SOLID STATE 9. Determination of Crystal Structures

Transcription

1 SOLID STATE 9 Determination of Crystal Structures In the diffraction experiment, we measure intensities as a function of d hkl. Intensities are the sum of the x-rays scattered by all the atoms in a crystal. The amount of scattering depends on: the phase difference, δ, between the x-rays the fractional co-ordinates, (x y z ) of the atoms the orientation of the scattering plane (hkl) the scattering factor of each atom, f 9.1 The atomic scattering factor, f For each atom, the electrons are mainly responsible for scattering the X-rays and so the atomic scattering factor, f, is proportional to the atomic number, Z. f also depends on the scattering angle, θ. At θ=0, f=z, but as θ increases, f decreases. This is known as the polarisation factor. The Lorentz factor, which depends on the geometry of the instrument, also influences f. These two effects are normally combined to give the L p factor. Other effects are due to multiplicities absorption by the sample under study thermal vibrations of the atoms within the crystal structure preferred orientation of powder samples extinction - reduced scattering powder in perfect single crystals 9.2 The Phase Difference, δ Each wave scattered from an atom has an amplitude proportional to f and a phase δ with respect to the origin of the unit cell. We thus need to calculate an expression for the phase in terms of the positions of the atoms in the unit cell and the indices of reflection. Consider a set of lattice planes: 1

2 There is a phase difference of 1 wavelength, i.e. 2π radians, between successive planes of any hkl (from Bragg s law). Hence the phase difference between A and A = 2π. For atom B, with fractional co-ordinate x along the a-axis: δ AB = 2πx for (100) planes For (200) planes For (h00) planes δ AB = 2πx.2 δ AB = 2πhx For any set of (h00) planes, the b and c fractional co-ordinates do not affect the δ values. In the general case, (hkl), all three fractional co-ordinates are essential. Hence the total phase difference in radians between the origin (A) and atom at (x y z ) is δ = 2π (hx + ky + lz ) δ is also sometimes known as the Geometric structure factor. 9.3 The Structure Factor The Structure Factor or Structure Amplitude F hkl is the resultant of N waves scattered in the direction of the reflection hkl by the N atoms in the unit cell. In general, waves can be represented by such expressions as: a = a o exp (2πi νt) where a o is the amplitude of the wave and 2πi νt represents the phase difference. Hence for an atom, with a diffracted wave of x-rays of amplitude f and phase difference δ then F = f exp(iδ ) The intensity of the wave is proportional to FF* (where F* is the complex conugate of F) FF* = (f e iδ ) (f e -iδ ) = f 2 and so I α f 2 Summing the diffracted waves over all atoms,, in the unit cell gives F hkl = f exp( iδ ) so F hkl = f exp 2π i( hx + ky + lz ) Structure Factor Equation This can also be written as F hkl = f (cosδ + isin δ ) 2

3 9.4 Centrosymmetric Structures Some structures have a centre of symmetry at the origin, and are called centrosymmetric structures. This means that for every atom at (x y z ) there is an identical atom at (-x -y -z ). These atoms will have phases +δ and -δ respectively, and since sin(-δ) = -sinδ, the sine terms for each pair of centrosymmetrically related atoms cancel. We thus derive a simplified structure factor equation for centrosymmetric structures: F hkl = f cos 2π ( hx + ky + lz ) 9.E1 Example 1 - Systematic Absences Evaluate F hkl for body centred cubic α-fe - See lectures Q1 Using the centrosymmetric structure factor equation, derive the reflection conditions for an F-centred and a C-centred lattice. 9.E2 Example 2 - Intensities in CsCl Evaluate F hkl for CsCl - See lectures. 9.5 Calculating structure factors and intensities In some cases we cannot simplify the equations as above, but we can use the structure factor equation to calculate the phases and relative intensities of specific reflections, (hkl). In hexagonal close-packed titanium metal, atoms are at (0, 0, 0) and ( 1 2,, l ) F hkl = f Ti exp 2πi (0) + f Ti exp 2πi ( h 2 + k + l ) = f Ti [1 + exp 2πi ( h 2 + k + l )] Evaluate F hkl for (002), (010), (100) and (101) F 002 = f Ti (1 + exp 2πi) = 2f Ti F 010 = f Ti (1 + exp 4 3 πi ) = f Ti (1 + cos 4 3 π + i sin 4 3 π ) = f Ti (0.5 - i 0.866) F 100 = f Ti (1 + exp 2 3 πi ) = f Ti (1 + cos 2 3 π + i sin 2 3 π ) = f Ti (0.5 + i 0.866) F 101 = f Ti (1 + exp 2πi ( ) = f Ti (1 + cos π + i sin 5 3 π ) = f Ti (1.5 - i 0.866) 3

4 Measured intensities are given by FF* so: I 002 = 4f 2 I 010 = f 2 I 100 = f 2 I 101 = 3f 2 This is of vital importance because experimentally we can measure the intensity, but the phase information is lost. Q2 Calculate the intensity of the (220) reflection from diamond in terms of f C. Atomic positions: C at 0,0,0; ½,½,0; ½,0,½; 0,½,½; ¼,¼,¼; ¾,¾,¼; ¾,¼,¾; ¼,¾,¾ 9.6 The Phase Problem Given a structure, we can calculate the diffraction pattern using the structure factor equation. Each F depends on (hkl) and f, which in turn depends on the number of electrons, Z (or the electron density) of atom. We want to be able to calculate the structure from the diffraction pattern. For this we need to locate the atoms, which are represented by electron density. Electron density is a Fourier transform of the structure factor: ρ uvw 1 = πi hu + kv + lw V F obs exp[ hkl 2 ( )] h k l where ρ uvw is the electron density at point uvw and V is the unit cell volume. Areas of high electron density correspond to atoms with many electrons. However, to measure the electron density we need to know F hkl - in fact we measure FF* so whilst we know the amplitude of F we don t know the phase. Several methods have been developed to help: Direct Methods (Hauptmann and Karle) Patterson Methods Heavy Atom Methods 4

