Materials Science and Engineering: Diffusion and Crystalline Materials, Exams of Engineering Economy

Download Materials Science and Engineering: Diffusion and Crystalline Materials and more Exams Engineering Economy in PDF only on Docsity! Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Chapter 1: Introduction to Materials Science and Engineering 1-1 Define materials science and engineering (MSE). Solution: Materials science and engineering (MSE) is an interdisciplinary field that studies and manipulates the composition and structure of materials across length scales to control materials properties through synthesis and processing. 1-2 What is the importance of the engineering tetrahedron for materials engineers? Solution: Structure, properties and performance all depend on the route in which a material is processed. We cannot predict the end properties for a material until we have specified a process to produce the component. Using the same material, but changing the way it is processed will result in different structure, properties and performance of that material. This is applicable to all material systems. 1-3 Define the following terms: (a) composition; (b) structure; (c) synthesis; (d) processing; and (e) microstructure. Solution: (a) The chemical make-up of a material. (b) The arrangement of atoms, seen at different levels of detail. (c) How materials are made from naturally occurring or man-made chemicals. (d) How materials are shaped into useful components. (e) The structure of an object at the microscopic scale. 1-4 Explain the difference between the terms materials science and materials engineering. Solution: Materials scientists work on understanding underlying relationships between the synthesis and processing, structure, and properties of materials. Materials engineers focus on how to translate or transform materials into useful devices or structures. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 1-5 The myriad materials in the world primarily fall into four basic categories; what are they? What are materials called that have one or more different types of material fabricated into one component? Give one example. Solution: Metals, polymers and ceramics. The addition of one or more of these to a single system is called a composite. An example of a composite material is fiberglass. 1-6 What are some of the materials and mechanical properties of metals and alloys? Solution: Metals and alloys have good electrical and thermal conductivity, high strength, ductility and formability, and high stiffness. 1-7 What is a ceramic, and what are some of the properties that you expect from a ceramic? Solution: Ceramics tend to have very high compressive strengths, but behave in a brittle (glass- like) manner. They have very high melting temperatures. Poor thermal conductivity and electrical conductivity make ceramics behave as an insulator instead of a conductor. 1-8 Make comparisons between thermoplastics and thermosetting polymers (a) on the basis of mechanical characteristics upon heating, and (b) according to possible molecular structures. Solution: Thermoplastics tend to soften with elevated temperature exposure with gradually decreasing viscosity. Thermosetting polymers do not soften with elevated temperature exposure; instead they will remain hard and will degrade, possibly charring with prolonged exposure. Thermoplastics consist of long chain molecular arrangements of covalently bonded carbon atoms with various side groups. Thermosetting polymers tend to be a complex 3- D arrangement usually deviating from the clearly defined long-chain molecular arrangement. 1-9 Give three examples of composites that can be fabricated. Solution: Metal matrix composites (MMC) – A metal matrix reinforced with a ceramic material in the form of particles, whiskers or fibers. Example: Cobalt alloy reinforced with tungsten- carbide particulates. Polymer matrix composites (PMC) – A polymer matrix reinforced with a ceramic material in the form of whiskers or fibers. Example: Kevlar or fiberglass. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 (c) Ball bearings in a bicycle’s wheel hub reduce friction between metal surfaces. Therefore, steel is used because it has high strength and hardness. (d) Polyethylene terephthalate is easily formed by a blow molding process and is recyclable, critical properties for mass-produced water bottles. (e) Glass has high chemical resistance; thus, glass bottles are used to preserve the taste of the wine contained in them. 1-13 What properties should an engineer consider for a total knee replacement of a deteriorated knee joint with an artificial prosthesis when selecting the materials for this application? Solution: Properties that should be considered are those relating to strength (fatigue, tensile and compressive) since the knee sustains static, dynamic and cyclic loads. Hardness will promote wear resistance. Modulus of elasticity similar to that of the human bone otherwise other problems will occur. Chemical stability in regards to corrosion resistance and cellular toxicity to prevent negative reactions to the material selected. The material needs to have the ability to bond with the residual bone material and have longevity in order to avoid frequent replacement. 1-14 Write one paragraph about why single-crystal silicon is currently the material of choice for microelectronics applications. Write a second paragraph about potential alternatives to single-crystal silicon for solar cell applications. Provide a list of the references or websites that you used. You must use at least three references. Solution: Answers will vary. 1-15 Coiled springs should be very strong and stiff. Silicon nitride (Si3N4) is a strong, stiff material. Would you select this material for a spring? Explain. Solution: Springs are intended to resist high elastic forces, where only the atomic bonds are stretched when the force is applied. The silicon nitride would satisfy this requirement; however, we would like to also have good resistance to impact and at least some ductility (in case the spring is overloaded) to ensure that the spring will not fail catastrophically. We also would like to be sure that all springs will perform satisfactorily. Ceramic materials such as silicon nitride have virtually no ductility, poor impact properties, and often are difficult to manufacture without introducing at least some small flaws that cause failure even for relatively low forces. The silicon nitride is NOT recommended. