Lecture 3 – August 25, 2010 Review Crystal = lattice + basis there are 5 non-equivalent 2D lattices: 5 Bravais lattices divided into 4 systems (symmetries; “point groups”) –oblique, square, rectangular (2), and hexagonal. Actual formation (self-assembly) of these lattices depends on details – boundary conditions (marble demo) 1

3D systems(symmetries) SystemNumber of Lattices Lattice SymbolRestriction on crystal cell angle Cubic3P or sc, I or bcc,F or fcc a=b=c α =β =γ=90° Tetragonal2P, Ia=b≠c α=β =γ=90° Orthorhombic4P, C, I, Fa≠b≠ c α=β =γ=90° Monoclinic2F, Ca≠b≠ c α=β=90 °≠β Triclinic1Pa≠b≠ c α≠β≠γ Trigonal1Ra=b=c α=β =γ <120°,≠90° Hexagonal1Pa=b≠c α =β =90° γ=120° Table 1. Seven crystal systems make up fourteen Bravais lattice types in three dimensions. P - Primitive: simple unit cell F - Face-centred: additional point in the centre of each face I - Body-centred: additional point in the centre of the cell C - Centred: additional point in the centre of each end R - Rhombohedral: Hexagonal class only from similar to Christman handouthttp://britneyspears.ac Key to understanding relationships: start with cube, break symmetries (point group) 48=3!x2 3 elements 2

Lecture 4 - Aug 27, 2010 Review Crystal = lattice + basis there are 5 non-equivalent (Bravais) lattices in 2D, 14 in 3D divided into 4 systems (symmetries; “point groups”), 7 in 3D –Cubic, tetragonal, orthorhombic, monoclinic, triclinic; hex.,trigonal Close-packed crystals C B A C B A Stacking of these planes: ABABAB... = hexagonal close packed (hcp) ABCABC... = face centered cubic (fcc) 3

Real crystals Simple cubic w/o basis: none. Unstable. Uncharged atoms prefer close-packed structures, many near neighbors (12 n.n. in hcp or fcc). Ions: NaCl structure (fcc with 2-ion basis, 6 n.n.) CsCl structure (sc with 2-ion basis) (“2-ion unit cell”, 8 n.n.) We will skip Miller indices here (they are easier to understand after we discuss the reciprocal lattice), and go on to: Covalently bonded structures: Diamond. fcc, 2 atoms/primitive unit cell = 8 atoms/conventional unit cell. Basis at (000) and (1/4,1/4,/1/4) (fcc translations (1/2,1/2,0), (1,0,0). 4

 Bragg Scattering of x-rays Extra path length must be integer number of wavelengths: 2 d sin  = n Bragg’s parallel plane picture is mnemonic, not derivation. PH 481/581 Lecture 6, Sept. 1, 2010 Figure Courtesy Wikipedia 5

Better treatment: x-rays are scattered by electron density n(r) In 1D, n(x) = n(x+a) (periodic) Expand in Fourier series n(x) = Σ n G e iGx where G=integer * 2  /a These G’s form a “reciprocal lattice” – is also {G: e iGx is periodic in the direct lattice} where “direct lattice” is..., -2a, -a, 0, a, 2a,... We want to generalize this to 3D. “lattice” means: closed under addition 6 k x-ray k' r n(x) will be real if n -G = n G * (complex conjugates)

Periodic functions in a lattice To calculate scattering from a periodic electron density n(r), we need to describe what it means to be periodic: n(r+R) = n(r) for R in lattice: R = u 1 a 1 + u 2 a 2 + u 3 a 3. Any periodic function n(r) can be expressed as a Fourier series n(r) = Σ n G e iG·r where each e iG·r is periodic in the lattice: e iG·(r+R) = e iG·r ◄▬► e iG·R = 1 ◄▬► G·R = 2  *integer. The set of such G’s is closed under addition – a lattice, called the “reciprocal lattice” of the “direct lattice” determined by a 1, a 2, and a 3. You can show that the vectors b 1, b 2, and b 3 determined by a i · b j = 2   ij are primitive translation vectors of this reciprocal lattice (RL). This means b 1 must be perpendicular to a 2 and a 3. Explicit formulas are: 7 [ Start by showing b 1 is in RL: b 1 ·R = 2  *integer]

Examples of reciprocal lattices Orthorhombic direct lattice: Reciprocal lattice is Easy to check that a i · b j = 2   ij Hexagonal direct lattice: Reciprocal lattice is a1a1 a2a2 b1b1 b2b2 8

Calculating FT of n(r) n G = Fourier transform of scattering density n(r). Only nonzero at points G of reciprocal lattice. Explicitly, 9 n(r) = Σ n G e iG·r where V is the volume of the unit cell.

