1. Theory of kinematical di ffraction: a practical recap by Carmelo Giacovazzo Istituto di Cristallografia, CNR Dipartimento Geomineralogico, Bari University

2. The Thomson scattering model A X-ray plane wave propagates along the x direction, the scatterer is in O. Q is the observation point, in the plane (x,y). F= e E ; a= F/m I eth = I i sin 2  [e 4 /(m 2 r 2 c 4 )]  is the angle between the acceleration direction of the electron and the observation direction.

3. §If the beam is non-polarised, we can divide the overall incident beam into two components with the same intensity( = 0.5I i ), the first polarized along the y axis, the second along the z axis. § For the first component φ=(π/2-2θ) and §I ey = 0.5I i [e 4 /(m 2 r 2 c 4 )] cos 2 2 . §For the second component φ=π/2 and §I ez =0.5 I i [e 4 /(m 2 r 2 c 4 )] §Therefore § I eTh =I i [e 4 /(m 2 r 2 c 4 )] [(1+cos 2 2  )/2]

4. X-Ray versus e-diffraction -- e-scattering is caused by interaction with the electrostatic field ( sum of nucleus and of electron cloud); --e-scattering is weakly dependent on the atomic number ( light atom localization) -- e- absorption is strong; --e-interaction is thousands stronger than X-ray interaction: nanocrystals, dynamical effects

5. The Compton effect §The corpuscolar interpretation of the quantomechanics lead Compton to consider an energy transfer from the photon to the electron. § The radiation is incoherent ( no phase relationship between the incident and the diffuse beam). That induces a wavelength change equal to §  =0.024 (1-cos2  )

6. The interference §We will investigate the interference mechanism which governs the diffusion from an extended body. §We will start with two charged particles, one in O and the other in O’, both interacting with a X-ray beam.

7. 4 §AO = -r.s o ; BO= r.s ; AO+BO = r.(s-s o ) §(AO+BO)/ = r.(s-s o )/ = r.r* where r*= (s-s o )/ §  = 2  r*.r r*=2sin  / §F(r*) = A o + A o’ exp(2  ir*.r) §F(r*) =  j A j exp(2  ir*.r j ) =  j f j exp(2  ir*.r j ) §where f 2 = I/I eth

8. §If the scatterer distribution is a continuoum then §You can easily recognize that the amplityde F(r*) is the Fourier transform of the electron density. §Viceversa §The above formulas are quite general: they will be applied to the cases in which  (r) represents the electron density of an electron, of an atom, of a molecule, of the unit cell, of the full crystal.

9. §The scattering amplitude of an electron §An example : §the atom C (2x1s;2x2s;2x2p) §(  e ) 1s = (c 1 3 /  ) exp(-2c 1 r); §(  e ) 2s = c 2 5 /(96  ) r 2 exp(-c 2 r) §I coe +I incoe =I eTh §  I incoe =I eTh - I coe =I eTh -I eTh f e 2 § =I eTh (1-f e 2 ) 6

10. I coe =I eTh f a 2 =I eTh (Σ j=1,..,Z f ei ) 2 I incoe =I eTh (Σ j=1,..,Z [1-f ej 2 ])

11. The atomic thermal factor  a (r) is the electron density for an atom isolated, located at the origin. Owing to the thermal agitation, the nucleus moves at the instant t into r’: then the electron density will be  a (r- r’). Let p(r’) the probability of that movement: then

13. Scattering amplitude of a molecule Let  j (r) be the electron density for an atom isolated, thermically agitated and located at the origin. Then  (r-r j ) will be the electron density of the same atom when its nucleus is in r j. Let  M (r ) be the electron density of a molecule:.

14. Scattering amplitude of an infinite crystal As we know the electron density of a crystal may be expressed as a convolution:

15. Scattering amplitude of a finite crystal A finite crystal may be represented as the product where  (r) is the function form which is equal to 1 inside the volume of the crystal, equal to 0 outside. Then

18. The structure factor

20. The Ewald sphere OP=r* H =1/d H = IO sin  = 2sin  /

21. The Bragg law 2d H sin  = n It may be written as 2(d H /n ) sin  = or also 2d H’ sin  = where H’= (nh, nk, nl). Saying r*= r* H is equivalent to state the Bragg law: r* = 2sin  / =1/d H

22. The symmetry in the reciprocal space The Friedel Law F H = A H + iB H ; F -H = A H - iB H  -H = -  H I H  (A H + iB H )(A H - iB H ) = A H 2 +B H 2 I -H  (A H - iB H )(A H + iB H ) = A H 2 +B H 2 The intensities show an inversion centre.

23. The effect of a symmetry operator in the reciprocal space The general expression of the structure factor is Let C  (R,T) be a symmetry operator. Consider the quantity

24. Find the Laue group Apply the Friedel law and the symmetry induced law | F HR | =| F H | to find the Laue groups. ---------------------------------------------------- Laue group P1 | F h k l |= |F -h -k -l | -1 P-1 “ “ ----------------------------------------------------------- P2 |F hkl | =| F -h k -l |= |F -h-k-l |=| F h -k l | 2/m Pm ‘’ ‘’ ‘’ ‘’ P2/m ------------------------------------------------------------- P222 | F hkl | = | F h -k -l | = |F -h k -l | = |F -h -k l | 2/m 2/m 2/m = |F -h -k -l | = |F -h k l | = | F h -k l | = |F h k -l |

25. Find the systematically absent reflections Find the set of reflections for which Once you have found it, check if the product HT is not an integral value. In this case the ( true) equation is violated, unless |F H | = 0. The reflection is systematically absent.

