Introduction to Materials ScienceGraduate students (Applied Physics)

Prof. Michael Roth

Chapter 2Reciprocal Lattice and X-ray Diffraction

Reciprocal Lattice - 1We can define a crystal structure by representing each lattice planeas a vector Ghkl in the direction nhkl (unit vector normal to the plane(hkl)) with a length determined by dhkl – this defines the reciprocallattice (also called dual or fundamental lattice). Like a Bravais(direct) lattice in position space, it is a periodic array of points in thewave vector space.

Ghkl = 2πnhkl/dhkl

From a chosen origin the reciprocal lattice comprises all points Ghklwith one point for each set of planes in the real space lattice.

Bravais lattice RLa1, a2, a3 b1, b2, b3

primitive lattice primitive latticevectors 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 = 2πa1⋅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 vectorsG = 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 b1and 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

n 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.

n “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× = × × × = =

× × × × i i ii

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

RLDirect lattice

SCSC

fccbcc

bccfcc

hexagonalhexagonal

Simple Cubica1 = 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 pointn 1st BZ is defined as a volume encompassed around a

lattice point without crossing any Bragg planes;n 2nd BZ is the volume obtained by crossing only one plane;n 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 StructureBefore 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 whichovercomes their electrostatic repulsion. The number of protons is the atomic # Z. The electrons (their # is alsoZ) 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 canhave only certain orbital radii to form different shells and subshells which obey rules for their occupancy. Theseshells and subshells are labelled using two sets of integers n and l called the principal and angular momentumquantum numbers repectively. n has the values 1, 2, 3, ... and l = 0, 1, 2,...,(n-1); i.e. l < n. The shellscorresponding 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 TableOuter electrons have the most important role in atomic Interactions, they are valenceelectrons. The columns in the Periodic Table are made up of atoms with similar numbersof 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

KLMN

KαKβ

Kγ

LαLβ

0,4 0,6 0,8 1,0 λ / Å

Irel

KβKα2

Kα1

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

1pm 1nm 1µm 1mm 1mγ - rays X - rays visible

UV IR micro - radiowaves

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 DiffractionA row of atoms has infinite rotational symmetry along the axes passingthrough it. Two-dimensional array of equally spaced atoms consequentlyproduces scattered waves which reinforce along the lines of the crosssection of two sets of corresponding cones oriented along the coordinateaxes. In three-dimensional case, the set of the cones oriented along thethird coordinate axes causes the reinforcement of scattered waves(constructive interference) to occur at certain locations. Those locationsare the points of cross section of all three sets of cones, oriented alongthree coordinate axes of the crystal.

Laue equationsLaue 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 LawShortly after Laue, Bragg gave a simple mathematicaldescription of the X-ray diffraction. He regarded thecrystal as built of lattice planes, which reflect X-rays.Lattice plane is a plane, which passes through latticepoints. Its orientation towards the lattice can bedefined by its Miller indices. Two examples of planesand their Miller indices are shown in the figure. Theblue palen cuts the a–axis at ⅓ , b at ¼ and c at ½, orthe 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. Adjacentplanes are separated by the interplanar distance d.

a

b

c

(342)

(110)

θθ

θ d

X X1

Y Y1

AB

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 X1and 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 - continuedNote: 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 rayscattered by atom O, etc. Phase differences between rays diffractedfrom 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, whichIs also the interplanar spacing, is measured along the normal to theplane and amounts to (a/h)Cosδ1, where δ1 is the angle between thea-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 2

h 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

X-ray Diffraction – practical examples - 2

1912 - von Laue, Knipping, and Friedrichthe 1st experiment on X-ray diffractionn el.-m. nature of X-raysn the inner structure of crystals

(Nobel prize, 1914)

1913 - W. H. Bragg and W. L. Braggdetermined the KCl, NaCl, KBr, KI crystal structuresn 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.

Laue Condition

We consider two scatterers separated by a lattice vector T. Let X-rays be incident from infinity, alongdirection with wavelength λ and wavevector k = 2π /λ . We assume that the scattering is elastic, i.e. theX-rays are scattered in direction with same wavelength λ, so that the wavevector k’ = 2π /λ . The pathdifference 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 conditionLaue condition

where G is, by definition, such a vector for whichG⋅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 thedirections 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− = πi

k k Tˆ ˆ( ' ) n− = λi

Laue Condition - continuedIt is sometimes more convenient to give a different formulation of the diffraction condition. In elasticscattering the photon energy is conserved, so that the magnitudes of k and k’ are equal, and thereforek2 = 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 representedby 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 suchinterceptions that

(*)As we know, any plane can be defined by two non-collinear vectorslying 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 thatthe reciprocal vector G is normal to the plane (hkl), it is sufficient toprove 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 andreciprocal lattices and the last equation follows from Eq. (*). In the same manner we can showthat 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 ConditionLet us come back to the Laue condition, G = k’ – k , underthe 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 conditionBragg 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

i

( )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 terminateson a point of RL;b) Draw sphere of a radius k = 2π/λabout the origin;c) Bragg peaks exist if the sphereintersects 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 theentire reciprocal lattice rotates about the axis of rotation of the crystal. TheBragg reflections occur whenever these circles intersect the Ewald sphere.Similarly we can introduce The Rotating-Crystal Method.