Unformatted text preview: NE 125: Lecture 5 NE
Crystal Structure; Crystallographic Directions and Planes Instructor: William K. O’Keefe, P.Eng. [email protected] “Office Hours”: Mondays 1:30 to 2:20 RCH 106 Week 2 Teaching Assistant (TA): Hua (Leanne) Wei [email protected] Reading: Shackelford 6th Ed, Chapters 3 & 4 Assignment 2 (handout) problem set due: 21 January 2007 Suggested (extra) problems: Shackelford 6th Ed ( Ch 3 practice problems, 2,4,5,9,10,14) NE 125: Introduction to Materials Science and Engineering Introduction Review: hcp and fcc close packed structures
FCC and HCP close packed structures A
B C A
C A
C CN=12 CN=12 (both FCC and HCP) FCC and HCP have highest APF of all structures 0.74 APF (i.e. close packed structures) A
B A
B C A
B A A A A FCC structures form pattern ABCABCABC etc HCP structures form sequence ABABAB etc NE 125: NE Introduction to Materials Science and Engineering Introduction Review: Polymorphism and Metastability Review:
In nature, systems will move spontaneously towards the state which minimizes the Gibbs energy (G) Depending on how the condition is approached, a locally stable state may be achieved The locally stable state is said to be a metastable state The metastable When a compound exhibits multiple forms, the phenomenon is known as polymorphism and the When polymorphism forms are referred to as polymorphs polymorphs When an element exhibits multiple morphological forms, the phenomenon is known as allotropy and When allotropy the morphological forms are referred to as allotropes allotropes G Global Minimum Local Minimum
(metastable state) (Thermodynamically Favoured State) ξ NE 125: NE Introduction to Materials Science and Engineering Introduction Metastability and crystal habit: Metastability
Example: Bill’s Problem at Uniroyal Chemical The objective was to separate a chemical product from synthesis via crystallization The from a supernatant liquid (product mixture). Two crystal habits possible: Equant (cubic) habit and acicular (needle like) habit The Equant habit (macroscopic crystals have cubic form) was desirable from a process engineering perspective. The acicular crystal habit created processing problems and required increased residence time ( in the crystallizer for digestion. (An extra 24 hours) residence A direct crystallization to the cubic form was desired NE 125: NE Introduction to Materials Science and Engineering Introduction Metastability and crystal habit: Metastability
Example: Bill’s Problem at Uniroyal Chemical Equant habit
(equidimensional) 2L reactor Batch Batch crystallization crystallization Saturated liquid cool Desired route Digestion (~24 hrs)
Crystals dissolve and recrystallize to assume more thermodynamically favourable form Isometric crystals were charged to the crystallizer to Isometric “seed” the process (i.e. create nucleation sites) “seed” As the temperature was lowered:  The solution became supersaturated The supersaturated  The product crystals precipitated (solubility) The The rate of crystallization is proportional to the The availability of adsorption sites availability Acicular habit
(metastable state) Rate of crystal growth unequal Rate for different crystal facets for dm ∝ m ⇒ m( t ) =k 1e k 2t dt
Molecules adsorb preferentially on certain facet facet Crystals much longer In one dimension In than other two (needle shaped) (needle NE 125: NE Introduction to Materials Science and Engineering Introduction Polymorphism and Metastability: Polymorphism Ref. Levin, I. and Brandon, D., J. Am. Ceram. Soc., 81(8), (1998), 1995 NE 125: NE Introduction to Materials Science and Engineering Introduction Polymorphism and Metastability: Polymorphism
Eg. Diamond and Graphite (covalent network solid; carbon) (above) structure of graphite The diamond structure is favoured for thermodynamic conditions of high temperature and pressure The structure of diamond from various perspectives1
[1] Illustration courtesy of the United States Naval Research Laboratory; Used by permission in accordance with NLR policy www.nrl.navy.mil NE 125: NE Introduction to Materials Science and Engineering Introduction Ceramic Crystal Structure: Ceramic
The cation is almost always smaller than the anion Many ceramic structures consist of anions in closed packed layers with the Many smaller cations located within the interstitial sites interstitial Eg. CsCl As with metallic bonding, we can define an ionic packing factor (IPF) utilizing the ionic radii IPF = (volume of ions in unit cell)/(unit cell volume) Illustration from Introduction to Materials Science for Engineers, 6th Ed; Used by permission. NE 125: Introduction to Materials Science and Engineering Introduction Web resource demonstration: Crystallographic structures of ceramic materials etc. NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic
Eg BCC structure 6 8 7 5 1 2 3 * 4 9