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 1-16 Temperature indicators are sometimes produced from a coiled metal strip that uncoils a specific amount when the temperature increases. How does this work; from what kind of material would the indicator be made; and what are the important properties that the material in the indicator must possess? Solution: Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change. In order for the material to perform well, the two materials must have very different coefficients of thermal expansion and should have high enough moduli of elasticity so that no permanent deformation of the material occurs. 1-17 What is the purpose of the classification for functional materials? Solution: It specifically categorizes the types of materials used with specific applications. For instance, aerospace materials use lightweight materials such as aluminum alloys or carbon-composites for flight applications instead of using heavy materials such as stainless steel. 1-18 Explain the difference between crystalline and amorphous materials. Give an example of each that you use in your daily life. Solution: Crystalline materials have long-range order arrangement of its atoms while amorphous materials have short-range order. One example of a crystalline material is metal and an example of an amorphous material is glass. 1-19 If you were given a material and were asked to determine whether it is crystalline or amorphous, how would you determine it? Solution: Crystalline materials can be delineated from amorphous materials using diffraction techniques. Crystalline materials will produce diffraction patterns while amorphous materials will not. 1-20 List six materials performance problems that may lead to failure of components. Solution: Excessive deformation (overload), fracture, wear, corrosion, fatigue and creep. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 1-21 Steel is often coated with a thin layer of zinc if it is to be used outside. What characteristics do you think the zinc provides to this coated, or galvanized, steel? What precautions should be considered in producing this product? How will the recyclability of the product be affected? Solution: The zinc provides corrosion resistance to the iron in two ways. If the iron is completely coated with zinc, the zinc provides a barrier between the iron and the surrounding environment, therefore protecting the underlying iron. If the zinc coating is scratched to expose the iron, the zinc continues to protect the iron because the zinc corrodes preferentially to the iron (see Chapter 23). To be effective, the zinc should bond well to the iron so that it does not permit reactions to occur at the interface with the iron and so that the zinc remains intact during any forming of the galvanized material. When the material is recycled, the zinc will be lost by oxidation and vaporization, often producing a “zinc dust” that may pose an environmental hazard. Special equipment may be required to collect and either recycle or dispose of the zinc dust. 1-22 The relationship between structure and materials properties can be influenced by the service conditions (environmental conditions). Name two engineering disasters that have had tragic results and why they happened. Solution: Answers will vary. Two sample answers: The Titanic sank when it hit an iceberg. Hull plate and rivets of the ship were made of the strongest material to use at that time, but at room temperature. The material failed due to its anisotropy and temperature sensitivity for when it was exposed to the frigid waters during its voyage, behaved like glass (brittle). The Space Shuttle Challenger tragic flight when the rocket boosters exploded due to the solid rocket booster (SRB) O-ring failure. The freezing temperatures of that morning had the O-rings behave like glass and the vibrations during lift-off cracked the O-rings and the boosters exploded. 1-23 What is the difference between physical and mechanical properties? List three examples for each one. Solution: Physical properties are the properties of a material as it is found in nature such as density, thermal conductivity, electrical conductivity, luster, color, corrosion resistance, oxidation resistance, coefficient of thermal expansion (CTE) and magnetic permeability. Mechanical properties are properties that can be tested for such as strength (fatigue, impact, tensile, yield), ductility, stiffness, creep rupture, toughness and hardness. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 The main body of the shell might be a light weight fiber-reinforced composite that would provide impact resistance (preventing penetration by dust particles) but would be transparent to radio signals. 1-29 What properties should the head of a carpenter’s hammer possess? How would you manufacture a hammer head? Solution: The head for a carpenter’s hammer is produced by forging, a metal working process; a simple steel shape is heated and formed in several steps while hot into the required shape. The head is then heat treated to produce the required mechanical and physical properties. The striking face and claws of the hammer should be hard—the metal should not dent or deform when driving or removing nails. These portions must also possess some impact resistance, particularly so that chips do not flake off the striking face and cause injuries. 