Lecture notes, Materials Science Fundamentals, Lecture 1-6, 13-15 - Prof. Ravi PDF

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Chapter 1: Introduction Background Materials are used in transportation, housing, clothing, communication, recreation, food production. Early materials: Stone, Wood, Clay, etc Materials evolved: Metals, Plastics, Glasses, and Fibers

Materials Science and Engineering Materials Science:  Relationship between structure and properties Materials Engineering:  Based on knowledge of materials science, designing the structure to result in required set of properties

Properties: Mechanical, Electrical, Thermal, Magnetic, Optical, Deteriorative (e.g. Corrosion)

Why Study Materials Science and Engineering? Choosing right material  In-service condition  Possible deterioration of properties (e.g. temperature, corrosion effects)  Cost Make judicious choices.

Materials Metals, Ceramics, Polymers, Composites, Semiconductors, Biomaterials

Advanced Materials High-tech applications  Electronics, Fiber Optics, Spacecraft, Aircraft, Missiles

Materials of the future Smart Materials  Shape memory alloys  Piezoelectric ceramics  Magnetostrictive materials  Electrorheological/ Magnetorheological fluids Nanotechnology  Building materials from atomic level and up.  “Bottom-up” approach 10-9 meters, 100 nanometers ≈500 atom diameters

Modern materials’ needs (Future) Nuclear Energy  High strength low density materials  Solar cells Pollution control/Non-renewable resources  Discover additional reserves  Alternate/ new materials with less adverse impact on environment  Recycling

Chapter 2: Atomic Structure and Interatomic Bonding Fundamental Concepts

Electrons: Positive

Protons : Negative

Neutrons: Neutral

Atomic Number (Z): Number of protons in nucleus or number of electrons for electrically neutral or complete atom Atomic Mass (A): Sum of the masses of protons and neutrons within nucleus Number of neutrons (N) may vary for different atoms of an element. So, some elements may have two or more atomic masses. “Isotopes”. Atomic weight: Weighted average of the atomic masses of the atom’s naturally occurring isotopes. Atomic mass unit (amu): 1⁄12 of the atomic mass of the carbon i.e., A=12.000 A≈Z+N One mole: 6.023 x 1023 Atoms or Molecules

Electrons in Atoms Bohr’s Model Quantum Mechanics Set of principles and laws governing system of atomic and subatomic entities Bohr’s Atom And Orbitals Nucleus surrounded by electrons in discrete orbitals

Orbital

Energy Levels/States  Quantization of energies of electrons. i.e, each electron has only specific level of energy.  If energy of an electron changes, it jumps to a lower or higher level of energy  Ground state: Lowest possible level of energy.

Wave Mechanical Model To overcome deficiencies in Bohr’s Model:  Electron is both wave-like and particle-like  Electron is no longer a particle in a discrete orbital  Electron can be at various locations in a discrete orbital (uncertainty of location)

Quantum Numbers    

To describe size, shape and distribution of electron-density Four Quantum numbers Bohr’s energy levels separate into electron subshells Quantum numbers dictate the number of states within each subshell

First or Principal Quantum Number (n)  Specifies shells Table 2.1  Integral numbers K L M N n 1 2 3 4 Second Quantum number (l)  Specific subshells

O 5

Table 2.2 Subshell s p Energy State 1 3

P 6

d 5

… …

f 7

Third Quantum number (ml)  Number of energy states for each subshell Fourth (Spin) Quantum number (ms)  Up +1/2  Down -1/2 Table 2.3 #1 Principal Quantum No. n 1

  

Shell K

2

L

3

M

4

N

#2 Subshell l s s p s p d s

Number of Electrons

#3 No. of States

Per Subshell

Per Shell

1 1 3 1 3 5 1

2 2 6 2 6 10 2

2 8

18 32

Smaller n, means smaller energy level Within each shell, energy of a subshell level increases with the value of l Overlap in energy of a state in one shell with states in adjacent shell e.g. 3d>3s

Electron configurations Pauli exclusion principle: No two electrons can have all the four Quantum numbers identical, i.e., at least the spins have to be opposite (+1/2, -1/2) s 2 -1/2

p 6

f 10

d 14

electrons ie, they have pairing

with +1/2 and

Ground state: All electrons in an atom are at the lowest energy level. Examples of notations: Hydrogen, H

Table 2.4 He 1s2 2 (Z*)

Na 1s2 2s2 2p6 3s1 11 (Z*)

