# Symmetry & Crystal Structures

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Symmetry & Crystal Structures
Solid State Physics 355 Topic 1 Symmetry & Crystal Structures

Crystals Atoms that are bound together, do so in a way that minimizes their energy. This most often leads to a periodic arrangement of the atoms in space. If the arrangement is purely periodic we say that it is crystalline. The stable bonding arrangement implies that the spatial arrangement of the positive ion cores and the outer electrons has less total energy than any other arrangement, including infinite separation. The difference in energy of a configuration with respect to that of individual atoms is the cohesive energy, which varies from 0.1 eV/atom for van der Waals bonding to more than 7 eV/atom for some covalent and metallic compounds.

Crystal = Lattice + Basis
A lattice is not an arrangement of atoms. It is a geometric arrangement of mathematical points. To describe a crystal structure, we must specify both a lattice, or Bravais lattice, and a basis associated with each point. The basis consists of the atoms, their spacings, and bond angles, such that every atom in the crystal is accounted for. The basis is one atom for argon and sodium, two atoms for silicon, 4 for Ge, etc.

Unit Cell A unit cell contains the basis and is repeated in 3 dimensions to fill a volume of space defined by the basis vectors a, b, and c. The volume is given by… V = a x b . C. The choice of unit cell is arbitrary. Cell A is the most obvious choice for a primitive unit cell, but B, C, and D are also primitive cells that will work. E is not primitive. For structures of lower symmetry, the choice of unit cell may not be so obvious.

Face-centered Cubic Lattice
The three interior angles formed between unit cell edges are called: a (alpha, between edges a2 & a3) b (beta, between edges a1 & a3) g (gamma, between edges a1 & a2) In the FCC rhombohedral standard reduced cell, it can be shown that a = b = g = 60.  Note that a cube is a just a special rhombohedron, with a = b = g = 90. In the Face Centered Cubic (FCC) unit cell there is one basis atom at each corner and one basis atom in each face.  There are 4 atoms per unit cell: (1/8)8 + (1/2)6 = 4. The simplest primitive cell is actually rhombohedral. Its volume is ¼ that of the cubic, as we might expect. In your book you’ll read about the 7 crystal systems for which there are 14 three dimensional bravais lattices. One of those crystal systems is cubic, which comes in 3 flavors, simple, fcc, and bcc. The volume of the rhombohedral cell is one-fourth that of the cube.

BaTiO3

Body-centered Cubic Lattice
In the Body Centered Cubic (BCC) unit cell there is one host atom (lattice point) at each corner of the cube and one host atom in the center of the cube: Z = 2

Symmetry Operations Translational Reflection at a plane
Rotation about an axis Inversion through a point Glide (=reflection + translation) Screw (=rotation + translation)

Rotations Although objects themselves may appear to have 5-fold, 7-fold, 8-fold, or higher-fold rotation axes, these are not possible in crystals.  The reason is that the external shape of a crystal is based on a geometric arrangement of atoms.  Note that if we try to combine objects with 5-fold and 8-fold apparent symmetry, that we cannot combine them in such a way that they completely fill space.

Mirror Symmetry The operation is done by imagining that you cut the object in half, then place a mirror next to one of the halves of the object along the cut.  If the reflection in the mirror reproduces the other half of the object, then the object is said to have mirror symmetry.  The plane of the mirror is an element of symmetry referred to as a mirror plane, and is symbolized with the letter m.  As an example, the human body is an object that approximates mirror symmetry, with the mirror plane cutting through the center of the head, the center of nose and down to the groin. Water molecule – 2 planes Benezene – 7 planes

Inversion Another operation that can be performed is inversion through a point.  In this operation lines are drawn from all points on the object through a point in the center of the object, called a symmetry center (symbolized with the letter "i"). The lines each have lengths that are equidistant from the original points.  When the ends of the lines are connected, the original object is reproduced inverted from its original appearance.  In the diagram shown here, only a few such lines are drawn for the small triangular face. The right hand diagram shows the object without the imaginary lines that reproduced the object.

