itb (institut teknologi brunei) engineering materials lecture 3 - structure of crystalline solids...

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13/8/2011 Eng. Mat/ITB/UHH 1 Structure of Crystalline Solids Types of Solids Crystalline Material Atoms self-organize in a periodic Atoms arranged in repetitive 3- Dimensional pattern, in long range order (LRO) give rise to crystal structure. Amorphous (Non-Crystalline) Lacks a systematic atomic arrangement Atoms are arranged in short range order Crystalline Structure Non-Crystalline Structure 2

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ITB (Institut Teknologi Brunei) Engineering Materials Lecture 3 - Structure of Crystalline Solids PCE1ENM 2013

TRANSCRIPT

13/8/2011

Eng. Mat/ITB/UHH 1

Structure of Crystalline Solids

Types of Solids

• Crystalline Material

• Atoms self-organize in a periodic

• Atoms arranged in repetitive 3-Dimensional pattern, in long rangeorder (LRO) give rise to crystalstructure.

• Amorphous (Non-Crystalline)

• Lacks a systematic atomic arrangement

• Atoms are arranged in short range order

Crystalline Structure

Non-Crystalline

Structure

2

13/8/2011

Eng. Mat/ITB/UHH 2

Single Crystals and Polycrystalline Materials

• Single Crystal

• Atoms are in a repeating or periodic array over the entire

extent of material

3

• Polycrystalline material

• Comprised of many small crystals or grains

• The grains have different crystallographic orientation

• Atomic mismatch where grains meet (grain boundaries)

Crystal Structure

• When describing crystalline structures, atoms (or

ions) are thought of as being solid spheres with well-

defined diameters.

• Atomic hard sphere model

• Spheres representing

nearest-neighbor atoms

touch one another.

4

13/8/2011

Eng. Mat/ITB/UHH 3

Unit Cell

• Space Lattice

• An imaginary network of lines, with atoms at intersection of

lines, representing the arrangement of atoms

• Unit Cell

• The smallest structural unit or building block that can

describe the crystal structure

• Repetition of the unit cell generates the entire crystal

Space Lattice

Unit Cell

5

Crystal Systems and Bravais Lattice

• Different types of unit cells can be constructed,

depending on values of the lattice parameters.

Lattice parameters

(or Lattice Constants)axial lengths: a, b, c

interaxial angles: α, β, γ

In general: a ≠ b ≠ c

α ≠ β ≠ γ6

13/8/2011

Eng. Mat/ITB/UHH 4

Crystal Systems and Bravais Lattice

• Only seven different types of unit cells are necessary to

create all space lattices.

• Many of the seven crystal systems have variation of the

basic unit cells:

• Bravais (1811 -1863) showed that fourteen standard unit cells

can describe all possible lattice networks

• These Bravais lattices are illustrated in the next few slides.

• The Four basic types of unit cells are:

1. Simple

2. Body Centered

3. Face Centered

4. Base Centered

7

Types of Unit Cells

1.

1.

8

13/8/2011

Eng. Mat/ITB/UHH 5

Types of Unit Cells

3.

4.

9

Types of Unit Cells

5.

6.

7.

10

13/8/2011

Eng. Mat/ITB/UHH 6

Principal Metallic Crystal Structures

• Metals tend to be densely packed

• Typically, only one element is present, so all atomic radii

are the same

• Metallic bonding is not directional

• No restriction on number or positions of nearest

neighbour atoms: large number of nearest neighbours

and dense atomic packing.

• The densely packed structures are in lower and more

stable energy.

• Energy is released as the atoms come closer together

and bond more tightly with each other

• Have simplest crystal structures

11

Principal Metallic Crystal Structures

• 90% of the metals have either Body Centered Cubic

(BCC), Face Centered Cubic (FCC) or Hexagonal Close

Packed (HCP) crystal structure

• HCP is denser version of simple hexagonal crystal

structure

12

The unit cell

is shown by

solid lines

13/8/2011

Eng. Mat/ITB/UHH 7

Body Centered Cubic (BCC) Crystal Structure

• Represented as one atom at each corner of cube and

one at the center of cube

• Each atom has 8 nearest neighbors

• Therefore, coordination number is 8

Atomic-site

unit cell

Hard-sphere

unit cell

Isolated

unit cell

13

Body Centered Cubic (BCC) Crystal Structure

• In isolated unit cell, each of these

cells has the equivalent of two

atoms per unit cell.

• Each unit cell has eight ⅛ atom at

corners and 1 full atom at the center

• (8 × ⅛) + 1 = 2 atoms

• Atoms contact each other at cube

diagonal.

