Introduction to Materials Science Graduate students (Applied Physics)

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1 Introduction to Materials Science Graduate students (Applied Physics) Prof. Michael Roth Chapter Reciprocal Lattice and X-ray Diffraction

2 Reciprocal Lattice - 1 The crystal can be viewed as made up of different sets of parallel planes Low index planes - less dense, more widely spaced High index planes more dense, more closely spaced We can define a crystal structure by representing each lattice plane as a vector G hkl in the direction n hkl (unit vector normal to the plane (hkl)) with a length determined by d hkl this defines the reciprocal lattice (also called dual or fundamental lattice). Like a Bravais (direct) lattice in position space, it is a periodic array of points in the wave vector space. G hkl = πn hkl /d hkl From a chosen origin the reciprocal lattice comprises all points G hkl with one point for each set of planes in the real space lattice. Reciprocal lattice b 3 G b Bravais lattice RL a 1, a, a 3 b 1, b, b 3 primitive lattice primitive lattice vectors vectors for RL b 1

3 Reciprocal Lattice - The basis set of the reciprocal lattice vectors is defined by the equation: a i b j = πδ ij (δ ιj = 0 if i j, δ ij = 1 if i = j), where a i are the basis vectors of the direct lattice and b j basis vectors of the RL. In two dimensions: a 1 b 1 = π, a b = π a 1 b = 0, a b 1 = 0 From these equations - b 1 must be perpendicular to a, and that b must be perpendicular to a 1. See the top figures for a two-dimensional rectangular lattice in the image below. Real lattice a a 1 Reciprocal lattice b b 1 If a 1 a (θ = 90 ), the conditions are automatically fulfilled and b 1 is in the same direction as a 1 ; and b is collinear with a. The reciprocal lattice is a set of points in reciprocal space which are connected to a given point by the vectors G = m 1 b 1 + m b, where m 1 and m are integers. It is also rectangular. The magnitudes of the vectors are given by b a a1θ 90 θ b 1 b 1 = π /a 1, b = π /a The bottom figures show a general two-dimensional lattice. If the angle between a 1 and a is θ then the angle between b 1 and b is θ. The magnitudes of the RL vectors are: b 1 = (π /a 1 ) Cos(90 - θ); b = (π /a ) Cos (90 - θ)

4 Reciprocal lattice - 4 In three dimensions we can define the RL vectors as: a a3 b 1 = π a a a ( ) 1 3 a3 a1 b = π a a a ( ) 1 3 a1 a b 3 = π a a a ( ) 1 3 It is easy to prove that b 1 = π/a 1, and b 1 a, b 1 a 3. Prove in class. G = m 1 b 1 + m b + m 3 b 3 (m 1, m, m 3 = 0, ±1, ±, ) Note: a 1 (a a 3 ) = a (a 3 a 1 ) = a 3 (a 1 a ) Volume of the primitive cell of RL is V G = [b 1 (b b 3 )]. π π ( ) ( ) ( ) ( ) ( ) ( 3 1) ( 1 ) a i( a a ) Reciprocal lattice of RL is the direct lattice. b 1 b 3 b ( ) ( ) π a a a a π b b 3 = i = = i a 3 a1 a1 a i a1 a a3 a1 a 3 a1 a a1 a a But a 1 (a a 3 ) = V C - volume of the primitive cell of the direct lattice V G = (π) 3 / V C.

1,0,0 a a a 1 [ ] b = π/a[0,1,0], b 3 = π/a[0,0,1], The primitive cell of the RL is a cube with an edge

5 Reciprocal Lattice Examples_1 Direct lattice SC bcc fcc hexagonal RL SC fcc bcc hexagonal Simple Cubic a 1 = a[1,0,0], a = a[0,1,0], a 3 = a[0,0,1] a a 3 = a [1,0,0] b a [ ] [ 1,0,0 ] [ 1,0,0 ] 1,0,0 π = π = 1,0,0 a a a 1 [ ] b = π/a[0,1,0], b 3 = π/a[0,0,1], The primitive cell of the RL is a cube with an edge π/a and a volume V G = (π) 3 /a 3. The Wigner-Seitz cell of the RL is called the 1 st Brillouin Zone. π/a

without crossing any Bragg planes; nd BZ is the volume obtained by crossing only one plane; 3

