engr 151: materials of engineering midterm 1 review material · 2017-03-02 · midterm 1 general...

MIDTERM 1 REVIEW MATERIAL

ENGR 151: Materials of Engineering

MIDTERM 1

General properties of materials

Bonding (primary, secondary and sub-types)

Properties of different kinds of bonds

Types of materials (metals, non-metals, ceramics,

polymers, semiconductors) and their characteristics

Different cell structures and properties (APF,

coordination number, relationship between a

and R, etc.)

Directions and planes

Geometric problems, sketching and interpreting

MIDTERM 1

Different coordinate systems (e.g. Miller-

Bravais)

Conversion from one system to the other

Density calculations (linear, planar)

Bragg’s law problems

Read Chapter 1 from textbook

CHAPTER 1 - INTRODUCTION

Metals used in structures & machinery

Plastics used in packaging, medical devices, consumer goods, clothing

Ceramics used in electronics (insulative)

Composites are novel materials for all applications listed above

4

CHAPTER 2 OBJECTIVES

Understand the elements used to make engineering materials

Review basic chemistry and physics principles

Overview of the materials classes:

Metals: good conductors, strong, lustrous

Polymers: organic, low densities, flexible

Ceramics: clay, cement, glass; insulators

Composites: fiberglass; strength and flexibility

5

ATOMIC STRUCTURE

Elements

Atomic number (Z, number of protons in nucleus)

Protons, neutrons, electrons (masses)

mp = mn = 1.67 x 10-27 kg, me = 9.11 x 10-31 kg

Atomic Mass (A) = sum of proton and neutron masses in nucleus

Isotope = same element, differing atomic masses E.g. Hydrogen (P = 1, N = 0), Deuterium (P = 1, N = 1), Tritium (P

= 1, N = 2).

Atomic Weight = Average atomic mass of all naturally-occurring isotopes.

amu (atomic mass unit) = 1/12 of atomic mass of carbon 12

One mole = 6.023 x 1023 (Avogadro’s number) atoms

6

ELECTRON CONFIGURATIONS

Rules of electron configuration (Table 2.1, pg.

23)

Electrons are quantized (have specific energies –

discrete energy levels)

Quantum numbers (4)

Principal: Position (n, distance of an electron from nucleus)

Azimuthal: Subshell (l)

Determines orbital angular momentum

s, p, d, or f (shape of electron subshell)

Magnetic: Number of energy states per subshell (ml)

s-1, p-3, d-5, f-7

Spin: Spin moment (ms)

+1/2, -1/2

7

ELECTRON CONFIGURATIONS – CONTD.

8

ELECTRON CONFIGURATIONS – CONTD.

Pauli Exclusion Principle:

No more than two electrons per electron state

Number of electron states per shell determined by magnetic quantum

number

Examples:

3p shell has 3 states (-1, 0, +1), therefore can

accommodate up to 6 electrons (2 electrons per state).

3d shell has 5 states (-2, -1, 0, +1, +2), therefore can

accommodate up to 10 electrons (2 electrons per state).

9

QUICK REVIEW (TABLE 2.2, PG. 22)

How many valence electrons do they have?

Hydrogen, 1s1

Aluminum, 1s22s22p63s23p1

Chlorine, 1s22s22p63s23p5

Answer: 1, 3, 7

10

ELECTRONEGATIVITY

Measures the tendency of an element to give up or accept

valence electrons

Electropositive elements (e.g. alkali metals)

Capable of giving up few valence electrons to become positively charged

(e-, negative charge)

Electronegative elements (e.g. halogens)

Readily accept electrons to form negatively charged ions.

Also share electrons (covalent bonding)

Electronegativity increases left to right, bottom to top

Atoms accept electrons if shells are closer to nucleus

Example: Na gives up one electron, Cl accepts the electron to

form NaCl

ATOMIC BONDING

To understand the physical properties behind

materials, we must have an understanding of

interatomic forces that bind atoms together.

At large distances, the interactions between

two atoms are negligible…BUT…as they come

closer to each other they start to exert a force

on each other.

