Engr Materials Science and Engineering TEST 4  Sample Solution


 Chastity Annabella Roberts
 2 years ago
 Views:
Transcription
1 Engr Materials Science and Engineering TEST 4  Sample Solution Part : Solve each of the following problems completely. 1. In this problem we are asked to show that the minimum cationtoanion radius ratio for a coordination number of four is 0.5. If lines are drawn from the centers of the anions, then a tetrahedron is formed. The tetrahedron may be inscribed within a cube as shown below. The spheres at the apexes of the tetrahedron are drawn at the corners of the cube, and designated as positions A, B, C, and D. (These are reduced in size for the sake of clarity.) The cation resides at the center of the cube, which is designated as point E. Let us now express the cation and anion radii in terms of the cube edge length, designated as a. The spheres located at positions A and B touch each other along the bottom face diagonal. Thus, AB r A But (AB) a + a a or AB a r A And a r A There will also be an anion located at the corner, point F (not drawn), and the cube diagonal AEF will be related to the ionic radii as AEF r A + r C ( ) (The line AEF has not been drawn to avoid confusion.) From the triangle ABF (AB) + (FB) (AEF) But,
2 FB a r A and from above. Thus, ( ) + r A r A AB r A [ ( r A + r C )] Solving for the r C /r A ratio leads to r C r A A hypothetical AX type or ceramic material is known to have a density of 3.55 g/cm 3, and a unit cell of cubic symmetry with a cell edge length of nm. The atomic weights of the A and X elements are 60. and 16.0 g/mol, respectively. On the basis of this information, which of the following crystal structures would you expect this ceramic compound to have: sodium chloride, cesium chloride, zinc blende, diamond cubic, fluorite, and perovskite? Justify your choice(s). (See table 1.4) Answer: None of the suggested crystal structure (n 3.0 formula units/units cell) Using Equation 1.1 and solving for n ρvc N A n' A + A C (3.55)(4.75 n ' A 8 ( ) n 3.0 formula units/unit cell. 3 ) (6.03 For the AX type ceramics, sodium chloride and zinc blende have n 4, while cesium chloride has n 1. Therefore, none of the suggested crystal structures would work. 3 ) 3. Consider two hypothetical ions A + and X  which have radii of 0.15 and nm, respectively. (See Table 1.) a. What is the coordination number for each A + in the compound A X? Answer: 8 b. Assuming that the crystal structure for this compound belongs to the cubic crystal system, describe or draw a unit cell for A X. Answer: A variant of fluorite crystal structure. A unit cell would have eight interconnected cubes with X  ions positioned at all cube corners; half of the cube centers would be occupied by A + ions. r + A 0.15 a) Compute the cationanion radius ratio: r X The coordination number is eight (Table 1.).
3 b) The crystal structure would be a variation of the fluorite crystal structure (Figure 1.5). A unit cell would have eight interconnected cubes with X  ions positioned at all cube corners; half of the cube centers would be occupied by A + ions. 4. The energy gap of Germanium is 0.66 ev. The electron mobility if m /Vs. The hole mobility is m /Vs. The intrinsic hole concentration at room temperature (300 K) is.3 19 holes/m 3. a) Calculate the electrical conductivity of germanium (intrinsic) at room temperature. b) Calculate the electrical conductivity of germanium (intrinsic) at 0 C. ( e C, and the k ev/k) c) Germanium is to be doped with Antimony to produce an extrinsic ntype semiconductor. What atomic density (in atoms/m 3 ) is required to yield a conductivity of 4 (Ωm) 1. (Neglect the conductivity of intrinsic germanium.) Answers: a).04 (Ωm) 1 ; b) 4.9 (Ωm) 1 a) σ n e ( μ + μ ) e h b) σ ( ( Ωm) 1 )(1.6 E g 1 19 σ 1 exp σ 1 k T T1 ln σ 0.66 σ ( ) σ 4.9 (Ωm) 1 )( ) Calculate the resistance of a 150m long, 3.1mm diameter copper wire at an ambient temperature of 0 C. The electrical conductivity of the wire is (Ωm) 1. How much current will result if a voltage of V is applied at the ends of the wire? Ans: 0.33 Ω. L 1 L R ρ A σ A R : 0.33 Ω. I V/R / A 6. Is it possible to have a random poly(ethylenepropylene) copolymer that has number and weightaverage degrees of polymerization of 500 and 000, respectively, and number and weightaverage molecular weights of 8,500 and 67,00 g/mol? Why or why not? Answer: no. Mer molecular weights computed from weightaverage and numberaverage data are different 7. Molecular weight data from some polymer are tabulated below. Compute the (a) numberaverage molecular weight, and (b) the weightaverage molecular weight. (c) If it is known