1-30 You would like to select a material for the electrical contacts in an electrical switching device that opens and closes frequently and forcefully. What properties should the contact material possess? What type of material might you recommend? Would Al2O3 be a good choice? Explain. Solution: The material must have a high electrical conductivity to ensure that no electrical heating or arcing occurs when the switch is closed. High purity (and therefore very soft) metals such as copper, aluminum, silver, or gold provide the high conductivity. The device must also have good wear resistance, requiring that the material be hard. Most hard, wear resistant materials have poor electrical conductivity. One solution to this problem is to produce a particulate composite material composed of hard ceramic particles embedded in a continuous matrix of the electrical conductor. For example, silicon carbide particles could be introduced into pure aluminum; the silicon carbide particles provide wear resistance while aluminum provides conductivity. Other examples of these materials are described in Chapter 17. Al2O3 by itself would not be a good choice—alumina is a ceramic material and is an electrical insulator; however, alumina particles dispersed into a copper matrix might provide wear resistance to the composite. 1-31 Aluminum has a density of 2.7 g/cm3. Suppose you would like to produce a composite material based on aluminum having a density of 1.5 g/cm3. Design a material that would have this density. Would introducing beads of polyethylene, with a density of 0.95 g/cm3, into the aluminum be a likely possibility? Explain. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: In order to produce an aluminum-matrix composite material with a density of 1.5 g/cm3, we would need to select a material having a density considerably less than 1.5 g/cm3. While polyethylene’s density would make it a possibility, the polyethylene has a very low melting point compared to aluminum; this would make it very difficult to introduce the polyethylene into a solid aluminum matrix—processes such as casting or powder metallurgy would destroy the polyethylene. Therefore polyethylene would NOT be a likely possibility. One approach, however, might be to introduce hollow glass beads. Although ceramic glasses have densities comparable to that of aluminum, a hollow bead will have a very low density. The glass also has a high melting temperature and could be introduced into liquid aluminum for processing as a casting. 1-32 You would like to be able to identify different materials without resorting to chemical analysis or lengthy testing procedures. Describe some possible testing and sorting techniques you might be able to use based on the physical properties of materials. Solution: Some typical methods might include: measuring the density of the material (may help in separating metal groups such as aluminum, copper, steel, magnesium, etc.), determining the electrical conductivity of the material (may help in separating ceramics and polymers from metallic alloys), measuring the hardness of the material (perhaps even just using a file), and determining whether the material is magnetic or nonmagnetic (may help separate iron from other metallic alloys). 1-33 You would like to be able to physically separate different materials in a scrap recycling plant. Describe some possible methods that might be used to separate materials such as polymers, aluminum alloys, and steels from one another. Solution: Steels can be magnetically separated from the other materials; steel (or carbon- containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around 2.7; that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators while aluminum has a particularly high electrical conductivity. 1-34 Some pistons for automobile engines might be produced from a composite material containing small, hard silicon carbide particles in an aluminum alloy matrix. Explain what benefits each material in the composite may provide to the overall part. What problems might the different properties of the two materials cause in producing the part? Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: Aluminum provides good heat transfer due to its high thermal conductivity. It has good ductility and toughness, reasonably good strength, and is easy to cast and process. The silicon carbide, a ceramic, is hard and strong, providing good wear resistance, and also has a high melting temperature. It provides good strength to the aluminum, even at elevated temperatures. There may be problems, however, producing the material—for example, the silicon carbide may not be uniformly distributed in the aluminum matrix if the pistons are produced by casting. We need to ensure good bonding between the particles and the aluminum—the surface chemistry must therefore be understood. Differences in expansion and contraction with temperature changes may cause debonding and even cracking in the composite. 1-35 Investigate the origins and applications for a material that has been invented or discovered since you were born or investigate the development of a product or technology that has been invented since you were born that was made possible by the use of a novel material. Write one paragraph about this material or product. Provide a list of the references or websites that you used. You must use at least three references. Solution: Answers will vary. 15 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 ( 3) 8.902 g Ni 1 mol Ni 6.022 × 1023 atoms 𝑁𝑎𝑡o𝑚𝑠 = 6.55 cm ( 1 cm3 ) ( 58.71 g Ni ) ( 1 mol ) (b) 𝑁 ( 3) 8.902 g Ni 1 mol Ni 𝑚o𝑙e𝑠 = 6.55 cm ( 1 ) ( 58.71 g Ni ) 2-9 Define electronegativity. Solution: Electronegativity is the tendency of an atom to accept an electron (which has a negative charge) and become an anion. 