Engr Materials Science and Engineering TEST 4 -- Sample Solution

Save this PDF as:

Size: px
Start display at page:

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 cation-to-anion 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 cation-anion 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 /V-s. The hole mobility is m /V-s. 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 n-type 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 150-m long, 3.1-mm 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(ethylene-propylene) copolymer that has number- and weight-average degrees of polymerization of 500 and 000, respectively, and number- and weight-average molecular weights of 8,500 and 67,00 g/mol? Why or why not? Answer: no. Mer molecular weights computed from weight-average and number-average data are different 7. Molecular weight data from some polymer are tabulated below. Compute the (a) numberaverage molecular weight, and (b) the weight-average molecular weight. (c) If it is known

4 that the material s number-average degree of polymerization is 477, which one of the polymers listed in Table 14.3 is this polymer? Why? Molecular wt.range x i 8,000-0, ,000-3, ,000-44, ,000-56, ,000-68, ,000-80, ,000-9, (a) From the tabulated data, we are asked to compute M n, the number-average molecular weight. This is carried out below. Molecular wt. Range Mean M i x i xm i i 8,000-0,000 14, ,000-3,000 6, ,000-44,000 38, ,000-56,000 50, ,000 56,000-68,000 6, ,160 68,000-80,000 74, ,000-9,000 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 8,000-0,000 14, ,000-3,000 6, ,000-44,000 38, ,000-56,000 50, ,500 56,000-68,000 6, ,60

5 68,000-80,000 74, ,840 80,000-9,000 86, M w w i M i 53,70 g/mol (c) We are now asked if the number-average 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 Polytetrafluoroethylene--0.0 g/mol Polypropylene g/mol Polystyrene g/mol Polymethyl methacrylate g/mol Phenol-formaldehyde 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.

Hydrocarbon Molecules Polymer Molecules The Chemistry of Polymer Molecules

14-1 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

Sample 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

Hydrocarbon Molecules Polymer Molecules The Chemistry of Polymer Molecules

14-1 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

The Crystal Structures of Solids

The Crystal Structures of Solids Crystals of pure substances can be analyzed using X-ray diffraction methods to provide valuable information. The type and strength of intramolecular forces, density, molar

Types 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 Ion-ion Metallic Na, Fe Metallic Molecular Ice, I 2 Dipole Ind. dipole Network Diamond Extended Graphite

HW 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;

Crystal Structure Metals-Ceramics

Crystal Structure Metals-Ceramics 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

0% (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

Imperfections 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

Unit 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

3a, 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

M 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 number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000. (a) The repeat unit

CHAPTER 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

MAE 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

(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,

Basic 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

CHAPTER 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 volume-unit cell volume ratio.

Materials Science and Engineering Department MSE , Sample Test #1, Spring 2010

Materials Science and Engineering Department MSE 200-001, 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

interstitial 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, 6-coordinate

CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS

4-1 CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and Self-Interstitials 4.1 In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation

MCEN 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

15.32 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

THE 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

Section 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

THE STRUCTURE OF CRYSTALLINE SOLIDS. IE-114 Materials Science and General Chemistry Lecture-3

THE STRUCTURE OF CRYSTALLINE SOLIDS IE-114 Materials Science and General Chemistry Lecture-3 Outline Crystalline and Noncrystalline Materials 1) Single, Polycrystalline, Non-crystalline solids 2) Polycrystalline

CHAPTER 19 THERMAL PROPERTIES PROBLEM SOLUTIONS

19-1 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

Chapter 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.

Electroceramics 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,

ELECTRICAL 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?

b) 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?

Module 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.

Condensed 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

Lecture 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........................

Solid 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

v kt = N A ρ Au exp (

4-2 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

This 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

Chapter 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 Face-centered cubic Body-centered cubic Hexagonal close-packed

Chapter 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..

12 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.

CHEM 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!

