BNQ 30503 Chapter 1 (S).pdf

1 1 1 1

BNQ 30503

BNQ 30503

MA

MA

TERIAL

TERIAL

ENGINE

ENGINE

ERING

ERING

TECHNOLOGY

TECHNOLOGY

CHAPTER 1:

CHAPTER 1:

SOL

SOL

ID STRUCTUR

ID STRUCTUR

E

E

By: Engr. Dr. Nasrul Fikry Che Pa By: Engr. Dr. Nasrul Fikry Che Pa

http://bit.ly/19o6LoL

Content’

Content’s s ReviewReview

•

• Chapter 1: Solid StructureChapter 1: Solid Structure

 –

 – What you will learn?What you will learn?

••

Crystal structures fundamental concepts, metallic crystalCrystal structures fundamental concepts, metallic crystal structures, crystallographic points, directions and

structures, crystallographic points, directions and planes, crys

planes, crystalline talline and noncrystand noncrystallinealline matematerials, poinrials, pointt defects, dislocations, importance of defects.

Content’

Content’s s ReviewReview

•

• Chapter 1: Solid StructureChapter 1: Solid Structure

 –

 – What you will learn?What you will learn?

••

Crystal structures fundamental concepts, metallic crystalCrystal structures fundamental concepts, metallic crystal structures, crystallographic points, directions and

structures, crystallographic points, directions and planes, crys

planes, crystalline talline and noncrystand noncrystallinealline matematerials, poinrials, pointt defects, dislocations, importance of defects.

3

3

Schematic representation of the most Schematic representation of the most

basic structural unit for quartz (all silicate basic structural unit for quartz (all silicate material). Each atom of Si surrounded by material). Each atom of Si surrounded by 4 O

4 O₂₂ atom whose center are located atatom whose center are located at the corner of a tetrahedron. Represented the corner of a tetrahedron. Represented as

as





₄₄4−4−

Sketch of a unit cell for Quartz, Sketch of a unit cell for Quartz,

composed of several interconnected composed of several interconnected





₄₄4−4− tetrahedral.tetrahedral.

Introduction

4

Schematic diagram showing a large interconnected



₄4− tetrahedral. Observe the shape resembles quartz.

Photograph of 2 single crystals of quartz.

Fundamental Concepts

•  Atoms pack in periodic, 3D arrays Crystalline materials...

metals, many ceramic & some polymers

crystalline SiO2

• Examples:

•  Atoms have no periodic packing Non-crystalline materials...

- complex structures - rapid cooling

- Lack systematic & regular arrangement of atoms over relatively large atomic distance

noncrystalline SiO2

" Amorphous" = Noncrystalline

Si Oxygen • occurs for:

-literally means ‘without form’

“Atoms in crystalline solid are position in orderly & repeated patterns that are in contrast to the random & disordered atomic distribution found in

Metallic Crystal Structures

• How can we stack metal atoms to minimize empty space?

2-dimensions

vs.

Imagine that these 2D layers are stacked to make 3D structures. Which one having a minimized empty spaces?

Metallic Crystal Structures

• Tend to be densely packed.

• Assumptions in 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

• Have the simplest crystal structures.

Next…. We _{will examine four such structures...}

Simple Cubic Structure (SC)

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

• Close-packed directions are cube edges.

• Coordination # = 6 (# nearest neighbors)

Click once on image to start animation (Courtesy P.M. Anderson)

Atomic Packing Factor (APF)

• APF for a simple cubic structure = 0.52

 APF = a3 4 3

p

(0.5a) 3 1 atoms unit cell atom volume unit cell volume  APF = Volume of atoms in unit cell* (Vs)

Volume of unit cell (V_c) *assume hard spheres

close-packed directions

a

R=0.5a

Body Centered Cubic Structure (BCC)

• Coordination # = 8

• Atoms touch each other along cube diagonals.

--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.

ex: Cr, W, Fe (), Tantalum, Molybdenum

2 atoms/unit cell:

1 center + 8 corners x 1/8

Click once on image to start animation (Courtesy P.M. Anderson)

Atomic Packing Factor: BCC

length = 4R =

Close-packed directions: 3 a • APF for a body-centered cubic structure = 0.68

a R a 3 a a 2  APF = atoms

unit cell _atom

volume 3 unit cell volume 4 3

p

( 3a/4)3 2 a

Face Centered Cubic Structure (FCC)

• Coordination # = 12

• Atoms touch each other along face diagonals.

--Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.