4 that the material s numberaverage degree of polymerization is 477, which one of the polymers listed in Table 14.3 is this polymer? Why? Molecular wt.range x i 8,0000, ,0003, ,00044, ,00056, ,00068, ,00080, ,0009, (a) From the tabulated data, we are asked to compute M n, the numberaverage molecular weight. This is carried out below. Molecular wt. Range Mean M i x i xm i i 8,0000,000 14, ,0003,000 6, ,00044,000 38, ,00056,000 50, ,000 56,00068,000 6, ,160 68,00080,000 74, ,0009,000 86, M n x i M i 47,70 g/mol (b) From the tabulated data, we are asked to compute M w, the weight average molecular weight. This determination is performed as follows: Molecular wt. Range Mean M i w i wm i i 8,0000,000 14, ,0003,000 6, ,00044,000 38, ,00056,000 50, ,500 56,00068,000 6, ,60
5 68,00080,000 74, ,840 80,0009,000 86, M w w i M i 53,70 g/mol (c) We are now asked if the numberaverage degree of polymerization is 477, which of the polymers in Table 14.3 is this material? It is necessary to compute m in Equation (14.4a) as m M n 47,70 g/mol n 477 n 0.04 g/mol The mer molecular weights of the polymers listed in Table 14.3 are as follows: Polyethylene g/mol Polyvinyl chloride g/mol Polytetrafluoroethylene0.0 g/mol Polypropylene g/mol Polystyrene g/mol Polymethyl methacrylate g/mol Phenolformaldehyde g/mol Nylon 6, g/mol PET g/mol Polycarbonate g/mol Therefore, polytetrafluoroethylene is the material since its mer molecular weight is closest to that calculated above.
For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2; possible m l. values are 0, ±1, and ±2; and possible m s
.6 Allowed values for the quantum numbers of electrons are as follows: n 1,,,... l 0, 1,,,..., n 1 m l 0, ±1, ±, ±,..., ±l m s ± 1 The relationships between n and the shell designations are noted in Table.1.
More informationHydrocarbon Molecules Polymer Molecules The Chemistry of Polymer Molecules
141 CHAPTER 14 POLYMER STRUCTURES PROBLEM SOLUTIONS Hydrocarbon Molecules Polymer Molecules The Chemistry of Polymer Molecules 14.1 The repeat unit structures called for are sketched below. (a Polychlorotrifluoroethylene
More informationSample Exercise 12.1 Calculating Packing Efficiency
Sample Exercise 12.1 Calculating Packing Efficiency It is not possible to pack spheres together without leaving some void spaces between the spheres. Packing efficiency is the fraction of space in a crystal
More informationHydrocarbon Molecules Polymer Molecules The Chemistry of Polymer Molecules
141 CHAPTER 14 POLYMER STRUCTURES PROBLEM SOLUTIONS Hydrocarbon Molecules Polymer Molecules The Chemistry of Polymer Molecules 14.1 The repeat unit structures called for are sketched below. (a Polychlorotrifluoroethylene
More informationThe Crystal Structures of Solids
The Crystal Structures of Solids Crystals of pure substances can be analyzed using Xray diffraction methods to provide valuable information. The type and strength of intramolecular forces, density, molar
More informationTypes of Solids. Metallic and Ionic Solids. Crystal Lattices. Properties of Solids. Cubic Unit Cells. Metallic and Ionic Solids
1 Metallic and Ionic Solids 1 Types of Solids Table 13.6 2 TYPE EXAMPLE FORCE Ionic NaCl, CaF 2, ZnS Ionion Metallic Na, Fe Metallic Molecular Ice, I 2 Dipole Ind. dipole Network Diamond Extended Graphite
More informationHW 10. = 3.3 GPa (483,000 psi)
HW 10 Problem 15.1 Elastic modulus and tensile strength of poly(methyl methacrylate) at room temperature [20 C (68 F)]. Compare these with the corresponding values in Table 15.1. Figure 15.3 is accurate;
More informationCrystal Structure MetalsCeramics
Crystal Structure MetalsCeramics Ashraf Bastawros www.public.iastate.edu\~bastaw\courses\mate271.html Week 3 Material Sciences and Engineering MatE271 1 Ceramic Crystal Structures  Broader range of chemical
More information0% (0 out of 5 correct) The questions marked with symbol have not been graded.