ATOMIC BONDING

There are two types of forces that are both

functions of the distance between two atoms:

1) Attractive Force (FA) – Depends on

bonding between atoms

2) Repulsive Force (FR) – Originates due

to repulsion between atoms’ individual

(negatively-charged) electron clouds

ATOMIC BONDING

Magnitude of an attractive force varies with

distance.

The Net Force (FN) is the sum of the attractive

and repulsive forces:

ATOMIC BONDING – CONTD.

When FA = FR the net force is zero:

(State of equilibrium)

In a state of equilibrium, the two atoms will remain separated by the distance, ro. Attractive force is the same as repulsive force at ro.

For many atoms, ro is approximately .3 nm or 3 angstroms (Å)


Another way to represent this relationship in attractive and repulsive forces is to look at potential energy relationships.

Force-energy relationships: Both force and energy are functions of distance r

Measure of amount of work done to move an atom from infinity (zero force) to a distance r.

Alternatively:


Energy relationships:

EN = net energy

EA = attractive energy

ER = repulsive energy


Energy relationships:

Why does zero force correspond to minimum energy?


The net potential energy curve has a trough

around its minimum. The potential energy

minimum is ro away from the origin.

Force is the derivative of energy.


The Bonding Energy, Eo, refers to the vertical

distance between the minimum potential

energy and the x-axis. This is the energy that

would be required to separate the atoms to an

infinite separation.

Force and Energy plots become more complex

in actual materials. Why?

ATOMIC BONDING ENERGY

Magnitude of bonding energy and shape of

energy-versus-interatomic separation curve

vary from material to material AND depend on

the type of bonding that is taking place

between atoms.

ATOMIC BONDING ENERGY (EO)

Materials with large Bonding Energies also have high MELTING POINTS

At room temperature, materials with high bonding energy are in the SOLID STATE; those with small bonding energy are in the GASEOUS STATE; LIQUID STATE for intermediate bonding energy

FLEXIBILITY: stiff materials have a steep slope at the r=ro position; the slope is less steep for more flexible materials. Flexibility or hardness of a material is gauged by the MODULUS of ELASTICITY.

How much a material expands upon heating or contracts upon cooling is related to the potential energy curve. This is called LINEAR COEFFICIENT of THERMAL EXPANSION.

TYPES OF ATOMIC BONDING

Three types of PRIMARY bonds: IONIC,

COVALENT, METALLIC.

Nature of the bond depends on the electron

structures of the bonding atoms AND type of

bond depends on the tendency of atoms to

assume stable electron structures (completely

filling outermost electron shell)

IONIC BONDING

Always found in compounds that are formed by reactions between metallic and nonmetallic elements (elements at the horizontal extremities of the periodic table).

Metallic elements easily give up valence electrons to nonmetallic atoms.

In this process, all atoms acquire stable or inert gas configurations and electrical charges (becoming ions)

IONIC BONDING

The attractive bonding forces within ionic bonds are

COULOMBIC (positive and negative ions attracting one

another).

E.g. Na + Cl Na+ + Cl- NaCl

Notice the crystalline structure in the diagrams below.

IONIC BONDING – CONTD.

Attractive energy is a function of interatomic distance (inversely

proportional):


Constant A is expressed as:


Repulsive energy:


Equation for attractive force:

COVALENT BONDING

Stable electron configurations are assumed by

the sharing of electrons between adjacent

atoms.

Atoms contribute at least one electron to the

bond, and the shared electrons are considered

to belong to both atoms

Covalent bonds are directional (between

specific atoms)

COVALENT BONDING – CONTD.

Examples include: Hydrogen (H2), Chloride (Cl2),

Fluoride (F2), Water (H20), diamond, silicon,

germanium, elemental solids located on right

hand side of the periodic table, Gallium

Arsenide (GaAs), indium antimonide (InSb), and

silicon carbide (SiC).

Covalent bond in Methane (CH4)

COVALENT BONDING

The number of covalent bonds for an atom is

determined by the number of valence

electrons.

