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September 28, 2010 Class 6 Ewald construction The Bragg or Laue condition is a rather restrictive one. A way to see the level of the coincidence that has to happen for diffraction to occur, notice that since k G v1b1 v 2 b2 v3 b3 , the dot product with our real space unit translation vectors gives, a1.k=2v1 a2. k=2v2 a3. k=2v3 These equations tell us that k lies on a cone around a1, but also on a cone around a2, and a3. So k must be along the single line where the three cones intersect. This is a very rigorous restriction and it is difficult to find out what that line is. So Ewald came up with a construction that allow to determine those wavevectors that will produce diffraction A k vector is drawn from any origin but ending in one of the reciprocal points. Then a sphere is drawn centered on the origin of the k vector. Diffraction will occur only if another point in the reciprocal lattice lies on the surface of the sphere then the reciprocal vector G that connects these two points determines k’=k+G In fact, any two points which lay on the surface of a sphere with radius k, will represent a set of planes which meet the diffraction condition. The figure above illustrates the severity of this condition in two dimensions. It is clear that if the k is fixed with respect to the reciprocal lattice, most of the time no diffraction will arise. This is equivalent to an experiment where a crystal is positioned on a holder fixed in place and monochromic x-rays are incident at a fixed angle. In order to measure something several techniques exist. 1) The Laue Method: The crystal is fixed in place, the incoming radiation has a fix angle, but the beam is not monochromatic. This is equivalent to draw a continuous of spheres of increasing radius all coincident in the tip of k. any reciprocal point in the area between the sphere with the minimum radius and the one with the maximum radius will produce diffraction 2) The Rotating Crystal Method: in this case monochromatic radiation is used but the crystal is placed on a rotating holder. As the crystal rotates both the direct and the reciprocal lattice rotate about the same axis. As the reciprocal points rotate the intersect the Ewald sphere 1 3) The Powder or Debye-Scherrer methods: A polycrystalline sample is used, thus small crystals in the sample are oriented in all ld sphere producing diffraction directions, this is equivalent to the rotating crystal method but the sample need not to be rotated. Brillouin Zones A geometrical interpretation of the Laue condition can also be obtained by dividing the Laue equation by two, then 2 k.G=G2 => k. (½G)=(½G)2 So all the waves with wavevectors k whose projection over G is equal to ½ of the length of G will produce diffraction. Consider all the G vectors from a lattice points and draw the set of planes that are perpendicular bisectors to these vectors, the intersection of all these planes will enclose portions of the space. The one with the smallest volume is known as the first Brillouin zone. Notice that this is the recipe to build the Weigner-Seitz primitive cell Any k that connects the center point and the surface of this volume is a wavevector that will produce diffraction. So giving a relationship between and . The Weigner-Seitz primitive cell of the reciprocal lattice is known as the first Brillouin Zone and has very particular implication in the crystal property. We will find the name Brillouin Zone applied to a number of properties in solid state, although we will not always be talking about the same thing, in general the Brillouin zones refer to the first, second, third, etc period on a periodical property. In this particular case, the first Brillouin zone contain ALL the wave vectors that will produce first order diffraction (corresponding to n=1 in the Bragg condition) the second zone will contain the wave vectors that satisfy Bragg for n=2, etc. Fourier analysis of the Basis Remember that F dV n( r ) exp ik r (scattering amplitude) When the diffraction condition is satisfied for a particular lattice vector G, then k=G and the scattering amplitude for a crystal of N cells is FG N dVcell n(r ) exp iG r NSG Where SG is called the structure factor. It is convenient to express the crystal electron density in term of atomic electron density nj as follows: s nr n j r rj Where s is the number of atoms in the atomic basis in the cell and rj is the j 1 vector position of each of the atoms with respect to origin in the cell. 2 With this substitution s s s iG j iGrrj rj s iGrj iGrj SG dV n j (r rj ) e iGr dV n j (r rj ) e e dV n e f e j j j j 1 j 1 j 1 j 1 iG j f j dV n j j e is the atomic factor. Notice that the j=r-rj (or r=+rj) is the coordinate with origin at the center of the atom and sampling outwards towards the electron cloud of each atom. SG is then the atomic factor for the basis There is no unique way to partition the atomic electrons to assign them to particular atoms, thus nj have no unique form. However the physics of the crystal, total atomic density, diffraction spectrum, etc, is not altered by the choice of nj as long as it is consistent. By writing rj, the position of each atom of the basis with respect to the origin in the cell in terms of the direct lattice vectors G.rj=(v1b1+v2b2+v3b3). (xja1+yja2+zja3)=2(xjv1+yjv2+zjv3) Where now xj, yj, and zj are not necessarily integer numbers and are relative to the lattice constants, then s SG 1 , 2 , 3 , f j e i 2 x j v1 y j v2 z j v3 j 1 The structure factor does not need to be real since the scatter intensity, which is what is actually measured, involves S2 Structure factor for the bcc lattice Consider the conventional lattice for a bcc lattice. The basis consists of two identical atoms, 1 at 000 and a second one at ½½½ So the structure factor for the bcc lattice is S G 1 , 2 , 3 , f 1 e i v1 v2 v3 What means that S=0 when v1+ v2+ v3=odd S=2f when v1+ v2+ v3=even This means that diffraction of planes such that (100), (300), (111), etc will not be present in the diffraction spectrum (where the index refer to a cubic lattice). For instance for the (100) line, the phase difference must be 2 (one full wave length) for a wave bouncing off planes at a distance a from each other. For the bcc lattice, there exists an intermediate layer between the two that exist on a cubic lattice. Waves bouncing off that layer are out of phase and cancel the other waves. For this to happen, the atoms must be of the same nature, the (100) does show in the CsCl 3 spectrum since Cs and Cl have different atomic factors but again, CsCl is NOT a bcc but a sc lattice Structure factor for the fcc lattice Consider the conventional lattice for a fcc lattice. The basis consist of four identical atom, 1 at 000; 0½½; ½0½; ½½0 So the structure factor for the bcc lattice is SG 1 , 2 , 3 , f 1 e i v2 v3 e i v1 v3 e i v1 v2 What means that if all the indexes are even integers or odd integers, then SG=4f. However if one has a different parity, SG=0 S=4f when v1, v2, v3 have the same parity S=0 when v1, v2, v3 do have the same parity 4