L16 - due to differences in pathlengths, there will be a...

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Unformatted text preview: 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 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 X-ray 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 X-ray 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 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 one-dimensional, 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.

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L16 - due to differences in pathlengths, there will be a...

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