For N' valence electrons, an atom can

covalently bond with at most 8 - N' other atoms.

METALLIC BONDING

Metallic materials have one, two, or at most three valence

electrons which are not bound to any atom within the solid and

can drift throughout the metal

Creates a “sea of electrons” (electrons that belong to the entire

metal)

Remaining nonvalence electrons and their atomic nuclei are

called ION CORES (have net positive charge equal in magnitude

to total valence charge per atom)

METALLIC BONDING – CONTD.

Free electrons shield ion cores from repulsive forces

and act as glue to hold ion cores together

Metallic bond is nondirectional

Free electrons allow metal to be good conductors of

heat and electricity

METALLIC BONDING ENERGIES

Can be strong or weak (Mercury, 68 kJ/mol, -

39 degrees Celsius; Tungsten, 850 kJ/mol,

3410 degrees Celsius)

SECONDARY BONDING

Very weak bonds (10 kJ/mol)

Exist between all atoms and molecules but obscured if other primary bonding is occurring

Found in:

inert gases because of their stable electron structures

between molecules that are covalently bonded

arise from dipoles (separation of positive and negative portions of an atom or molecule).

Also known as van der Waals bonding or physical bonding.

Physical, as opposed to chemical bonding – weaker bonds.

SECONDARY BONDING – CONTD.

+ -

+ -

Atomic or Molecular Dipoles

FLUCTUATING INDUCED DIPOLE BONDS

Atom structure is not symmetric and positively

charged nucleus creates dipole with electron cloud on

opposite side of atom.

Instantaneous and short-lived distortions of electric field.

+ -

Electrically Symmetric Atom

Induced Atomic Dipole

Atomic Nucleus

Electron Cloud

Atomic Nucleus

Electron Cloud

POLAR MOLECULE-INDUCED DIPOLE BONDS

Polar molecules: asymmetrical arrangements of

positively and negatively charged regions within

molecules.

Can also induce dipoles in adjacent nonpolar

molecules to form a bond.

H

Cl

+ -

PERMANENT DIPOLE BONDS

Hydrogen Bonding: a type of secondary

bonding occurring in molecules that have

hydrogen covalently bonded to fluorine, oxygen,

and nitrogen.

Stronger than induced dipole bond

Adjacent polar molecules attracting each other

H

F H

F

SUMMARY OF BONDING TYPES

Bonding

Primary

Secondary

Ionic

Covalent

Metallic

Fluctuating Induced

Dipole Bonds

Polar Molecule-Induced

Dipole Bonds

Permanent Dipole

Bonds

CRYSTAL STRUCTURE

Material properties depend on crystal structure of the material

Atoms are thought of as being solid spheres having well-defined diameters (atomic hard sphere model, atoms touching each other)

Lattice: three-dimensional array of points coinciding with atom positions (sphere centers)

43

CRYSTAL STRUCTURE – CONTD.

44

UNIT CELLS

Subdivide the crystal structure into small

repeating entities called UNIT CELLS

These cells are mainly in cubes, prisms, six-

sided figures – regular, repeating geometric

structures

Geometric symmetry

The unit cell is the basic structural unit or

building block of crystal structure

45

METALLIC CRYSTAL STRUCTURES

No restrictions as to the number and position

of nearest-neighbor atoms (dense atomic

packing)

Each sphere in atomic hard sphere model

equates to an ion core

46

METALLIC CRYSTAL STRUCTURES CONTD.

Four simple crystal structures are found in

metals:

Simple Cubic (SC)

Face-Centered Cubic (FCC)

Body-Centered Cubic (BCC)

Hexagonal Close-Packed (HCP)

47

ANALYSIS TECNIQUES

Analysis of crystal structures gives insight into

properties of material such as strength, density,

how the material may behave under physicals

stress, etc.

Analysis steps:

Identify spatial geometries associated with unit cell

Relate dimensions of unit cell to atomic radius

Characterize/calculate required properties of

crystal structure

48

49

• Tend to be densely packed.

