Introduction to crystallography The unitcell The resiprocal space and unitcell Braggs law Structure factor Fhkl and atomic scattering factor fzθ

Introduction to crystallography The specimen Introduction to crystallography We divide materials into two categories: Amorphous materials The atoms are ”randomly” distributed in space Not quite true, there is short range order Examples: glass, polystyrene (isopor) Crystalline materials The atoms are ordered Short range and long range order Deviations from the perfect order are of importance for the properties of the materials. Why do we see facets on the surface

Introduction to crystallography The specimen Introduction to crystallography

Basic aspects of crystallography The specimen Basic aspects of crystallography Crystallography describes and characterise the structure of crystals Basic concept is symmetry Translational symmetry: if you are standing at one point in a crystal, and move a distance (vector) a the crystal will look exactly the same as where you started. a 2D b a a a 1D a

The lattice described as a set of mathematical points in space The specimen The lattice described as a set of mathematical points in space each of these points represents one or a group of atoms, basis Basis + Lattice = crystal structure a a

The specimen Axial systems x y z a b c α γ β The point lattices can be described by 7 axial systems (coordinate systems) Axial system Axes Angles Triclinic a≠b≠c α≠β≠γ≠90o Monoclinic α=γ=90o ≠ β Orthorombic α= β=γ=90o Tetragonal a=b≠c Cubic a=b=c Hexagonal a1=a2=a3≠c α= β=90o γ=120o Rhombohedral α= β=γ ≠ 90o Enhetscellen defineres av tre ikke planære vektorer a,b og c som er aksene i enhetscellen. -kan også beskrives ved lengdene til a, b og c samt vinklene mellom dem. Vektorene definerer et aksesystem. I eksempel 1 var…, 2 var… Dette er eksempler på hva vi kaller kubisk aksesystem I eks. 3 var var aksene a=b men ikke lik c, vinkl. 90 gr.…Dette er et eksempel på hva vi kaller et tetragonalt aksesystem. (Kommer tilbake til andre mulige aksesyst.) Gjentakende like grupper av atomer i enhetscellen kan erstattes med gitterpunkt. 1 eller flere atomer pr. pkt.

Bravais lattice The specimen The point lattices can be described by 14 different Bravais lattices Hermann and Mauguin symboler: P (primitiv) F (face centred) I (body centred) A, B, C (bace or end centred) R (rhombohedral)

The specimen Unit cell The crystal structure is described by specifying a repeating element and its translational periodicity The repeating element (usually consisting of many atoms) is replaced by a lattice point and all lattice points have the same atomic environments. The unit cells are the smallest building blocks. A primitive unit cell has only one lattice point in the unit cell. a c b α β γ Repeating element, basis Lattice point

Exaples of materials with a face centered cubic lattice The specimen Exaples of materials with a face centered cubic lattice Copper

Exaples of materials with a face centered cubic lattice The specimen Exaples of materials with a face centered cubic lattice Silicon

Exaples of materials with a face centered cubic lattice The specimen Exaples of materials with a face centered cubic lattice ZnS

What about other symmetry elements? The specimen What about other symmetry elements? We have discussed translational symmetry, but there are also other important symmetry operations: Mirror planes Rotation axes Inversion Screw axes Glide planes The combination of these symmetry operations with the Bravais lattices give the 230 space groups

Space groups The specimen Crystals can be classified according to 230 space groups. Details about crystal description can be found in International Tables for Crystallography. Criteria for filling Bravais point lattice with atoms. A space group can be referred to by a number or the space group symbol (ex. Fm-3m is nr. 225) Structural data for known crystalline phases are available in “books” like “Pearson’s handbook of crystallographic data….” but also electronically in databases like “Find it” or f.ex. “Crystallography open database”: http://www.crystallography.net/cod/ . Pearson symbol like cF4 indicate the axial system (cubic), centering of the lattice (face) and number of atoms in the unit cell of a phase (like Cu). 13

Lattice planes The specimen z y x Miller indexing system Crystals are described in the axial system of their unit cell Miller indices (hkl) of a plane is found from the interception of the plane with the unit cell axis (a/h, b/k, c/l). The reciprocal of the interceptions are rationalized if necessary to avoid fraction numbers of (h k l) and 1/∞ = 0 Planes are often described by their normal (hkl) one single set of parallel planes {hkl} equivalent planes y z x c/l a/h b/k Z Y X (110) Når vi skal beskrive posisjonene til gitterpunkter og atomer bruker vi posisjonsvektorer av typen-……. Hvor a, b og c er enhetsvektorene I de ulike krystall systmene, u, v, w er lengden langs hver av disse vektorene. Hva ville uvw være for et atom på en sideflate? Plan I enhetcelle kan angiv ved normalen til planet eller ved Miller-indeksene hkl Z Y X (010) (001) (100) (111) Z Y X

Directions The specimen uh+vk+wl= 0 z wc ua a b x z c y vb wc [uvw] The indices of directions (u, v and w) can be found from the components of the vector in the axial system a, b, c. The indices are scaled so that all are integers and as small as possible Notation [uvw] one single direction or zone axis <uvw> geometrical equivalent directions [hkl] is normal to the (hkl) plane in cubic axial systems Zone axis [uvw] (hkl) uh+vk+wl= 0

Reciprocal vectors, planar distances The specimen Reciprocal vectors, planar distances The normal of a plane is given by the vector: Planar distance between the planes {hkl} is given by: Planar distance (d-value) between planes {hkl} in a cubic crystal with lattice parameter a: The reciprocal lattice is defined by the vectors : a * = ( b ´ c ) / V b * = ( c ´ a ) / V c * = ( a ´ b ) / V a*=(bcsinα)/V b*=(casinβ)/V c*=(absinγ)/V