2-10 Write the electronic configuration of the following elements (a) tungsten, (b) cobalt, (c) zirconium, (d) uranium, and (e) aluminum. Solution: (a) W: [Xe] 4f145d46s2 (b) Co: [Ar] 3d74s2 (c) Zr: [Kr] 4d24s2 (d) U: [Rn] 5f36d17s2 (e) Al: [Ne] 3s23p1 2-11 Write the electron configuration for the element Tc. Solution: Since Technetium is element 43: [Tc] = 1s22s22p63s23p64s23d104p65s24d5 2-12 Assuming that the Aufbau Principle is followed, what is the expected electronic configuration of the element with atomic number Z = 116? Solution: Using the Aufbau diagram produces: [116] = 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p4 Or in shorthand: 𝑁𝑚o𝑙e𝑠 = 0.99 mol Ni 𝑁𝑎𝑡o𝑚𝑠 = 598 × 1021 atoms Ni cm3 16 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 [116] = [Rn] 5f146d107s27p4 2-13 Using the Aufbau Principle, what is the expected electronic configuration of the hypothetical element with atomic number Z = 123? Solution: Using the Aufbau diagram produces: [123] = 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p68s25g3 Or in shorthand: [123] = [Rn] 5f145g36d107s27p68s2 Or assuming that [118] is another inert gas: [123] = [118] 5g38s2 2-14 Suppose an element has a valence of 2 and an atomic number of 27. Based only on the quantum numbers, how many electrons must be present in the 3d energy level? Solution: We can let x be the number of electrons in the 3d energy level. Then: 1s22s22p63s23p63dx4s2 (must be 2 electrons in 4s for valence = 2) Since 27 - (2 + 2 + 6 + 2 + 6 + 2) = 7 = x there must be 7 electrons in the 3d level. 2-16 Bonding in the intermetallic compound Ni3Al is predominantly metallic. Explain why there will be little, if any, ionic bonding component. The electronegativity of nickel is about 1.8. Solution: The electronegativity of Al is 1.5, while that of Ni is 1.9. These values are relatively close, so we wouldn’t expect much ionic bonding. Also, both are metals and prefer to give up their electrons rather than share or donate them. 2-17 Plot the melting temperatures of elements in the 4A to 8–10 columns of the periodic table versus atomic number (i.e., plot melting temperatures of Ti through Ni, Zr through Pd, and Hf through Pt). Discuss these relationships, based on atomic bonding and binding energies: (a) as the atomic number increases in each row of the periodic table and (b) as the atomic number increases in each column of the periodic table. 17 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Melting temperature v. atomic number 4000 3500 3000 2500 2000 1500 1000 500 0 Ti - Ni Zr - Pd Hf - Pt Solution: The melting temperatures are taken from Figure 2-9 and are plotted below: For each row, the melting temperature is highest when the outer “d” energy level is partly full. In Cr, there are 5 electrons in the 3d shell; in Mo, there are 5 electrons in the 4d shell; in W there are 4 electrons in the 5d shell. In each column, the melting temperature increases as the atomic number increases—the atom cores contain a larger number of tightly held electrons, making the metals more stable. 2-18 Plot the melting temperature of the elements in the 1A column of the periodic table versus atomic number (i.e., plot melting temperatures of Li through Cs). Discuss this relationship, based on atomic bonding and binding energy. Solution: Using data from Figure 2-9: M el ti n g te m p er at u re ( °C ) 20 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: Starting with Equation 2-1, we determine the fraction of ionic bonding with this equation: ƒio𝑛i𝑐 = 1 − e(−0.25∆𝐸 2) The electronegativities of magnesium and oxygen are 1.2 and 3.5, so their difference is 2.3. Inserting: Math time! ƒio𝑛i𝑐 = 1 − e(−0.25[2.3]2) The compound is held together mostly by ionic bonding. 2-24 Calculate the fraction of bonding that is covalent for silica (SiO2). Solution: Using Equation 2-1, we determine the fraction of covalent bonding: ƒ𝑐o𝑣𝑎𝑙e𝑛𝑡 = e(−0.25∆𝐸 2) The electronegativities of silicon and oxygen are 1.8 and 3.5, so their difference is 1.7. Inserting: ƒ𝑐o𝑣𝑎𝑙e𝑛𝑡 = e(−0.25[1.7]2) 2-25 Calculate the fraction of bonding that is ionic in nature for zirconia (ZrO2)? Solution: ƒio𝑛i𝑐 = 1 − e(−0.25∆𝐸 2) The electronegativities of zircon and oxygen are 1.4 and 3.5, so their difference is 2.1. Inserting: ƒio𝑛i𝑐 = 1 − e(−0.25[2.1]2) 2-26 What is the type of bonding in diamond? Are the properties of diamond commensurate with the nature of the bonding? ƒio𝑛i𝑐 = 0.668 ƒ𝑐o𝑣𝑎𝑙e𝑛𝑡 = 0.486 ƒio𝑛i𝑐 = 0.734 21 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: In diamond, the carbon atoms are covalently bonded. Diamond is electrically insulating, which makes sense: each carbon is bonded to four other carbon atoms thus leaving no free valence electrons available to conduct electricity. 2-27 What are the bonding mechanisms in thermoplastics? Solution: The primary bond in thermoplastics are the covalent bonds that hold the individual carbon atoms together along the polymer chains while van der Waals secondary bonds hold the polymer chains close together. 2-28 Why are covalently bonded materials generally less dense than those that are bonded ionically or metallically? Solution: Covalently bonded materials are typically less dense than metallically or ionically bonded materials due to the nature of their bonding. The bonding in covalent materials is directional in nature which doesn’t allow the atoms to pack together in a dense manner like the ionic or metallic bonded materials. This results in a lower mass and lower density for covalently bonded materials. 2-31 Calculate the fractions of ionic bonds in silicon carbide (SiC) and silicon nitride (Si3N4). Solution: We use Equation 2.1 and take the electronegativities from Figure 2-9: ƒio𝑛i𝑐 = 1 − e(−0.25∆𝐸 2) The electronegativities of silicon and carbon are 1.8 and 2.5, so their difference is 0.7. Inserting: Math time! ƒio𝑛i𝑐 = 1 − e(−0.25[0.7]2) Repeating this for silicon nitride: ƒio𝑛i𝑐 = 1 − e(−0.25∆𝐸 2) The electronegativities of silicon and nitrogen are 1.8 and 3.0, so their difference is 1.2. Inserting: ƒio𝑛i𝑐 = 1 − e(−0.25[1.2]2) ƒio𝑛i𝑐 = 0.115 22 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 2-32 One particular form of boron nitride (BN) known as cubic boron nitride is a very hard material and is used in grinding applications. Calculate the fraction of bonding that is covalent in this material. Solution: Using Equation 2-1, we determine the fraction of covalent bonding: ƒ𝑐o𝑣𝑎𝑙e𝑛𝑡 = e(−0.25∆𝐸 2) The electronegativities of boron and nitrogen are 2.0 and 3.0, so their difference is 1.0. Inserting: ƒ𝑐o𝑣𝑎𝑙e𝑛𝑡 = e(−0.25[1.0]2) 2-34 Is there a trend in the number of electrons in the outermost energy shell of atoms or ions that have formed bonds? Solution: Yes. They try to fill their valence shells. 