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L16 - X-ray diffraction Summary 1 due to differences in...

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due to differences in pathlengths, there will be a phase difference between x-rays scattered from P relative to the origin O O P p q ˆ s 0 ˆ s r r the phase shift term is e " 2 # i $ p + q ( ) = e " 2 # i $ r r % ˆ s o " r r % ˆ s ( ) = e 2 # i r r % ˆ s " ˆ s o ( ) $ = e 2 # i r r % r S X-ray diffraction Summary 1
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Summary 2 F ( r S ) = " r r ( ) # e 2 $ i r r % r S d r r = F ( r S ) e i & r S " r r ( ) = 1 V F ( r S ) # e 2 $ i r r % r S d r S The electron density ρ (r) and the scattering pattern F(S) are related by Fourier transformations: F(S) is a complex number with amplitude |F| and phase φ can measure |F| 2 experimentally (intensity of scattered wave) but φ is generally lost and must be obtained indirectly “The phase problem” J.P. Allen, Chap. 15 “X-ray diffraction and EXAFS”
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Applications solution scattering (radial distribution function) For problems with spherical symmetry F ( r S ) = " r x ( ) # e 2 $ i r S % r x d r x reduces to F ( r S ) = " r r ( ) # e 2 $ i r S % r r r 2 sin & d d & dr = d 0 2 $ # r 2 0 ( # " r ( ) dr sin & e 2 $ iSr cos & d & 0 $ # = 2 S r 0 ( # " r ( ) sin 2 $ Sr ( ) dr
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for a uniform sphere, ρ (r) = 1, r < R F ( S ) = 4 " 3 R 3 # $ % & ( 3 sin 2 " SR ( ) ) 2 " SR cos 2 " SR ( ) ( ) 2 " SR ( ) 3 * + , , - . / / first zero, SR = 0.72
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Applications scattering from an oriented sample (fibers, crystals) x 0 a 2a 3a F ( r S ) = G ( r S ) e 2 " i ( n # 1) r a $ r S n = 1 N % = G ( r S ) e 2 " i ( n # 1) r a $ r S n = 1 N % N=1 N=2 N=5 N=20 G(S) = scattering from a single object a•S
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F ( r S ) = G ( r S ) e 2 " ij r b # r S j = $ N N % = G ( r S ) e 2 " ij r b # r S j = $ N N % Fourier transform of a lattice of objects G ( r S ) e 2 " ij r b # r S j = $ N N % 2N+1=3 2N+1=21 F ( r S ) " " = = “convolution theorem”
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Atlas of Optical Transforms, Harburn, Taylor and Welberry
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lattice spacing “a” a= λ /2sin θ ~ λ /2 θ = λ d / d=2400 mm λ = 6.5 x 10 -4 mm =23/6=3.8 mm (vert.) =21/6=3.5 mm (horiz.) calculate: a=0.41 mm (vert.) a=0.45 mm (horiz.) (60 mesh = 0.42 mm) optical diffraction of 60 mesh wire sieve h 0 1 2 3 4 5 6 7 8
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one-dimensional, axial “electron density” profile lattice spacing = 0.42 mm opening = 0.246 mm (0.246/0.42 = 2x0.293 ) F ( S ) = F ( h ) = cos2 " hx dx # 0.293 0.293 $ = sin2 " h (0.293) " h h 0 1 2 3 4 5 6 7 8 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 0 1 2 3 4 5 6 7 8 F(h) h -0.5 -0.293 0 0.293 0.50 fractional 0 0.123 0.21 mm ρ (x) 1 0.246 mm 0.42 mm x
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Resolution and Bragg s Law ˆ s 0 ˆ s θ θ θ d dsin θ dsin θ the minimum spacing d corresponding to constructive interference between scattered waves occurs when 2 d sin " = # or 2sin " # = 1 d = r S (Bragg s Law) S
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