Page 1 of 15 0% (0 out of 5 correct) The questions marked with symbol have not been graded. Responses to questions are indicated by the symbol. 1. Which of the following materials may form crystalline
More informationImperfections in atomic arrangements
MME131: Lecture 8 Imperfections in atomic arrangements Part 1: 0D Defects A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Today s Topics Occurrence and importance of crystal defects Classification
More informationUnit 12 Practice Test
Name: Class: Date: ID: A Unit 12 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) A solid has a very high melting point, great hardness, and
More information3a, so the NN spacing is. = nm. SOLUTIONS: ECE 305 Homework 1: Week 1 Mark Lundstrom Purdue University
ECE 305 SOLUTIONS: ECE 305 Homework 1: Week 1 Mark Lundstrom Purdue University 1) Ge has the same crystal structure (diamond) as Si, with a lattice constant of a = 5.6 Angstroms = 0.56 nm. Find the atomic
More informationM n = (DP)m = (25,000)(104.14 g/mol) = 2.60! 10 6 g/mol
14.4 (a) Compute the repeat unit molecular weight of polystyrene. (b) Compute the numberaverage molecular weight for a polystyrene for which the degree of polymerization is 25,000. (a) The repeat unit
More informationCHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS
HPTER THE STRUTURE OF RYSTLLINE SOLIDS PROBLEM SOLUTIONS Fundamental oncepts.1 What is the difference between atomic structure and crystal structure? tomic structure relates to the number of protons and
More informationMAE 20 Winter 2011 Assignment 2 solutions
MAE 0 Winter 0 Assignment solutions. List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal structure (Figure.6). In Figure.6, the barium ions are
More information(10 4 mm 2 )(1000 mm 3 ) = 10 7 mm = 10 4 m = 6.2 mi
CHAPTER 8 DEFORMATION AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS Basic Concepts of Dislocations Characteristics of Dislocations 8.1 To provide some perspective on the dimensions of atomic defects,
More informationBasic laws and electrical properties of metals (I) Electrical properties. Basic laws and electrical properties of metals (II)
Electrical properties Electrical conduction How many moveable electrons are there in a material (carrier density)? How easily do they move (mobility)? Semiconductivity Electrons and holes Intrinsic and
More informationCHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS
CHAPTER THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS Fundamental Concepts.6 Show that the atomic packing factor for HCP is 0.74. The APF is just the total sphere volumeunit cell volume ratio.
More informationMaterials Science and Engineering Department MSE , Sample Test #1, Spring 2010
Materials Science and Engineering Department MSE 200001, Sample Test #1, Spring 2010 ID number First letter of your last name: Name: No notes, books, or information stored in calculator memories may be
More informationinterstitial positions
Interstitial Sites In the spaces between the sites of the closest packed lattices (planes), there are a number of well defined interstitial positions: The CCP (FCC) lattice in (a) has 4 octahedral, 6coordinate
More informationCHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS
41 CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and SelfInterstitials 4.1 In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation
More informationMCEN Fall 2003.
Basic types of solid materials. Overview The theory of bands provides a basis for understanding the classification and physical properties of solid materials such as electrical conductivity, optical behavior
More information15.32 Of those polymers listed in Table 15.2, which polymer(s) would be best suited for use as ice cube trays? Why?
