lorentz and Polarization Factors Temperature Factors • Neutron diffraction • Electron diffraction 1 Interaction of X-rays with Matter These are the diffracted X-rays Elastic Scattering by an Electron • Charged particles (electron) scatter electromagnetic radiation (x-rays) – The varying electric field of the X-ray induces an oscillation of the electron – The oscillating electron then acts as a source of electromagnetic radiation – In this way the x-rays are scattered in all directions • JJ Thompson analyzed the scattering and found that: I = I0[(μ0/4p)2(e4/m2r2)sin2α] I = I0(K/r2) sin2α α is the angle between the scattering direction and the direction in which the electron is accelerated r is the distance from the scattering electron K is a constant 2 Scattering by an Atom We can consider an atom to be a collection of electrons. The electrons around an atom scatter radiation in the manner described by Thompson. However, due to the coherence of the radiation we need to consider interference effects from different electrons within an atom. This leads to a strong anglular dependence of the scattering. We express the scattering power of an atom by its form factor (f). X-ray Form Factors 25 The form factor is equivalent to the atomic number at θ = 0 Form Factor, f 20 Ca Ca2+ 15 10 The form factor drops rapidly as a function of (sin θ)/λ 5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 Sin(θ)/λ 3 The Effect of Form Factors on Diffraction Patterns 4500 TbBaFe2O5 - 70 K Neutron Data TbBaFe2O5 - 300 K Synchrotron X-ray 4000 Intensity (Arb. Units) 3500 Intensity 3000 2500 2000 1500 1000 500 0 5 25 45 65 2-Theta (Degrees) The peak intensities drop off at high angles in an X-ray diffraction pattern because the form factor decreases 10 30 50 70 90 110 130 2-Theta (Degrees) Neutrons are scattered from the nucleus and the form factor is not angle dependent. Intensities do not drop off at high angle. Diffraction Intensities The integrated intensity (peak area) of each powder diffraction peak is given by the following expression: I(hkl) = |S(hkl)|2 × Mhkl × LP(θ) × TF(θ) – – – – S(hkl) = Structure Factor Mhkl = Multiplicity LP(θ) = Lorentz & Polarization Factors TF(θ) = Temperature factor (more correctly referred to as the displacement parameter) This does not include effects that can sometimes by problematic such as absorption, preferred orientation and extinction. 4 Structure Factor The structure factor reflects the interference between atoms in the basis (within the unit cell). All of the information regarding where the atoms are located in the unit cell is contained in the structure factor. The structure factor is given by the following summation over all atoms (from 1 to j) in the unit cell: S(hkl) = Σj fj exp {-i2π(hxj+kyj+lzj)} – fj = form factor for the jth atom – h, k & l = Miller indices of the hkl reflection – xj, yj & zj = The fractional coordinates of the jth atom To evaluate the value of the imaginary term in the exponential function, remember Euler’s equation: exp (-ix) = cos(x) – i sin(x) The value of this function is a real number when x is a multiple of 2π. It is equal to 1 for even multiples of 2π and -1 for odd multiples of 2π. Structure Factors: Example CsCl • Let’s calculate the structure factors for the first 6 peaks CsCl. To do this we need to know the atomic positions and the Miller Indices. Cl at (0,0,0) Cs at (½,½,½) 5 Structure Factors: Example CsCl S(hkl) = Σj fj exp {-i2π(hxj+kyj+lzj)} S(hkl) = fCl exp {-i2π([h×0]+[k×0]+[l×0]} + fCs exp {-i2π([h×½]+[k×½]+[l×½]} S(hkl) = fCl exp {0} + fCs exp {-iπ(h+k+l)} S(hkl) = fCl – fCs when h+k+l is an odd number S(hkl) = fCl + fCs when h+k+l is an even number S(100) = fCl - fCs S(111) = fCl - fCs S(210) = fCl - fCs S(110) = fCl + fCs S(200) = fCl + fCs S(211) = fCl + fCs Systematic