26. Space group determination: the single crystal scheme a)First step : lattice analysis, which provides the Unit Cell parameters. E.g. a = 11.10(2), b = 14.27(3), c = 9.72(2)  = 90.25(37),  = 89.48(30),  = 90.32(35) Is the unit cell orthorhombic ? b)Second step: symmetry analysis In the example above, three two-fold axes should be present to confirm the orthorhombic nature of the crystal.

27. The symmetry analysis 1) First step: identification of the Laue group via the recognition of the “ symmetry equivalent reflections”. The Laue group generally identifies without any doubt the crystal system ( pseudo-symmetries, twins etc. are not considered). b)Second step: identification of the systematically absent reflections.

28. The Laue Group Symmetry operators : Then are symmetry equivalent positions in direct space. In the reciprocal space we will observe the symmetry equivalent reflections If the space group is n.cs., the Friedel opposites should be added: the complete set is

29. The Laue group The equivalent reflections are

30. The systematically absent reflectio ns Since we have If hR s =h and hT s =n the relation (1) is violated unless the reflection h is a systematically absent reflection. The combination of the information on the Laue Group with that on the systematically absent reflections allows the determination of the Extinction Symbol.

31. The extinction symbols (ES) In the first position: cell Centric type ( e.g., P - - a, in orth.) Then the reflection conditions for each symmetry direction are given. Symmetry directions not having conditions are represented by a dash. A symmetry direction with conditions is represented by the symbol of the screw axis or glide plane. Table of ES are given in IT per crystal system. There are 14 ES for monoclinic, 111 for orthorhombic, 31 for tetragonal, 12 for trigonal-hexagonal, 18 for cubic system

32. Extinction symbol and compatible space groups ES P - - - P222, Pmmm, Pm2m, P2mm, Pmm2 (in orth.) P - - a Pm2a, P2 1 ma, P mma (in orth.) P - - - P4, P-4, P4/m, (in tetrag.) P422, P4mm, P-42m, P-4m2 P4/mmm P6 1 -- P6 1 P6 5 P6 1 22 P6 5 22

33. Find the reflections with symmetry restricted phase values Look for the set of reflections for which Since  HR =  H - 2  HT for that set of reflections it will be  -H =  H -2  HT, or 2  H =2  HT or  H =  HT+ n 

34. Some exercises §Find the rotation and translation matrices for the space group P4 1, and P3 1 ; §Find the equivalent reflections for the same space groups; §Find the systematically absent reflections for the space group P2 1 /c, P4 1,P3 1, P2 1 2 1 2 1

35. The diffraction intensities I H = K 1 K 2 I 0 L P T E |F H | 2 where I 0 is the intensity of the direct beam, I H is the integrated intensity of the reflection H, K 1 = e 4 /(m 2 c 4 ) K 2 = 3  /V P  polarisation factor (e.g., [1+cos 2 2  ]/2 L  Lorentz factor T  Transmission fctor ( e.g., I/I oss )

36. Why integrated intensity? The diffraction conditions are verified on a domain owing to: a) the finitness of the crystal b)non vanishing divergence of the incident radiation; c) non-monochromatic incident radiation; d) crystal defects

37. Electron density calculation

38. Electron density map and finitness of the diffraction data The limiting sphere (and/or other factors) limits the number of measurable diffraction intensities. Let  (r*) be the form function of the measured reciprocal space( equal to 1 in the volume containing the measured reflections, equal to 0 elsewere).

39. CodeSGASUCOLLRES/N RE S COMP RES R int aker a P -4 2 1 mO 14 Mg 2 Al 6 Si 4 Ca 4 P0.40/339238.8 anatase b I 4 1 /a m dO 8 Ti 4 P0.25/2672213.8 barite c P n m aO 16 S 4 Ba 4 P-A0.77/3558215.3 charoite d P 2 1 /mCa 24 K 14 Na 10 Si 72 O 186 P-A1.18/28789713.3 cnba e P 2 1 /cC 29 N H 17 A0.76/39758321.9 gann f P 6 3 m cGa 2 N 2 P0.25/170324.5 mayenite g I -4 3 dO 88 Al 28 Ca 24 P0.54/2284421.5 mfu41 h F m -3 mZn 2 Cl N 2 C 3 OP-A1.09/6559629.2 mullite i P b a mAl 4.56 Si 1.44 O 9.72 P-A0.76/2138630.9 natio j C 2/mO 3 Ti 1.5 Na 0.5 P-A0.76/5177314.2 natrolite k F d d 2Na 2 (Al 2 Si 3 O 10 ) (H 2 O) 2 P-A0.75/7439920.6 srtio3 l P m -3 mO 3 Ti 1 Sr 1 P0.26/1804318.8 srtio3_s m P b n mO 12 Ca 4 Ti 4 P0.26/464613.8 zn8sb7 n P -1Zn 32 Sb 28 P-A0.77/39435417.0

40. CodeORD L EX T ORD7ORD8 aker1P -4 21 m33 anatase1I 41/a m d11 barite1P n m a11 charoite1P 21/m11 cnba1P 21/c11 gann1P 63 m c11 mayenite1I -4 3 d12 mfu411F m -3 m1>100 mullite1P b a m33 natio1C 2/m11 natrolite1F d d 211 srtio31P m -3 m31 srtio3_s1P b n m6036 zn8sb74P -1--