2D - Point Groups

Point and Space Groups 7 crystal systems 14 Bravais lattices
230 non-Bravais lattices

Space Groups plus the Bravais lattices yields
73 simple 3D Space Groups 32 point symmetries 2 triclinic 3 monoclinic 3 orthorhombic 7 tetragonal 5 cubic 5 trigonal 7 hexagonal plus compound operations (glide and screw operations) yields 157 more = 230 space groups

Crystal Directions

Crystal Planes

Crystal Planes: Miller Indices

Find the intersection t of the plane with the c axis.
Procedure for finding Miller indices in four-index notation: Find the intersections, r and s, of the plane with any two of the basal plane axes. Find the intersection t of the plane with the c axis. Evaluate the reciprocals 1/r, 1/s, and 1/t. Convert the reciprocals to smallest set of integers that are in the same ratio. Use the relation i = -(h+ k), where h is associated with a1, k is associated with a2, and i is associated with a3. Enclose all four indices in parentheses: (h k i l) The angle between the a1 a2 and a3 axes is 120 degrees. The c axis is perpendicular to these three.

Example: What is the designation, using four Miller indices, of the lattice plane shaded pink in the figure?      The plane intercepts the a1, a3, and c axes at r = 1, u = 1, and t = ∞, respectively. The reciprocals are 1/r = 1 1/u = 1 1/t = 0 These are already in integer form. We can write down the Miller indices as (h k i l) = (1 k 1 0), where the k index is undetermined so far. Use i = (h + k). That is, k = 2. The designation of this plane is (h k i l) =

Diffraction covered in detail in the next chapter.
Why are planes in a lattice important? (A) Determining crystal structure Diffraction methods directly measure the distance between parallel planes of lattice points. This information is used to determine the lattice parameters in a crystal and measure the angles between lattice planes. (B) Plastic deformation Plastic (permanent) deformation in metals occurs by the slip of atoms past each other in the crystal. This slip tends to occur preferentially along specific lattice planes in the crystal. Which planes slip depends on the crystal structure of the material.       (C) Transport Properties In certain materials, the atomic structure in certain planes causes the transport of electrons and/or heat to be particularly rapid in that plane, and relatively slow away from the plane. Example: Graphite Conduction of heat is more rapid in the sp2 covalently bonded lattice planes than in the direction perpendicular to those planes. Example: YBa2Cu3O7 superconductors Some lattice planes contain only Cu and O. These planes conduct pairs of electrons (called Cooper pairs) that are responsible for superconductivity. These superconductors are electrically insulating in directions perpendicular to the Cu-O lattice planes. Diffraction covered in detail in the next chapter. Thermal conductivity is 400 times greater in a-b plane than in the c direction of graphite. Planes and chains structure was determined by my co-workers at Argonne in We will make some of this later in the semester when we study superconductors.

(GPa)

The interplanar spacing (aka d-spacing)
perpendicular distance between planes in a given family (hkl) symbolically designated as dhkl       The d-spacing decreases as the Miller indices increase The density of lattice points in a plane---i.e. the number of lattice points per unit area in the plane--- decreases as the Miller indices increase

d spacing Example The lattice constant for aluminum is angstroms. What is d220? Answer Aluminum has an fcc structure, so a = b = Å

Interstitials FCC BCC Within any crystal structure, there are void spaces between the atoms These are called interstices, or interstitial voids, or interstitial sites. Each crystal structure has specific types of voids geometry determined by the nearest neighbors size determined by atomic/ionic radius and void geometry Interstitial sites are important because they are possible sites for other types of atoms - formation of compounds (NaCl, CsCl) - alloying agents (hydrogen occupies interstitial sites to make metal hydrides) - impurity atoms

What types/sizes of atoms or ions can fit in a given interstitial site?
Answer: Calculate the effective radius ρ of the void space using trigonometry AND the hard-sphere model with the radius r for the host atom (for metals) or ions (for ionic crystals). Any element with atomic/ionic radius less than or equal to ρ can occupy that interstitial site. If the element has radius larger than ρ, then it will cause some distortion to the crystal structure The increased energy due to distortion will limit the number of interstitial sites that can be occupied (example: carbon in iron).