2-35 In order to increase the operating temperature of an engine, it is suggested that some of the aluminum components be coated with a ceramic. What kinds of problems could this pose? How could you overcome these problems? Solution: Creating a mechanical bond between the ceramic and the metallic component could pose a problem since the ceramic is ionic in nature and the component is metallically bonded. This can be overcome by creating a slightly roughened surface and choosing a ceramic that has a limited degree of chemical reactivity with the metal which would enhance bonding. Another problem that we face is likely to spallation or debonding of the coating due to the differences in the coefficients of thermal expansion. To overcome this problem, it would be possible to select a ceramic that could have a compatible coefficients of thermal expansion with the aluminum for the given operating temperature range. 2-36 Aluminum and silicon are side by side on the periodic table. Compare the melting temperatures of the two elements and explain the difference in terms of atomic bonding. ƒ𝑐o𝑣𝑎𝑙e𝑛𝑡 = 0.779 ƒio𝑛i𝑐 = 0.302 25 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Chapter 3: Atomic and Ionic Arrangements 3-1 What is a crystalline material? Solution: A uniform substance made up of a three dimensional, grid-like repeating pattern. 3-2 What is a single crystal? Solution: A crystalline material that is composed of only one crystal. 3-3 State two applications in which single crystals are used. Solution: Silicon integrated circuit (computer) chips and lithium niobate optoelectric devices. 3-4 What is a polycrystalline material? Solution: A crystalline material which is composed of many smaller crystals or grains, randomly oriented. 3-5 What is a liquid crystal material? Solution: A material that behaves as a liquid in one state, but forms small crystalline regions when an external stimulus is applied. 3-6 What is an amorphous material? Solution: A material that displays only short range order. For solids, amorphous is antonymic to crystalline. 3-7 Why do some materials assume an amorphous structure? Solution: The kinetics of the manufacturing process do not allow a regular or crystalline pattern to form. 26 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 3-8 Explain how it is possible for a substance to exhibit short-range, but not long-range order. Solution: A substance with only short-range order is similar to a scattered ream of paper or deck of cards. Each piece of paper represents a short-range organization of wood fiber, but the pieces have no order with respect to each other. If they were collected into a stack, they would have both short- and long-range order. 3-9 Can an alloy exist in both crystalline and amorphous forms? Solution: Not at the same time. Forming amorphous alloys is frequently very difficult and requires extraordinary cooling rates. 3-10 Approximately how many grains are shown in Figure 3-2(b)? Solution: By rough count, about 50. Count the areas separated by the darker lines, which are the grain boundaries. 3-11 Define the terms lattice, unit cell, basis, and crystal structure. Solution: A lattice is a collection of points arranged in a spatial pattern so each point has a relation its surrounding points that is valid for all other points as well. A unit cell is the smallest part of a lattice that shows the repeating pattern of the lattice. The basis is composed of atoms with a certain relationship spatial relationship. The crystal structure is the combination of the lattice and basis and shows how the atoms of a crystal are arranged with respect to each other throughout the entire thing. 27 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 𝑟 = 1.426 × 10−8 cm 𝑟 = 1.4447 × 10−8 cm 3-12 List the seven different crystal systems and the types of Bravais lattices that are associated with their groups. Solution: They are cubic (simple, face-centered, and body-centered), tetragonal (simple and body- centered), orthorhombic (simple, body-centered, base-centered, and face-centered), rhombohedral/trigonal, hexagonal, monoclinic (simple and base-centered), and finally triclinic. The order of this list is random. 3-13 Explain why there is no face-centered tetragonal Bravais lattice. Solution: The facial lattice points would not have identical surroundings, which is necessary for a unit cell. 3-14 Calculate the atomic radius in cm for the following: (a) BCC metal with a0 = 0.3294 nm; and (b) FCC metal with a0 = 4.0862 Å. Solution: For BCC metals: 𝑟 = √3 𝑎 4 0 𝑟 = √3 (0.3294 nm) = 0.1426 nm 4 For FCC metals: 𝑟 = √2 𝑎 4 0 𝑟 = √2 (4.0862 Å) = 1.4447 Å 4 3-15 Calculate the lattice parameter of each of the following elements using their atomic radii: (a) iron, (b) aluminum, (c) copper, and (d) magnesium. Solution: Iron is BCC and the a0 to r relationship is 30 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 0 mol ) 𝑟 = 1.7980 × 10−8 cm Density 𝜌 = (number of atoms/cell)(atomic mass) (volume of unit cell)(𝑁Æ) 11.72 g cm3 (2 atom ) (232 g ) = cell mol (𝑎3) (6.022 × 1023 atom ) 0 𝑎3 = 1.315297 × 10−22 cm3 mol For the second part, (b), we use the relationship between the atomic radius and the lattice parameter: √2 𝑟 = 𝑎0 4 𝑟 = (5.0856 × 10−8 cm) √2 4 3-19 A metal having a cubic structure has a density of 2.6 g/cm3, an atomic weight of 87.62 g/mol, and a lattice parameter of 6.0849 Å. One atom is associated with each lattice point. Determine the crystal structure of the metal. Solution: Using Equation 3-5: Density 𝜌 = (number of atoms/cell)(atomic mass) (volume of unit cell)(𝑁Æ) 2.6 g cm3 (𝑥) (87.26 g ) = mol (6.0849 × 10−8 cm)3 (6.022 × 1023 atom 𝑥 = 4 atom cell The crystal must be face-centered cubic (FCC). 3-20 A metal having a cubic structure has a density of 1.892 g/cm3, an atomic weight of 132.91 g/mol, and a lattice parameter of 6.13 Å. One atom is associated with each lattice point. Determine the crystal structure of the metal. Solution: Using Equation 3-5: 𝑎0 = 5.0856 × 10−8 cm 31 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 mol ) mol ) 0 𝑉 = 2.1226 × 10−22 cm3 Density 𝜌 = (number of atoms/cell)(atomic mass) (volume of unit cell)(𝑁Æ) 1.892 g cm3 (𝑥) (132.91 g ) = mol (6.13 × 10−8 cm)3 (6.022 × 1023 atom 𝑥 = 2 atom cell The crystal must be body-centered cubic (BCC). 