Introduction

Question 1

Define amorphous and crystalline solids.

Answer 1

Amorphous solids - Locally ordered but no long range order.

Crystalline solids - Periodic structure repeating over long distances.

Question 2

Name the 9 important Bravais lattices and describe their differences.

Answer 2

Cubic (all edges the same length): Primative (P), Body (I) and Face (F)

Tetragonal (square base but different length): Primative(/simple) and Body

Orthorhombic (all different lengths): Primative, Body, Base-centered and Face

Question 3

How do you calculate the density of a unit cell?

Answer 3

Find the side that defines the lattice parameter (where the atoms touch) and use the atomic radius to find the lattice parameter. Then find the volume.

Calculate the number of atoms in a cell and find the weight of the cell from the atomic mass. Density can then be found.

Question 4

Draw each of the miller planes for (100) to (222) for all variations and give the distance between planes for each.

Answer 4

Question 5

What is the difference between allotropy and polymorphism?

Answer 5

Allotropy is the ability of single atom to form more than one structure in a particular state (e.g. O₂ and O₃).

Polymorphism is the ability of a compound to form more than one crystal form.

Question 6

What must be considered when using a metal/compound which has multiple allotropes?

Answer 6

How the volume of the lattice may change when changing between allotropes. The volume per atom of each form can be found to show the percentage volume change.

The properties of the structure will change as well. This is especially true for tin.

Question 7

Describe the transition between β and α tin.

Answer 7

β tin is a BCT structure stable above 13 degrees and is a good metallic conductor, coordination 8.

α tin is stable below 13 degrees and is a non metallic insulator. The material fractures as the volume changes, coordination 4.

This is problem where tin is being used however the transition occurs over 100s of years.

Question 8

What is the equation for Braggs law?

How is the difference between planes calculated?

What are systematic absences?

Answer 8

2d sinθ = nλ (n = 1 when using miller indices)

Cubic structures: d² = a²/(h² + k² + l²) where h, k and l are the lattice parameters.

Orthorhombic structures: d^-2 = h²/a² + k²/b² + l²/c²

If there is a layer of atoms between the planes which are being studied for diffraction then the diffaction angle expected will be absent (100 cancelled by 200).

Question 9

What planes cause the first angles seen in the diffraction of the primitive, body and face cubic structures?

How does the angle of the diffraction change when the distance between planes increases?

Answer 9

PC: 100

BCC: 110

FCC: 111

When the distance between planes increases, the angle of diffraction angle becomes smaller.

Question 10

Calculate the angle of diffraction for the planes of Na (BCC) from (100) to (200) and determine which are seen.

λ = 154 pm, a = 423 pm.

Answer 10

(100) θ = 10.488º not seen
(110) θ = 14.918º seen
(111) θ = 18.378º not seen (centre atom isn’t part of the plane)
(200) θ = 21.350º seen

Question 11

Name and briefly describe the 8 types of defects and state their dimensionality.

Answer 11

Interstitial impurity (0D) - impurity between atoms in the lattice

Self interstitial (0D) - extra atom between atoms in the lattice

Vacancy (0D) - atom removed from the lattice

Substitutional impurity (0D) - atom substituted for an impurity

Group of impurities (1/2D) - multiple substitutional impurities grouped together

Edge dislocation (1D) - extra half plane of atoms added/removed

Vacancy dislocation loop (1/2D) - set of vacancies as a group

Interstitial dislocation loop (1/2D) - interstital atoms forming as a group between the lattice

Question 12

Describe the two types of defect lines.

Answer 12

Edge - plane is correct when completing a square movement but start location is not the same as the end point.

Screw - Following the defect in a square takes you to the correct site, but in the incorrect plane.

Question 13

Define grains and grain boundaries.

Answer 13

Grains - areas of well formed crystal

Grain boundaries - defects in a line between grains

Question 14

Define Schottky and Frenkel defects for ionic lattices.

Answer 14

Schottky - ion and cation pair removed from lattice, the volume is the same but the density is reduced as mass is lost. 2 vacancies are formed.

Frenkel - An ion is moved to form an interstitial. Cation interstitials are much more common as anions are usually larger.

Question 15

How do defects affect the lattice atoms surrounding it?

Answer 15

Vacancies cause the surrounding atoms to be pulled towards it. Intertitials push away surrounding atoms.

Edge dislocations stress the top and bottom in different ways:

Where an additional atom is present the atoms are compressed - they want to push outwards.

Where an atom is missing there is tension - the atoms want to pull inwards.

Question 16

What is the equation for the number of defects?

What are the main steps towards its derivation?

Answer 16

n_d = Ne^-ε​_d/kT

Where n_d = number of defects, N = number of sites, ε_d = energy of defect

k = boltzmann constant, T = temperature

Using G = H - TS, some defects are preffered as entropy favours it.