4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8

Click once on image to start animation (Courtesy P.M. Anderson)

Atomic Packing Factor: FCC

• APF for a face-centered cubic structure = 0.74

maximum achievable APF

 APF =

4

3

p

( 2a/4)3 4

atoms

unit cell _atom

volume a3 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

 A sites B _B B B B B B C sites C _C C  A B B sites

• ABCABC... Stacking Sequence • 2D Projection • FCC Unit Cell

FCC Stacking Sequence

B _B B B B B B B sites C _C C  A C _C C  A  A B C

Hexagonal Close-Packed

Structure (HCP)

• Coordination # = 12

• ABAB... Stacking Sequence

• APF = 0.74

• 3D Projection • 2D Projection

6 atoms/unit cell

ex: Cd, Mg, Ti, Zn • c/a = 1.633 (long over short branch)

c

a

 A sites B sites

 A sites _{Bottom layer} 

Middle layer  Top layer 

Exercise:

1. If the atomic radius of aluminum (FCC) is 0.143

nm, calculate the volume of its unit cell in cubic

meters.

Theoretical Density,

where n = number of atoms/unit cell

 A = atomic weight

V_C = Volume of unit cell = a3 _{for cubic}

N _A = Avogadro’s number 

= 6.022 x 1023 _atoms/mol Density =

r

= V_CN _A n A

r

= Cell Unit of  Volume Total Cell Unit in  Atoms of  Mass

• Ex: Cr (BCC)  A _{= 52.00 g/mol} R _{= 0.125 nm} n _{= 2 atoms/unit cell}

Theoretical Density,

a = 4R/ 3 = 0.2887 nm a  R 

r

= a3 52.00 2 atoms unit cell mol g unit cell volume atoms mol 6.022x1023

r

_theoretical

r

_actual = 7.18 g/cm3 = 7.19 g/cm3

Exercise:

20

Exercise:

1) Zirconium has an HCP crystal structure and a density of 6.51 g/cm3_.

 – (a) What is the volume of its unit cell in cubic meters?

 – (b) If the c/a ratio is 1.593, compute the values of c and a.

V_C N _A n A =

Exercise

2. Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your

determination. A simple cubic unit cell is shown in Figure 3.24.

Alloy Atomic Weight (g/mol) Density (g/cm3) Atomic Radius (nm) A 77.4 8.22 0.125

B 107.6 13.42 0.133 C 127.3 9.23 0.142

Fig. 3.24: Hard sphere unit cell representation of the simple cubic crystal structure.

Exercise

3. Rhenium has an HCP crystal structure, an atomic

radius of 0.137 nm, and a c/a ratio of 1.615.

Densities of Material Classes

r

_{metals >}

r

_{ceramics >}

r

_polymers

Why?       r    (  g    /  c  m    )    3 Graphite/ Ceramics/ Semicond Metals/  Alloys Composites/ fibers Polymers 1 2 20 30

Based on data in Table B1, Callister *GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers

in an epoxy matrix). 10 3 4 5 0.3 0.4 0.5 Magnesium  Aluminum Steels Titanium Cu,Ni Tin, Zinc Silver, Mo Tantalum Gold, W Platinum Graphite Silicon Glass -soda Concrete Si nitride Diamond  Al oxide Zirconia HDPE, PS PP, LDPE PC PTFE PET PVC Silicone Wood  AFRE* CFRE* GFRE* Glass fibers Carbon fibers  Aramid fibers Metals have... • close-packing (metallic bonding)

• often large atomic masses

Ceramics have...

• less dense packing • often lighter elements

Polymers have...

• low packing density (often amorphous)

• lighter elements (C,H,O)

Composites have...

• intermediate values

Crystals as Building Blocks

• Some engineering applications require single crystals:

• Properties of crystalline materials often related to crystal structure.

-- Ex: Quartz fractures more easily along some crystal planes than others.

-- diamond single

crystals for abrasives -- turbine blades

(Courtesy Martin Deakins, GE Superabrasives,

Worthington, OH. Used with permission.)

Polycrystals

• Most engineering materials are polycrystals.

• Each "grain" is a single crystal. • If grains are randomly oriented,

overall component properties are not directional. • Grain sizes typically range from 1 nm to 2 cm

(i.e., from a few to millions of atomic layers).

 Adapted from Fig. K, color inset pages of Callister 5e.

(Fig. K is courtesy of Paul E. Danielson, Teledyne Wah Chang  Albany)

1 mm

Isotropic  Anisotropic

Single crystal vs Polycrystals

• Single Crystals

-Properties vary with direction: anisotropic. -Example: the modulus

of elasticity (E) in BCC iron:

Data from Table 3.3, Callister & Rethwisch 8e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.)

• Polycrystals

-Properties may/may not vary with direction.