Interaction with sample Elastic scattering Why do we want a monochromatic wave in diffraction studies? X-rays are scattered by the electrons in a material Electrons are scattered by both the electron and the nuclei in a material The electrons are directly scattered and not by an field to field exchange as in the case for X-rays The diffraction theory is the same for electrons, X-rays and neutrons. Based on Braggs law

Braggs law Interaction with sample Coherent incoming wave Gives the angle when the scattered wave is in phase. Elastic scattering

Deduction of Braggs law: Interaction with sample Deduction of Braggs law: ψi Sum ψ scattered in phase θ θ d x x In order for the scattered waves to be in phase (resulting in constructiv interference), the difference in the travelled distance of the waves must be a multiple of λ (i.e 2X = nλ). X= d sinθ and hence 2d sinθ= nλ.

Effect of adding a scattering plane with d/2 Interaction with sample Effect of adding a scattering plane with d/2 Sum ψ001 = 0 ψi θ (001) (002) d/2 x’ x’ θ d x x Imaging that the red planes represents the (001) planes in a cubic structure with the same atom (=electron) density on the (002) plane (like in fcc and bcc structures). The (002) plane will now scatter the incoming wave and travel 1/2nλ shorter/longer than the waves scattered from (001). The amplitude of the sum of the waves in this situation is 0 (x’=d/2 sinθ, 2x’= d sinθ and this is equal to 1/2nλ. In this example “d“ represented the planar distance of the (001) planes and “d/2” the planar distance of the (002) plane. This is commonly written as d001 and d002. For a cubic unitcell d001 and d002 is equivalent to d100 and d200. The latter ones are the ones commonly tabulated in d-value tables if the intensity is not 0.

Example: Bragg angle for (002) relative to (001) Interaction with sample Example: Bragg angle for (002) relative to (001) Sum ψ001 = 0 ψi θ (001) (002) d/2 x’ x’ θ d x x Imaging that the wave length and the crystal are the same as in the previous examplered planes represents the (001) planes in a cubic structure with the same atom (=electron) density on the (002) plane (like in fcc and bcc structures). The (002) plane will now scatter the incoming wave and travel 1/2nλ shorter/longer than the waves scattered from (001). The amplitude of the sum of the waves in this situation is 0 (x’=d/2 sinθ, 2x’= d sinθ and this is equal to 1/2nλ. In this example “d“ represented the planar distance of the (001) planes and “d/2” the planar distance of the (002) plane. This is commonly written as d001 and d002. For a cubic unitcell d001 and d002 is equivalent to d100 and d200. The latter ones are the ones commonly tabulated in d-value tables if the intensity is not 0.

Atomic scattering factor (X-ray) Z Intensity of the scattered X-ray beam

Structure factors X-ray: rj uja a b x z c y vjb wjc The coordinate of atom j within the crystal unit cell is given rj=uja+vjb+wjc. h, k and l are the miller indices of the Bragg reflection g (and represents the normal to the plane (hkl). N is the number of atoms within the crystal unit cell. fj(n) is the x-ray scattering factor, or x-ray scattering amplitude, for atom j. The intensity of a reflection is proportional to: The structure factors for x-ray, neutron and electron diffraction are similar. For neutrons and electrons we need only to replace by fj(n) or fj(e) .

Example: Cu, fcc eiφ = cosφ + isinφ enπi = (-1)n eix + e-ix = 2cosx Atomic positions in the unit cell: [000], [½ ½ 0], [½ 0 ½ ], [0 ½ ½ ] Fhkl= f (1+ eπi(h+k) + eπi(h+l) + eπi(k+l)) What is the general condition for reflections for fcc? If h, k, l are all odd then: Fhkl= f(1+1+1+1)=4f What is the general condition for reflections for bcc? If h, k, l are mixed integers (exs 112) then Fhkl=f(1+1-1-1)=0 (forbidden)

Example: Cu, fcc Atomic positions in the unit cell: [000], [½ ½ 0], [½ 0 ½ ], [0 ½ ½ ] enπi = (-1)n The coordinate of atom j within the crystal unit cell is given rj=uja+vjb+wjc. (hkl): a plane Fhkl= fCu (exp2πi(0+0+0) + exp2πi(1/2h+1/2k+0) + exp2πi(1/2h+0+1/2l) + exp2πi(0+1/2k+1/2l)) Fhkl= fCu (exp0πi + expπi(h+k) + expπi(h+l) + expπi(k+l)) : General expression for fcc F100=fCu(1 + expπi(1+0) + exp πi(1+0) + exp πi(0+0))= fCu(1-1-1+1)= 0 F200=fCu(1 + expπi(2+0) + exp πi(2+0) + exp πi(0+0))= fCu(1+1+1+1)= 4fCu F110=fCu(1 + expπi(1+1) + exp πi(1+0) + exp πi(1+0))= fCu(1+1-1-1)= 0 F111=fCu(1 + expπi(1+1) + exp πi(1+1) + exp πi(1+1))= fCu(1+1+1+1)=4fCu If h, k, l are all odd (ex. 111) or even then: Fhkl= f(1+1+1+1)=4f (intensity) If h, k, l are mixed integers (exs 110) then: Fhkl=f(1+1-1-1)=0 (no intensity) The fcc real space lattice results in a bcc reciprocal lattice 002 111 c* b* a* 200