15.31 Name the following polymer(s) that would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why? This
More informationTHE STRUCTURE OF CRYSTALLINE SOLIDS
CHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS Fundamental Concepts 3.1 What is the difference between atomic structure and crystal structure? Atomic structure relates to the number of
More informationSection A5: Current Flow in Semiconductors
Section A5: Current Flow in Semiconductors Conductive behaviors in materials, defined by the parameter conductivity, are a primary factor in the development of electronic and optoelectronic devices. Electrical
More informationTHE STRUCTURE OF CRYSTALLINE SOLIDS. IE114 Materials Science and General Chemistry Lecture3
THE STRUCTURE OF CRYSTALLINE SOLIDS IE114 Materials Science and General Chemistry Lecture3 Outline Crystalline and Noncrystalline Materials 1) Single, Polycrystalline, Noncrystalline solids 2) Polycrystalline
More informationCHAPTER 19 THERMAL PROPERTIES PROBLEM SOLUTIONS
191 CHAPTER 19 THERMAL PROPERTIES PROBLEM SOLUTIONS Heat Capacity 19.1 The energy, E, required to raise the temperature of a given mass of material, m, is the product of the specific heat, the mass of
More informationChapter Number of Atoms. 2.2 Atoms and Density. 2.3 Mass and Volume. 2.4 Plating Thickness. 2.4 Electronegativity and Bonding
Chapter 2 2.1 Number of Atoms 2.2 Atoms and Density 2.3 Mass and Volume 2.4 Plating Thickness 2.4 Electronegativity and Bonding 2.1 Number of Atoms Calculate the number of atoms in 100 grams of silver.
More informationElectroceramics Prof. Ashish Garg Department of Material Science and Engineering Indian Institute of Technology, Kanpur.
Electroceramics Prof. Ashish Garg Department of Material Science and Engineering Indian Institute of Technology, Kanpur Lecture  3 Okay, so in this third lecture we will just review the last lecture,
More informationELECTRICAL CONDUCTION
Chapter 12: Electrical Properties Learning Objectives... How are electrical conductance and resistance characterized? What are the physical phenomena that distinguish conductors, semiconductors, and insulators?
More informationb) What is the trend in melting point as one moves up the group in alkali metals? Explain.
Problem Set 2 Wed May 25, 2011 1. What is the rough trend in the number of oxidation states available to the elements (excluding the noble gases) as you move a) to the right and b) down the periodic table?
More informationModule 6 : PHYSICS OF SEMICONDUCTOR DEVICES Lecture 34 : Intrinsic Semiconductors
Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES Lecture 34 : Intrinsic Semiconductors Objectives In this course you will learn the following Intrinsic and extrinsic semiconductors. Fermi level in a semiconductor.
More informationCondensed Matter Physics Prof. G. Rangarajan Department of Physics Indian Institute of Technology, Madras. Lecture  36 Semiconductors
Condensed Matter Physics Prof. G. Rangarajan Department of Physics Indian Institute of Technology, Madras Lecture  36 Semiconductors We will start a discussion of semiconductors one of the most important
More informationLecture 3: Electron statistics in a solid
Lecture 3: Electron statistics in a solid Contents Density of states. DOS in a 3D uniform solid.................... 3.2 DOS for a 2D solid........................ 4.3 DOS for a D solid........................
More informationSolid State Theory Physics 545
Solid State Theory Physics 545 CRYSTAL STRUCTURES Describing periodic structures Terminology Basic Structures Symmetry Operations Ionic crystals often have a definite habit which gives rise to particular
More informationv kt = N A ρ Au exp (
42 4.2 Determination of the number of vacancies per cubic meter in gold at 900 C (1173 K) requires the utilization of Equations 4.1 and 4.2 as follows: N v N exp Q v N A ρ Au kt A Au exp Q v kt (6.023
More informationThis is Solids, chapter 12 from the book Principles of General Chemistry (index.html) (v. 1.0).
This is Solids, chapter 12 from the book Principles of General Chemistry (index.html) (v. 1.0). This book is licensed under a Creative Commons byncsa 3.0 (http://creativecommons.org/licenses/byncsa/
More informationThis is the 11th lecture of this course and the last lecture on the topic of Equilibrium Carrier Concentration.
Solid State Devices Dr. S. Karmalkar Department of Electronics and Communication Engineering Indian Institute of Technology, Madras Lecture  11 Equilibrium Carrier Concentration (Contd.) This is the 11th
More informationChapter Outline. How do atoms arrange themselves to form solids?
Chapter Outline How do atoms arrange themselves to form solids? Fundamental concepts and language Unit cells Crystal structures Simple cubic Facecentered cubic Bodycentered cubic Hexagonal closepacked
More informationChapter 3. 1. 3 types of materials amorphous, crystalline, and polycrystalline. 5. Same as #3 for the ceramic and diamond crystal structures.
Chapter Highlights: Notes: 1. types of materials amorphous, crystalline, and polycrystalline.. Understand the meaning of crystallinity, which refers to a regular lattice based on a repeating unit cell..
