Where are atoms located in crystals?
•Remember crystal structure = lattice + basis. So far we have only treated the measurement of diffraction peak angular positions, which provides information to specify the lattice (lattice parameters and Bravais lattice). •Now we treat the measurement of diffraction peak intensities, which allows the positions/arrangements of the atoms in the basis to be determined. •Consider the figure below (left), 2 atoms/unit cell, either is derivable by shift of one atom by the vector ½ c. •Consider reflections from the (001) planes, in (a) C base-centered lattice suppose Bragg law is satisfied for particular values of q and l used. This means that the path difference ABC between rays 1’ and 2’ is 1 wavelength and in phase. Similarly in I body-centered lattice in (b), rays 1’ and 2’ are also in phase, however additional plane of atoms with path difference DEF at ½ wavelength. Thus rays 1’ and 3’ are completely out of phase and annul each other. Similarly ray 4’ if shown in next plane down would annul ray 2’, and so on. Thus, you see no (001) reflections for any body-centered (I) lattice (I-cubic, I-tetragonal, or I-orthorhombic):
1 l difference Diffraction from between (001) planes A and A’:
However, the (002) reflection will be present since the ray from B.D. Cullity “Elements of XRD” from B plane is displaced by 1 l and B plane is in-phase with A and A’ rays: See handout, Ex. 5.6 : solves this
Diffraction from geometrically/ (002) planes mathematically class25/1 Diffraction from (001) planes Recall - Example of BCC XRD Pattern
•The unit cell size and geometry may be resolved from the angular positions of the diffraction peaks. •When the distance between the peak spacings are all pretty much the same (2q) it is likely cubic. •The arrangement of atoms within the unit cell is associated with the relative intensities of these peaks z z z (200) c (110) c c (211)
y (110) y y a b a b a b x x x (211) Diffraction d-spacing decreases pattern for according to Bragg’s Law polycrystalline (200) Intensity (relative) Intensity a-iron (BCC)
Diffraction angle 2q h+k+l even are only allowed according to previously shown class25/2 selection rules for BCC. Intro to Structure Factor in Crystals
•This example shows how a simple rearrangement of atoms within unit cell can eliminate reflection completely. The directions of these rays are fixed by Bragg’s law, however if Bragg’s Law is not satisfied, no diffracted ray will occur. However, as we just saw, Bragg’s law may be satisfied for a certain set of atomic planes (001) and yet no diffraction occurs because annulled by (002) planes. •More generally, the intensity of diffracted beam is changed, not necessarily to 0, by any change in atomic positions, and conversely we can determine atomic positions only by diffracted intensities. •Now we consider this relation between atom position and intensity….. •Have to take this step by step: x-ray scattering by a single electron (oscillatory motion), then by an atom (atomic scattering factor), and finally by all the atoms in a unit cell. •Assuming that Bragg’s law is satisfied, we want to find the intensity of the beam diffracted by a crystal as a function of atom position function of structure factor….. •As in the case of phase differences occurring in waves scattered by the individual electrons, individual atoms in unit cell also have waves scattered. This phase difference depends on the arrangement of the atoms. •Phase difference (f) = 2p(hu+kv+lw) [1] between wave scattered by atom B and atom A (at origin) for given hkl reflection. uvw are atom coordinates in x,y,z directions for atom B. (0,0,0) for values are tabulated atom A. for atoms and ions •These two waves may differ, not only in phase, but also in amplitude if atom B and atom A are of different kinds. These amplitudes of the waves are given, relative to amplitude of the wave scattered by a single electron, by the appropriate values class25/3 of f, the atomic scattering factor.See table (Appendix 12) handout. Intro to Structure Factor in Crystals (continued) •We now see that the problem of scattering from a unit cell resolves itself into one of adding waves of different phase and amplitude in order to find the resultant wave. •Waves scattered by all the atoms of the unit cell including the one at the origin must be added. •The most convenient way of carrying out this summation is by expressing each wave as a complex exponential function, which contains a real and an imaginary part (i=√-1). •For a wave vector in the complex plane it can be found that eix cos x i sinx or Ae if Acosf Ai sinf •Now return to problem of adding the scattering waves from each of the atoms in the unit cell. •The amplitudes of each wave is given by the appropriate value of f for the scattering atom considered and the value of sinq/l involved in the hkl reflection. •The phase of each wave is given by [1] in terms of hkl reflection considered and the uvw coordinates of the atom. •Using previous relations, any scattered wave in the complex exponential form is: Ae if fe2pi(hukvlw) •The resultant wave scattered by all the atoms of the unit cell is called the structure factor, because it describes how the atom arrangement, given by uvw for each atom, affects the scattered beam. •The structure factor (F) is obtained by simply adding together all the waves scattered by the individual atoms in the basis.
