SOLID STATE 18. Reciprocal Space
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1 SOLID STATE 8 Reciprocal Space Wave vectors and the concept of K-space can simplify the explanation of several properties of the solid state. They will be introduced to provide more information on diffraction which involves the concept of the reciprocal lattice. In the following text the vector is given in bold text, its modulus (length) in normal text.. Reciprocal lattice vectors Any set of planes in a crystal can be defined by: () their orientation in the crystal - by the Miller index (h k l) (2) their d-spacing The orientation of a plane in a crystal can also be defined by the direction of a normal - very important. This is how we generate the reciprocal lattice, which is defined such that a.a* = b.b* = c.c* = where a*, b* and c* are the reciprocal lattice dimensions. This will be proved later. Take two sets of planes in the real lattice. Here we will consider two dimensions, but of course the concept can be extended to include a third. We can draw directions normal to these planes as shown: It is thus a simple process to define the orientation, but we also need to define the length. We use d to define the length of the vector; the longer the vector, the shorter the d-spacing. These are called reciprocal lattice vectors, G (see note below) which have dimensions /length. The general equation for a reciprocal lattice vector is: G hkl NB Sometimes instead of a factor of 2 is used in definitions of G hkl. The dimensions remains the same, but absolute values of distances are then changed by a factor of 2. The system adopted here is most commonly used in crystallography, but the alternative is often used in solid state physics. In addition sometimes d* is used instead of G. The basics are the same! (The change is related to the structure factor) d hkl /7 JMSS
2 2. Reciprocal Lattice (or Reciprocal Unit Cell) Take a monoclinic unit cell (in two dimensions) in real space, and define some lattice planes (h k l): Now look at the reciprocal lattice vectors for these planes, as defined above: To get the reciprocal lattice, we need only concentrate on 00 and G 00, 00 and G 00, since these define the unit cell. In the real lattice, the unit cell vectors a and c were equivalent to d 00 and d 00, respectively. In the reciprocal lattice, the reciprocal unit cell is given by the vectors G 00 and G 00. We call the reciprocal unit cell vectors a* and c*, respectively So a* = G 00 and a* = c* = G 00 and c* = and of course b* = G 00 and b* = d 00 d 00 d 00 * is the complement of. It also follows that a.a* = and a*.b = 0 and so on (9 equations). The cross product (b x c) defines a vector parallel to a* with modulus of the area defined by b and c. The volume of the unit cell is thus given by a.(bxc) We can define the reciprocal lattice, thus, as follows: b c c a a b a* b* c* V V V 2/7 JMSS
3 We can express reciprocal lattice vectors in terms of the reciprocal lattice unit cell vectors a* b* c* G hkl = ha* + kb* + lc* Also from the above diagram we can see that each corner point of the reciprocal unit cell represents a family of planes, (hkl). 3. The K-vector and the Laue formulation Within the Bragg construction we define the incident and reflected X-rays as k o and k, respectively. Each has modulus /. k o and k are of equal length, and so the length of K is given by K = 2 sin hkl = 2 sin hkl K is perpendicular to the (hkl) plane, so can be defined as K = [ 2 sin hkl ] n where n is a vector of unit length perpendicular to (hkl) G is also perpendicular to (hkl), so n G hkl Ghkl and K = 2 G hkl sin hkl G hkl Substitute G hkl = to give K = 2d hkl sin hkl G hkl But Bragg s Law states that 2dsin = so K=G hkl The Laue condition 3/7 JMSS
4 PX The Ewald Sphere In reality, the Laue condition implies that incident radiation impinges on the crystal then radiates out in all directions. Constructive interference only occurs when the scattering vector, K, coincides with a reciprocal lattice vector, G. We can illustrate this concept by use of the Ewald construction. The steps are as follows: Draw the reciprocal lattice. Choose a reciprocal lattice point as the origin For a fixed wavelength and direction of incident radiation, draw (-)k o with length equal to /. Draw a sphere with radius / centred on the end of k o. Note if the sphere intersects a reciprocal lattice point - this reflection is observed. The centre of the Ewald sphere is at O, which is the origin of real space. In a diffraction experiment, we can either change k o, i.e. change the direction/wavelength of the incident radiation, and hence move the Ewald sphere, or we can move the crystal(s), and hence move the reciprocal lattice and vectors G hkl. In single crystal methods, the crystal is rotated or moved so that each G is brought to the surface of the sphere. In powder methods, the crystallites are randomly oriented so that G takes up all orientations at once. 4/7 JMSS
