15677511
Description
Flashcards by Viktoria W, updated more than 1 year ago
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Created by Viktoria W
over 5 years ago
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| Question | Answer |
| 1.Sketch the fcc and bcc unit cells. Which one has the higher atomic packing? (APF) Which one has the higher density? | The atomic packaging factor of the fcc structure is 0,74 and bcc 0,68. Hence th fcc has the higher density |
| Make a sketch of the cubic crystal plane with miller indices (¨6¨26) (Prickarna är streck över siffran) | |
| 1.A cubic plane has the following axial intercepts: a = -2/3, b = 6, c = 1. What are the Miller indices of this plane? Give also the normal direction to this plane! | |
| 1.Draw the following directions in a cubic unit cell: [00¨1 ], [113], [ ¨112], [ ¨52¨1 ]. (Prickarna är streck över kommande siffra) | |
| 1.Write down the Miller indices of all cubic face plans. | Cubic faces: (100), (010), (001), ('100), (0'10), (00'1) Streck över ettorna där tecknet är. |
| 1.Make a sketch of a hcp unit cell. | |
| 1.How can you determinate if a crystal structure is a fcc or bcc type? | |
| 1. Make a sketch of the potential energy-versus-interatomic separation curves for a strongly and a weakly bonded material. | |
| 1. Make sketches of all planes of the type {110}! (You may want to use more than one sketch.) | |
| When the material is exposed to higher temperatures, first, recovery sets in and at even higher temperatures (ca 0.4 x Tm) recrystallization occurs. New, dislocation-free grains nucleate and start to grow. After recrystallization a new microstructure has formed that has consumed the deformation microstructure. As a result, the tensile strength and yield strength go down and the ductility (here expressed as % elongation) increases. Yield strength is defined as the intersection of the deformation curve with a straight line paralleln to the elastic portion and offset 0.2 % on the strain axis. It describes the transition from elastic to plastic behavior. | |
| 2. There are different methods of strengthen metal. Name them! | Strengthening methods for metal: Precipitation s. Solute s. Grain size s. Work hardening |
| 2. How can you strengthen a polymeric material? | To strengthen a polymeric material you can do that by adding fibers or particles. |
| 2. You are supposed to deform both copper (Cu) and magnesium oxide (MgO. How are dislocations moving in these material and what type of problem may you encounter (especially in MgO)? describe in words or make a sketch for illustration. | Copper is a metal that comes in a fcc structure. A dislocation can move in specific crystallographic planes and directions (glide systems). As seen in (a). There are many combinations of such combiantions planes/directions(12st). But this is not the case in MgO since it is not possible to move in many crystallographic planes/directions because ions of the same charge will be placed next to each other. This leading to fracture in the material hence there are a limited amount of glide systems in ceramic materials which makes deformation more difficult. |
| 2.Draw a stress-strain curve of a metal and explain in which regime of the curve you should deform/shape a metal! | |
| Should a metal be shaped at room temperature or elevated temperature? | |
| 2. What is the strenghtening relate to/how does it work? | The strengthening is achieved by hindering the dislocations in their motions (put obstacles in front of them). |
| 3.Make a sketch of an edge dislocation and explain/sketch how it's gliding. | When stress is applied on an edge dislocation, the dislocation can glide (breaking of bonds and establishing of new bonds along the glide plane bring the dislocation one atomic step forward. |
| 3.What is required/what are prerequisites for diffusion to happen? | For diffusion to happen you have to have high energy (high temperature) and space available. (Displacement and in case of substitutional diffusion the formation of vacancies) |
| 3. The diffusion coefficients for self-diffusion and carbon diffusion in α-iron at 500ºC are 3.0 10-21 m2/sec and 2.4 10-12 m2/sec, respectively. Explain the huge difference! | Self-diffusion is the exchange of an atom in a free place in the lattice position (gitterposition). Carbon diffusion is an exchange in interstitial places (not in ideal positions). There are far more interstitial places and therefore C-diffusion is much faster. |
