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Unformatted text preview: due to differences in pathlengths, there will be a phase difference between xrays scattered from P relative to the origin O O P p q s s r r the phase shift term is e " 2 # i $ p + q ( ) = e " 2 # i $ r r % s o " r r % s ( ) = e 2 # i r r % s " s o ( ) $ = e 2 # i r r % r S Xray diffraction Summary 1 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 (r) and the scattering pattern F(S) are related by Fourier transformations: F(S) is a complex number with amplitude F and phase can measure F 2 experimentally (intensity of scattered wave) but is generally lost and must be obtained indirectly The phase problem 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 (r) and the scattering pattern F(S) are related by Fourier transformations: F(S) is a complex number with amplitude F and phase can measure F 2 experimentally (intensity of scattered wave) but is generally lost and must be obtained indirectly The phase problem J.P. Allen, Chap. 15 Xray diffraction and EXAFS 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 2 $ # r 2 ( # " r ( ) dr sin & e 2 $ iSr cos & d & $ # = 2 S r ( # " r ( ) sin 2 $ Sr ( ) dr for a uniform sphere, (r) = 1, r < R F ( S ) = 4 " 3 R 3 # $ % & ( 3 sin 2 " SR ( ) ) 2 " SR cos 2 " SR ( ) ( ) 2 " SR ( ) 3 * + , , . / / Frst zero, SR = 0.72 Applications scattering from an oriented sample (Fbers, crystals) x 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 aS 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 Atlas of Optical Transforms, Harburn, Taylor and Welberry lattice spacing a a= /2sin ~ /2 = d / d=2400 mm = 6.5 x 104 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 onedimensional, axial...
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This note was uploaded on 10/27/2010 for the course BI 110 taught by Professor Richards,j during the Winter '08 term at Caltech.
 Winter '08
 Richards,J
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