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Diffraction maxima include diffraction from All atoms in the crystal Diffraction in Uniform Direction h=1 k=0 • “Planes” divide unit cell in hkl divisions • Each maxima is from planes with spacing dhkl h=1 k=1 • Smaller spacing yields higher angle maxima: 2 dhkl sin θ = λ 1 h=2 k=1 • “Planes” divide unit cell in hkl divisions h=2 • Each maxima is from planes with spacing dhkl k=3 • Smaller spacing yields higher angle maxima: 2 dhkl sin θ = λ h=2 k = -1 Fhkl is the Fourier Transform of ρxyz Fhkl = V∑∑∑ρxyz cos [2π(hx+ky+lz)] + ρxyz sin [2π(hx+ky+lz)] xyz Structure Factor Electron Density Diffraction in direction hkl is from constructive interference of diffraction by all electrons xyz 2 Waves and Diffraction Fhkl α √Ihkl Fhkl = Fhklcos αhkl + iFhklsin αhkl Phase lost Diffraction by Atoms Fhkl = V∑∑∑ρxyz cos [2π(hx+ky+lz)] + ρxyz sin [2π(hx+ky+lz)] xyz Fhkl = ∑ fi cos [2π(hx+ky+lz)] + fi sin [2π(hx+ky+lz)] N Atomic Scattering Factor Note: the volume (V) is used in the top equation to account for ρ being expressed in electrons per unit volume, whereas F is in electrons per unit cell. 3 Argand Diagram Vector Addition of Atomic Scattering Factors Fhkl 4 Calculating Electron Density ρxyz = 1/V∑∑∑ Fhkl cos [2π(hx+ky+lz) + αhkl] hkl - Fhkl sin [2π(hx+ky+lz) + αhkl] A question: Crystal A is identical to crystal B, except crystal B contains one less atom in each molecule. How will the diffraction patterns of the two crystals differ? (Hint: consider the Fourier summation for Fhkl) 5