6 Reciprocal Lattice Examples_ Higher order Brillouin Zones General: BZ are defined in the RL around a lattice point 1 st BZ is defined as a volume encompassed around a lattice point without crossing any Bragg planes; nd BZ is the volume obtained by crossing only one plane; 3 rd BZ continue on to higher orders 3-D SC Lattice 1st Brillouin Zone nd Brillouin Zone 3rd Brillouin Zone

) a 1 π a = a( x + z ), b = ( x y + z ) a 1 π a3 = a( x + y ), b3 = ( x + y z ) a Upper Fig.

7 Reciprocal Lattice Examples_3 (bcc, fcc) bcc: fcc: 1 π a1 = a( x + y + z ), b1 = ( y + z ) a 1 π a = a( x y + z ), b = ( x + z ) a 1 π a3 = a( x + y z ), b3 = ( x + y ) a 1 π a1 = a( y + z ), b1 = ( x + y + z ) a 1 π a = a( x + z ), b = ( x y + z ) a 1 π a3 = a( x + y ), b3 = ( x + y z ) a Upper Fig.: bcc lattice with its 1 st BZ (WS cell of the fcc RL) Lower Fig.: fcc lattice with its 1 st BZ (WS cell of the bcc RL)

8 Reciprocal Lattice 1st BZ of the BCC lattice

9 Reciprocal Lattice 1st BZ of the FCC lattice

10 Atomic Structure Before QM, according to the simple Bohr model (Niels Bohr Nobel prize 19): Atoms have nucleus of protons (+ q) and neutrons (0 q) with similar mass held together by strong force which overcomes their electrostatic repulsion. The number of protons is the atomic # Z. The electrons (their # is also Z) orbit the nucleus at relatively large distances. The smallest electron-nucleus separation is nm for H. The electrons orbit the nucleus quickly and effectively and form a cloud (shell) around the nucleus. They can have only certain orbital radii to form different shells and subshells which obey rules for their occupancy. These shells and subshells are labelled using two sets of integers n and l called the principal and angular momentum quantum numbers repectively. n has the values 1,, 3,... and l = 0, 1,,...,(n-1); i.e. l < n. The shells corresponding to n = 1,, 3, 4,...are labelled K, L, M, N, and the subshells l = 0, 1,, 3,.. s, p, d, f,.. etc. 1s s p or [He]s p For a multi-electron atom, the shells are filled up with the lowest quantum numbers first. The number of electrons in a subshell is indicated by a superscript on the subshell symbol. For example, the carbon (C) atom has 6 electrons therefore its electron arrangement is 1s s p (see the Table and figure above).

11 Atomic Structure Periodic Table Outer electrons have the most important role in atomic Interactions, they are valence electrons. The columns in the Periodic Table are made up of atoms with similar numbers IA of electrons in their outer shells; therefore, they have similar chemical properties. 0 H IIA IIIB IVB VB VIB VIIB VIIIB IB IIB IIIA IVA VA VIA VIIA Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr He Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac** Unq Unp Unh Uns * Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu ** Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

12 X-ray Diffraction Origin of X-rays pm 1nm 1µm 1mm 1m γ - rays UV IR micro - radiowaves X - rays visible I rel Kα 1 Lα Lβ Kβ Kα 0,4 0,6 0,8 1,0 λ / Å K L M N Kα Kβ Kγ

13 X-ray Diffraction X-ray diffraction must behave like optical phenomena. - Crystals must be composed of periodic arrays of atoms. Diffraction and interference of light If an X-ray beam is directed at a row of equally spaced atoms (picture above), each atom becomes a source of scattered waves spreading spherically and reinforce in certain directions to produce the zero-, first-, second-, and higher order diffracted beams.