More information12 x 1/6 + 2 x 1/2 + 3 = 6 atoms per unit cell for HCP.
FCC. BCC and HCP Metals Introduction The majority of common metals have either a Face Center Cubic Structure, fig la, a Body Centered Cubic Structure, fig.lb or an Hexagonal Close Packed structure fig.lc.
More informationCHEM 10113, Quiz 7 December 7, 2011
CHEM 10113, Quiz 7 December 7, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, show charges as appropriate, and please box your answers!
More informationUNIT 1 THE SOLID STATE VERY SHORT ANSWER TYPE QUESTIONS (1 MARKS)
UNIT 1 THE SOLID STATE VERY SHORT ANSWER TYPE QUESTIONS (1 MARKS) Q1. How many spheres are in contact with each other in a single plane of a close packed structure? A1. Six(6). Q2.Name the two closest
More information100 Practice Questions for Chem 1C Midterm 1  Joseph
100 Practice Questions for hem 1 Midterm 1  Joseph 1. Which of the following statements is incorrect? A) Ionic bonding results from the transfer of electrons from one atom to another. B) Dipole moments
More informationDefinition : Characteristics of Metals :
Metallic Bond Definition : It may be defined as, 1. The force that binds a metal ion to a number of electrons with in its sphere of influence. 2. The attractive force which holds the atoms of two or more
More informationLecture 25 The Solid State: types of crystals, lattice energies and solubility s of ionic compounds.
2P32 Principles of Inorganic Chemistry Dr.M.Pilkington Lecture 25 The Solid State: types of crystals, lattice energies and solubility s of ionic compounds. 1. Types of crystals: metalic, covalent, molecular
More informationChapter 5. Second Edition ( 2001 McGrawHill) 5.6 Doped GaAs. Solution
Chapter 5 5.6 Doped GaAs Consider the GaAs crystal at 300 K. a. Calculate the intrinsic conductivity and resistivity. Second Edition ( 2001 McGrawHill) b. In a sample containing only 10 15 cm 3 ionized
More informationCoordination and Pauling's Rules
Page 1 of 8 EENS 2110 Tulane University Mineralogy Prof. Stephen A. Nelson Coordination and Pauling's Rules This document last updated on 24Sep2013 The arrangement of atoms in a crystal structure not
More informationChem 106 Thursday Feb. 3, 2011
Chem 106 Thursday Feb. 3, 2011 Chapter 13: The Chemistry of Solids Phase Diagrams  (no BornHaber cycle) 2/3/2011 1 Approx surface area (Å 2 ) 253 258 Which C 5 H 12 alkane do you think has the highest
More informationPolymers: Introduction
Chapter Outline: Polymer Structures Hydrocarbon and Polymer Molecules Chemistry of Polymer Molecules Molecular Weight and Shape Molecular Structure and Configurations Copolymers Polymer Crystals Optional
More informationAnalog & Digital Electronics Course No: PH218
Analog & Digital Electronics Course No: PH218 Lecture 1: Semiconductor Materials Course Instructors: Dr. A. P. VAJPEYI Department of Physics, Indian Institute of Technology Guwahati, India 1 Semiconductors
More informationLiquids and Solids. AP Chemistry Chapter 10. 9/20/2009 Jodi Grack; Wayzata High School; images used with permission from Zumdahl
Liquids and Solids AP Chemistry Chapter 10 Liquids and Solids Gases are much easier to study because molecules move independent of each other. In liquids and solids forces between molecules become very
More informationCHAPTER  45 SEMICONDUCTOR AND SEMICONDUCTOR DEVICES
1. f = 101 kg/m, V = 1 m CHAPTER  45 SEMCONDUCTOR AND SEMCONDUCTOR DEVCES m = fv = 101 1 = 101 kg No.of atoms = 101 10 6 10 = 64.6 10 6. a) Total no.of states = N = 64.6 10 6 = 58.5 = 5. 10 8 10 6 b)
More informationNext, solid silicon is separated from other solid impurities by treatment with hydrogen chloride at 350 C to form gaseous trichlorosilane (SiCl 3 H):
University Chemistry Quiz 5 2014/12/25 1. (5%) What is the coordination number of each sphere in (a) a simple cubic cell, (b) a bodycentered cubic cell, and (c) a facecentered cubic cell? Assume the
More informationPhysics Notes Class 12 Chapter 14 Semiconductor Electronics, Materials, Devices and Sample Circuits
1 P a g e Physics Notes Class 12 Chapter 14 Semiconductor Electronics, Materials, Devices and Sample Circuits It is the branch of science which deals with the electron flow through a vacuum, gas or semiconductor.
