Introduction to Materials Science

Prof. Michael Roth

Chapter 2 Reciprocal Lattice and X-ray Diffraction

Reciprocal Lattice - 1 We can define a crystal structure by representing each lattice plane as a vector Ghkl in the direction nhkl (unit vector normal to the plane (hkl)) with a length determined by dhkl – 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.

Ghkl = 2πnhkl/dhkl

From a chosen origin the reciprocal lattice comprises all points Ghkl with one point for each set of planes in the real space lattice. Bravais lattice RL a1, a2, a3 b1, b2, b3

primitive lattice primitive lattice vectors vectors for RL

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

Reciprocal lattice

G

b3

b2

b1

Reciprocal Lattice - 2

θ 90−θa2

a1

a1

a2

b1

b2

b1

b2

Real lattice Reciprocal lattice

The basis set of the reciprocal lattice vectors is defined by the equation: ai⋅bj = 2πδij (διj = 0 if i ≠ j, δij = 1 if i = j),

where ai are the basis vectors of the direct lattice and bj – basis vectors of the RL.

In two dimensions: a1⋅b1 = 2π, a2⋅b2 = 2π a1⋅b2 = 0, a2⋅b1 = 0

From these equations - b1 must be perpendicular to a2, and that b2 must be perpendicular to a1. See the 2 top figures for a two-dimensional rectangular lattice in the image below.

If a1 ⊥ a2 (θ = 90°), the conditions are automatically fulfilled and b1 is in the same direction as a1; and b2 is collinear with a2. The reciprocal lattice is a set of points in reciprocal space which are connected to a given point by the vectors G = m1b1 + m2b2 , where m1 and m2 are integers. It is also rectangular. The magnitudes of the vectors are given by

b1 = 2π /a1, b2 = 2π /a2 The bottom figures show a general two-dimensional lattice. If the angle between a1 and a2 is θ then the angle between b1 and b2 is 180° - θ. The magnitudes of the RL vectors are:

b1 = (2π /a1)⋅Cos(90° - θ); b2 = (2π /a2)⋅Cos (90° - θ)

Reciprocal lattice - 4

In three dimensions we can define the RL vectors as: It is easy to prove that b1 = 2π/a1, and b1 ⊥ a2, b1 ⊥ a3. Prove in class.

G = m1b1 + m2b2 + m3b3 (m1, m2, m3 = 0, ±1, ±2, …)

Note: a1⋅(a2×a3) = a2⋅(a3×a1) = a3⋅(a1×a2)

Volume of the primitive cell of RL is VG = [b1⋅(b2×b3)].

But |a1⋅(a2×a3)| = VC - volume of the primitive cell of the direct lattice VG = (2π)3 / VC. “Reciprocal lattice” of RL is the direct lattice.

b3

b2

b1

( )

( )

( )

×= π

×

×= π

×

×= π

×

2 31

1 2 3

3 12

1 2 3

1 23

1 2 3

a ab 2a a a

a ab 2a a a

a ab 2a a a

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

( )( )

π × × × ππ π× = × × × = =

× × × ×

2 23 1 1 2

2 3 3 1 1 2 122 3 1 3 1 2 1 2 31 2 3

2 a a a a 22 2b b a a a a aa a a a a a a a aa a a

Reciprocal Lattice – Examples_1

RL Direct lattice

SC SC

fcc bcc

bcc fcc

hexagonal hexagonal

Simple Cubic a1 = a[1,0,0], a2 = a[0,1,0], a3 = a[0,0,1]

a2×a3 = a2[1,0,0]

b2 = 2π/a[0,1,0], b3 = 2π/a[0,0,1],

The primitive cell of the RL is a cube with an edge 2π/a and a volume VG = (2π)3/a3.

The Wigner-Seitz cell of the RL is called the 1st Brillouin Zone.

[ ][ ] [ ] [ ]⋅ π

= π =⋅

2

1 2

1,0,0 22 1,0,01,0,0 1,0,0a

a a ab

2π/a

Reciprocal Lattice – Examples_2

Higher order Brillouin Zones General: BZ are defined in the RL around a lattice point 1st BZ is defined as a volume encompassed around a

lattice point without crossing any Bragg planes; 2nd BZ is the volume obtained by crossing only one plane; 3rd BZ – continue on to higher orders

1st Brillouin Zone 2nd Brillouin Zone 3rd Brillouin Zone

3-D SC

Lattice

Reciprocal Lattice – Examples_3 (bcc, fcc)

bcc: fcc: Upper Fig.: bcc lattice with its 1st BZ (WS cell of the fcc RL)

Lower Fig.: fcc lattice with its 1st BZ (WS cell of the bcc RL)