14 X-ray Diffraction A row of atoms has infinite rotational symmetry along the axes passing through it. Two-dimensional array of equally spaced atoms consequently produces scattered waves which reinforce along the lines of the cross section of two sets of corresponding cones oriented along the coordinate axes. In three-dimensional case, the set of the cones oriented along the third coordinate axes causes the reinforcement of scattered waves (constructive interference) to occur at certain locations. Those locations are the points of cross section of all three sets of cones, oriented along three coordinate axes of the crystal. a(cosα cosα) = hλ, h = 0, 1,, 3, Laue equations a(cosα cosα) = hλ b(cosβ cosβ) = kλ c(cosγ cosγ) = lλ

15 X-ray diffraction Bragg Law Shortly after Laue, Bragg gave a simple mathematical description of the X-ray diffraction. He regarded the crystal as built of lattice planes, which reflect X-rays. Lattice plane is a plane, which passes through lattice points. Its orientation towards the lattice can be defined by its Miller indices. Two examples of planes and their Miller indices are shown in the figure. The blue palen cuts the a axis at, b at ¼ and c at ½, or the fractional coordinates are, ¼, ½ = 1/h, 1/k, 1/l. Their reciprocal values are the Miller indices hkl = 34. Each plane is in fact one of a stack of planes. Adjacent planes are separated by the interplanar distance d. c In order to explain X-ray diffraction he assumed that each lattice plane acts as a semitransparent mirror. When a crystal is irradiated with X rays, a fraction of the radiation is reflected at an angle = to the angle of incidence. The rest is transmitted to the next plane, where it is subsequently reflected, and so on, as shown below. The beams XX 1 and YY 1 are reflected by adjacent planes. Diffraction occurs only if the reflected beams X 1 and Y 1 are in phase, which means that the optical difference ABC must be equal to a whole number of wavelengths: = nλ = AB + BC = dsinθ. X X 1 Y Y 1 θ θ (34) (110) b A θ d C B n λ = d sinθ (n = 1,, 3...) a This is Bragg Law

16 X-ray diffraction Bragg Law - continued Note: All higher order reflections can be regarded as first order reflections from imaginary sets of planes. If the spacing of these is d = d/n, where d is the true spacing, then the Bragg equation can be simplified to λ = d sinθ. In the figure: The ray scattered by atom A is h wavelengths ahead of the ray scattered by atom O, etc. Phase differences between rays diffracted from atoms O and A,B,C are πh, πk and πl respectively. Alternatively, the rays are in phase ( φ = π) at points A, B, C at distances a/h, b/k and c/l from the origin. The distance d between the plane A B C (hkl) and the origin, which Is also the interplanar spacing, is measured along the normal to the l plane and amounts to (a/h)cosδ 1, where δ 1 is the angle between the a-axis and the (hkl) plane normal. Similarly: d=(b/k)cosδ, d=(c/l)cosδ 3. h k l Cos δ + Cos δ + Cos δ = d + + a b c λ h k l Orthorhombic crystal: Sin θ = a b c 1 3 Tetragonal crystal: Sin λ h + k l θ = + 4 a c. Since Cos δ + Cos δ + Cos δ =, for orthogonal crystals: Cubic crystal: Hexagonal crystal: Sin Sin λ 4a ( h k l ) θ = + + λ 4 h + k + hk l θ = a c These formulas are special cases of a more general expression derived within the reciprocal lattice concept.

17 X-ray Diffraction practical examples

18 X-ray Diffraction practical examples - Atomic form factor (f) efficiency of an atom in scattering X-rays (f gives of intensity scattered by an atom to the corresponding intensity from an electron). The Structure factor (F) amplitude of the sum of waves (sine waves of different amplitude and phase, but identical wavelength) scattered by each of the atoms in the unit cell. F is the intensity of the observed reflection von Laue, Knipping, and Friedrich the 1st experiment on X-ray diffraction el.-m. nature of X-rays the inner structure of crystals (Nobel prize, 1914) W. H. Bragg and W. L. Bragg determined the KCl, NaCl, KBr, KI crystal structures X-ray crystallography (Nobel prize, 1915)

19 Laue Condition Laue regarded a 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. Sharp peaks are observed only in the directions and at wavelengths for which the x-rays scattered from all lattice points interfere constructively. We consider two scatterers separated by a lattice vector T. Let X-rays be incident from infinity, along direction k with wavelength λ and wavevector k = π k /λ. We assume that the scattering is elastic, i.e. the X-rays are scattered in k' direction with same wavelength λ, so that the wavevector k = π k' /λ. The path difference between the X-ray scattered from the two atoms should be an integer number of wavelengths. Therefore, as seen from the Fig., the condition of constructive interference is ( k ' k ) it = nλ (n - an integer). Multiplying both sides by π/λ leads to a condition on incident and scattered wave vectors: ( k ' k) it = πn Defining the scattering wave vector k = k - k, the diffraction condition can be written as k = G, Laue condition where G is, by definition, such a vector for which G T = πn A set of vectors G which satisfies this condition form a reciprocal lattice.