Stable electronic configuration: Outermost shell is completely filled with electrons

Ne 1s 2s2 2p6 10 (Z*) 2

Table 2.5: Inert Gases Ar Kr 2 2 6 2 6 2 2 6 2 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p6 3d10 4s2 4p6 18 (Z*) 36 (Z*)

Reason for inertness: No incompletely filled outermost shell

The periodic table of elements • Arranged in order of increasing atomic number • Seven horizontal rows: periods • Column(s) or group(s): similar valence electron structure. Similar chemical and physical properties. • Across each period (e.g., left to right) the properties changes gradually and systematically • Group 0 (zero): Right-most group, Inert gases: He, Ne, Ar, Kr, Xe, Rn Groups VII A and VI A: Electrons have one and two electrons deficient from having a stable structure. Oxygen (O) : 8 1s2 2s2 2p4 : 2 deficient Fluorine (F) : 9 1s2 2s2 2p5 : 1 deficient Group VII A: Halogens Group I A: Li, Na, K, Rb, Cs, Fr

Alkali

Group II A: Be, Mg, Ca, Sr, Ba, Ra Alkaline earth Group III B and II B: Transition metals • Partially filled d electron states • In some cases two electrons in the next higher energy shell Example: (Group III B) Sc: 21 1s2 2s2 2p6 3s2 3p6 3d1 4s2 Group III A, IV A, V A: Intermediate between metals and non-metals LEFT : Electropositive (e.g, Alkalis)

RIGHT: Electronegative (e.g, Halogens)

Atomic bonding in solids • Net force between two atoms: FN = FA + FR where: FA: attractive force FR: repulsive force • No net force if FA + FR = 0 (state of equilibrium) • At a state of equilibrium (r0: equilibrium spacing), the two atoms will counteract any attempt to separate them by a repulsive force, or to push them together by an attractive force (E0: equilibrium net potential energy) Energy,

𝐸 = ∫ 𝐹 ∗ 𝑑𝑟

where, 𝐹 = 𝑓𝑜𝑟𝑐𝑒

𝐸𝑁 = ∫ 𝐹𝑁 ∗ 𝑑𝑟 = ∫𝐹 𝐴 ∗ 𝑑𝑟 + ∫ 𝐹𝑅 ∗ 𝑑𝑟 = 𝐸𝐴 + 𝐸𝑅 **Integrals are from ‘r’ to infinity.

Net energy,

Net energy = Attractive energy + Repulsive energy Then, Eo = Bonding energy (Energy at minimum point)

Primary Interatomic Bonds Ionic (Electrovalent Bonds) • In compounds which have metallic and non-metallic elements • Due to transfer of electrons Left Extreme: Right Extreme:

Na (IA) Cl (VIIA)

Na: 2, 8, 1: 11

1s2 2s2 2p6 3s1 1s2 2s2 2p6

-1 electron Neon, Ne

Cl: 2, 8, 7: 17

1s2 2s2 2p6 3s2 3p5 1s2 2s2 2p6 3s2 3p6

+1 electron Argon, Ar

Attractive Bonding Forces: Coulombic forces • Positive and negative ions attract each other Na+ Cl- Na+ Cl• Ionic bonding is non-directional (magnitude of the bond is equal in all directions around the ion) Example: Ceramic materials exhibit ionic bonding Covalent Bonding • Sharing of electrons between adjacent atoms. (Example: CH4)

C: 4 valence electrons H: 1 valence electron

Directional: Between special atoms and only in the direction of participation. Example: H2O, HNO3, HF N: no. of valence electrons • An atom can bond with (8-N) other atoms Cl: C:

N=7 N=4

8-N=1 8-N=4

Cl + Cl = Cl2 carbon can bond with four hydrogen atoms

Attractive Bonding Forces • Covalent bonds are strong e.g, Diamond melting point>3550°C (6400°F) or, covalent bonds can be weak e.g, Bismuth melting point: 270°C (518°F) • Polymers: Covalent bonds • Partially ionic + partially covalent: possible • Wider separation in the periodic table: Ionic • Closer together in the periodic table: Covalent

Metallic Bonds • Metals have 1, 2 or 3 valence electrons • Valence electrons are not bound to an atom, but are free to drift through the entire metal. i.e, forming an “electron cloud” (-ve) • Remaining non-valence electrons and the atomic nuclei form the ionic core (+ve) • Free valence electrons shield the +ve ion core • Non-directional • Free valence electrons hold the ion core together • Good conductors • Bonding may be weak. e.g, Hg: 6.8 KJ/mole (0.7 ev/atom) • Bonding may be strong e.g, W: 850 KJ/mole (8.8 ev/atom)