• Reasons for dense packing:

- Typically, only one element is present, so all atomic

radii are the same.

- Metallic bonding is not directional.

- Nearest neighbor distances tend to be small in

order to lower bond energy.

- Electron cloud shields cores from each other.

• Metals have the simplest crystal structures.

We will examine four such structures...

METALLIC CRYSTAL STRUCTURES

50

• Rare due to low packing density (only Po has this structure)

• Close-packed directions are cube edges.

• Coordination # = 6

(# nearest neighbors)

SIMPLE CUBIC STRUCTURE (SC)

Fig. 3.3, Callister & Rethwisch 9e.

51

• APF for a simple cubic structure = 0.52

APF =

a 3

4

3 π (0.5a) 3 1

atoms

unit cell atom

volume

unit cell

volume

ATOMIC PACKING FACTOR (APF)

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

Adapted from Fig. 3.3 (a),

Callister & Rethwisch 9e.

close-packed directions

a

R = 0.5a

contains 8 x 1/8 = 1 atom/unit cell

FACE-CENTERED CUBIC (FCC)

The UNIT CELL is CUBIC

Atoms located at each of the corners and the

centers of all the cubic faces

Aluminum, Copper, Gold, Lead, Nickel,

Platinum, Silver

52

FACE-CENTERED CUBIC

The cube edge length a and the atomic radius

are related through:

a = 2R√2

53

54

• APF for a face-centered cubic structure = 0.74 ATOMIC PACKING FACTOR: FCC

maximum achievable APF

APF =

4

3 π ( 2 a/4 ) 3 4

atoms

unit cell atom

volume

a 3

unit cell

volume

Close-packed directions:

length = 4R = 2 a

Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

a

2 a

Adapted from

Fig. 3.1(a),

Callister &

Rethwisch 9e.

FCC

Each corner atom is shared by eight unit cells

(Eight 1/8 portions per unit cell)

Each face atom is shared by two unit cells (six

½ portions per unit cell)

Grand total of four whole atoms per unit cell

55

COORDINATION NUMBER

The number of nearest neighbor or touching

atoms

The coordination number for FCC is ??

56

COORDINATION NUMBER

The number of nearest neighbor or touching

atoms

The coordination number for FCC is 12

57

ATOMIC PACKING FACTOR (APF)

The fraction of solid sphere volume in a unit

cell:

Volume of atoms in a unit cell

Total unit cell volume

Using the APF equation and the volume of an

FCC unit cell, find the APF of a FCC crystal

structure

APF =

58

BODY-CENTERED CUBIC (BCC)

59

BCC

Atoms located in all eight corners and a single

atom at the cube center

Derive an expression for the unit cell edge

length (a) using the atomic radius (R)

60

61

ATOMIC PACKING FACTOR: BCC

APF =

4

3 π ( 3 a/4 ) 3 2

atoms

unit cell atom

volume

a 3

unit cell

volume

length = 4R =

Close-packed directions:

3 a

• APF for a body-centered cubic structure = 0.68

Adapted from

Fig. 3.2(a), Callister &

Rethwisch 9e.

a

a 2

a 3

a R

BCC

8 x 1/8 = 1 (each corner)

1 atom in center

2 atoms per unit cell

What is the coordination number and APF?

62

BCC

8 x 1/8 = 1 (each corner)

1 atom in center

2 atoms per unit cell

What is the coordination number and APF?

C.N. = 8

APF = .68

63

HEXAGONAL CLOSE-PACKED (HCP)

Unit cell is hexagonal

Six atoms form a regular hexagon and surround

a single atom

Another plane is situated in between top and

bottom plane (provides three additional atoms)

64

HCP

Six atoms altogether within unit cell

65

HCP

Six atoms altogether within unit cell

1/6 x 12 (top and bottom plane portions

½ x 2 (center face atoms)

3 midplane interior atoms

C.N. = 12

APF = .74

c/a = 1.633

66

OVERVIEW

Crystal

Structure

Relationship

between a

and R (cubic

structures)