$Laue Condition - continued It is sometimes more convenient to give a different formulation of the diffraction condition.$

20 Laue Condition - continued It is sometimes more convenient to give a different formulation of the diffraction condition. In elastic scattering the photon energy is conserved, so that the magnitudes of k and k are equal, and therefore k = k. Therefore, it follows from k = G that k = (G + k) 0 = G + k G. By replacing G to -G, which is also a reciprocal lattice vector, we arrive at another Laue condition: k G = G (1) We show that the reciprocal lattice vector G = h b 1 + k b + l b 3 is orthogonal to the plane represented by Miller indices (hkl). Consider the plane (hkl) which intercepts axes at points x,y, and z given in units a 1, a and a 3 : By the definition of the Miller indices we can always find such interceptions that ( h, k, l) =,, x y z (*) As we know, any plane can be defined by two non-collinear vectors lying within this plane. We can choose vectors u and v as shown. They are given by u = ya xa 1 and v = ya za 3. To prove that the reciprocal vector G is normal to the plane (hkl), it is sufficient to prove that this vector is orthogonal to u and v, i.e. u G = 0 and v G = 0. We have u G = (ya xa 1 ) (h b 1 + k b + l b 3 ) = π (yk - xh) = 0, where the second equation follows from the orthogonality condition of the vectors of the direct and reciprocal lattices and the last equation follows from Eq. (*). In the same manner we can show that G is orthogonal to v. We have proved, therefore, that vector G is orthogonal to the plane (hkl).

21 Bragg Condition Let us come back to the Laue condition, G = k k, under the assumption of elastic scattering: k k = k. Then G = k k Cosθ = 4k Sin θ G = ksinθ = (π/λ) Sinθ G = m 1 b 1 + m b + m 3 b 3 = n (h b 1 + k b + l b 3 ) k k θ (n positive integer; h,k,l have no common division). So: n G(h,k,l) = (π/λ) Sinθ d(h,k,l) Sinθ = n λ, Bragg condition where d(h.k.l) = π G( h, k, l ) spacing of adjacent lattice planes specified by Miller indices (h.k.l) The convenient quantity 1 / d (h,k,l) is given by the relation ( ) ( ) 1 G( h, k, l) = = 1 ( hb1 + kb + lb3) i hb1 + kb + lb 3 d ( h, k, l ) ( π) π ( ) = 1 ( h b1 + k b + l b3 + hkb1b Cos γ * + klbb3cos α * + lhb3b1cos β * π where α*, β*, γ* are the tabulated angles between the axes of the reciprocal lattice, which can be expressed in terms of the interaxial angles of the direct crystal lattice. For crystals of the orthorhombic, tetragonal and cubic systems α* = β* = γ* = 90 (Cos... = 0), and the result fits the former calculations.

22 Ewald Construction Laue condition: k k = G a) Pick origin such that k terminates on a point of RL; b) Draw sphere of a radius k = π/λ about the origin; c) Bragg peaks exist if the sphere intersects some other RL points.

Amendment to experimental methods The Rotating-Crystal Method Ewald sphere determined by the incident k-vector is fixed in k-space, while the entire reciprocal lattice rotates

23 Amendment to experimental methods The Rotating-Crystal Method Ewald sphere determined by the incident k-vector is fixed in k-space, while the entire reciprocal lattice rotates about the axis of rotation of the crystal. The Bragg reflections occur whenever these circles intersect the Ewald sphere. Similarly we can introduce The Rotating-Crystal Method.

PART 1 Introduction to Theory of Solids

Elsevier UK Job code: MIOC Ch01-I044647 9-3-2007 3:03p.m. Page:1 Trim:165 240MM TS: Integra, India PART 1 Introduction to Theory of Solids Elsevier UK Job code: MIOC Ch01-I044647 9-3-2007 3:03p.m. Page:2