Secondary or Vander Waal’s bonding Example: Hydrogen bond Fluctuating induced dipole bonds

Summary Atomic Structure • Bohr and wave mechanical models • Electron energy states  Quantum number • Periodic table Primary Bonds • Ionic • Covalent • Metallic Secondary Bonds

Chapter 3: The Structure of Crystalline Solids Fundamental Concepts Crystalline: Repeating/periodic array of atoms; each atom bonds to nearest neighbor atoms. Crystalline structure: Results in a lattice or three-dimensional arrangement of atoms

Unit cells  Smallest repeat unit/entity of a lattice.  Represents symmetry of the crystal structure.  Basic structure unit/building block of crystal structure  Defines the crystal structure by its geometry and atom positions

Co-ordination number For each atom, it is the number of nearest-neighbors or touching atoms Example: FCC:12, HCP:12, BCC:8

Atomic packing factor (APF): 𝐴𝑃𝐹 =

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑖𝑛 𝑎 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 𝑉𝑠 = 𝑇𝑜𝑡𝑎𝑙 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑉𝑐

= 0.74 (𝑓𝑜𝑟 𝐹𝐶𝐶 𝑜𝑟 𝐻𝐶𝑃) = 0.68 (𝑓𝑜𝑟 𝐵𝐶𝐶) Atoms per unit cell FCC Face atoms= 6 x ½ =3 4 Corners atoms = 8 x 1/8 =1 Examples: Al, Ni, Cu, Au, Ag, Pb, Gamma (γ)-Iron Body atom=1 2 Corners atoms = 8 x 1/8 =1 Examples: Cr, W, Alpha (α)-Iron, Delta (δ)- Iron, Mo, V, Na BCC

SC

Corners atoms = 8 x 1/8 =1

1

Metallic crystal structure

SC (Simple Cubic)

BCC (Body-Centered Cubic) FCC (Face-Centered Cubic)

Where: R = Radius of the atom a = cube edge a2 + a2 = (4R)2 2a2 = (4R)2 = 16R2 a = 2R √2 APF = Unit cell volume (Vc ) = a3

3

= (2R√2) = 16 √2 R3

4 π R3 ∗ 4 3 16 = π R3 3

Vs =

4 atoms per unit cell

Total cell voleume (Vc ) = 16 √2 R3 APF =

Vs

Vc

=

16 3

(

)πR3

16√2 R3

= 0.74

Volume of atoms in a unit cell Vs = Total unit cell volume Vc

Body Centered Cubic All sides are equal to dimension “a” a2 + a2 = 2a2 2

2

2

(a√2) + a = 3a = (4R)

2

a√3= 4R

The Hexagonal Close-Packed

12

6 Atoms at top 6 Atoms at bottom 2 Centre face atoms 3 Midplane atoms 12 x 1/6 = 2 2 x 1/2 = 1 Midplane 3

6 atoms/unit cell

Co-ordinate number: 12 (HCP or FCC) Atomic packaging factor (APF): 0.74 Examples: Cd, Zn, Mg, Ti

Density Computations Density,

𝑛𝐴

𝜌 = 𝑉𝑁

𝑐 𝐴

Where, N = No. of atoms/unit cell A = Atomic weight Vc = Volume of unit cell NA = Avogadro’s number (6.023 x 1023/mole)

a√3 a√2

Example: Copper has an atomic radius of 0.128 nm, an FCC crystal structure, and an atomic weight of 63.5 g/mol. Compute its theoretical density and compare the answer with its measured density. Given: Atomic radius = 0.128 nm (1.28 Ǻ) Atomic weight = 63.5 g/mole n=4 ACU = 63.5 g/mol Solution: Unit cell volume = 16 R3√2 R = Atomic Radius ρ

4 63.5g/mole [16 2(1.28 108 cm)3 /unit cell](6.0231023)atoms

= 8.89 g/cm3 Close to 8.94 g/cm3 in the literature

Crystal system x, y, z : Coordinate systems a, b, c : Edge lengths α, β, γ : Inter axial angles Cubic system: a=b=c α=β=γ=90° Lattice parameter (e.g., a,b,c, α, β, γ) determine the crystal system. There are seven crystal systems which are Cubic, Tetragonal, Hexagonal, Rhombohedral (Trigonal), Monoclinic, Triclinic.