Absences: Body Centering Recall that earlier we said that for a body centered space group all of the peaks where h+k+l is an odd number will be missing. Consider the structure factors for sodium which has the BCC structure with Na atoms at (0,0,0) and (½,½,½). The equations will be just like CsCl, but the two form factors will be equal. S(hkl) = fNa exp {-i2π([h×0]+[k×0]+[l×0]} + fNa exp {-i2π([h×½]+[k×½]+[l×½]} S(hkl) = fNa exp {0} + fNa exp {-iπ(h+k+l)} S(hkl) = fNa – fNa = 0 when h+k+l is an odd number S(hkl) = fNa + fNa = 2fNa when h+k+l is an even number 6 Int. Na (BCC) (110) 12000 10000 4000 (220) (200) 6000 (211) 8000 2000 Simulated X-ray powder diffraction patterns [Calculated with Diamond] 0 20 30 40 2Theta 50 60 Int. (210) (220) (110) 20000 (200) 30000 (111) 40000 (100) 50000 (211) CsCl 60000 10000 0 20 30 40 2Theta 50 60 Multiplicity Factor In a powder diffraction experiment the d-spacings for related reflections are often equivalent. Consider the examples below: 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 In general for a cubic system where the Miller indices are n1, n2 and n3 (all unequal) the multiplicity factors Mhkl are: n100 (i.e. 100) → M = 6 n1n10 (i.e. 110) → M = 12 n1n1n2 (i.e. 221) → M = 24 n1n1n1 (ie 111) → M = 8 n1n20 (ie 210) → M = 24 n1n2n3 (ie 321) → M = 48 The multiplicities are lower in lower symmetry systems. For example in a tetragonal crystal the (100) is equivalent with the (010), (-100) and (0-10), but not with the (001) and the (00-1). 7 Lorentz Factor There are a number of factors that lead to a theta dependence of the peak intensities (integrated intensities). • Diffraction can occur for angles slightly different from the value predicted by Bragg’s Law → I α 1/sin(2θ) • The number of crystals oriented in such a way as to satisfy Bragg’s Law is highest for low angles → I α cos(θ) • The fraction of the diffraction cone that intersects the detector is highest at low angles → I α 1/sin(2θ) When combine these considerations and do some trigonometric manipulation we get the Lorentz Factor: I α 1/(4sin2θ cosθ) Polarization and the LP Factor An X-ray propagating in the x-direction will have an electric vector oriented in the yz plane. The y and z components of the X-ray will be scattered differently because the angle between the scattered beam and the electric field gradient will differ, as derived by Thompson. This leads to the polarization factor: I α (1+cos22θ)/2 Typically the Lorentz and Polarization terms are combined to give the Lorentz-Polarization (LP) factor. I α (1+cos22θ)/(8sin2θ cosθ) 8 Temperature Factor The vibrations of atoms in a crystal lead to an angle dependent effect on the diffracted peak intensities. The more an atom vibrates the scattering power is decreased, because the scattering power of the atom is smeared out. As an approximation we can assume that all atoms vibrate equally. In that case the TF can be expressed by the equation: TF(θ) = exp {B(sin θ/λ)2} Where the coefficient B is the isotropic temperature factor. – B is proportional to the mean squared displacement of the atoms – Typical values for inorganic extended solids range from 0.5 to 1.5. – The temperature factor has its biggest impact at high angles. More sophisticated treatments assume – Different values of B for each crystallographically independent atom – Anisotropic vibrations (elliptical in shape rather than spherical). 