-If grains are randomly oriented: isotropic.

(Epoly iron = 210 GPa)

-If grains are textured, anisotropic.

200 mm  Adapted from Fig._{4.14(b), Callister &}

Rethwisch 8e.

(Fig. 4.14(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC [now the National Institute of Standards and

Technology,

Gaithersburg, MD].)

E (diagonal) = 273 GPa

Polymorphism

• Two or more distinct crystal structures for the same material (allotropy/polymorphism) titanium



,



-Ti carbon diamond, graphite BCC FCC BCC 1538ºC 1394ºC 912ºC -_Fe liquid iron system -_Fe -_Fe

Crystal Systems

Fig. 3.4, Callister & Rethwisch 8e.

7 crystal systems

14 Bravais crystal lattices

Unit cell: smallest repetitive volume which contains

the complete lattice pattern of a crystal.

Three edge lengths a, b, and c and three interaxial angles α, β, and γ are the lattice constants

4. Below is a unit cell for a hypothetical metal.

(a) To which crystal system does this unit cell belong?

(b) What would this crystal structure be called?

(c)

Calculate the density of the material, given that its atomic weight is 141 g/mol.

Point Coordinates

•Point coordinates for unit cell center are

a_/2, b_/2, c_{/2 ½ ½ ½}

•Point coordinates for unit cell corner are 111 z x y a b c 000 111

The manner in which the q, r and s

coordinates at point P within the unit cell are determined. The q coordinate

(which is a fraction) corresponds to the distance qa along the x axis, where a is the unit cell edge length. The respective r and s coordinates for the y and z axes are determined similarly.

Exercise

5. List the point coordinates for all atoms that

are associated with the FCC unit cell.

Crystallographic Directions

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

families of directions <uvw> (crystallographically equivalent [spacing btw atoms is the same] e.g in cubic system)

z

x

 Algorithm

where overbar represents a negative index

[111] =>

Linear Density

Linear Density of Atoms



LD =

ex: linear density of Al in [110] direction

a = 0.405 nm

a

[110]

Unit length of direction vector 

No. of atoms centered on direction vector 

# atoms length 1 3.5 nm a 2 2 LD

=

=

-Exercise

6. What are the indices for the directions

indicated by the two vectors in the sketch

below?

Exercise

7. For tetragonal crystals, cite the indices of

directions that are equivalent to each of the

following directions:

(a) [001]

(b) [110]

(c) [010]

HCP Crystallographic Directions

1. Vector repositioned (if necessary) to pass through origin.

2. Read off projections in terms of unit cell dimensions a₁, a₂, a₃, and c 3. Adjust to smallest integer values

4. Enclose in square brackets, no commas [uvtw]

[1120]

ex: ½, ½, -1, 0 =>

dashed red lines indicate

projections onto a₁ and a₂ axes a₁

a₂ a₃ -a₃ 2 a₂ 2 a₁ z  Algorithm

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

-a₃

a₁ a₂

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

Exercise

8. Determine indices for the directions shown in

the following hexagonal unit cells:

Crystallographic Planes

Crystallographic Planes

 Adapted from Fig

 Adapted from Fig. 3.10,. 3.10, Callister & Rethwisch 8e. Callister & Rethwisch 8e.

Crystallographic Planes

Crystallographic Planes

•

•

Miller Indices: Reciproc

Miller Indices:

Reciprocals of

als of the (thr

the (three) a

ee) axial

xial

inter

intercepts for a plane, cl

cepts for a plane, cleared of fractions & common

eared of fractions & common

multiples.

multiples. All par

All parallel pl

allel planes ha

anes have

ve same Miller

same Miller indices.

indices.

•

•

Algorithm

Algorithm

1.

1. Read Read off ioff intenterceprcepts of ts of plane plane with awith axes xes inin terms of

terms of aa_,, bb_,, cc

2. Take reciprocals of intercepts 2. Take reciprocals of intercepts 3.

3. ReduReduce to smallece to smallest intst integer vaeger valueslues 4.

4. EncloEnclose in se in parenparenthesestheses, no, no commas i.e.,

Crystallographic Planes

Crystallographic Planes

z z x x y y a a bb c c 4.

4. Miller Miller Indices Indices (1(110)10)

example example aa bb cc zz x x y y a a bb c c 4.

