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Molecular Biophysics Diffraction theory (2)

Molecular Biophysics Diffraction theory (2). F cryst ( S ) = n a n b n c F 0 ( S ). if and only if 2 p a . S = 2 p b . S = 2 p c . S = 0 or 2 p or 4 p etc. }. h , k , l are integers, called the Miller indices of the scattered beam. Diffraction from a crystal.

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Molecular Biophysics Diffraction theory (2)

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  1. Molecular Biophysics Diffraction theory (2)

  2. Fcryst(S) = nanbncF0(S) if and only if2pa.S = 2p b.S = 2p c.S = 0 or 2p or 4p etc. } h, k,l are integers, called the Miller indices of the scattered beam Diffraction from a crystal Fcryst(S) = F0(S) Sexp {2pi ta.S} Sexp {2pi ub.S} S exp {2pi vc.S} a.S = h b.S = k c.S = l These conditions are called the Laue conditions

  3. Diffraction from a crystal

  4. a.S = h b.S = k c.S = l If , then c* c b* b a g b a* a Diffraction from a crystal S = ha*+ kb* + lc* a.a*= 1 a.b*= 0a.c*= 0 b.a*= 0 b.b*= 1b.c*= 0 c.a*= 0 c.b*= 0c.c*= 1 a*, b*and c*are the reciprocal lattice vectors

  5. Diffraction from a crystal: the reciprocal lattice

  6. b b* a* a Diffraction from a crystal

  7. Diffraction from a crystal: the reciprocal lattice

  8. |F| |S| A perfect crystal

  9. exp{-Bsin2q/l2} |F| |S| A real crystal: (1) translational disorder

  10. The temperature (B) factor Debye-Waller temperature factor B = 8p<u>2

  11. Diffraction from a real crystal

  12. Bragg‘s law Bragg (1913): diffraction as reflection from crystal planes path difference = 2dsinq For constructive interference, nl = 2dsinq q q d |S| = 2sinq/l = 1/d If a diffraction pattern fades out at an angle of 2qmax, then dmin = l / 2sinqmax This is termed the resolution of the pattern

  13. High angle reflections = high resolution

  14. A perfect crystal |F| |S|

  15. |F| |S| A real crystal: (2) local disorder

  16. A perfect crystal |F| |S|

  17. |F| |F| e |S| |S| A real crystal: (3) rotational disorder e is called the mosaicity

  18. Fcryst(S) = nananaF0(S) if and only if2pa.S = 2p b.S = 2p c.S = 0 or 2p or 4p etc. } h, k,l are integers, called the Miller indices of the scattered beam Diffraction from a crystal Fcryst(S) = F0(S) Sexp {2pi ta.S} Sexp {2pi ub.S} S exp {2pi vc.S} a.S = h b.S = k c.S = l These conditions are called the Laue conditions

  19. S s s0 q a* b* Ewald‘s construction What does the diffraction pattern of a 3d-crystal look like? S = s - s0 = ha*+ kb* + lc* For a reflection to occur, the circle (sphere) must intersect with a point on the reciprocal lattice. This sphere is the Ewald sphere. |s0| = |s| = 1/l

  20. Diffraction from a crystal

  21. rj=xa+ yb + zc x,y and z are termed fractional coordinates Diffraction from a crystal Fcryst(S) = nananaFcell(S) Fcell (S) = cellr(r)exp (2pir.S) d3r S = ha*+ kb* + lc* Fcell (hkl) = cellr(xyz)exp (2pi{hx+ ky+lz}) d3r

  22. F(hkl) a(hkl) a(-h-k-l) F (-h-k-l) Friedel‘s law F(hkl) = cellr(xyz)exp (2pi{hx+ ky+lz}) d3r F(-h-k-l) = cellr(xyz)exp (2pi{-hx+ -ky+-lz}) d3r |F (hkl)| = |F (-h-k-l)| F (-h-k-l) and F (hkl) are complex conjugates F(-h-k-l) = F*(hkl) I(hkl) = F(hkl) . F*(hkl) = |F(hkl)|2

  23. The electron density equation F(hkl) = cellr(xyz)exp (2pi{hx+ ky+lz}) d3r r(xyz)= ShklF(hkl) exp (-2pi{hx+ ky+lz}) But we can only measure the intensity I(hkl) = F(hkl) . F*(hkl) = |F(hkl)|2 We have lost the phase information: this is the fundamental problem in X-ray crystallography – The PHASE PROBLEM

  24. The phase problem

  25. The phase problem

  26. Influence of intensities Influence of phases The phases are more important than the amplitudes!!!!

  27. so we have to use diffraction

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