More informationA. Xray diffraction B. elemental analysis C. band gap energy measurement based on absorption of light D. none of the above
LED Review Questions 1. Consider two samples in the form of powders: sample A is a physical mixture comprising equal moles of pure Ge and pure Si; sample B is a solid solution of composition Si0.5Ge0.5.
More informationIntroduction to Materials Science, Chapter 13, Structure and Properties of Ceramics. Chapter Outline: Ceramics
Chapter Outline: Ceramics Chapter 13: Structure and Properties of Ceramics Crystal Structures Silicate Ceramics Carbon Imperfections in Ceramics Optional reading: 13.6 13.10 Chapter 14: Applications and
More informationDoped Semiconductors. Dr. Katarzyna Skorupska
Doped Semiconductors Dr. Katarzyna Skorupska 1 Doped semiconductors Increasing the conductivity of semiconductors by incorporation of foreign atoms requires increase of the concentration of mobile charge
More informationLiquids and Solids. 1. Are liquids closer in physical properties to solids or gases? Why?
Liquids and Solids 1. Are liquids closer in physical properties to solids or gases? Why? Liquids are more similar to solids. There are many intermolecular forces experienced by solids and liquids and very
More information13. In the circuit shown the amount of charge on the plates of capacitor is 5 V
CAPACITORS 1. A parallel plate air capacitor consists of two circular plates of 2m 2 area separated by 1mm. If the gap between plates is doubled then its capacitance will be 1. Halved 2. Doubled 3. 4 times
More informationPart IA Paper 2: Structures and Materials MATERIALS Examples Paper 1 TEACH YOURSELF MICROSTRUCTURE Mike Ashby and Hugh Shercliff. 1.
Engineering Tripos FIRST YEAR Part IA Paper 2: Structures and Materials MATERIALS Examples Paper 1 TEACH YOURSELF MICROSTRUCTURE Mike Ashby and Hugh Shercliff 1. Introduction 2. Atomic structure. Atomic
More informationSemiconductors, diodes, transistors
Semiconductors, diodes, transistors (Horst Wahl, QuarkNet presentation, June 2001) Electrical conductivity! Energy bands in solids! Band structure and conductivity Semiconductors! Intrinsic semiconductors!
More information(ii) Relative sizes of tetrahedral and octahedral sites A general guideline for the relative size of sites:
(ii) Relative sizes of tetrahedral and octahedral sites A general guideline for the relative size of sites: tetrahedra < octahedra < polyhedra of higher coordination number In ionic structures, cations
More informationUniversity of Toronto Department of Electrical and Computer Engineering. ECE 330F SEMICONDUCTOR PHYSICS Eng. Annex 305
University of Toronto Department of Electrical and Computer Engineering ECE 330F SEMICONDUCTOR PHYSICS Eng. Annex 305 Experiment # 1 RESISTIVITY AND BAND GAP OF GERMANIUM TA: Iraklis Nikolalakos OBJECTIVE
More informationRESISTIVITY OF A SEMICONDUCTOR BY THE FOURPROBE METHOD
1 Experiment322 A RESISTIVITY OF A SEMICONDUCTOR BY THE FOURPROBE METHOD Dr Jeethendra Kumar P K, Tata Nagar, Bengaluru560 092. INDIA. Email: labexperiments@rediffmail.com, labexperiments@kamaljeeth.net
More informationEVERYDAY ENGINEERING EXAMPLES FOR SIMPLE CONCEPTS
EVERYDAY ENGINEERING EXAMPLES FOR SIMPLE CONCEPTS Thermal Properties ENGR 3350  Materials Science Dr. Nedim Vardar Copyright 2015 Thermal Properties of Materials Engage: MSEIP Engineering Everyday Engineering
More informationSemiconductor Physics
10p PhD Course Semiconductor Physics 18 Lectures NovDec 2011 and Jan Feb 2012 Literature Semiconductor Physics K. Seeger The Physics of Semiconductors Grundmann Basic Semiconductors Physics  Hamaguchi
More informationUniversity of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 7 Solutions to Purcell problems by P.