3-21 Indium has a tetragonal structure with a0 = 0.32517 nm and c0 = 0.49459 nm. The density is 7.286 g/cm3, and the atomic weight is 114.82 g/mol. Does indium have the simple tetragonal or body-centered tetragonal structure? Solution: Using Equation 3-5: Density 𝜌 = (number of atoms/cell)(atomic mass) (volume of unit cell)(𝑁Æ) 7.286 g cm3 (𝑥) (114.82 g ) = mol (3.2517 × 10−8 cm)2(4.9459 × 10−8 cm) (6.022 × 1023 atom 𝑥 = 2 atom cell Therefore, the crystal is body-centered tetragonal (BCT). 3-22 Bismuth has a hexagonal structure, with a0 = 0.4546 nm and c0 = 1.186 nm. The density is 9.808 g/cm3, and the atomic weight is 208.98 g/mol. Determine (a) the volume of the unit cell; and (b) the number of atoms in each unit cell. Solution: The volume, (a), of the unit cell is: 𝑉 = 𝑎2𝑐0 cos 30° 𝑉 = (0.4546 nm)2(1.186 nm)(cos 30°) Reprising Equation 3-5 for part (b): Density 𝜌 = (number of atoms⁄cell)(atomic mass) (volume of unit cell)(𝑁Æ) 32 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 mol ) mol Packing factor = 0.387 9.808 g cm3 (𝑥) (208.98 g ) = mol (2.1226 × 10−22 cm3) (6.022 × 1023 atom 𝑥 = 6 atom cell 3-23 Gallium has an orthorhombic structure, with a0 = 0.45258 nm, b0 = 0.45186 nm, and c0 = 0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm3, and the atomic weight is 69.72 g/mol. Determine (a) the number of atoms in each unit cell; and (b) the packing factor in the unit cell. Solution: The volume of the unit cell is: 𝑉 = 𝑎0𝑏0𝑐0 𝑉 = (0.45258 nm)(0.45186 nm)(0.76570 nm) = 0.1556 nm3 𝑉 = 1.566 × 10−22 cm3 Using Equation 3-5 for part (a): Density 𝜌 = (number of atoms⁄cell)(atomic mass) (volume of unit cell)(𝑁Æ) 5.904 g cm3 (𝑥) (69.72 g ) = mol (1.566 × 10−22 cm3) (6.022 × 1023 atom ) 𝑥 = 8 atom cell The packing factor (PF) is calculated using Equation 3-4: Packing factor = (number of atoms⁄cell)(volume of each atom) (volume of unit cell) (8 atom ) (4𝜋/3)(0.1218 nm)3 Packing factor = cell (0.1566 nm3) 3-24 Beryllium has a hexagonal crystal structure, with a0 = 0.22858 nm and c0 = 0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine (a) the number of atoms in each unit cell; and (b) the packing factor in the unit cell. 35 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 0 Using Equation 3-5 for part (a): Density 𝜌 = (number of atoms⁄cell)(atomic mass) (volume of unit cell)(𝑁Æ) (2 atom ) (22.99 g ) Density 𝜌 = cell mol atom (1.7321 × 10−8 cm)3 (6.022 × 1023 𝜌 = 14.69 g cm3 This was already done: mol ) 3-28 Calculate the density for zinc (HCP) if the c/a ratio is 1.85 and r = 1.332 Å. Solution: 𝑎0 = 2(1.332 Å) = 2.664 Å 𝑐 = 1.85(2.664 Å) = 4.928 Å 𝑉 = 𝑎2𝑐0 cos 30° 𝜌 = 𝑚 𝑉 2 atom (65.39 g ) ( mol ) 6.022 × 1023 atom 𝜌 = mol (2.664 × 10−8 cm)2(4.928 × 10−8 cm)(cos 30°) 𝜌 = 7.169 g cm3 3-29 Thoria or thorium dioxide can be described as an FCC lattice with a basis of Th (0, 0, 0) and O (1/4, 1/4, 1/4) and O (1/4, 1/4, 3/4). Thorium and thorium dioxide are radioactive materials. Thorium dioxide is commonly used in tungsten electrodes for arc welding. Thoria improves the high temperature properties of the electrodes and promotes emission of electrons from the electrode tip. (a) How many atoms of each type are there in the conventional unit cell of thorium dioxide? (b) Is this consistent with the chemical formula of thorium dioxide? Explain. (c) What is the coordination number (number of nearest neighbors) for each Th atom? (d) What is the coordination number (number of nearest neighbors) for each O atom? (e) What type of interstitial site do the oxygen atoms occupy? 𝑎0 = 1.7321 × 10−8 cm 36 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 0 Packing factor = 0.65 Solution: The Th atom at the origin contributes one-eighth of an atom, but it recurs at each corner. There are eight corners per unit cell, so ( 1 ) 8 Th atom corner × 8 corner unit cell Th atom = 1 unit cell The two oxygen atoms occur wholly within the unit cell and contribute two atoms each. The total then is three atoms per unit cell. Yes. For each unit cell there is one thorium atom and two oxygen atoms, for a molecular formula of ThO2. Each thorium atom has four oxygen atoms equally close to it, so its coordination number is 4. Each oxygen atom has one atom closer to the others: the thorium atom. It has a coordination number of 1. 3-30 Zinc has the hexagonal close-packed crystal structure. The lattice parameters for zinc are a = 0.26648 nm and c = 0.49470 nm, and the atomic radius is 0.1332 nm. Note that zinc does not have the ideal atomic packing factor. (a) What is the number of atoms per unit cell in the hexagonal close-packed structure? (b) Determine the atomic packing factor of the Zn unit cell. (c) Is the c/a ratio for zinc greater or less than the ideal HCP c/a ratio? Will slip be harder or easier in zinc compared to the ideal HCP structure? Explain your answer fully. Solution: Examining Table 3-2, there are two atoms per unit cell of HCP crystal. For this, we repair to Equation 3-4: Packing factor = (number of atoms⁄cell)(volume of each atom) (volume of unit cell) First we find the volume of the unit cell: 𝑉 = 𝑎2𝑐0 cos 30° 𝑉 = (0.26648 nm)2(0.49470 nm)(cos 30°) (2 atom ) (4𝜋/3)(0.1332 nm)3 Packing factor = cell (0.0304 nm3) The ideal c/a ratio for HCP is 1.663, the c/a ratio for zinc is: ( 𝑐 ) = 0.49470 nm = 1.856 𝑎 Zn 0.26648 nm 37 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 ( 𝑐 ) 𝑎 Zn 𝑎 HCP ideal > ( 𝑐 ) So then: 3-31 Rutile is the name given to a crystal structure commonly adopted by compounds of the form AB2, where A represents a metal atom and B represents oxygen atoms. One form of rutile has atoms of element A at the unit cell coordinates (0, 0, 0) and (1/2, 1/2, 1/2) and atoms of element B at (1/4, 1/4, 0), (3/4, 3/4, 0), (3/4, 1/4, 1/2), and (1/4, 3/4, 1/2). The unit cell parameters are a = b ≠ c and a = b = γ = 90°. Note that the lattice parameter c is typically smaller than the lattice parameters a and b for the rutile