• Therefore, lattice constant:

a =4R

3

14

Ra 43 = or

13/8/2011

Eng. Mat/ITB/UHH 8

Atomic Packing Factor of BCC Structure

• If the atoms in the BCC are considered to be spherical:

• Since there are two atom per BCC unit cell:

• Volume of the BCC unit cells is:

• Therefore:

Atomic Packing Factor =Volume of atoms in unit cell

Volume in unit cell

15

Face Centered Cubic (FCC) Crystal Structure

• Represented as one atom each at the corner of cube

and at the center of each cube face

• Coordination number for FCC structure: 12

• Atomic Packing Factor is 0.74

Atomic-site

unit cell

Hard-sphere

unit cell

Isolated

unit cell

16

13/8/2011

Eng. Mat/ITB/UHH 9

Face Centered Cubic (FCC) Crystal Structure

• In isolated unit cell, each of these

cells has the equivalent of four

atoms per unit cell.

• Each unit cell has eight ⅛ atom at

corners and six ½ atoms at the

center of six faces.

• (8 × ⅛) + (6 × ½) = 4 atoms

• Atoms contact each other across

cubic face diagonal

• Therefore, lattice constant:

2

4Ra =

17

Ra 42 = Or

Hexagonal Close-Packed (HCP) Crystal Structure

• Represented as an atom at each of 12 corners of a

hexagonal prism, 2 atoms at top and bottom face and 3

atoms in between top and bottom face

• Atoms attain higher APF (0.74) by attaining HCP Structure

than simple hexagonal structure

• The coordination number: 12

Atomic-site

unit cell

Hard-sphere

unit cellIsolated

unit cell

18

13/8/2011

Eng. Mat/ITB/UHH 10

• In isolated unit cell, each of thesecells has the equivalent of six atomsper unit cell.

• Each unit cell has six ⅙ atoms ateach of top and bottom layer, twohalf atoms at top and bottom layerand 3 full atoms at the middle layer.

• (2 × 6 × ⅙) + (2 × ½) + 3 = 6 atoms

• Ideal c/a ratio is 1.633

• Ratio of the height c of the hexagonalprism of the HCP structure to itsbasal side a.

Hexagonal Close-Packed (HCP) Crystal Structure

19

HCP Crystal Structure (Example)

• Calculate the volume of the zinc crystal structure unit cell by

using the following data: pure zinc has the HCP structure with

lattice constant a = 0.2665 nm and c = 0.4947.

Solution:

Volume of Zn HCP unit cell = Area of the base × height

20• Area of base =

Area of ABDEFG

• Area of ABDEFG:

Total areas of 6 equivalent triangles of area

ABC

13/8/2011

Eng. Mat/ITB/UHH 11

Atom Positions in Cubic Unit Cells

• Cartesian coordinate system is use to locate atoms (x, y, z)

• In a cubic unit cell

• y-axis: direction to the right

• x-axis: direction coming out of the paper

• z-axis: direction towards top

• Negative directions are to the opposite of positive directions

• Atom positions are

located using unit

distances along the

axes.21

Directions in Cubic Unit Cells

• In cubic crystals, Direction Indices [uvw] are vector

components of directions resolved along each axes,

resolved to smallest integers.

• Direction indices are position coordinates of unit cell

where the direction vector emerges from cell surface,

converted to integers.

• Some directions in cubic unit cells:

22

13/8/2011

Eng. Mat/ITB/UHH 12

Procedure to Find Direction Indices

Produce the direction vector till it

emerges from the surface of cubic cell

Determine the coordinates of point of

emergence and origin

Subtract coordinates of point of

emergence by that of origin

Are all are integers?

Are any of the direction vectors negative?

Represent the indices in a square

bracket without comas with a bar

(_) over a negative index

Represent the indices in a square

bracket without comas

Convert them to smallest

possible integer by

multiplying by an integerYES

YES

NO

NO

23

For Cubic Unit

Cells

For Cubic Unit

Cells

Direction Indices Cubic Unit Cells(Example)

• Determine direction indices of the given vector.

• Origin coordinates are (3/4, 0, 1/4)

• Emergence coordinates are (1/4, 1/2, 1/2)

Solution:

• Subtracting origin coordinates

from emergence coordinates:

• Multiply by 4 to convert all

fractions to integers:

24

13/8/2011

Eng. Mat/ITB/UHH 13

Miller Indices in Cubic Unit Cells

• Used to refer to specific lattice planes of atoms

• They are reciprocals of the fractional intercepts (with

fractions cleared) that the plane makes with the

crystallographic x, y and z axes of three non-parallel

edges of the cubic unit cell.

25

Procedure to Find Miller Indices

Choose a plane that does not pass

through origin

Determine the x, y, and z intercepts of

the plane

Find the reciprocals of the intercept

Fractions?

