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Wave Diffraction and Reciprocal Lattice. Most methods for determining the atomic structure of crystals are based of the idea of scattering of radiation. The wavelength of the radiation should be comparable with the distance between atoms ( ).

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## Wave Diffraction and Reciprocal Lattice

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**Wave Diffraction and Reciprocal Lattice**Most methods for determining the atomic structure of crystals are based of the idea of scattering of radiation. The wavelength of the radiation should be comparable with the distance between atoms ( ) E=2-10 keV, X rays X-rays interact with electronic shells of atoms in a solid. Electrons absorb and re-radiate x-rays which can then be detected. Nuclei are too heavy to respond. The reflectivity of X-rays is of the order of 10-3-10-5, so that the penetration in the solid is deep. Therefore, X-rays serve as a bulk probe.**Bragg’s law (1913):**crystalline solids have remarkably characteristic patterns of reflected X-ray radiation. In crystalline materials, for certain wavelengths and incident directions, intense peaks of scattered radiation were observed. Mirror reflection from the planes, Path difference between the planes Condition for constructive interference: There are a number of various setups for studying crystal structure using X-ray diffraction. In most cases, the wavelength of radiation is fixed, and the angle is varied to observe diffraction peaks corresponding to reflections from different crystallographic planes. Using the Bragg law one can then determine the distance between the planes. The Bragg law is greatly oversimplified (but it works!). It (i) says nothing about intensity and width of X-ray diffraction peaks; (ii) neglects differences in scattering from different atoms; (iii) neglects distribution of charge around atoms.**Typical numbers:**Cubic crystal, lattice constant 4.00 Å. Line CuKα1 has wavelength of 1.54 Å. Then reflection from the (100) plane corresponds to For γ-rays the angles are less (sliding beams)**Max von Laue:**crystal as composed of identical atoms placed at the lattice sites T and assumed that each atom can reradiate the incident radiation in all directions Scattering is elastic Interference is constructive if or If we define the scattering wave vector as and introduce a new vector by the definition , then the Laue condition can be written as Vector is called the reciprocal lattice vector, a set of such vectors forms the reciprocal lattice.**How one can construct the reciprocal lattice?**Let be the primitive vectors of direct lattice. Then we define: By definition, The reciprocal lattice is then defined as Therefore, are primitive vectors of the reciprocal lattice. It is easy to prove this statement: since we get: Examples: reciprocal lattices for 1D and 2D-rectangular structures**Since scattering is elastic, we can write:**Since –G is also a reciprocal lattice vector this is equivalent to We will show that this condition is equivalent to the Bragg reflection Some 3D examples We will come back to the examples later, but now let us assume that we are able to draw a reciprocal lattice for a given direct one. We will now proceed to the Laue equations.**By the definition of the Miller indices we can always find**such interceptions that The plane is defined by two vectors, We have: and similarly . Thus, G is perpendicular to the (hkl) plane. 2. The distance between the adjacent planes is then 1. First, let us map a plane with Miller indices (hkl) to a reciprocal vector G. For this let is choose and consider the plane (hkl) which intercepts axes at points x,y, and z given in units a1, a2 and a3:**3. Then, from it follows the**Bragg’s condition: Scattered Ewald construction Center Incident Origin of the reciprocal lattice • The reciprocal vector G(hkl) is associated with the crystal planes (hkl) and is normal to these planes. • The separation between these planes is 2π times the inverse of G.**In a similar way we can draw other lines (planes), which**satisfy the diffraction condition. Brillouin zones One can rewrite the diffraction condition as That leads to the following picture in the reciprocal space: Any vector k satisfies the diffraction condition So the Brillouin construction exhibits all the wave vectors k which can be Bragg-reflected by the crystal.**The central part of in the reciprocal lattice is of special**importance in the theory of solids. It is the first Brillouin zone. The first Brillouin zone is the smallest volume entirely enclosed by the planes that are perpendicular bisectors of the reciprocal lattice vectors. In this construction it is the rectangle about the origin. The first Brillouin zone is the Wigner-Seitz primitive cell in the reciprocal lattice. Taking dot product of the relationship with aiand choosing we derive the Laue equations in their original form**BCC lattice:**Reciprocal lattice Examples SC lattice: Reciprocal lattice is also s SC lattice with the lattice constant (2π/a)**Shortest G’s (12):**One primitive cell of the reciprocal lattice is the parallelepiped described by . The volume of this cell in reciprocal space The cell contains one reciprocal lattice point, because each of the eight corner points is shared among eight parallelepipeds. Each parallelepiped contains one-eighth of each of eight corner points. General reciprocal lattice vector:**First Brillouin zone of the body centered cubic lattice.**The figure is a regular rhombic dodecahedron. The first Blillouin zone: Each such cell contains one lattice point at the central point of the cell. This zone (for the bcc lattice) is bounded by the planes normal to the 12 shortest G’s of at their midpoints. The zone is a regular 12-faced solid, a rhombic dodecahedron. Homework: make a similar analysis for the FCC and HCP lattices.**Diffraction amplitude**So far we have not discussed the amplitude and the width of diffraction peaks, which play an important role in the interpretation of x-ray diffraction data. Scattering occurs due to the interaction of the incident X-rays with the electron charge distributed in a solid with charge density**The amplitude of the scattered wave is proportional to**with The total amplitude is then the intensity being Now, where is the Fourier component of the electron density Check: Result:**Scattering from a lattice with basis**If the crystal structure represents a lattice with a basis, then we should take into account scattering by the atoms which have non-equivalent positions in a unit cell. The intensity of radiation scattered in a given Bragg peak will depend of the extent to which the rays scattered from these basis sites interfere with one another. To analyze this issue let us assume that , and then split the integral over the volume as: At the integral is equal to the volume V, the amplitude being , otherwise the integral is very small Therefore, the intensity of a Bragg peak is determined by a respective Fourier component of the charge density**Let us assume that a cell has s atoms located at the points**rj, Then where Atomic form-factor we have: Since Here N is the total number of cells, while is the structure factor.