Crystallographic Direction Steps: 1. Choose a vector of convenient length 2. Obtain vector projection on each of three axes (for the direction to be drawn, if necessary) 3. Divide the three numbers by a common factor (if the indices are to be assigned) to reduce to the smallest integer values 4. Use square brackets [ ]

Crystallographic Planes Miller Indices (hkl)

Steps: 1. Obtain lengths of planar intercepts for each axis. 2. Take reciprocals 3. Change the three numbers into a set of smallest integers (use a common factor ) 4. Enclose within parenthesis e.g., (012) Tips: 1. Parallel planes have the same indices 2. An index 0(zero) implies the plane is parallel to that axis.

Cubic Crystal system

( ) { } [ ] < >

Plane Family of planes Direction Family of directions

Example: {111}: (111) (111) (111) (111) (111) (111) (111) (111) Hexagonal Crystal system

[u’v’w’] -------> [u v t w] [0 1 0] -------> [1210] u = n/3 (2u’ – v’) e.g., u = n/3 (2 x0 – 1) Where, n=factor to convert into indices = 3 u= 1

= n/3 (0 -1)

v = n/3 (2v’ – u’) Example: v = n/3 (2 x 1 -0) = n/3 (2) Where: n = factor to convert into indices = 3 v=2 t = - (u’ + v’) Example: t = -(0 + 1) = -1 = 1 w = w’

u v t w = 1210

a1, a2, a3 axes: Z-axis:

all in basal plane (at 120° to each other) Perpendicular to basal plane

[u’v’w’] -------> [u v t w] abc abzc Miller -------> Miller-Bravais u = n/3 (2u’ – v’)

[0 1 0] -------> [1210]

v = n/3 (2v’ – u’) u’v’w’ ----> u v t w t = - (u’ + v’) u = (0 -1), t = -(1), v = 2, w = 0 w = nw’ n=factor to convert into indices

Linear and Planar Atomic Densities Linear density 4R = a√3 BCC a = 4R/√3

a√3 N a√2

Distance occupied distance available 2R = a 2R = 4R √2

BCC LD [100] =

= 0.866

M

a

X- Ray Diffraction

Interplanar spacing

𝑛𝜆 = SQQT Where, n = an integer, order of reflection = 1 (unless stated otherwise) Bragg’s law of diffraction nλ = d

hkl

sinθ + d sinθ

= 2d

hkl

sinθ

hkl

For cubic system,

a2/d2 = h2 + k2 + l2

X-Ray Diffraction nλ = SQQT = path difference = dhkl sinθ + dhklsinθ = 2dhkl sinθ a2/d2 = h2 + k2 + l2

where n = integer = 1

(h + k + l) must be even: BCC 2, 4, 6, 8, 10, 12…… h k l: all odd or all even FCC 3, 4, 8, 11, 12, 16…….. If the ratio of the sin2θ values of the first two diffracting planes is 0.75, it is FCC structure. If it is 0.5, it is BCC structure. 𝜆 = 2𝑑 𝑠𝑖𝑛𝜃 =

2𝑎 𝑠𝑖𝑛𝜃

√ℎ2 + 𝑘 2 + 𝑙2

sin2 𝜃 = (𝜆2 (ℎ2 + 𝑘 2 + 𝑙2 ))/(4𝑎 2 )

(sin2 𝜃1 )/(sin2 𝜃2 ) =

ℎ21 + 𝑘12 + 𝑙12 ℎ22 + 𝑘22 + 𝑙22

Example Givens: {211} Planes aFe = 0.2866 nm (2.866Å) λ= 0.1542 nm (1.542Å) n=1 Determine dhkl, 2θ (diffraction angle) a) dhkl = a/ √ (h2 + k2 + l2) = 0.2866 nm /√ (22 + 12 + 12) = 0.1170 nm (1.170Å) 𝑛𝜆

b) sin 𝜃 = 2𝑑 𝜃 = 𝑠𝑖𝑛

ℎ𝑘𝑙

=

1∗0.1542 2∗0.1170

−1 (0.659)

2𝜃 = 82.44°

= 41 .22°

λ 𝑎𝑛𝑑 𝑎 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠

Crystalline and Non-crystalline materials Single crystal: Polycrystalline: Anisotropy: Isotropy:

No grain boundary Several crystals Directionality in properties No directionality Modulus of elasticity (E), psi x 106 (MPa x 103) FCC Al FCC Cu