structure. (a) How many atoms of element A are there per unit cell? (b) How many atoms of element B are there per unit cell? (c) Is your answer to part (b) consistent with the stoichiometry of an AB2 compound? Explain. (d) Draw the unit cell for rutile. Use a different symbol for each type of atom. Provide a legend indicating which symbol represents which type of atom. (e) For the simple tetragonal lattice a = b ≠ c and α = β = γ = 90°. There is one lattice point per unit cell located at the corners of the simple tetragonal lattice. Describe the rutile structure as a simple tetragonal lattice and a basis. Solution: There are two atoms of element A per unit cell: one at the corners of the unit cell and one at the body–centered position. There are four atoms of element B per unit cell: two atoms on faces and two located in the interior of the unit cell. Yes, the answer to (b) is consistent with the stoichiometry of an AB2 compound. The ratio of A:B atoms is 2:4 or 1:2. The unit cell is shown below. 40 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 base 3 The volume of a unit cell is 𝑉cell = 𝐴base𝑐 Where Abase is the area of the base of the unit cell and c the height of the unit cell. The geometry of the unit cell is shown here: 𝑉cell = 𝐴base𝑐 𝐴 = 2 ( 1 × 𝑎 × 2 𝑎√3 ) = 2 𝑎2√3 2 𝑉cell = ( 𝑎2√3 2 ) 𝑐 𝑉cell = ( [3.24 × 10−10 m]2√3 2 ) (5.19 × 10 −10 m) 𝑉cell = 4.718 × 10−29 m3 𝑁 𝑉total atom atoms = ( 𝑉cell ) (4 cell ) 3-34 Calculate the atomic packing fraction for the hexagonal close-packed crystal structure 8 3 . Remember that the base of the unit cell is a parallelogram. Solution: This is an exercise in studying the geometry of the HCP unit cell. The base of the HCP unit cell can be treated as two equilateral triangles each of side a as shown below. The B atom in the HCP structure is located at the (2/3, 1/3, 1/2) location. The distance from the coordinate (1, 0, 0) to (2/3, 1/3, 0), which is the centroid of the triangle in the base, is 𝑎√3 . 𝑁atoms = 1.33 × 108 atom for which √𝑐 = 41 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 3 2 ) The c/a ratio is found by applying the Pythagorean Theorem to the triangle shown below. It is necessary to recognize that each B atom touches the three A atoms in the layer below it at a distance a = 2R, where R is the radius of the atom. 2 ( 𝑎√3 ) 3 + ( 𝑐 2 = 𝑎2 Solving for c/a: 𝑎2 𝑐2 2 3 + 4 = 𝑎 𝑐2 8𝑎2 = 3 𝑐 8 𝑎 = √ ∎ 3-35 Magnesium has the hexagonal close-packed crystal structure. The lattice parameters for magnesium are a = 0.32087 nm and c = 0.5209 nm, and the atomic radius is 0.1604 nm. Note that magnesium does not have the ideal atomic packing factor. (a) What is the number of atoms per unit cell in the hexagonal close-packed structure? (b) Determine the atomic packing factor of the Mg unit cell. (c) Is the c/a ratio for Mg greater or less than the ideal HCP c/a ratio? Will slip be harder or easier in Mg compared to the ideal HCP structure? Explain your answer fully. Solution: Examining Table 3-2, there are two atoms per unit cell of HCP crystal. 42 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 0 Packing factor = 0.744 ( 𝑐 ) 𝑎 Mg 𝑎 HCP ideal > ( 𝑐 ) For this, we repair to Equation 3-4: Packing factor = (number of atoms⁄cell)(volume of each atom) (volume of unit cell) First we find the volume of the unit cell: 𝑉 = 𝑎2𝑐0 cos 30° 𝑉 = (0.32087 nm)2(0.5209 nm)(cos 30°) = 0.04645 nm3 (2 atom ) (4𝜋/3)(0.1604 nm)3 Packing factor = cell (0.04645 nm3) The ideal c/a ratio for HCP is 1.663, the c/a ratio for magnesium is: ( 𝑐 ) = 0.5209 nm = 1.623 𝑎 Mg 0.32087 nm So then: Slip in Mg will be slightly harder to produce than in ideal HCP structures because the basal close packed planes where slip generally occurs in HCP metals will be marginally more closely packed. 3-36 What is the difference between an allotrope and a polymorph? Solution: Allotropy describes the property of having multiple crystals forms when pure elements are described. Polymorphism is the same property, but occurring in compounds of elements. 3-37 What are the different polymorphs of zirconia? Solution: Monoclinic, tetragonal, cubic and orthorhombic. These occur at different temperatures and pressures. 3-38 A number of metals undergo an allotropic transformation from one crystal structure to another at a specific temperature. There is generally a volume change that accompanies the transformation. What is the practical significance of such data? Solution: 45 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 mol ) mol 𝜌 = 0.657 𝜌tetragonal ZrO2 (1 atom ) (123.22 g ) = cell mol (0.1376 × 10−21 cm3) (6.022 × 1023 atom 𝜌tetragonal ZrO2 = 1.487 g cm3 For cubic, there is 1 atom per unit cell. The volume is: 𝑉cubic unit cell = 𝑎3 𝑉cubic unit cell = (5.124 Å) 3 𝑉cubic unit cell = 0.1345 nm3 (1 atom g 𝜌cubic ZrO2 = cell ) (123.22 mol) (0.1345 × 10−21 cm3) (6.022 × 1023 atom ) 𝜌cubic ZrO2 = 1.521 g cm3 3-42 From the information in this chapter, calculate the volume change that will occur when the cubic form of zirconia transforms into a tetragonal form. Solution: From problem 3-41, we have the densities of the two forms: 𝜌cubic ZrO2 𝜌 = 1.521 g cm3 = 1.487 g tetragonal ZrO2 cm3 By assuming a constant mass of 1 g, we can take the inverse of the densities and take their differences: −1 cubic ZrO2 −1 cm3 g cm3 𝜌tetragonal ZrO2 = 0.672 g ( ) −1 ( ) cm3 ∆𝑉 = 1 g ∆𝜌 = 1 g (−0.015 g ) 3-43 Monoclinic zirconia cannot be used effectively for manufacturing oxygen sensors or other devices. Explain. ∆𝑉 = −0.015 cm3 46 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: Monoclinic zirconia is not volume stable and may fail in use. 3-44 What is meant by the term stabilized zirconia? Solution: Dopants (deliberate impurities) are added to the crystal structure to make the cubic crystal form preferred energetically, so it does not change. 3-45 State any two applications of stabilized zirconia ceramics. Solution: Thermal barrier coats for turbine blades and electrolytes for oxygen sensors. 3-46 Explain the significance of crystallographic directions using an example of an application. Solution: Metal defamation is easier in directions of crystal orientation. 