**Example: BCC lattice, conventional cell**Two identical atoms: Reciprocal lattice vector: Thus, diffraction peaks will be observed, e.g, from the (110), (200), (211) planes, but not from the (100), (111), (210) planes. The later fact is due to the destructive interference from the basis atoms which cancel some peaks.**Structure factor of FCC lattice.**The basis of the FCC structure referred to the cubic cell has identical atoms at r1 = 000; r2 = ½½0; r3 = 0½½; r4 = ½0½. Therefore which is non-zero only if all the indices are even or all the indices are odd. Allowed peaks are, e.g., (111), (200), (222), (220), (131). Example: (100) plane Here the phase difference between successive planes is π . Therefore, the structure factor vanishes**Atomic form factor**• The atomic form factor is defined as • It depends on the • number and distribution of atomic electrons, • wavelength and angle of scattering of the radiation. • It measures the scattering power of j-th atom in the unit cell. Example from Kittel’s book: KCl vs. KBr, both having sodium chloride structure In KCl the numbers of electrons of K+ and Cl- ions are equal. The scattering amplitudes, f(K+) and f(CI-), are almost exactly equal, so that the crystal looks to X-rays as if it were a monatomic simple cubic lattice of lattice constant a/2. Only even integers occur in the reflection indices when these are based on a cubic lattice of lattice constant a. In KBr the form factor of Br ionis quite different to that of Kl- , and all reflections of the FCC lattice are present.**Let us for simplicity assume that the charge distribution,**n(r), is spherically symmetric. Then putting we get Would all the density be concentrated at a point r=0, then This relationship is exact for forward scattering, G=0. The overall electron distribution in a solid as seen in X-ray diffraction is fairly close to that of the appropriate free atoms. This statement does not mean that the outermost or valence electrons are not redistributed somewhat in forming the solid; it means only that the X-ray reflection intensities are represented well by the free atom values of the form factors and are not very sensitive to small redistributions of the electrons.**Few words about experimental methods**The Laue Method Both the positions of the crystal and the detector are fixed, a broad X-ray spectrum. So, it is possible to find diffraction peaks according to the Ewald picture. This method is mainly used to determine the orientation of a single crystal with a known structure. The Rotating Crystal Method The crystal is placed in a holder, which can rotate with a high precision. The X-ray source is fixed and monochromatic. At some angle the Bragg conditions are met and the diffraction takes place. In the Ewald picture it means the rotating of reciprocal basis vectors. As long as the X-ray wave vector is not too small one can find the intersection with the Ewald sphere at some angles.**The Bragg spectrometer. A sufficiently monochromatic beam,**of wavelength of the order of 1 Å, is collimated by a system of slits and then falls on the large extended face of a single crystal as shown by figure. The crystal is turned at constant angular speed through the Bragg law position, and the total diffracted energy received by the ionization chamber during this process is measured. Similar readings with the chamber set on either side of the peak give a background correction. How many reciprocal lattice points can we see?**Powder (or Debye-Scherrer) method**The powder method involves the diffraction of a collimated monochromatic beam from a sample containing an enormous number of tiny crystals having random orientation. X-rays diverging from a target at T fall on the sample at O, the sample being a flat-faced briquette of powder. Diffracted radiation from the sample passes through the receiving slit at S and enters the Geiger counter. During the operation the sample turns at angular velocity ω and the counter at 2ω. The distances TO and OS are made equal to satisfy approximate focusing conditions. A filter F before the receiving slit gives the effect of a sufficiently monochromatic beam. A chart recording of the amplified output of the Geiger counter gives directly a plot of intensity versus scattering angle 2Θ.**Double Crystal Diffraction**This is a very powerful method which uses one very high-quality crystal to produce a beam acting upon the specimen When the Bragg angles for two crystals are the same, the narrow diffraction peaks are observed. This method allows, in particular, study epitaxial layers which are grown on the substrate.**Temperature Dependent Effects**The positions of atoms vibrate in time, , that results in an additional phase shift Since the vibrations are fast and random, we have to evaluate an average, Assuming that that u(t) obey Gaussian distribution we get Since the scattered intensity is Now, from classical considerations, where m is the atomic mass while ω is the typical lattice vibrations frequency (1013-1014 s-1). Therefore, According to quantum mechanics, at T=0 there exist zero-point vibrations. Their intensity is that leads to the law For T=0, G=109 cm-1, ω/2π=1014 s-1, m=10-22 g the exponential factor is 0.997.**To take home (1)**• Various statements for the Bragg condition: • Laue conditions: • Definition of primitive vectors of reciprocal lattice: • The reciprocal vector has a form: • The scattered amplitude in the direction k’=k+Gis proportional to the geometrical structure factor**To take home (2)**• Any function invariant under a lattice translation T may be expanded in a Fourier series of the form • The first Brillouin zone is the Wigner-Seitz primitive cell of the reciprocal lattice. Only waves whose wave vector k drawn from the origin terminates on a surface of the Brillouin zone can be diffracted by the crystal. First Brillouin zone Cube Rhombic dodecahedron Truncated octahedron Crystal lattice Simple cubic Body-centered cubic Face-centered cubic

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