Often described as 2 interpenetrating FCC lattices
Examples of common structures: (1) The Sodium Chloride (NaCl) Structure (LiH, MgO, MnO, AgBr, PbS, KCl, KBr) The NaCl structure is FCC The basis consists of one Na atom and one Cl atom, separated by one-half of the body diagonal of a unit cube There are four units of NaCl in each unit cube Atom positions: Cl : 000 ; ½½0; ½0½; 0½½ Na: ½½½; 00½; 0½0; ½00 Each atom has 6 nearest neighbours of the opposite kind Often described as 2 interpenetrating FCC lattices

NaCl structure Crystal a LiH 4.08 Å MgO 4.20 MnO 4.43 NaCl 5.63 AgBr
5.77 PbS 5.92 KCl 6.29 KBr 6.59 a

(2) The Cesium Chloride (CsCl) structure
(CsBr, CsI, RbCl, AlCo, AgZn, BeCu, MgCe, RuAl, SrTl) The CsCl structure is BCC The basis consists of one Cs atom and one Cl atom, with each atom at the center of a cube of atoms of the opposite kind There is on unit of CsCl in each unit cube Atom positions: Cs : 000 Cl : ½½½ (or vice-versa) Each atom has 8 nearest neighbours of the opposite kind

CsCl structure Crystal a BeCu 2.70 Å AlNi 2.88 CuZn 2.94 CuPd 2.99
AgMg 3.28 LiHg 3.29 NH4Cl 3.87 TlBr 3.97 CsCl 4.11 TlI 4.20 a Why are the a values smaller for the CsCl structures than for the NaCl (in general)?

Closed-packed structures
(or, what does stacking fruit have to do with solid state physics?)

Closed-packed structures
There are an infinite number of ways to organize spheres to maximize the packing fraction. The centres of spheres at A, B, and C positions (from Kittel) There are different ways you can pack spheres together. This shows two ways, one by putting the spheres in an ABAB… arrangement, the other with ACAC…. (or any combination of the two works)

(3) The Hexagonal Closed-packed (HCP) structure
Be, Sc, Te, Co, Zn, Y, Zr, Tc, Ru, Gd,Tb, Py, Ho, Er, Tm, Lu, Hf, Re, Os, Tl The HCP structure is made up of stacking spheres in a ABABAB… configuration The HCP structure has the primitive cell of the hexagonal lattice, with a basis of two identical atoms Atom positions: 000, 2/3 1/3 ½ (remember, the unit axes are not all perpendicular) The number of nearest-neighbours is 12. Conventional HCP unit cell

(looking along [111] direction
The FCC and hexagonal closed-packed structures (HCP) are formed from packing in different ways. FCC (sometimes called the cubic closed-packed structure, or CCP) has the stacking arrangement of ABCABCABC… HCP has the arrangement ABABAB…. [1 1 1] [0 0 1] FCC (CCP) (looking along [111] direction HCP ABAB sequence ABCABC sequence

HCP and FCC structures Crystal c/a He 1.633 Be 1.581 Mg 1.623 Ti 1.586 Zn 1.861 Cd 1.886 Co 1.622 Y 1.570 Zr 1.594 Gd 1.592 Lu The hexagonal-closed packed (HCP) and FCC structures both have the ideal packing fraction of 0.74 (Kepler figured this out hundreds of years ago) The ideal ratio of c/a for this packing is (8/3)1/2 = 1.633

Amorphous Materials Glass
The continuous random network structure of amorphous silicon dioxide, notice that each Si atom (gold spheres) has 4 bonds, and each oxygen atom (red spheres) has 2 bonds.