π= + = − + +

π= + = − +

π= + = + −

1 1

2 2

3 3

1 2ˆ ˆ ˆ ˆ ˆ( ), ( )21 2ˆ ˆ ˆ ˆ ˆ( ), ( )21 2ˆ ˆ ˆ ˆ ˆ( ), ( )2

aa

aa

aa

a y z b x y z

a x z b x y z

a x y b x y z

π= − + + = +

π= − + = +

π= + − = +

1 1

2 2

3 3

1 2ˆ ˆ ˆ ˆ ˆ( ), ( )21 2ˆ ˆ ˆ ˆ ˆ( ), ( )21 2ˆ ˆ ˆ ˆ ˆ( ), ( )2

aa

aa

aa

a x y z b y z

a x y z b x z

a x y z b x y

Reciprocal Lattice – 1st BZ of the BCC lattice

Reciprocal Lattice – 1st BZ of the FCC lattice

Atomic Structure Before QM, according to the simple Bohr model (Niels Bohr 1919 - Nobel prize 1922):

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 0.053 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, 2, 3, ... and l = 0, 1, 2,...,(n-1); i.e. l < n. The shells corresponding to n = 1, 2, 3, 4,...are labelled K, L, M, N,… and the subshells l = 0, 1, 2, 3,.. s, p, d, f,.. etc.

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 1s22s22p2 (see the Table and figure above).

1s22s22p2 or [He]2s22p2

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 of electrons in their outer shells; therefore, they have similar chemical properties. IA 0

IIA IIIA IVA VA VIA VIIA

IIIB IVB VB VIB VIIB ⌐ VIIIB ¬ IB IIB

H He

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

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

X-ray Diffraction – Origin of X-rays

K L M N

Kα Kβ

Kγ

Lα Lβ

0,4 0,6 0,8 1,0 λ / Å

Irel

Kβ Kα2

Kα1

10-12 10-10 10-8 10-6 10-4 10-2 100 102

1pm 1nm 1µm 1mm 1m γ - rays X - rays visible

UV IR micro - radiowaves

Production of X-rays

X-rays are produced whenever high-speed electrons collide with a metal target. A source of electrons – hot W filament, a high accelerating voltage between the cathode (W) and the anode and a metal target, Cu, Al, Mo, Mg. The anode is a water-cooled block of Cu containing desired target metal.

Cross section of sealed-off filament X-ray tube

X-ray Diffraction - 1

Diffraction and interference of light

- X-ray diffraction must behave like optical phenomena. - Crystals must be composed of periodic arrays of atoms.

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.

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.

Laue equations

a(cosα’ – cosα) = hλ,

h = 0, 1, 2, 3, …

a(cosα’ – cosα) = hλ

b(cosβ’ – cosβ) = kλ

c(cosγ’ – cosγ) = lλ

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 ⅓ , b at ¼ and c at ½, or the fractional coordinates are ⅓, ¼, ½ = 1/h, 1/k, 1/l. Their reciprocal values are the Miller indices hkl = 342. Each plane is in fact one of a stack of planes. Adjacent planes are separated by the interplanar distance d.

a

b

c

(342)

(110)

θ θ θ d

X X1

Y Y1

A B

C

n λ = 2d sinθ (n = 1, 2, 3...)

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 XX1 and YY1 are reflected by adjacent planes. Diffraction occurs only if the reflected beams X1 and Y1 are in phase, which means that the optical difference ABC must be equal to a whole number of wavelengths: ∆ = nλ = AB + BC = 2dsinθ.

This is Bragg Law

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 = 2d’ 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 2πh, 2πk and 2πl respectively. Alternatively, the rays are in phase (∆φ = 2π) 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 plane and amounts to (a/h)Cosδ1, where δ1 is the angle between the a-axis and the (hkl) plane normal. Similarly: d=(b/k)Cosδ2, d=(c/l)Cosδ3. . Since , for orthogonal crystals:

Cos Cos Cos2 2 21 2 3 1δ + δ + δ =Cos Cos Cos d

a b c

2 2 22 2 2 2

1 2 3 2 2 2h k l

δ + δ + δ = + +

Sina c

2 2 2 22

2 24

4 3h k hk l λ + +

θ = +

( )h k l2

2 2 2 2

4λ

θ = + +2Sina

Sina b c

2 2 2 22

2 2 24h k l λ

θ = + +

2l

Sina c

2 2 2 22

2 24h k l λ +

θ = +

Orthorhombic crystal: Cubic crystal:

Tetragonal crystal: Hexagonal crystal:

These formulas are special cases of a more general expression derived within the reciprocal lattice concept.