University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 7 Solutions to Purcell problems by P. Pebler 1 Purcell 4.8 A copper wire 1 km long is connected across a 6
More informationGen Chem I Exam 1 Review (Chapters 1 & 2)
Gen Chem I Exam 1 Review (Chapters 1 & 2) 1 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. All of the following are properties of antimony. Which one
More informationMetals, Semiconductors, and Insulators
Metals, Semiconductors, and Insulators Every solid has its own characteristic energy band structure. In order for a material to be conductive, both free electrons and empty states must be available. Metals
More information(a) Diamond is an electrical insulator and is one of the hardest substances known. Interpret these properties in terms of its bonding.
MSE 200A Survey of Materials Science Fall, 2008 Problem Set No. 1 Problem 1: In the diamond modification of solid carbon each atom has exactly four nearest neighbors that are configured in a tetrahedral
More informationCrystalline solids. A solid crystal consists of different atoms arranged in a periodic structure.
Crystalline solids A solid crystal consists of different atoms arranged in a periodic structure. Crystals can be formed via various bonding mechanisms: Ionic bonding Covalent bonding Metallic bonding Van
More informationIn Problems #1  #4, find the surface area and volume of each prism.
Geometry Unit Seven: Surface Area & Volume, Practice In Problems #1  #4, find the surface area and volume of each prism. 1. CUBE. RECTANGULAR PRISM 9 cm 5 mm 11 mm mm 9 cm 9 cm. TRIANGULAR PRISM 4. TRIANGULAR
More informationThe atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C. = 2(sphere volume) = 2 = V C = 4R
3.5 Show that the atomic packing factor for BCC is 0.68. The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C Since there are two spheres associated
More informationChapter 12. Intermolecular Forces and Liquids and Solids
Chapter 12 Intermolecular Forces and Liquids and Solids Intermolecular Forces Intermolecular Forces Intermolecular forces will affect: Boiling point Melting point States of Matter  phases Intermolecular
More informationChapter 10  Liquids and Solids
Chapter 10  Liquids and Solids 10.1 Intermolecular Forces A. DipoleDipole Forces 1. Attraction between molecules with dipole moments a. Maximizes (+)  (  ) interactions b. Minimizes (+)  (
More informationCHAPTER 1: Semiconductor Materials & Physics
Chapter 1 1 CHAPTER 1: Semiconductor Materials & Physics In this chapter, the basic properties of semiconductors and microelectronic devices are discussed. 1.1 Semiconductor Materials Solidstate materials
More informationChapter 13 The Chemistry of Solids
Chapter 13 The Chemistry of Solids Jeffrey Mack California State University, Sacramento Metallic & Ionic Solids Crystal Lattices Regular 3D arrangements of equivalent LATTICE POINTS in space. Lattice
More informationStructure of materials
Structure of materials The atomic number is the number of protons for each element. Atoms of the same element have the same number of protons in the nucleus but may differ by one or more neutrons forming
More informationElectrical Properties
Electrical Properties Outline of this Topic 1. Basic laws and electrical properties of metals 2. Band theory of solids: metals, semiconductors and insulators 3. Electrical properties of semiconductors
More information1.5 Light absorption by solids
1.5 Light absorption by solids BlochBrilloin model L e + + + + + allowed energy bands band gaps p x In a unidimensional approximation, electrons in a solid experience a periodic potential due to the positively
More informationConsider a onedimensional chain of alternating positive and negative ions. Show that the potential energy of an ion in this hypothetical crystal is
Chapter 11 The Solid State. Home Wor Solutions 11.1 Problem 11.5 Consider a onedimensional chain of alternating positive and negative ions. Show that the potential energy of an ion in this hypothetical
More informationNuclear reactions determine element abundance. Is the earth homogeneous though? Is the solar system?? Is the universe???