| 3. What is happening in carburization and why is it used? | During carburization Carbon atoms are supposed to diffuse(sprida sig i) a material ex steel. Since carbon is a interstitial atom it will harden the surface. (by solute strengthening) |
| 3. In order for two components to have complete solid solubility in each other, they usually have to follow certain conditions. Name them! | The diameter of the atoms can not differ more than by 8% to achieve full solubility. Between 8%-15% partial solubility will be achieved. Same crystal structure is preferred. A big difference in electronegativity should be avoided. The soulate is preffered to have higher valance. (valans -> bindningar mellan atomer) |
| 3. Which hardening mechanisms can be achieved in an alloy (legering) with two elements that are totally soluble (löslig) in eachother? | Solute (löst ämne) Grain boundary (korngräns) Deformation hardening (deformationshärdning) |
| 3. Why is substitutional diffusion slower than interstitial diffusion? Explain briefly! | Substitutional diffusion is slower since the moving atoms need to have a vacancy nearby while interstitial does not need a vacancy and can choose from many other places to diffuse to. |
| 3. How can you distinguish between an edge and a screw dislocation? Describe briefly! | A dislocation is expressed by the Burgers vector. This by making a Burgers circuit in for the crystal an edge dislocation is when the Burgers vector is perpendicular (vinkelrät) to the dislocation line and a screw dislocation when the Burgers vector is parallell to the dislocation line. |
| 4. Draw cooling curves of a single component and a binary system! | |
| (i) The microstructure is eutectic (α + β) containing primary α (distributed). (ii) The first liquid forms at 183ºC (iii) The first liquid consists of about 62% Sn and 38% Pb (iv) Complete melting occurs at ~252ºC. (v) The last solid prior to complete melting consists of 13% Sn and 87% Pb. | |
| 4. Depending on temperature, pure Fe occurs in different crystal structures. Give the name of the different types of Fe, their crystal structure, their temperature range of existence, and the highest amount of carbon they may contain. The Fe-C phase diagram is given below! | δ-ferrite: BCC, 1394°C - 1534°C (Tm) (contains low amount of C (similar to ferrite) Austenite (γ-Fe): FCC, 910°C - 1394°C, contains up to 2.1 wt.%C Ferrite (α-Fe): BCC, RT - 910°C, contains up to 0.035 wt.%C |
| 4. What is the difference between a eutectic and a eutectoid reaction? Explain and give examples for both reactions. | Eutectic: A liquid transforms into two solid phases. Ex liquid into austenite and cementite. Eutectoid: A solid transforms into two other solid phases. Ex austenite into ferrite and cementite |
| 4.For a Fe – 0.5 wt% C alloy at a temperature at 725ºC, determine the phase amounts of primary ferrite, total ferrite and of the cementite phase. | For a Fe – 0.5 wt% C alloy at 725ºC: Phase amount of primary ferrite: (0.76-0.5)/(0.76-0.022)=0.352 Phase amount of total ferrite: (6.7-0.5)/(6.7-0.022)=0.928 Phase amount of cementite: (0.5-0.022)/(6.7-0.022)=0.072 |
| 4.Make sketches of the microstructure of the Fe – 0.5 wt% C alloy at 750 and 650ºC! | |
| 5.What is a martensitic transformation? Explain! Where to find martensite in the Fe-C phase diagram? | A martensitic transformation is a diffusion-less transformation. It occurs by a sudden re-orientation of C and Fe from the Fcc solid solution to a Bcc tetragonal structure. The maretensitic transformation does not exist in an equilibrium phasediagram since it is a metastable phase. |
| 5. Explain the difference between homogeneous and heterogeneous nucleation and state which one is more advantageous and why! | Homogenous: Occurs in liquid Hetrogenous: Occurs at a surface (mold) Hetorgenous is advatageous as a part of the surface energy which has to be brought up can be saved. |
| 5. You have a lamellar structure, for example pearlite. How can you determine if the microstructure was achieved at higher or at lower temperature? | The width of the lamellar pearalite tells about the materials thermal history. Higher temperatures: Not so many nucleation sites(kärnbildningsställen) but the atoms can diffuse long distances which results in a coarse(grov) structure. Low temperatures: There are more nucleation and the atoms diffuses short distances which leads to a fine structure and a narrow lamellae width. |