Calculating x-ray scattering intensity 10 Incoming amplitude (electric field?) Lecture 7 Sept. 3, 2010 PH 481/581 k x-ray r n(r) = Σ n G e iG·r r’ detector Amplitude reaching detector = Re n(r)d 3 r number of scatterers Phase change where F is called the scattering amplitude (in Kittel’s book) r’ - r

.. G Conclusion: x-ray scattering probes reciprocal lattice (RL) If incoming wavevector is k, amplitude for outgoing wavevector k’ ~ Fourier component n k’-k. In a perfect crystal, this is nonzero only if k’ – k is a reciprocal lattice vector, call it G: k’ k You have to rotate the crystal, powder it,... Also, this is elastic scattering: |k|=|k’|, so k’ must lie on a sphere. The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness) Powder pattern: crystallites are in all possible orientations, which rotates each G over the surface of a sphere: Sphere of possible G’s intersects sphere of possible k’ s in a circle (ring), seen on screen. G k k’ screen

12 Powder pattern Pattern for the mineral olivine, mirrors/armel/egypte/conf2/theory.html Ball-milled Hand-crushed

d Relating k’ – k = G to the Bragg equation 13 G k’ k   Bragg picture is k’ k  leading to 2 d sin  = n In reciprocal lattice picture, k’ = k + G, so elastic scattering condition is Geometrically, we have Consistent only if (so smallest G goes with n=1: )

Predicting angles of powder diffraction rings 14 G k’ k   Recall picture of k’ = k + G & from top triangle, 

Lecture 8 PH 481/581, So the Bragg planes are related to reciprocal lattice vectors, and the plane spacing d is related to the RLV magnitude G = |G| by d = 2  /G 15 Schematically, RLV G ◄▬► lattice planes or more precisely, a plane and a vector G each determine a direction in space. There are many Gs along this direction, and many planes normal to it, so the relation is really between families of G’s and families of planes: family {0,± G, ± 2G,...} of parallel RLVs ◄▬► family of parallel planes. Only some of these planes go through lattice points, and these are spaced d = 2  /G apart. (G must be the smallest length in the family.)

Direct lattice Reciprocal Lattice Drawing low-index planes (far apart) high-index planes (close together)

Calculating scattering amplitude from atomic form factors 17 Consider a crystal with a basis, r1r1 r2r2 Plot density along dashed line: n(r) = n 1 (r-r 1 ) + n 2 (r-r 2 ) = n 1 (   ) + n 2 (  2 ) rxrx xx  2x   r 2x r 1x where   = r - r 1 and   = r – r 2 n1n1 n2n2  r The scattering amplitude F G is defined by Lecture 10 PH 481/581, Sept. 13, 2010 But the integrand is periodic, so the result is proportional to the number N of unit cells. The amplitude per unit cell is called the structure factor S G = F G / N

Calculating scattering amplitude (structure factor) 18 Structure factor r1r1 r2r2  r where Example: CsCl: simple cubic, r 1= = (0,0,0), r 2 = (a/2,a/2,a/2) Some algebra gives In limit f 1  f 2,(i.e., CsCl  FeFe bcc) this becomes fcc RL.

Another example (NaCl structure) 19 KCl: isoelectronic, f 1 = f 2, lose reflections due to extra symmetry NaCl is fcc, basis is r 1= = (0,0,0), r 2 = (a/2,0,0) (if some even and some odd, e.g., (100), these points are not even ON the reciprocal lattice so we don’t calculate S G.) Example in Fig. 2-17, p. 42: KBr has 111, 200, 220,... as predicted.

Chapter 3: Binding Types of bonding: van der Waals Ionic Covalent metallic 20 Start with van der Waals because all pairs of atoms have vdW attraction. For neutral atoms (so there is no electrostatic force) it always dominates at long distances. Ionic: as a Na and a Cl atom approach each other, the extra electron on Na tunnels to Cl. Covalent: as 2 atoms approach, each with extra electron, bonding and anti-bonding orbitals are formed Metallic: long-wavelength plane waves play the role of the bonding orbitals; shorter-wavelength (higher energy) waves are like anti-bonding orbitals. Lecture 12 Sept. 17, 2010 PH 481/581

21 Comparison of different core potentials V=r -n for n = 1,2,3,6, and12 n=1 is said to be “soft”, n=12 is “hard”. n=infinity

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