Number of

atoms per

unit cell

Coordination

Number APF

FCC 4 12 0.74

BCC 2 8 0.68

HCP ------ 6 12 0.74

67

THEORETICAL DENSITY, R

68

where n = number of atoms/unit cell

A = atomic weight

VC = Volume of unit cell = a3 for cubic

NA = Avogadro’s number

= 6.022 x 1023 atoms/mol

Density = =

VC NA

n A =

Cell Unit of Volume Total

Cell Unit in Atoms of Mass

THEORETICAL DENSITY, R

Ex: Cr (BCC)

A (atomic weight) = 52.00 g/mol

n = 2 atoms/unit cell

R = 0.125 nm

69

theoretical

a = 4R/ 3 = 0.2887 nm

actual

a R

= a 3

52.00 2

atoms

unit cell mol

g

unit cell

volume atoms

mol

6.022 x 1023

= 7.18 g/cm3

= 7.19 g/cm3

• Coordinates of selected points in the unit cell.

• The number refers to the distance from the origin in terms

of lattice parameters.

Point Coordinates

• Each unit cell is a reference or basis.

• The length of an edge is normalized as a unit of

measurement.

• E.g. if the length of the edge of the unit cell along the

X-axis is a, then ALL measurements in the X-direction

are referenced to a (e.g. a/2).

Point Coordinates – Contd.

Point coordinates for unit cell center are

a/2, b/2, c/2 ½ ½ ½

Point coordinates for unit cell corner are 111

Translation: integer multiple of lattice constants identical position in another unit cell

72

z

x

y a b

c

000

111

y

z

2c

b

b

Point Coordinates

EXAMPLE PROBLEMS

73

EXAMPLE PROBLEMS CONTD.

74

EXAMPLE PROBLEMS CONTD.

75

Determine the Miller indices of directions A, B, and C.

Miller Indices, Directions

(c) 2003 Brooks/Cole Publishing /

Thomson Learning™

GENERAL APPROACH FOR MILLER INDICES

77

SOLUTION

Direction A

1. Two points are 1, 0, 0, and 0, 0, 0

2. 1, 0, 0, -0, 0, 0 = 1, 0, 0

3. No fractions to clear or integers to reduce

4. [100]

Direction B

1. Two points are 1, 1, 1 and 0, 0, 0

2. 1, 1, 1, -0, 0, 0 = 1, 1, 1

3. No fractions to clear or integers to reduce

4. [111]

Direction C

1. Two points are 0, 0, 1 and 1/2, 1, 0

2. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1

3. 2(-1/2, -1, 1) = -1, -2, 2

2]21[ .4

(c) 2003 Brooks/Cole Publishing

/ Thomson Learning™

CRYSTALLOGRAPHIC DIRECTIONS

79

1. Vector repositioned (if necessary) to pass

through origin.

2. Read off projections in terms of

unit cell dimensions a, b, and c

3. Adjust to smallest integer values

4. Enclose in square brackets, no commas

[uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]

-1, 1, 1

z

x

Algorithm

where overbar represents a

negative index

[ 111 ] =>

y

Problem 3.6, pg. 58

EXAMPLE PROBLEMS

80

EXAMPLE PROBLEMS, CONTD.

81


82

Exam problems – make

sure you translate and

scale vector to fit WITHIN

unit cube if asked!

83

HCP CRYSTALLOGRAPHIC DIRECTIONS

Hexagonal Crystals

4 parameter Miller-Bravais lattice coordinates are

related to the direction indices (i.e., u'v'w') as follows.

=

=

=

' w w

t

v

u

) v u ( + -

) ' u ' v 2 ( 3

1 -

) ' v ' u 2 ( 3

1 - =

] uvtw [ ] ' w ' v ' u [

Fig. 3.8(a), Callister & Rethwisch 8e.

- a3

a1

a2

z

84


Fig. 3.8(a), Callister & Rethwisch 8e.

- a3

a1

a2

z

85


1. Vector repositioned (if necessary) to pass

through origin.