| 6. Give estimates of the used energy over life time of a civil airplane and a vacuum cleaner. Motivate your answer! | Extremely much energy will be used for an airplane therefor this energy use dominates. For the vacuum the energy use is still a major part but materials will too become more important but since it is made of polymeric material energy disposal is relatively low. |
| 6. With respect to recyclability compare metals and polymers. Recyclability of which of the materials is more economic and why? For comparison, how would you rank fiberreinforced polymers? Motivate! | Recycling metals is both economic and helps save energy since it takes less energy to produce a product from recycled metal than to make new one. This is not the case for polymers and fiberreinforced polymers is not possible to recycle. At least to the fibers. |
| 6. The total energy use of a product is given by 4 different energies. Name them! | (i) material’s embodied energy (kroppsliga engerin) (ii) energy of processing as well as (skapandets energi) (iii) use-energy, and (energin för användning) (iv) energy of disposal. (energin för undandröandet) |
| 6. Briefly assess energy-use-over-life for a mobile phone! | (i) embodied energy: quite high when it comes to producing some of the materials for example battery (ii) energy of processing: not too high (iii) use-energy: high (battery) (iv) energy of disposal: relatively high is taking the parts apart again to recycle some of the materials |
| 7.Explain – in words and with a sketch – the difference in electrical conductivity (or electron concentration) upon temperature of an intrinsic semiconductor and an extrinsic semiconductor | Intrinsic: Needs high temperature to overcome the band gap. Extrinsic: Has a defect level to which the electrons can get exited. Therefore it needs lesser temperature. |
| 7. Sketch the electron energy band structure of a p-type semiconductor and explain how conductivity can be achieved. | Defektnivån ligger i bandgapet precis ovanför den över änden av valensbandet. En elekton kan vara excited in defektnivån och ett hål bildas då i valensbandet. Which can move in an electrical field. This brings electric conduction. |
| 7. What are the major charge carriers in (i) an intrinsic semiconductor, (ii) an n-type semiconductor, (iii) a p-type semiconductor, and a (iv) metal? | (i) (inre) intrinsic semiconductor: electrons and holes (ii) n-type semiconductor: electrons (iii) p-type semiconductor: holes (iv) metal: electrons |
| 7. Ceramic are usually bad conductors or in fact insulators. Describe a case where ceramicmaterial can be an excellent electrical conductor. What is required to achieve that? | At a critical temperature the resistivity(resistansen) can drop to zero which makes the material to a superconductor which will when set in motion continue in a loop forever in the with electric current. |
| 7. Matthiesen’s rule describes what is affecting conductivity in metals? Name the effects! | Matthiesen’s rule describes the total resistivity in metals which is the sum of contributions from thermal vibrations (ρt), impurities (ρi), and plastic deformation (ρd): ρtot = ρt + ρi + ρd with ρt: temperature effect, ρi: impurity effect ρd: effect due to deformation |
| In metals the electrical conductivity decreases with higher temperatures In semiconductors the electrical conductivity increases with the temperature. This is related to the bandgap. | |
| Shows 3 different areas. T<100K: Freeze out - thermal energy is not enough to excite the electrons. 50 K < T < 450 K: extrinsic behavior; the defects provide charge carriers enough. number; this i the regime for our computers T >> 450 K: Intrinsic behavior, energy is enough to excite the electrons and can overcome bandgap To achive conduction in our computers it should be heaten up to 450K | |
| 7. Knowing the electron energy band structure, why can you predict if a material is a good or bad thermal conductor? | Thermal conductivity relies on phonons and electrons and are responsible for the major part in thermal conductivity. When knowing the electron energy band structure, you know if you have free electrons available which then tells you if the material is a good/bad thermal conductor. |