4. Miller Miller Indices Indices (100)(100) 1. Intercepts 1. Intercepts 1 1 11  2. Reciprocals 2. Reciprocals 1/1 1/1 1/1 1/1 1/1/ 1 1 1 1 00 3. Reduction 3. Reduction 1 1 1 1 00 1. Intercepts 1. Intercepts 1/21/2   2. Reciprocals 2. Reciprocals 1/½ 1/½ 1/1/ 1/1/ 2 2 0 0 00 3. Reduction 3. Reduction 2 2 0 0 00 example example aa bb cc

Crystallographic Planes

z x y a b c    4. Miller Indices (634) 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),

9. Determine the Miller indices for the planes shown in the

following unit cell:

10. Cite the indices of the direction that results from the

intersection of each of the following pair of planes

within a cubic crystal: (a) (100) and (010) planes, (b)

(111) and (111) planes, and (c) (101) and (001) planes.

Crystallographic Planes (HCP)

•

In hexagonal unit cells the same idea is used

example a₁ a₂ a₃ c 4. Miller-Bravais Indices (1011) 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), Callister & Rethwisch 8e.

a1 = h a3 = i = -(h+k) a2 = k z = l

Planar Density

•

Used to examine the atomic packing of crystallographic

planes

•

E.g: Iron foil can be used as a catalyst. The atomic

packing of the exposed planes is important.

a) Draw (100) crystallographic planes for Fe assuming that the structure is BCC.

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 = 2D repeat unit = Planar Density = a2 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

11. (a) Derive planar density expressions for FCC (100) and (111) planes in terms of the atomic radius R.

(b) Compute the planar density values for these same two planes for nickel when R = 0.125 nm.

Summary from part 1

• Atoms may assemble into crystalline or amorphous structures.

• We can predict the density of a material, provided we know the atomic weight, atomic radius, and crystal geometry (e.g., FCC, BCC, HCP).

• Common metallic crystal structures are FCC, BCC, and HCP. Coordination number and atomic packing factor  are the same for both FCC and HCP crystal structures.

• Crystallographic points, directions and planes are specified in terms of indexing schemes.

Crystallographic directions and planes are related to atomic linear densities and planar densities.

Summary from part 1 (Cont.)

• Some materials can have more than one crystal structure. This is referred to as polymorphism (or allotropy).

• Materials can be single crystals or polycrystalline.

Material properties generally vary with single crystal orientation (i.e., they are anisotropic), but are generally non-directional (i.e., they are isotropic) in polycrystals with randomly oriented grains.

• Solidification- result of casting of molten material

 – 2 steps

• Nuclei form

• Nuclei grow to form crystals – grain structure

• Start with a molten material – all liquid

Imperfections in Solids

• Crystals grow until they meet each other 

grain structure crystals growing

nuclei

Polycrystalline Materials

• Grain Boundaries

• regions between crystals

• transition from lattice of one region to that of

the other

• slightly disordered

• low density in grain boundaries

 – high mobility

 – high diffusivity

 – high chemical reactivity

• Grains can be - equiaxed (roughly same size in all directions) - columnar (elongated grains)

Solidification

Columnar in area with less undercooling Shell of equiaxed grains due to rapid cooling (greater T) near wall

Grain Refiner - added to make smaller, more uniform, equiaxed grains.

heat flow

~ 8 cm

heat flow

Types of Imperfections

• Vacancy atoms

• Interstitial atoms

• Substitutional atoms

Point defects

• Dislocations

Line defects

Point Defects in Metals

• Vacancies:

-vacant atomic sites in a structure.

• Self-Interstitials:

-"extra" atoms positioned between atomic sites.

Vacancy

distortion of planes

self-interstitial

distortion of planes

Equilibrium Concentration:

Point Defects



Boltzmann's constant (1.38 x 10-23 J/atom-K) (8.62 x 10-5 eV/atom-K)

N

_v

N

=

exp

-

_Q

_v

kT





 

 



No. of vacancies Total No. of atomic sites  Activation energy Temperature

Each lattice site is a potential

vacancy site

Estimating Vacancy Concentration

• Find the equil. # of vacancies in 1 m3 _{of Cu at 1000}



_C.

• Given:

 A_Cu = 63.5 g/mol

r

= 8.4 g/cm3

Qv = 0.9 eV/atom N A = 6.02 x 1023 atoms/mol

For 1 m3_{, N =} N A  A Cu

r

x x 1 m3_{= 8.0 x 10}28 _sites

= 2.7 x 10

-4 8.62 x 10-5 _eV/atom-K 0.9 eV/atom 1273 K



N

_v

N

=

exp

-

Q

_v

kT





 

 

• Answer: Nv = (2.7 x 10-4_{)(8.0 x 10}28_{) sites = 2.2 x 10}25 _vacancies

Exercise (Tutorial)

1.

Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K).

Assume an energy for vacancy formation of 0.55 eV/atom.