5 So routine for solving a structure: Measure I hkl for many hkl Extract F hkl Fourier Transform - Electron Density Locate Atoms - Structure 9.7 The Agreement Factor, R To assess the validity of the crystal structure solution, we compare calculated F values (F calc ) with the observed pattern (F obs ). A common measure of agreement is the R value: R = F obs F F obs calc If R<0.1 then the structural model is probably correct. 9.8 Improvements Once a structural model has been deduced, several improvements can be made: 1. Refinement Positions of atoms are varied slightly so as to minimise R 2. Thermal Vibrations If we assume isotropic atomic vibrations then f exp2πi(hx + ky + lz ) can be multiplied by a factor: exp 2 B K 16 B is the isotropic thermal factor. Anisotropic thermal factors can also be calculated. 9.9 Limitations of X-ray Structure Determination gives average structure light atoms are difficult to detect difficult to distinguish atoms of similar atomic number need to grow single crystals long time for data collection and analysis 5

6 9.10 Neutron Diffraction For neutrons, the scattering factor has a very small dependence on atomic number; the maor contribution is Z independent. This means that light atoms and atoms of similar Z can be distinguished. There are problems, however, with atoms and isotopes which are good neutron absorbers - e.g. Cd, Gd, 6 Li atoms and isotopes with low scattering - e.g. vanadium Diffraction with neutrons assumed a fully elastic collision - no energy is lost. However, it is also possible to design neutron experiments where energy is lost (inelastic) which gives information on vibrations in the lattice and stretching of bonds. 9.10a Time of Flight Neutron Diffraction This method uses constant angle, θ and variable wavelength, λ, and takes advantage of the full white spectrum. Using the equations λ = h L and v = mv t where m,v = mass and velocity of neutron L = length of flight path of neutron and t= time of flight of neutron ht then λ = = 2d sin θ and so t = L 2dm sinθ i.e. t α d ml h Errors in t decrease with the length, L, of the flight path. Q3 Calculate the velocity of a neutron that would have the same wavelength as CuKα radiation (λ=1.54å); mass of neutron = x kg Q4 Silicon has a cubic unit cell. A neutron diffraction experiment using a detector at 10m and θ=45 reveals that the (111) reflection of silicon has a time of flight of microseconds. What is the unit cell of silicon? (h=6.626 x Js) 6

7 TUTORIAL PROBLEMS CONCEPTS 9.1 Explain the term atomic scattering factor, f. 9.2 How does f vary with θ? How does f vary with atom number Z? 9.3 What is the probable lattice type of crystalline substances that give the following observed reflections: (i) 100, 110, 200, 220, , 200, 112, 202, Explain the term structure factor, F. 9.6 What is the relation between F, f and δ? 9.7 How does this relationship simplify if the cell is centrosymmetric? 9.8 How are observed F values obtained? 9.9 What is meant by the R-value in crystal structure determination? 9.10 What are the limitations in using X-ray diffraction for structure determination? 9.11 What are the advantages and disadvantages of neutron diffraction? 9.12 What is meant by the time-of-flight method in neutron diffraction? 9.13 Explain the origin of nuclear and magnetic neutron scattering Explain the difference between elastic and inelastic neutron scattering. PROBLEMS Speed of light c = ms -1 Mass of neutron M N = kg Boltzmann constant, k B = J K -1 Planck constant, h = J s 7.1 Using the structure factor equation given below, derive the systematic absences for a facecentred lattice and a body centred lattice. F hkl = f exp 2π i( hx + ky + lz ) 7.2 The CsCl structure has Cs at positions (0, 0, 0) and Cl at positions (1/2, 1/2, 1/2). Calculate the structure factor for reflections with (h+k+l) even and (h+k+l) odd. 7.3 The nickel arsenide (NiAs) structure is hexagonal with atoms at: As (0, 0, 0) (1/3, 2/3, 1/2) Ni (2/3, 1/3, 1/4) (2/3, 1/3, 3/4) Calculate the intensities of the 002, 010 and 100 reflections. 7.4 The calcium fluoride (CaF 2 ) structure is cubic with atoms at: Ca (0, 0, 0) (½, ½, 0) (½, 0, ½) (0, ½, ½) F(¼, ¼, ¼) (¼, ¾, ¼) (¼, ¼, ¾) (¾, ¼, ¼) (¾, ¾, ¼) (¾, ¼, ¾) (¼, ¾, ¾) (¾, ¾, ¾) Calculate the intensities of the 202, 111, 420 and 311 reflections. 7.5 Calculate the velocity of a neutron which would have a wavelength of 1.3Å. How fast would this neutron travel a flight path of 20m? 7.6 A neutron diffraction experiment has a detector at θ=30 at a length of 50m. A reflection is detected at microseconds. What is the corresponding d-spacing of this reflection? 7.7 A neutron diffraction experiment has a detector at θ=90 at a length of 18m. The 110 reflection of a cubic substance is detected at microseconds. What is the unit cell of the substance? 7

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