2. Read off projections in terms of unit

cell dimensions a1, a2, a3, or c


4. Enclose in square brackets, no commas

[uvtw]

[ 1120 ] ex: ½, ½, -1, 0 =>

Adapted from Fig. 3.8(a),


dashed red lines indicate

projections onto a1 and a2 axes a1

a2

a3

-a3

2

a 2

2

a 1

- a3

a1

a2

z

Algorithm

86

REDUCED-SCALE COORDINATE AXIS

87

PROBLEM 3.8

88

PROBLEM 3.8, CONTD.

89

PROBLEM 3.8, CONTD.

90

PROBLEM 3.8, CONTD.

91

PROBLEM 3.8, CONTD.

92

PROBLEM 3.8, CONTD.

93

PROBLEM 3.9

94

PROBLEM 3.9, CONTD.

95

DRAWING HCP CRYSTALLOGRAPHIC

DIRECTIONS (I)

1. Remove brackets

2. Divide by largest integer so all values

are ≤ 1

3. Multiply terms by appropriate unit cell

dimension a (for a1, a2, and a3 axes)

or c (for z-axis) to produce

projections

4. Construct vector by placing tail at

origin and stepping off these

projections to locate the head

Algorithm (Miller-Bravais coordinates)

Adapted from Figure 3.10,


96

DRAWING HCP CRYSTALLOGRAPHIC

DIRECTIONS (II)

Draw the direction in a hexagonal unit cell.

[1213]

4. Construct Vector

1. Remove brackets -1 -2 1 3

Algorithm a1 a2 a3 z

2. Divide by 3 1 3

1

3

2

3

1

3. Projections

proceed –a/3 units along a1 axis to point p

–2a/3 units parallel to a2 axis to point q

a/3 units parallel to a3 axis to point r

c units parallel to z axis to point s

[1 2 13]

p

q r

s

start at point o

Adapted from p. 72,

Callister &

Rethwisch 9e.

[1213] direction represented by vector from point o to point s

97

1. Determine coordinates of vector tail, pt. 1:

x1, y1, & z1; and vector head, pt. 2: x2, y2, & z2.

in terms of three axis (a1, a2, and z)

2. Tail point coordinates subtracted from head

point coordinates and normalized by unit cell

dimensions a and c


4. Enclose in square brackets, no commas,

for three-axis coordinates

5. Convert to four-axis Miller-Bravais lattice

coordinates using equations below:

6. Adjust to smallest integer values and

enclose in brackets [uvtw]

Adapted from p. 72, Callister &

Rethwisch 9e.

Algorithm

DETERMINATION OF HCP CRYSTALLOGRAPHIC

DIRECTIONS (II)

98

4. Brackets [110]

1. Tail location 0 0 0

Head location a a 0c

1 1 0 3. Reduction 1 1 0

Example a1 a2 z

1/3, 1/3, -2/3, 0 => 1, 1, -2, 0 => [ 1120 ]

6. Reduction & Brackets

Adapted

from p. 72,

Callister &

Rethwisch

9e.

DETERMINATION OF HCP CRYSTALLOGRAPHIC

DIRECTIONS (II)

Determine indices for green vector

2. Normalized

FAMILIES OF DIRECTIONS <UVW>

For some crystal structures, several

nonparallel directions with different

indices are crystallographically equivalent;

this means that atom spacing along each

direction is the same.

99

100

CRYSTALLOGRAPHIC PLANES

Adapted from Fig. 3.10,


101

CRYSTALLOGRAPHIC PLANES

Miller Indices: Reciprocals of the (three) axial intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices.