| 7. Make sketches of the electron energy band structure of a p-type and an n-type semiconductor (incl. defect level) and briefly explain for both cases how conductivity can be achieved. | Addition of an atom of higher valence; non-bonding electron can be exited and be a carrier of a negativ charge. Addition of an electron with a lower valens; Defect level can accept an electron from the valance band leading to a hole in the valance band and in this way to positive charged carrier. |
| 8. How is heat transported? Explain! | Heath is transported by phonons and electrons. Traveling by lattice (gitter) ways |
| 8. Make a sketch of the potential energy versus interatomic separation curve for a strongly bonded and a weakly bonded material. | Weakly and strongly bonded material (example polymer versus ceramic). The curve for the ceramic is more deep and narrow, while the weaker bonded material (polymer) has a broader and more shallow curve |
| 8. Why don’t you burn yourself at your steel sink when you are doing the dishes? Explain briefly! | The sink is made of stainless steel, i.e. it contains a number of solutes. They are efficient scattering sites for travelling electrons. Heat is not transported so well by phonons, hence, the sink is not getting hot even though the water inside may be hot. |
| 8. Why are ceramics sensitive to thermal shock? Explain (case to be chosen). | Ceramics are prone to thermal shock as they have a low thermal conductivity. When you change the external temperature, the material tries to expand. As the thermal conductivity is low, surface of the material will be at a different temperature than the interior. This will lead to stresses and as the material is brittle, this can lead to failure. |
| 8. What is thermal expansion and how is it related to bonding? Explain! | Thermal expansion describes the fact that material is expanding when it gets warmer (and is contracting upon cooling). It occurs because the bonds are stretched when heated (asymmetrical shape of the potential energy versus interatomic distance curve – when energy is increased, the atoms are moving apart). |
| Explain ball bonding and name the wire which has to be used! | Ball bonding is possible for Au-wires only!(Al cannot be ball bonded, since Al does not form balls upon melting; Au has better conductivity, anyhow ) Use of hydrogen flame (heat) → melted wire end forms into a small ball (due to high surface tension of Au) |
| Explain what is meant with electromigration! | Very high current densities – collisions between electrons and atoms in the metallic film → drift of atoms in the direction of electron flow; due to divergences voids and hillocks or whisker extrusion are formed. Failure: opening and/or shorting. Higher temperatures accelerate failure! |
| What are direct and expanded contacts? Explain advantages! | Metallization direct on Si; wire is directly on top of metallization. Metallization extended to the chip edge –Bond is at periphery of the chip Advantage: contact between bare wires and metallization patterns can be avoided |
| Describe ball bonding! | |
| 9. What is epitaxy? Explain briefly! | In epitaxial growth, a layer is mimicking(imiterar) the crystal structure of the layers below. |
| What is happening when you join a p-type and an n-type semiconductor? Explain! | When joining an n- and p-type semiconductor material, electrons from the n-type will tend to diffuse across the junction into the p-type material. Likewise, holes will tend to diffuse from the p-type to the n-type material. This creates a potential difference between the two types of material, which will act against the diffusion of holes and electrons, pulling them back to where they started. This opposing flow of charge under a potential difference is called drift. |
| What is happening in case of forward bias and reverse bias? Explain briefly! | Forward biasing: The positive terminal of the battery is connected to the p-type material and the negative terminal to the n-type material. The effective electric field in the depletion region is reduced. Hence, electrons and holes now cross the junction and constitute a current. Reverse biasing: The positive terminal of the battery is connected to the n-type material and the negative terminal to the p-type material. The electric field induced by the DC voltage source will reinforce the internal potential. Hence, depletion zone is increased and there is no flow of current due to majority carriers when the diode is reverse biased |
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