2

. Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of vacancies at 500



C (773 K) is 7.55



1023 _m-3_{. The atomic weight and density (at 500}



_{C) for aluminum are,}

Imperfections in Metals

Two outcomes if impurity (B) added to host (A):

• Solid solution of B in A (i.e., random dist. of point defects)

• Solid solution of B in A plus particles of a new phase (usually for a larger amount of B)

OR

Substitutional solid soln. (e.g., Cu in Ni)

Interstitial solid soln. (e.g., C in Fe)

Second phase particle -- different composition -- often different structure.

Impurities in Solids

•

Specification of composition

weight percent x 100 2 1 1 1 m m m C  = m₁ = mass of component 1 100 x 2 1 1 ' 1 m m m n n n C  =

n_m1 = number of moles of component 1 ,

atom percent

Derive an equation that relate C’₁ & C₁ ?

 Ans: C’₁ = [(C₁ A₂)/ [(C₁ A₂) + (C₂ A₁) ] ] x 100

n_m1 =   ()

Additional Equation 64 _′ =  __+__ x 100 ₂′ =  __+__ x 100

Conversion of weight percent to atom percent (for two-element alloy)

_ =  _  __+__ x 100 ₂ =  _  __+__ x 100

Conversion of atom percent to weight percent (for two-element alloy)

_  ₂ = 100 _′  ₂′ = 100

_ & ₂ = weight percent

 _ & ₂ = Atomic weight 

′ _{ }

2′ = Atom percents

65 Additional Equation _"= _ +   X 103 ₂"= _  +   X 103

Conversion of weight percent to Mass per unit volume (for 2 element alloy)

_" , = Concentration (kg/3) ρ_, ρ₂ = density in (g/3 ) ρ_= _ +   ρ_ = +     +   

Computation of density (for two-Element metal alloy)

 _= _ +     _ = +  

Computation of atomic weight (for a two-element metal alloy)

Line / Linear Defects

• are line defects,

• slip between crystal planes result when dislocations move, • produce permanent (plastic) deformation.

Dislocations:

Schematic of Zinc (HCP):

• before deformation • after tensile elongation

Imperfections in Solids

•

Linear Defects (Dislocations)

Are one-dimensional defects around which atoms are misaligned • Edge dislocation:

extra half-plane of atoms inserted in a crystal structure b perpendicular () to dislocation line

• Screw dislocation:

spiral planar ramp resulting from shear deformation b parallel () to dislocation line

Imperfections in Solids

•

Edge Dislocation

• Dislocation motion requires the successive bumping of a half plane of atoms (from left to right here).

• Bonds across the slipping planes are broken and remade in succession.

Imperfections in Solids

Burgers vector b Dislocation line b (a) (b)

Screw Dislocation

Imperfections in Solids

Edge Screw Mixed

Imperfections in Solids

•

Dislocations are visible in electron micrographs

TEM micrograph of a titanium alloy in which the dark lines are

dislocations. 51,450x.

TEM micrograph of dislocations in austenitic stainless steel

Dislocations & Crystal Structures

• Structure: close-packed planes & directions

are preferred.

view onto two close-packed planes.

close-packed plane (bottom) close-packed plane (top)

close-packed directions • Specimens that were tensile tested. Mg (HCP)  Al (FCC) tensile direction

Significance of Dislocations

1.

It provide a mechanism for plastic deformation.

2.

It provide ductility in metals.

3.

We control the mechanical properties of a metal or

alloy by interfering with the movement of

dislocations.

•

Strain hardening

•

Solid solution strengthening

•

Grain size strengthening

4.

It influence the electronic and optical properties of

materials.

Interfacial Defects in Solid (Area)

-

Grain Boundaries

• Boundary separating two small grains or crystals having different orientations.

• Atoms are bonded less regularly along the grain boundary, result in boundary energy (similar to surface energy).

• High angle boundaries have higher energy.

• Grain boundaries are

chemically reactive compared to grains because of the

boundary energy.

• Impurity atom often segregate along grain boundaries

Interfacial Defects in Solids

•

Phase Boundaries

 – Exist in multiple phase materials.

 – The different phases exist on each side of the boundary and it have its own relative physical and/or chemical characteristics.

 – Important in determining

the mechanical characteristic of multiphase metal alloys.

500x Ni

Cu Solder

Interface IMC

Optical micrograph of intermetallic layer between Nickel-gold surface finish and lead free solder.

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Twin boundary (plane)

 – Essentially a reflection of atom positions across the twin plane. (mirror-image)

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Stacking faults

 – For FCC metals an error in ABCABC packing sequence

 – Ex: ABCABABC