•If a unit cell contains atoms 1,2,3,…N (atoms/unit cell) with coordinates u1v1w1, u2v2w2, etc. and atomic scattering factors f1,f2,etc. then the structure factor for the hkl reflection is given by: N This equation tells us what reflections (hkl) to expect in a 2pi(hu kv lw ) F f e n n n [2] diffraction pattern from a given crystal structure with atoms hkl n located at positions u,v,w. KNOW THIS RELATION. class25/4 1 Calculating Structure Factors
•F is in general a complex number and it expresses both the amplitude and phase of the resultant wave. •Its absolute value |F| gives the amplitude of the resultant wave in terms of the amplitude of the wave scattered by a single electron: |F|=amplitude of the wave scattered by all the atoms of u.cell (see Cullity handout, p.120) amplitude of the wave scattered by one electron •The intensity (I) of the beam diffracted by all the atoms of the unit cell in a direction predicted by Bragg law is proportional to |F|2, the square of the amplitude of the diffracted beam, and |F|2 is
obtained by multiplying equation [2] for Fhkl by its complex conjugate (see handout, p.228)….. *Thus, equation 2 permits calculation of the intensity of any hkl reflection from a knowledge of the atomic positions.* •In calculating structure factors by complex exponential functions, many particular relations often occur and verified by means of eix cos x i sinx pi 3pi 5pi e e e 1 In general enpi (1)n where n is any integer e2pi e4pi e6pi 1 In general enpi enpi where n is any integer eix eix 2cos x Example 1: simplest case is that of unit cell containing only one atom at origin, uvw=000 basis. F fe2pi(hukvlw) fe2pi(0) f and I F 2 f 2 F2 is thus independent of h,k, and l and is the same for all reflections, e.g. Primitive cubic, a-Po (all reflections allowed, as we discussed last class.). Any primitive cell with this elemental basis will also have all hkl’s present, e.g. As, Bi (Rhombohedral lattice). Reason is F is independent of the shape and size of the unit cell. The lattice parameters never entered into Eq. [2]. class25/5 Calculating Structure Factors (continued) Example 2: consider previous body-centered (I) unit cell containing two atoms of the same kind located at origin, uvw=000 and uvw=½, ½, ½ basis. F fe2pi (hukvlw) fe2pi (0) fe2pi (h/ 2k / 2l / 2) F f [1 epi (hkl ) ] S= 2 4 6 etc. F 2 f when (h+k+l) is even: (110) (200) (211) etc. are allowed Selection F 2 4 f 2 rules for hkl S= 1 3 5 etc. reflections when (h+k+l) is odd: F 0 (100) (111) (210) etc. F 2 0 are not allowed This also agrees with slide 1 when (001) planes are out of phase and cancel, and (002) planes are allowed. This F and selection rules are for elemental BCC metals (W, Fe,...) & BCT metals (In and Pa)
Previous HW Q#1: base (A and C) centered and FC lattice: 1. (a) Give a fully simplified representation of the structure factors for (1) face-centered (F), (2) A- base centered, and (3) C-base centered lattices. (b) Discuss if anything in this calculation changes if the crystal is cubic, orthorhombic, or tetragonal and why/why not. Answers given in class handout. class25/6 Crystal Structure Centric or Not?
•As examples, FCC Cu and BCC W are centric (or centro-symmetric/center of symmetry) crystal structures (11 of 32 point groups=Laue class, class 19/slide 9) where for every point at (x,y,z) there was an identical (equi)point at (-x,-y,-z) [where -x=1-x, -y=1-y, and -z=1-z] = inversion •Ex. Cu (P.G. m3m) at (0,0,0)(½, ½, 0)(½, 0, ½) (0, ½, ½) (1,1,1)(½ ,½, 1)(½ ,1 ,½) (1, ½, ½) “Equivalent points thus centric since they are all identical points”
•Thus when determining the structure factor (F) and calculating the Ihkl, for Laue class point groups, the sine term will cancel out and you are only left with the cosine term to calculate Ihkl ; for example in FCC Cu (next slide for next class). eix cos x
•However, for non-centric (no center of symmetry) crystal structures (other 21 non-Laue point groups) there will not be an identical (equi)point at (-x,-y,-z) and thus the sine term must be included with the cosine term in I calculation. hkl eix cos x i sinx
•E.g. Te atoms in ZnTe (zincblende, P.G. 43m) at 4T+ sites (¼,¼,¼) (¼,¾,¾) (¾,¼,¾) (¾,¾,¼) (¾,¾,¾) (¾,¼,¼) (¼,¾,¼) (¼,¼,¾) “Non-Equivalent points thus non-centric since they are not identical points”
class25/7