5 1.2 Lorentz-Polarization 4 3.5 1 0.8 3 2.5 0.6 2 0.4 1.5 1 Temperature Factor LP LP Inc. Beam Mono Temp. Factor (B=1) Temp. Factor (B=2) 4.5 0.2 0.5 0 0 50 100 2-Theta 150 0 200 9 Int. (220) (210) 20000 (200) 30000 (111) 40000 (211) CsCl (110) 50000 (100) Intensity Calculations for CsCl 60000 10000 0 20 h k l 0 0 1 0 0 1 0 0 1 1 0 1 1 2 1 1 1 2 2 2 2 2Theta [°] d-spacing [Å] 21.54 30.64 37.76 43.88 49.39 54.47 63.80 4.123 2.915 2.380 2.062 1.844 1.683 1.458 30 40 2Theta S LP 34.66 58.72 31.54 51.98 29.32 47.68 44.64 6.80 3.23 2.05 1.47 1.12 0.90 0.63 50 60 Mult. Relative Int. 6 12 8 6 24 24 12 37 100 12 18 17 37 11 Neutron Diffraction Diffraction from crystals can also be carried out with neutron and electrons. The basic concept are the same though there are important distinctions. Let’s begin with the DeBroglie relationship. λ = h/p = h/mv but the kinetic energy, E, and velocity, v, of a particle are related by the expression E = (1/2)mv2 → v = sqrt(2E/m) combining the two relationships gives λ = h/sqrt(2mE) finally for thermal neutrons from a reactor source the energy of an electron is E = kT so that λ = h/sqrt(2kT) Typical values for a neutron reactor source are 300<T<400K and 1.0 A< λ < 2.5 A. 10 Reactor Diffractometers Thermal Neutron Instruments at NIST BT1 Powder Diffractometer BT1 Powder Diffractometer at NIST 11 X-rays vs. Neutrons • Neutron beams are highly penetrating (penetration depths of centimeters-decimeters), whereas X-rays are highly attenuated by matter (penetration depths of microns-millimeters). – Larger samples are needed for neutron diffraction • Neutrons are scattered from nuclei, whereas X-rays are scattered by electrons – Neutron scattering factors do not drop off with increasing θ • Neutron scattering power varies irregularly with atomic number and mass number, whereas X-ray scattering power varies smoothly with atomic number. – Neutrons are more sensitive to light elements (H, N, O, F, C, etc.) and isotope distributions – Some elements strongly absorb neutrons and cannot be easily studied (B, Cd, Sm, Eu, Gd, Dy, Hf, Ir, Hg), some are essentially transparent to neutrons (V) or have negative scattering lengths (Ti, Mn) • Neutrons have an intrinsic magnetic moment, whereas X-rays have no magnetic moment – Neutrons are used to determine ordered magnetic structures Electron Diffraction Diffraction from crystals can also be carried out with neutron and electrons. The basic concept are the same though there are important distinctions. Let’s begin with the DeBroglie relationship. λ = h/p = h/mv but the kinetic energy, E, and velocity, v, of a particle are related by the expression E = (1/2)mv2 → v = sqrt(2E/m) combining the two relationships gives λ = h/sqrt(2mE) The energy of an electron is related to the accelerating voltage, V, by the relationship E = eV. Upon correcting for relativistic effects we get an approximate relationship between the wavelength (in Angstroms) and the voltage λ = sqrt(150/V) Typical values for a electron microscope are V ~ 100 kV and λ ~ 0.04 Angstroms 12 Transmission Electron Microscope This is the CM300 UltraTwin Field Emission Gun TEM. It is located in the Campus Center for Electron Optics (CEOF) in the MSE department X-rays vs. Electrons • Electrons are highly attenuated by matter (penetration depth of nanometers), whereas X-rays are strongly attenuated by matter (penetration depths of microns-millimeters). – Data collection done under high vacuum – Data collection is usually a transmission measurement (TEM) on a thin sample or crystal • Electrons beams can be highly focused, whereas X-rays are difficult to focus – The electron beam is usually focused onto a very small micro-single crystal • Electron scattering power drops off strongly as a function of 2theta, whereas X-ray scattering power drops off smoothly as a function of 2-theta – Data sets are often collected over a very limited angular range (± 4° 2-theta) • Diffracted electrons can interact strongly with the lattice making it difficult to get reliable intensities, whereas X-ray intensities and peak positions can be determined with a great deal of accuracy – Structure refinement and solution is challenging 13