BCC

Fe

BCC

W

[100]

[110]

[111]

9.2

10.5

11.0

(63.7)

(72.6)

(76.1)

9.7

18.9

27.7

(66.7)

(130.3)

(191.1)

18.1

30.5

39.6

(125)

(210.5)

(272.7)

55.8

55.8

55.8

(384.6)

(384.6)

(384.6)

Non-Crystalline • Amorphous • No systematic arrangement (regular) of atoms

Summary • Crystalline –lattice • Crystal system: BCC, FCC, HCP • Planes, directions, packing • X-Ray diffraction

Chapter 4: Imperfections in Solids Introduction  Metals  Alloys  Solid solutions  New/second phase  Solute (the guest)  Solvent (the host)

Solid solutions Substitutional solid solutions Interstitial solid solutions Cu Dia: 1.28Å

Ni Dia: 1.25 Å

+1

+2

FCC

FCC

Substitutional Atomic size factor (difference in atomic radii: ±15%) Electrochemical factor (must be close in periodic table otherwise intermetallic compound) Relative valencies factor (higher valence metal dissolves in lower valence metal) Same crystal structure, solid solution

Defect: Irregularity Point defects:  Vacant lattice site: vacancy  Addition atom crowding  into an interstitial site  Self-interstitial: same type of atom 𝑄

No. of vacancies, 𝑁𝑣 = 𝑁 exp (− 𝑣) 𝑘𝑇 Where: N = no. of atomic sites QV = Activation energy to form vacancy T = Absolute temperature, K k = Boltzmann constant = 1.38 x 10-23 J/atom-K = 8.62 x 10-5 ev/atom-K R = 8.31 J/mol-K (gas constant) = 1.987 calories/mol-K

Example: Calculate the number of vacancies per cubic meter of copper at 1000°C. The activation energy for vacancy formation is 0.9 ev/atom; the atomic weight and density (at 1000°C), for copper are 63.5 gm/mol and 8.4 gm/cm3 respectively. Solution N = No. of atomic sites/ cubic meter of copper ACu = Atomic weight of copper = 63.5 gm/mol ρCu = Density of copper = 8.4 gm/cm3 NA = Avogadro’s no. = 6.023 x 1023 atoms/mol N = NA ρCu/ACu = 8.0 x 1028 atoms/m3 No. of vacancies at 1000 °C (1273 K) 𝑄𝑣 𝑁𝑣 = 𝑁 exp (− ) 𝑘𝑇 QV N k T

= 0.9 ev/atom = 8.0 x 1028 atom/m3 = 8.62 x 10-5 ev/K = 1273 K

Therefore, NV = 2.2 x 1025 vacancies/m3

Interstitial solid solutions  Fill voids/ interstices  For high APF, interstitial positions are small. So, atomic diameter of interstitial impurity must be very small (relative to host atoms)  Max. allowable concentration: lattice strains  Carbon atom: interstitial in Fe max 2% Atomic radius of carbon, C: 0.71 Å Atomic radius of iron, Fe: 1.24 Å

Specification of Composition A and B atoms 𝑚𝐴 𝑊𝑡%𝐴 = 𝑥 100 𝑚𝐴 + 𝑚𝐵 𝑚𝐴 𝐴 𝐴𝑡𝑜𝑚%𝐴 = 𝑚 𝐴 𝑚 𝐵 𝐴 𝐴𝐴 + 𝐴𝐵

𝑚,

𝑤𝑒𝑖𝑔ℎ𝑡 𝑚𝑎𝑠𝑠

𝑚𝐴 : 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴

𝐴𝐴 : 𝑎𝑡𝑜𝑚𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐴

Dislocations- Line defects • Edge dislocation • Screw dislocation Edge dislocation: Extra half plane of atoms terminates within the crystal • Dislocation line/extra half plane causes lattice to pull apart. • Where the extra half plane is not there, the lattice is squeezed together Screw dislocation: • Shear stress to distort the lattice • Upper region is shifted one atomic distance relative to bottom • Associated spiral or helical ramp Mixed dislocation: Neither edge perpendicular () or screw (

)

Burger’s vector Close-failure of a dislocation, b (extent of distortion). Dislocation and b are perpendicular in edge. They are parallel in screw.

Extra half plane of atoms b: Burger’s vector : Edge Dislocation

Edge Dislocation

b points in a closed-packed direction and is equal to atomic spacing(s)

Sc...