Nuclear reactions determine element abundance Is the earth homogeneous though? Is the solar system?? Is the universe??? Earth = anion balls with cations in the spaces View of the earth as a system of anions
More informationSolution for Homework #1
Solution for Homework #1 Chapter 2: Multiple Choice Questions (2.5, 2.6, 2.8, 2.11) 2.5 Which of the following bond types are classified as primary bonds (more than one)? (a) covalent bonding, (b) hydrogen
More informationStructures and properties
Crystal structures Structure plays an important role in understanding properties Need to be introduced to different ways of describing structures Models play an important role in communicating structural
More informationChapter 3: The Structure of Crystalline Solids
Sapphire: cryst. Al 2 O 3 Insulin : The Structure of Crystalline Solids Crystal: a solid composed of atoms, ions, or molecules arranged in a pattern that is repeated in three dimensions A material in which
More informationChapter 18 Electric Current and Circuits
Chapter 18 Electric Current and Circuits 3. When a current flows down a wire: A. electrons are moving in the direction of the current. B. electrons are moving opposite the direction of the current. C.
More informationIntrinsic and Extrinsic Semiconductors, FermiDirac Distribution Function, the Fermi level and carrier concentrations
ENEE 33, Spr. 09 Supplement I Intrinsic and Extrinsic Semiconductors, FermiDirac Distribution Function, the Fermi level and carrier concentrations Zeynep Dilli, Oct. 2008, rev. Mar 2009 This is a supplement
More informationStrength and Stiffness
Strength and Stiffness Stress = load/area is applied to a material by loading it Strain = deformation/length a change of shape (dimensions and twist angles) is its response Stiffness = stress/strain is
More informationSemiconductor Detectors Calorimetry and Tracking with High Precision
Semiconductor Detectors Calorimetry and Tracking with High Precision Applications 1. Photon spectroscopy with high energy resolution. Vertex detection with high spatial resolution 3. Energy measurement
More informationANSWERS to CIRCUITS V
1 ANSWERS to CIRCUITS 1. The speed with which electrons move through a copper wire is typically 104 m s 1. a. Explain why is it that the electrons cannot travel faster in the conductor? b. Explain why
More informationCovalent and Metallic Bonding
Covalent and Metallic Bonding Chemistry for Earth Scientists, DM Sherman University of Bristol Failures of Ionic Model: Ionic radius of S 2 NaCl structure CaS is highly ionic. Cell constant 5.70 Å Ca
More informationLecture 4.1: Thermoplastics and Thermosets
Lecture 4.1: Thermoplastics and Thermosets The word plastic comes from the Greek word Plastikos, meaning able to be shaped and molded. Plastics can be broadly classified into two major groups on the basis
More information8/19/2011. Periodic Trends and Lewis Dot Structures. Review PERIODIC Table
Periodic Trends and Lewis Dot Structures Chapter 11 Review PERIODIC Table Recall, Mendeleev and Meyer organized the ordering the periodic table based on a combination of three components: 1. Atomic Number
More informationCURRENT ELECTRICITY OHM S LAW KIRCHHOFF S LAW
CURRENT ELECTRICITY OHM S LAW KIRCHHOFF S LAW 1 01. A current of 16mA flows through a conductor. The number of electrons flowing per second is, (1) 10 14 (2) 10 15 (3) 1017 (4) 10 17 2 Solution (1):
More informationChapter 3: Structure of Metals and Ceramics. Chapter 3: Structure of Metals and Ceramics. Learning Objective
Chapter 3: Structure of Metals and Ceramics Chapter 3: Structure of Metals and Ceramics Goals Define basic terms and give examples of each: Lattice Basis Atoms (Decorations or Motifs) Crystal Structure
More informationIonic Bonding Pauling s Rules and the Bond Valence Method
Ionic Bonding Pauling s Rules and the Bond Valence Method Chemistry 754 Solid State Chemistry Dr. Patrick Woodward Lecture #14 Pauling Rules for Ionic Structures Linus Pauling,, J. Amer. Chem. Soc. 51,,
More informationResins for Optics. 1. LightCuring Resin 2. HeatCuring Resin. Introduction
Three Bond Technical News Issued July 1, 2004 63 Resins for Optics 1. LightCuring Resin 2. HeatCuring Resin Introduction As the word "optoelectronics is popular," the fields of electronics and optics
More informationResistance is defined as a measure of the ease of current flow, and is the ratio of the potential difference to the current  Ohm s law.
Resistance and Ohms Law When we have a fixed potential difference then the electric field will also be fixed and therefore there will be a steady electric force on the charge carriers (electrons)  this
More informationLecture 8: Extrinsic semiconductors  mobility
Lecture 8: Extrinsic semiconductors  mobility Contents Carrier mobility. Lattice scattering......................... 2.2 Impurity scattering........................ 3.3 Conductivity in extrinsic semiconductors............
More information