UNIT 1 THE SOLID STATE VERY SHORT ANSWER TYPE QUESTIONS (1 MARKS)

UNIT 1 THE SOLID STATE VERY SHORT ANSWER TYPE QUESTIONS (1 MARKS) Q-1. How many spheres are in contact with each other in a single plane of a close packed structure? A-1. Six(6). Q-2.Name the two closest

100 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

Definition : 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

Lecture 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

Chapter 5. Second Edition ( 2001 McGraw-Hill) 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 McGraw-Hill) b. In a sample containing only 10 15 cm -3 ionized

Coordination 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 24-Sep-2013 The arrangement of atoms in a crystal structure not

Chem 106 Thursday Feb. 3, 2011

Chem 106 Thursday Feb. 3, 2011 Chapter 13: -The Chemistry of Solids -Phase Diagrams - (no Born-Haber cycle) 2/3/2011 1 Approx surface area (Å 2 ) 253 258 Which C 5 H 12 alkane do you think has the highest

Polymers: 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

Analog & Digital Electronics Course No: PH-218

Analog & Digital Electronics Course No: PH-218 Lecture 1: Semiconductor Materials Course Instructors: Dr. A. P. VAJPEYI Department of Physics, Indian Institute of Technology Guwahati, India 1 Semiconductors

Liquids 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

CHAPTER - 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)

Next, 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 body-centered cubic cell, and (c) a face-centered cubic cell? Assume the

Physics 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.

A. X-ray 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.

Introduction 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

Doped 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

Liquids 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

13. 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

Part 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

Semiconductors, diodes, transistors

Semiconductors, diodes, transistors (Horst Wahl, QuarkNet presentation, June 2001) Electrical conductivity! Energy bands in solids! Band structure and conductivity Semiconductors! Intrinsic semiconductors!

(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

University 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

RESISTIVITY OF A SEMICONDUCTOR BY THE FOUR-PROBE METHOD

1 Experiment-322 A RESISTIVITY OF A SEMICONDUCTOR BY THE FOUR-PROBE METHOD Dr Jeethendra Kumar P K, Tata Nagar, Bengaluru-560 092. INDIA. Email: labexperiments@rediffmail.com, labexperiments@kamaljeeth.net

EVERYDAY 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

Semiconductor Physics

10p PhD Course Semiconductor Physics 18 Lectures Nov-Dec 2011 and Jan Feb 2012 Literature Semiconductor Physics K. Seeger The Physics of Semiconductors Grundmann Basic Semiconductors Physics - Hamaguchi

University 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

Gen 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

Metals, 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

(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

Crystalline 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

In 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

The 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

Chapter 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

Chapter 10 - Liquids and Solids

Chapter 10 - Liquids and Solids 10.1 Intermolecular Forces A. Dipole-Dipole Forces 1. Attraction between molecules with dipole moments a. Maximizes (+) ----- ( - ) interactions b. Minimizes (+) ----- (

CHAPTER 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 Solid-state materials

Chapter 13 The Chemistry of Solids

Chapter 13 The Chemistry of Solids Jeffrey Mack California State University, Sacramento Metallic & Ionic Solids Crystal Lattices Regular 3-D arrangements of equivalent LATTICE POINTS in space. Lattice

Structure 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

Electrical 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

1.5 Light absorption by solids

1.5 Light absorption by solids Bloch-Brilloin 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

Consider a one-dimensional 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 one-dimensional chain of alternating positive and negative ions. Show that the potential energy of an ion in this hypothetical

Nuclear 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

Solution 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

Structures 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

Chapter 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

Chapter 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.

Intrinsic and Extrinsic Semiconductors, Fermi-Dirac Distribution Function, the Fermi level and carrier concentrations

ENEE 33, Spr. 09 Supplement I Intrinsic and Extrinsic Semiconductors, Fermi-Dirac Distribution Function, the Fermi level and carrier concentrations Zeynep Dilli, Oct. 2008, rev. Mar 2009 This is a supplement

Strength 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

Semiconductor 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

1 ANSWERS to CIRCUITS 1. The speed with which electrons move through a copper wire is typically 10-4 m s -1. a. Explain why is it that the electrons cannot travel faster in the conductor? b. Explain why

Covalent 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

Lecture 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

8/19/2011. 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

CURRENT 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) 10-17 (4) 10 17 2 Solution (1):-

Chapter 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

Ionic 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,,

Resins for Optics. 1. Light-Curing Resin 2. Heat-Curing Resin. Introduction

Three Bond Technical News Issued July 1, 2004 63 Resins for Optics 1. Light-Curing Resin 2. Heat-Curing Resin Introduction As the word "optoelectronics is popular," the fields of electronics and optics