H = n_dε_d and S = klnω where ω = the number of microstates which is found by pascals triangle and ^NC_nd. We use the general formula for ^NC_nd.

An approximation is used that is d(Aln(A))/dA = - 1 - lnA.

Assuming N>>n_d gives the equation in the form we use.

Question 17

What happens if 2 Burgers vectors in opposite directions meet?

Answer 17

They annihilate, leaving no defects.

Question 18

Define an edge and a screw dislocation in terms of Burgers vectors.

Answer 18

Edge dislocation - Burgers vector perpendicular to dislocation/defect (all one plane)

Screw dislocation - Burgers vector is parallel to dislocation/defect (vectors cover one plane

Question 19

How do dislocation lines minimse the lattice energy? Why does this occur?

Answer 19

They bring lines closer in together to try close the gaps the defects leave behind. Defects also repel each other if the compressions and tensions are on the same side. If they are on opposite sides then they will annihilate.

The elastic energy of a defect is proportional to the Burgers vector, b, squared so 2 seperate defects are better than 2 together.

Question 20

What is the significance of defects for crystals?

Answer 20

For diamonds, defects occur naturally but when they are synthetically made there are very few. For lasers and electronics, perfect crystals are required.

Question 21

Define elastic and plastic deformation in terms of sheering a lattice.

Answer 21

Elastic - the bonds are stretched under the load but are not permanently deformed. The lattice appears to lean.

Plastic - bonds are broken and reformed as the planes slide across each other. The structure is permanently deformed.

During sheering, the vertical bonds lengthen until they are closer to the adjacent atom, at which point, the bonds will break and reform with the closer atom.

Question 22

Define the following in terms of stess and strain: ductile, brittle, toughness, stiffness and strength.

Answer 22

Ductile - the ability of a material to draw wires under strain

Brittle - materials that break under a small strain

Toughness - the ability to absorb energy before fracture, the area under a stress-strain curve

Stiffness - the ability to resist deformation

Strength - the ability to withstand a load

Question 23

Give full definitions of stress and strain, including equations.

Answer 23

Stress, σ = Force, F/Cross sectional area, A. Units are Nm^-2.

Strain, ε = extension, x/original length, L. This is dimensionless.

These are also known as engineering or nominal stress and strain.

True stress uses the actual area which will change as the load is applied. True strain is defined as ln(actual length/original length).

Question 24

How does the diameter change when a material is under compression and tension?

Answer 24

Tension = reduction in diameter

Compression = increase in diameter

Question 25

Define a stress strain plot and sketch an example with each of the regions of the line described.

Answer 25

A strain is applied and the stress is measured.

O-A: Linear region, elastic/proportional, Hooke’s law obeyed

A-B: No longer linear but still elastic region, B marks the end of elasticity

C: Proof stress (aka. offset yield stress, 0.2 proof stress) is the point where if released, the material will be deformed by 0.2% strain

D: UTS - ultimate tensile strength/stress, maximum stress that can be taken

E: Fracture point/final instability point, where the material breaks

Question 26

How in the Young’s modulus defined? What are its units? What actually determines the Young’s modulus?

Answer 26

It is defined by Hooke’s law from the linear section of the stress-strain graph.

E = σε^-1, units are Nm^-2

The Young’s modulus is physically determined by the resistance to changing chemical bond length of adjacent atoms. This depends on the plane that material is being stretched as well as the atoms.

Question 27

Draw 3 rods that have been stretched until they break, one which is brittle, one which is ductile and one inbetween. Then draw a stress strain graph showing each material. Which material is the toughest?

Answer 27

Question 28

What is necking and when does it occur?

Answer 28

Necking is where the rod non-uniformally changes diameter, making a point where the rod breaks. This occurs at the UTS.

Question 29

How can the properties of a material be changed without altering the composition? What effects do these have?

Answer 29

Work-Hardening - heating and working with force gives a more brittle material with a higher UTS

Annealing - heating to a high temperature and letting the grains recrystalise, this makes the material more ductile and workable

Question 30

How can aluminiums compostion be changed to be significantly stronger?

Answer 30

Adding 4% by weight of copper to form Dural makes the material more brittle but means it can take a signifcantly higher stress. This is because CuAl₂ precipitates forms especially at grain boundaries which adds resistance to the movements of the grains and dislocations.

Question 31

Draw a stress-strain graph for an elastic band with lines for the stretch and the release. Describe the cause and results of its shape.

Answer 31

Rubber is cross-linked so the polymers all tangle together when loose. When stretched they untangle and allign to the limit of the cross links. When you allign the elastic band you give the rubber energy as you give it a higher entropy so when it relaxes, there is an energy gain. This is represented by the area between the curves, called hysteresis, and is given out as heat during relaxation.

Question 32

Draw the stress-strain graph for a brittle polymer, a ductile polymer, a typical polymer and rubber.

