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Symmetry & Crystal Structures Solid State Physics 355 Topic 1.

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1 Symmetry & Crystal Structures Solid State Physics 355 Topic 1

2 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.

3 Crystal = Lattice + Basis

4 Unit Cell

5 Face-centered Cubic Lattice 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. The three interior angles formed between unit cell edges are called:  alpha, between edges a 2 & a 3   (beta, between edges a 1 & a 3 )  (gamma, between edges a 1 & a 2 ) In the FCC rhombohedral standard reduced cell, it can be shown that  =  =  = 60 . Note that a cube is a just a special rhombohedron, with  =  =  = 90 .

6 BaTiO 3

7 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

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

9 Rotations

10 Mirror Symmetry

11 Inversion

12 2D - Point Groups

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

14  32 point symmetries –2 triclinic –3 monoclinic –3 orthorhombic –7 tetragonal –5 cubic –5 trigonal –7 hexagonal Space Groups

15 Crystal Directions

16 Crystal Planes

17 Crystal Planes: Miller Indices



20 Procedure for finding Miller indices in four-index notation: 1.Find the intersections, r and s, of the plane with any two of the basal plane axes. 2.Find the intersection t of the plane with the c axis. 3.Evaluate the reciprocals 1/r, 1/s, and 1/t. 4.Convert the reciprocals to smallest set of integers that are in the same ratio. 5.Use the relation i = -(h+ k), where h is associated with a 1, k is associated with a 2, and i is associated with a 3. 6.Enclose all four indices in parentheses: (h k i l)

21 Example: What is the designation, using four Miller indices, of the lattice plane shaded pink in the figure? The plane intercepts the a 1, a 3, 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) =

22 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 sp 2 covalently bonded lattice planes than in the direction perpendicular to those planes. Example: YBa 2 Cu 3 O 7 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.

23 (GPa)


25 d spacing Example The lattice constant for aluminum is angstroms. What is d 220 ? Answer Aluminum has an fcc structure, so a = b = Å

26 Interstitials BCCFCC

27 What types/sizes of atoms or ions can fit in a given interstitial site? Answer: 1. 1.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 4. 4.The increased energy due to distortion will limit the number of interstitial sites that can be occupied (example: carbon in iron).

28 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 Na: ½½½; 00½; 0½0; ½00  Each atom has 6 nearest neighbours of the opposite kind Often described as 2 interpenetrating FCC lattices

29 NaCl structure Crystala LiH4.08 Å MgO4.20 MnO4.43 NaCl5.63 AgBr5.77 PbS5.92 KCl6.29 KBr6.59 a

30 (2) The Cesium Chloride (CsCl) structure  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 (CsBr, CsI, RbCl, AlCo, AgZn, BeCu, MgCe, RuAl, SrTl)

31 CsCl structure Crystala BeCu2.70 Å AlNi2.88 CuZn2.94 CuPd2.99 AgMg3.28 LiHg3.29 NH4Cl3.87 TlBr3.97 CsCl4.11 TlI4.20 a Why are the a values smaller for the CsCl structures than for the NaCl (in general)?

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

33 Closed-packed structures  There are an infinite number of ways to organize spheres to maximize the packing fraction. 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) The centres of spheres at A, B, and C positions (from Kittel)

34 (3) The Hexagonal Closed-packed (HCP) structure  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. Be, Sc, Te, Co, Zn, Y, Zr, Tc, Ru, Gd,Tb, Py, Ho, Er, Tm, Lu, Hf, Re, Os, Tl Conventional HCP unit cell

35 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…. HCP ABAB sequence FCC (CCP) (looking along [111] direction ABCABC sequence [1 1 1] [0 0 1]

36 HCP and FCC structures  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 = Crystalc/a He1.633 Be1.581 Mg1.623 Ti1.586 Zn1.861 Cd1.886 Co1.622 Y1.570 Zr1.594 Gd1.592 Lu1.586

37 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.

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