X-ray Diffraction – practical examples

XRD Patterns of Simple Cubic and FCC

SC, e.g. BaTiO3 – T > 130°C

fcc

h2 + k2 + l2

simple cubic (any

combination)

FCC (either all odd

or all even)

BCC (h + k + l) is

even 1 100 - - 2 110 - 110 3 111 111 - 4 200 200 200 5 210 - - 6 211 - 211 7 - - - 8 220 220 220 9 300, 221 - -

10 310 - 310 11 311 311 - 12 222 222 222

The Structure Factor •Any hkl is from diffraction of all atoms in the lattice. •Each reflection has a different phase relative to its neighbor. •The overall diffraction from a plane hkl is the sum (Σ) over numerous wavelets of amplitude fj and phase φj.

•The sum is called the structure factor F(hkl) with magnitude F and phase α.

Powder XRD on PhotographicFilm

X-Rays

Sample

2Θ = 180° 2Θ = 0°

X-ray Diffraction – practical examples - 2

1912 - 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)

1913 - W. H. Bragg and W. L. Bragg determined the KCl, NaCl, KBr, KI crystal structures X-ray crystallography

(Nobel prize, 1915)

Atomic form factor (f) – efficiency of an atom in scattering X-rays (f2 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|2 is the intensity of the observed reflection.

X-Ray Powder Diffraction Why 2Θ ?

Laue Condition

We consider two scatterers separated by a lattice vector T. Let X-rays be incident from infinity, along direction with wavelength λ and wavevector k = 2π /λ . We assume that the scattering is elastic, i.e. the X-rays are scattered in direction with same wavelength λ, so that the wavevector k’ = 2π /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 (n - an integer). Multiplying both sides by 2π/λ leads to a condition on incident and scattered wave vectors: 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 = 2πn A set of vectors G which satisfies this condition form a reciprocal lattice.

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.

k'ˆ k'ˆk̂ k̂

k k T( ' ) 2 n− = π

k k Tˆ ˆ( ' ) n− = λ

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 k2 = k’2 . Therefore, it follows from ∆k = G that

k’2 = (G + k)2 0 = G2 + 2k⋅G. By replacing G to -G, which is also a reciprocal lattice vector, we arrive at another Laue condition:

2k⋅G = G2

(1) We show that the reciprocal lattice vector G = h b1 + k b2 + l b3 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 a1, a2 and a3: By the definition of the Miller indices we can always find such interceptions that (*) 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 = ya2 – xa1 and v = ya2 – za3 . 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 = (ya2 – xa1)⋅(h b1+ k b2+ l b3) = 2π (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).

1 1 1( , , ) , ,h k lx y z

=

Bragg Condition Let us come back to the Laue condition, G = k’ – k , under the assumption of elastic scattering: |k| ≈ |k’| = k. Then

G2 = 2k2 – 2k2Cos2θ = 4k2Sin2θ

G = 2kSinθ = 2 (2π/λ) Sinθ

G = m1b1 + m2b2 + m3b3 = n (h b1 + k b2 + l b3) (n – positive integer; h,k,l have no common division). So: n |G(h,k,l)| = 2 (2π/λ) Sinθ 2⋅d(h,k,l)⋅Sinθ = n⋅λ , Bragg condition where d(h.k.l) = ≡ spacing of adjacent lattice planes specified by Miller indices (h.k.l)

2θk’’

k

G2

( , , )h k lπ

The convenient quantity 1 / d2(h,k,l) is given by the relation

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.

( ) ( ) ( )Gb b b b b b

22

1 2 3 1 2 322

( , , )1 1 ( )2( , , ) 2

h k lh k l h k l

d h k l= = + + + +ππ

( )22 2 2 2 2 2

1 2 3 1 2 2 3 3 11 ( b b b 2 b b Cos * 2 b b Cos * 2 b b Cos *2 h k l hk kl lh= + + + γ + α + βπ

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 = 2π/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 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.

Sample 1

04-008-9822 (A) - Whewellite - Ca(C2O4)(H2O) - WL: 1.5406 - Monoclinic - a 10.11831 - b 7.29452 - c 6.29136 - alpha 90.000 - beta 109.465 - gamma 90.00Sample 1 - File: Sample 1.raw - Start: 6.013 ° - End: 66.011 ° - Step: 0.020 ° - Step time: 1. s - Anode: Cu - WL1: 1.5406 - WL2: 1.54439 - Creation: 07-Nov-1

Lin

(Cou

nts)

0

1000

2000

2-Theta - Scale6 10 20 30 40 50 60

Sample name: Sample 1 04-008-9822 Whewellite, Calcium Oxalate Hydrate ----- Ca(C2O4)(H2O) Crystallite Size (Scherrer): 1657.8 A System: Monoclinic Space group: C2/m (12) Cell param.: Initial Final a: 10.11720 10.11831 b: 7.30000 7.29452 c: 6.29000 6.29136 beta: 109.517 109.465