Answer 32

Question 33

When drawing a polymer, explain why there is such a long drawing region.

Answer 33

The chains are extending with almost no resistance called cold-drawing, this is where the diameter will decrease linearly as more and more of the polymer chains are alligned.

Question 34

Aside from the Young’s modulus, what other elastic moduli are used? How can these be combined for measuring strain in 1 direction? What is a typical range of values for this?

Answer 34

When a material is put under tension, a transverse strain occurs in the perpendicular directions. Poisson’s ratio, ν, is the ratio of transverse strain over longitudinal strain.

ν = -ε_t/ε_l

If stress is applied in 3 directions, the strain in the x direction is:

ε_x = E^-1(σ_x - νσ_y - νσ_z)

ν ≤ 0.5 and can even be negative in high tech materials used in trainers.

Question 35

Define specifc strength and give its units.

Answer 35

Specific strength is a materials strength (force per unit area) at failure, divided by the density. Units are N m kg^-1 or the Yuri.

Question 36

How do we define how a material deforms?

Answer 36

A slip system needs to be defined which consists of a slip plane and a slip direction. During plastic deformation there tends to be only one slip system but at high temperatures and high stress others may occur. The slip doesn’t occur all at once, they glide through a system rather than moving all at once.

Question 37

What determines how easy a dislocation moves? Why do they only move in certain directions?

Answer 37

The ease of movement depends on the degree of distortion around the dislocation core. If the distortion is spread over a larger region, it is easier to move. Wide dislocations are common in ductile materials.

The energy of the dislocation is proportional to the burgers vector so the dislocation will only move over close packed planes.

Question 38

What are the slip planes and directions of an FCC and a BCC crystal? Why do they slip this way? How many slip systems are possible for each?

Answer 38

FCC: Slip along close packed plane {111} in the <110> direction, 12 possible systems as the dot product must be zero so there are negatives.

BCC: Slip along the nearest neighbours plane since there is no close packed plane which is {110} in the <111> direction. There is also 12 possible slip systems.

Question 39

How do you determine which slip system will slip when undergoing stress in a perticular direction? Show all the steps.

Answer 39

OILS rule: O (zero) I (intermediate) L (lowest) S (sign change)

For FCC: 1. Write tensile axis [U V W], [2 1 -4]

2. Ignoring signs, identify the highest, intermediate and lowest number, [I L H]
3. The slip direction is the < 1 1 0 > direction where the 0 is the intermediate position and the signs of the other 2 positions are kept the same, < 0 1 -1 >
4. The slip plane is { 1 1 1 } with the signs of the highest and intermediate perserved but the lowest switched, { 1 -1 1 }

For BCC the same occurs but with the slip plane in steps 2 and 3 and the slip direction in step 4.

Question 40

Describe in terms of slip how carbon strengthans iron.

Answer 40

The carbon is a smaller atom which interupts the regular structure, this makes the material harder to slip. Low carbon steel (up to 0.3% carbon) is malleable and ductile but with lower tensile strength. High carbon steel (0.3 - 1.7%) carbon is much stronger and used for its higher tensile strength.

Question 41

What are the componants and units of a magnetic field?

Answer 41

A magnetic field is a vector field that points from south to north. The units are the Tesla (T) = kg s^-2 A^-1.

Question 42

Describe and give three examples of ferromagnetic materials.

Answer 42

Ferromagnetic materials have many domains which each have an alligned spin direction. In their equilibrium the domains point in random directions but when exposed to a magnetic field they all become alligned and become magnetic, and when the magnet is removed they stay magnetic. Soft materials lose magnetism over time but hard materials such as ferrite form permanent magnets when processed in magnetic fields.

The soft magnetic elements are iron, nickel and colbalt, hard materials include rare earth metals.

Question 43

Define ferro, antiferro and ferri.

Answer 43

Ferro: spin states alligned

Anti-Ferro: regular pattern of alternating spins

Ferri: can be weakly magnetised as alligning spin is favourable but has anti spin atoms often found where materials have different components

Question 44

Why do some magnets not retain their domain orientation?

Answer 44

Energy is stored in a magnetic field, known as magnetocrystalline energy. This makes the domains easier to align in specific directions in a crystal, but the boundaries are random.

Question 45

How does the induced magnetism in a material change when the applied field is increased? How does this depend on the axis the field is applied? Sketch a graph the situations.

Answer 45

Aligning in some directions is easier due to the domains in the material. But the boundaries are random.

Question 46

Draw a curve showing how the induced magnetisation changes in a material as a field is applied.

Answer 46

Question 47

How does applied magnetisim change when a field is turn off in soft and hard materials? Why is this important?

Answer 47

Question 48

What is the Curie temperature?

Answer 48

As temperature increases, the ablity to magnetise a material decreases. At the Curie temperature, no magnetism occurs even at high temperatures.