Algorithm 1. Read off intercepts of plane with axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce to smallest integer values 4. Enclose in parentheses, no commas i.e., (hkl)

102

CRYSTALLOGRAPHIC PLANES z

x

y a b

c

4. Miller Indices (110)

example a b c z

x

y a b

c


1. Intercepts 1 1

2. Reciprocals 1/1 1/1 1/

1 1 0 3. Reduction 1 1 0

1. Intercepts 1/2

2. Reciprocals 1/½ 1/ 1/

2 0 0 3. Reduction 1 0 0

example a b c

103

CRYSTALLOGRAPHIC PLANES z

x

y a b

c


example 1. Intercepts 1/2 1 3/4

a b c

2. Reciprocals 1/½ 1/1 1/¾

2 1 4/3

3. Reduction 6 3 4

(001) (010),

Family of Planes {hkl}

(100), (010), (001), Ex: {100} = (100),

FAMILY OF PLANES

Planes that are crystallographically equivalent

have the same atomic packing.

Also, in cubic systems only, planes having the

same indices, regardless of order and sign,

are equivalent.

Ex: {111}

= (111), (111), (111), (111), (111), (111), (111), (111)

104

(001) (010), (100), (010), (001), Ex: {100} = (100),

_ _ _ _ _ _ _ _ _ _ _ _

FCC UNIT CELL WITH (110) PLANE

105

BCC UNIT CELL WITH (110) PLANE

106

107

CRYSTALLOGRAPHIC PLANES (HCP)

In hexagonal unit cells the same idea is used

example a1 a2 a3 c

4. Miller-Bravais Indices (1011) (hkil)

1. Intercepts 1 -1 1 2. Reciprocals 1 1/

1 0

-1

-1

1

1

3. Reduction 1 0 -1 1

a2

a3

a1

z

Adapted from Fig. 3.8(b),


108

DISTINCTION BETWEEN CRYSTALLOGRAPHIC

DIRECTIONS AND MILLER INDICES

Crystallographic Direction Crystallographic Planes

Associated with point

coordinates in 3D space

Associated with planar

structures in 3D space

Denoted by enclosing in

square brackets as [abc]

Denoted by enclosing in

regular parentheses (hkl)

EXAMPLE PROBLEMS

109


110

Example of

family of

planes


111

Don’t

forget to

slide plane

back within

unit cell!


112

Strategy

Find answer in

3 coordinates

and then

convert to 4-

coordinate

representation


113

114

ex: linear density of Al in [110]

direction

a = 0.405 nm

LINEAR DENSITY

Linear Density of Atoms LD =

a

[110]

Unit length of direction vector

Number of atoms

# atoms

length

3.5 atoms/nm a 2

2 LD = =

Adapted from Fig. 3.1(a),


2 atoms per line, sharing computed

along vector length

115

PLANAR DENSITY OF (100) IRON Solution: At T < 912ºC iron has the BCC structure.

(100)

Radius of iron R = 0.1241 nm

R 3

3 4 a =

Adapted from Fig. 3.2(c), Callister & Rethwisch 8e.

2D repeat unit

= Planar Density = a 2

1

atoms

2D repeat unit

= nm2

atoms 12.1

m2

atoms = 1.2 x 1019

1

2

R 3

3 4 area

2D repeat unit

116

PLANAR DENSITY OF (111) IRON

Solution (cont): (111) plane 1 atom in plane/ unit surface cell

3 3 3

2

2

R 3

16 R

3

4

2 a 3 ah 2 area =

= = =

atoms in plane

atoms above plane

atoms below plane

a h 2

3 =

a 2

1

= = nm2

atoms 7.0

m2

atoms 0.70 x 1019

3 2 R 3

16 Planar Density =

atoms

2D repeat unit

area

2D repeat unit

CONSTRUCTIVE INTERFERENCE

Occurs at angle θ to the planes, if path length

difference is equal to a whole number, n, of

wavelengths (n equals order of reflection):

Bragg’s Law

BRAGG’S LAW

If Bragg’s Law is not satisfied then we get non-

constructive interference, which will yield a very

low-intensity diffracted beam.

For cubic unit cells:

BRAGG’S LAW

Specifies when diffraction will occur for unit

cells having atoms positioned only at the cell

corners

Atoms at different locations act as extra

scattering centers and results in the absence of

some diffracted beams.

EXAMPLE PROBLEM

engr 151: materials of engineering midterm 1 review material · 2017-03-02 · midterm 1 general...

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