Place a ‘bar’ over the negative indices

Enclose in parenthesis (hkl) where

h, k, l are miller indices of cubic

crystal place for x, y and z. Eg: (111)

Clear fractions by

multiplying by an integer to

determine smallest set of

whole numbers

No

Yes

26

For Cubic Unit

Cells

For Cubic Unit

Cells

13/8/2011

Eng. Mat/ITB/UHH 14

Miller Indices for Cubic Unit Cells(Example)

• Intercepts of the plane at x, y and

z are 1, ∞ and ∞

• Taking reciprocals we get (1, 0, 0)

• Miller indices are (100):

• one-zero-plane

1.

2. • Intercepts are 1/3, 2/3 and 1

• Taking reciprocals we get (3,

3/2, 1)

• Multiplying by 2 to clear

fractions, we get (6,3,2)

• Miller indices are (632)

27

Miller Indices for Cubic Unit Cells (Example)

• Taking reciprocals of the

indices we get (1, ∞, 0)

• The intercepts of the plane

are x = 1, y = ∞ (parallel to y)

and z = 1

3. Plot the plane (101)

4. Plot the plane (221)

• Taking reciprocals of the

indices we get (1/2, 1/2, 1)

• The intercepts of the plane are

x = ½, y = ½ and z = 1

28

13/8/2011

Eng. Mat/ITB/UHH 15

Miller Indices for Cubic Unit Cells(Example)

• Taking reciprocals of the indices we get (1, -1, ∞)

• The intercepts of the plane are x = 1, y = -1 and z = ∞

(parallel to z axis)

5. Plot the plane (1 1 0)_

• To show this plane in a

single unit cell, the origin

is moved along the

positive direction of y

axis by 1 unit

29

Miller Indices (Important Relationship)

• This is only for the CUBIC SYSTEM:

• Direction indices of a direction perpendicular to a crystal

plane are same as miller indices of the plane.

• Example:

• Interplanar spacing (dhkl) between parallel closest

planes with same miller indices is given by

dhkl =a

h2 + k 2 + l2

30a = lattice constant

h, ,k, l = miller indices of cubic

planes being considered)

13/8/2011

Eng. Mat/ITB/UHH 16

Indices for Crystal Planes in

Hexagonal (HCP) Unit Cells

• Four indices are used (hkil) called as Miller-Bravals

indices.

31

• Four axes are used (a1, a2,a3 and c)

• Reciprocal of the intercepts

that a crystal plane makes

with the a1, a2, a3 and caxes give the h, k, i and l

indices respectively.

Indicesfor Crystal Planesin HCP Unit Cells (Examples)

• Basal Planes:

• Prism Planes: For plane ABCD

32

13/8/2011

Eng. Mat/ITB/UHH 17

Comparison of FCC and HCP Crystals

• Both FCC and HCP are close packed and have APF 0.74

• FCC crystal is close packed in (111) plane while HCP is

close packed in (0001) plane.

FCC Crystal Structure

showing a close-packed

(111) plane

HCP Crystal Structure

showing a close-packed

(0001) plane

33

Structural Difference between HCP and FCC

34

13/8/2011

Eng. Mat/ITB/UHH 18

Volume Density

• Since the entire crystal can be generated by the

repetition of the unit cell, the density of a crystalline

material

• Example:

• Copper (FCC) has atomic mass of 63.54 g/mol and

atomic radius of 0.1278 nm.

• Volume of unit cell,

ρv =Mass /Unit cell

Volume /Unit cell

35

Densities of Material Classes

• In general

ρmetals > ρceramics > ρpolymers

• Reasons:

• Metals have close-packing (metallic bonding) and often

large atomic masses.

• Ceramics have less dense packing and often lighter

elements

• Polymers have low packing density (often amorphous)

and lighter elements (C, H, O)

36

13/8/2011

Eng. Mat/ITB/UHH 19

Polymorphism or Allotropy

• Polymorphism

• Materials exist in more than one crystalline form under

different conditions of temperature and pressure.

• If the material is an elemental solid, it is called allotropy

• Example: Iron exists in both BCC and FCC form

depending on the temperature.

37

• Another example: Carbon, which can exist as diamond,

graphite and amorphous carbon.

Polymorphism or Allotropy

38

13/8/2011

Eng. Mat/ITB/UHH 20

Non-Crystalline (Amorphous) Solids

• Solids lack of a systematic and regular arrangement of

atoms over relatively large atomic distances.

• Two-dimensional schematic diagrams for:

• Rapid cooling of metals (108 K/s) can give rise to

amorphous structure since little time is allowed for

ordering process.

39

Crystalline SiO2Non Crystalline SiO2

Much more

disordered and

irregular