5 PX302 In a real experiment: From the geometry of this arrangement, 2 = tan - (x/r) where x is the distance to the spot on the film and R is the distance from the crystal to the film. From previously, K = 2 sin hkl, so K = 2 sin [½ tan - (x/r)] Hence we can convert measured distances on a film to lengths of reciprocal lattice vectors. Q The atoms of an element form a layer which is only one atom thick. In this layer, the atoms are packed together, in a square primitive lattice, so that they just touch. The radius of an atom is r. (a) Draw a sketch which shows the arrangement of atoms. (b) What is the length of its unit cell side, in terms of r? (c) Sketch the two-dimensional lattice showing its dimensions. (d) Show how the two-dimensional reciprocal lattice can be generated from this lattice and show that a coincides with a* and a* = b* =. (e) What is the value of *, the angle between a* and b*?. (f) Sketch the two-dimensional reciprocal lattice indicating the positions of the 0,0;,0; 0,; -,0; 0,- ;,; -,;,- and -,- reciprocal lattice points. (g) A beam of radiation, wavelength Å, is used to record a diffraction pattern from this layer. The pattern is recorded on a flat film which is perpendicular to the transmitted beam of radiation and a distance 40.0 mm from the point where the incident beam encounters the specimen. The 0 reflection is recorded at a distance of 2.0 mm from the 00 reflection. Calculate the radius of the atoms. Q2 In another element, the atoms in the layer pack together in a 2-d hexagonal close packed arrangement. The radius of an atom is r. (a) Draw a sketch which shows the arrangement of atoms. (b) What is the length of its unit cell side, in terms of r? (c) Sketch the two-dimensional lattice showing its dimensions. (d) Show how the two-dimensional reciprocal lattice can be generated from this lattice and show that a does not coincide with a* but a*=b*=. (e) What is the value of *, the angle between a* and b*?. 5/7 JMSS
6 PX302 Tutorial Questions Concepts. How is the reciprocal lattice vector G hkl related to the d-spacing d hkl? 2. How is the reciprocal unit cell vector a* related to the real unit cell vector a? 3. How is the reciprocal unit cell angle * related to the real unit cell angle? 4. Define the reciprocal lattice vector G hkl in terms of the Miller indices hkl and the reciprocal unit cell vectors a b and c. 5. Show how the scattering vector K is defined in terms of incident and scattered radiation. 6. State the Laue condition. Explain the terms. 7. Explain how the Laue condition leads to the Ewald sphere construction. Problems. The atoms of an element form a layer which is only one atom thick. In this layer, the atoms are packed together, in an oblique primitive lattice, so that they just touch. The radius of an atom is r, a=b and the angle =75. (i) Draw a sketch which shows the arrangement of atoms. (ii) What is the length of its unit cell side, in terms of r? (iii) Sketch the two-dimensional lattice showing its dimensions. (iv) Show how the two-dimensional reciprocal lattice can be generated from this lattice and show that a* = b* =. (v) What is the value of *, the angle between a* and b*? 2. A primitive lattice of unit cell a = (r=radius of atoms) has reciprocal unit cell a* =. Radiation of wavelength 0.7Å is used to record a diffraction pattern. The pattern is recorded on a flat film which is perpendicular to the transmitted beam of radiation at a perpendicular distance 50mm from the sample. The 00 reflection is recorded at a distance 8.85mm from the 000 reflection. Calculate the radius, r, of the atoms. 3. Using the same equipment, a different sample with the same type of crystal structure but different r is measured. The 200 reflection is recorded at distance 25.46mm from the 000 reflection. Calculate the radius, r, of the atoms. [Clue: calculate d hkl in terms of r and hence G hkl for the 200 reflection] 4. Using the same equipment, a different sample with the same type of crystal structure but different r is measured. The 0 reflection is recorded at distance 2.66mm from the 000 reflection. Calculate the radius, r, of the atoms. 6/7 JMSS
7 PX The following is an electron diffraction picture taken of a high temperature superconductor. The lattice is primitive tetragonal, and the a*b* (here with notation a * a 2 *) is shown. Fine aluminium powder has been used as a calibrant, and produces the rings shown. Aluminium is face-centred cubic with a=4.05å, and so the two circles shown correspond to the () and (200) planes. From the diagram, calculate a (and hence a 2 ) for the material. Note any problems which could arise from this method. 7/7 JMSS
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