No 1 2 3 4 5 6 7 8 9 Fractional Lengths x y z 0 0 0 1 0 0 1 1 0 0 1 0 1/2 1/2 1/2 0 0 1 1 0 1 1 1 1 0 1 1 Coordinates (000) (100) (110) (010) (1/2 1/2 1/2) (001) (101) (111) (011) Coordinates for locations of atoms in bcc structure * If an atom is located at point 5, the structure is bcc; if there is a vacancy at point 5, The structure is simple cubic (SC); The location of any atom within a lattice may be expressed in terms of fractional multiples of the lattice The parameters (i.e. unit cell edge lengths) parameters
Illustration courtesy of Prof. L. Simon NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic
z y x If a = 0.48 nm, b = 0.46 nm, c = 0.40 nm; locate the point within the unit cell with coordinates (1/4 1 ½) To which crystal system would this unit cell belong? NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic
z (1/4 1 ½) y x Monoclinic a≠ b≠ c α =β =90° ≠ γ =90° NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic
The crystallographic direction between two points within a crystal lattice may be defined according to the following rules: defined 1. 1. 1. 1. 1. 1. Position the vector such that it passes through the origin of the coordinate system Determine the length of the vector’s projection onto the three coordinate axes Normalize the lengths by the respective lattice parameters (i.e. divide by the lattice Normalize parameter) parameter) Eliminate fractions by multiplying through by the lowest integer in the denominators Eliminate of the fractions of The vector is represented by the three coordinates enclosed with square brackets The and without using commas eg. [111] and A bar above a coordinate indicates a negative value instead of using a negative sign NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic 113 002 z y x NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic A family of equivalent directions family are denoted with angled brackets 〈 111〉 Illustration from Introduction to Materials Science for Engineers, 6th Ed; Used by permission. NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic From symmetry, often several nonparallel directions may be equivalent Eg. In cubic system 〈100〉 denotes the family: [100], [ 1 00], [010], [0 1 0], [001], [00 1 ] NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Planes: Crystallographic
The orientation of planes within crystals may be defined using Miller Indices The Miller
1. Define a coordinate system using the unit cell as a basis (as done previously) 1. If plane passes through origin, select another equivalent parallel plane or chose If new origin new 1. Determine points where plane intersects the 3 axes in terms of lattice parameters Determine a,b,c a,b,c 1. Take reciprocals of the values determined in #3. A plane parallel to axis has Take infinite intercept and its reciprocal is 0 infinite 1. Multiply through a common factor to eliminate fractions. Use multiplication Multiply factor which gives smallest set of integers (lowest values) factor 1. The plane is represented by the values calculated in #5 enclosed in parentheses The (hkl) (hkl) NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Planes: Crystallographic
Example of Miller Indices z c/2 a x b y NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Planes: Crystallographic
Example of Miller Indices z z z’ c/2 a x b y O’ y x x’ With respect to new origin O’, the plane intercepts the axis at x = ∞ y=b z= c/2 Reciprocals: 0, 1, 2 Miller index ( 0 1 2) NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic
Hexagonal Crystals Mapping (u’v’w’)→(uvtw)
where u’, v’ and w’ are projections on the a1, a2 and z axes z a2 a3 a1 1 u = ( 2u′ − v′) 3 1 v = ( 2v′ − u ′) 3 t = −( u + v ) w = w′ For example the directional vector [010] becomes [ 1 2 1 0] For crystals with hexagonal symmetry, we need to utilize the MillerBravais 4 axes scheme For MillerBravais axes NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic
Hexagonal Crystals z [0001] 1 ( 2u′ − v′) = 1 (2(1) − 1) = 1 3 3 3 1 1 1 v = ( 2v′ − u ′) = ( 2(1) − 1) = 3 3 3 2 t = −( u + v ) = − 3 w = w′ = 0 u=
a2
a2 a3 [11 2 0]
a1
For crystals with hexagonal symmetry, we need to utilize the MillerBravais 4 axes scheme For MillerBravais axes a1 NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Directions: Crystallographic
Hexagonal Crystals
Mapping (u’v’w’)→(uvtw) z 1 ( 2u′ − v′) 3 1 v = ( 2v′ − u′) 3 t = −( u + v ) w = w′ u= Consider direction indicated by green vector a2 a3 a1 u’=1; v’=1; w’= 1 ∴ u=1/3 v=1/3 u=1/3 t=2/3 w=1 [1123] For crystals with hexagonal symmetry, we need to utilize the MillerBravais 4 axes scheme For MillerBravais axes NE 125: NE Introduction to Materials Science and Engineering Introduction Crystallographic Planes: Crystallographic MillerBravais Indices for Hexagonal Crystals z (hkil) i = (h+k) ( 1 010) (10 1 1)
a2 a3 Basal plane (0001) a1 For crystals with hexagonal symmetry, we need to utilize the MillerBravais 4 axes scheme For MillerBravais axes ...
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This note was uploaded on 10/08/2010 for the course NE 125 taught by Professor Simon during the Spring '10 term at Waterloo.
 Spring '10
 SIMON

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