3-47 Why are Fe-Si alloys used in magnetic applications “grain oriented”? Solution: The direction of the grains will facilitate or hinder the magnetization of the alloy. 3-49 Determine the Miller indicies of the directions for the following points: (a) from (1, 0, 2) to (2, 4, 1); (b) from (2, 1, 3) to (5, 4, 2); and (c) from (3, 1, 3) to (9, 1, 5). Solution: [ 141 ], [ 331 ], [602] 3-50 Indicate the directions (a) [111], (b) [025], and (c) [414] within a unit cell. Solution: 3-51 Determine the Miller indices for the directions in the cubic unit cell shown in Figure 3- 37. 47 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: A: 0, 1, 0 - 0, 1, 1 = 0, 0, -1 = [ 001 ] B: ½, 0, 0 - 0, 1, 0 = ½, -1, 0 = [120 ] C: 0, 1, 1 - 1, 0, 0 = -1, 1, 1 = [ 111 ] D: 1, 0, ½ = 0, ½, 1 = 1, -½, -½ = [ 211 ] 3-52 Determine the indices for the directions in the cubic unit cell shown in Figure 3-38. Solution: A: 0, 0, 1 - 1, 0, 0 = -1, 0, 1 = [ 101 ] B: 1, 0, 1 - ½, 1, 0 = ½, -1, 1 = [122 ] C: 1, 0, 0 - 0, ¾, 1 = 1, -¾, -1 = [ 434 ] D: 0, 1, ½ = 0, 0, 0 = 0, 1, ½ = [021] 3-53 Indicate the planes (a) (100), (b) (134), and (c) (101) within a unit cell. 50 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 3-56 Determine the indices for the directions in the hexagonal lattice shown in Figure 3-41, using both the three-digit and four-digit systems. Solution: A: (1, -1, 0) – (0, 0, 0) = (1, -1, 0) = [110 ] h = ⅓(2 + 1) = 1 k = ⅓ (-2 - 1) = -1 i = -⅓ (1 - 1) = 0 l = 0 Therefore, [1100 ] B: (1, 1, 0) – (0, 0, 1) = (1, 1, -1) = [111 ] h = ⅓(2 - 1) = ⅓ k = ⅓(2 - 1) = ⅓ i = -⅓(1 + 1) = -⅔ l = -1 Therefore, [1123 ] C: (0, 1, 1) – (0, 0, 0) = (0, 1, 1) = [011] 51 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 h = ⅓(0 - 1) = -⅓ k = ⅓(2 - 0) = ⅔ i = -⅓(0 + 1) = -⅓ l = 1 Therefore, [ 1213 ] 3-57 Determine the indices for the directions in the hexagonal lattice shown in Figure 3-42, using both the three-digit and four-digit systems. Solution: A: (0, 1, 1) – (½, 1, 0) = (-½, 0, 1) = [ 102 ] h = ⅓(-2 - 0) = -⅔ k = ⅓(0 + 1) = ⅓ i = -⅓(-1 + 0) = ⅓ l = 2 Therefore, [ 2116 ] B: (1, 0, 0) – (1, 1, 1) = (0,-1, -1) = [ 011 ] h = ⅓(0 + 1) = ⅓ k = ⅓(-2 + 0) = -⅔ i = -⅓(0 - 1) = ⅓ 52 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 l = -1 [ 1213 ] C: (0, 0, 0) – (1, 0, 1) = (-1, 0, -1) = [ 101 ] h = ⅓(-2 + 0) = -⅔ k = ⅓(0 + 1) = ⅓ i = -⅓(-1 + 0) = ⅓ l = -1 [ 2113 ] 3-58 Determine the indices for the planes in the hexagonal lattice shown in Figure 3-43. Solution: A: a1 = 1; 1/a1 = 1 a2 = -1; 1/a2 = -1 a3 = ∞; 1/a3 = 0 c = 1; 1/c = 1 Therefore, ( 1101 ) with origin at a2 = 1. B: a1 = ∞; 1/a1 = 0 55 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: 3-61 Sketch the following planes and directions within a cubic unit cell. (a) [ 110 ] (b) [ 221 ] (c) [410] (d) [ 012 ] (e) [ 321 ] (f) [ 111 ] (g) ( 111 ) (h) ( 011 ) (i) (030) (j) ( 121 ) (k) ( 113 ) (l) ( 041 ) 56 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: 3-62 Sketch the following planes and directions within a hexagonal unit cell. (a) [ 0110 ] (b) [ 1120 ] (c) [ 1011 ] (d) (0003) (e) ( 1010 ) (f) ( 0111 ) Solution: 57 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 3-63 Sketch the following planes and directions within a hexagonal unit cell. (a) [ 2110 ] (b) [ 1121 ] (c) [ 1010 ] (d) ( 1210 ) (e) ( 1122 ) (f) ( 1230 ) Solution: 3-64 What are the indices of the six directions of the form <110> that lie in the ( 111 ) plane of a cubic cell? Solution: [ 110 ] [101] [011] [110 ] [ 101 ] [ 011 ] 3-65 What are the indices of the four directions of the form <111> that lie in the ( 101 ) plane of a cubic cell? Solution: [111] [ 111 ] [ 111 ] [ 111 ] 60 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: 𝜌 = 𝑁atoms = 2𝑟 = √2 = 0.707 〈100〉 Distance ( 4𝑟 ) 2 √2 𝜌 = 𝑁atoms = 4𝑟 = 1.000 〈110〉 Distance √2 ( 4𝑟 ) √2 𝜌〈111〉 = 𝑁atoms 2𝑟 = 0.408 Distance √3 ( 4𝑟 ) √2 So <110> is the most closely packed in a FCC unit cell. 3-71 Calculate and compare the planar densities for the {100}, {110} and {111} planes in a BCC unit cell. Which plane is the most close-packed (dense)? Solution: 1 𝜌 = (𝑁atoms in plane)(𝐴atom) = 4 (4) (𝜋𝑟2) = 0.59 {100} Area ( 4𝑟 ) 2 3 𝜌{110} = (𝑁atoms in plane)(𝐴atom) = Area √ 2𝜋𝑟2 4𝑟 2 = 0.83 √2 ( 3 ) 1 𝜌 = (𝑁atoms in plane)(𝐴atom) = (2) √ (𝜋𝑟2) = 0.34 {111} Area 1 (2) ( 2 4𝑟√2 ) √3 sin 60° So {110} is the most dense in a BCC unit cell. 3-72 Calculate and compare the linear densities for the {100}, {110} and {111} planes in an FCC unit cell. Which plane is the most densely packed? Solution: (𝑁atoms in plane)(𝐴atom) 2(𝜋𝑟2) 𝜌{100} = Area = 4𝑟 2 = 0.785 ( 2 ) 𝜌{110} = (𝑁atoms in plane)(𝐴atom) = Area √ 2𝜋𝑟2 4𝑟 2 = 0.555 √2 ( 3 ) (𝑁atoms in plane)(𝐴atom) √ 2𝜋𝑟2 𝜌{111} = Area = 1 (2) ( 2 4𝑟√2 ) √2 sin 60° = 0.907 61 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 So {111} is the most dense in a FCC unit cell. 3-73 Determine the repeat distance, linear density, and packing fraction for FCC nickel, which has a lattice parameter of 0.35167 nm, in the [100], [110], and [111] directions. Which of these directions is close packed? Solution: For [100]: 𝑟 = (√2)(0.35167 nm) = 0.1243 nm 4 Repeat distance = a0 = 0.35167 nm Linear density = 1/a0 = 2.84 nm-1 Linear packing fraction = (2)(0.1243 nm)(2.84 nm-1) = 0.707 For [110]: Repeat distance = 20.5a0/2 = 0.2487 nm Linear density = (2/20.5)a0 = 4.02 nm-1 Linear packing fraction = (2)(0.1243 nm)(4.02 nm-1) = 1.0 For [111]: Repeat distance = 30.5a0 = 0.6091 nm Linear density = 1/(30.5a0) = 1.642 nm-1 Linear packing fraction = (2)(0.1243 nm)(1.642 nm-1) = 0.408 Only the [110] is close packed; it has a linear packing fraction of 1. 62 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 3-74 Determine the repeat distance, linear density, and packing fraction for BCC lithium, which has a lattice parameter of 0.35089 nm, in the [100], [110], and [111] directions. Which of these directions is close packed? Solution: For [100]: 𝑟 = (√3)(0.35089 nm) = 0.1519 nm 4 Repeat distance = a0 = 0.35089 nm Linear density = 1/a0 = 2.85 nm-1 Linear packing fraction = (2)(0.1519)(2.85) = 0.866 For [110]: Repeat distance = 20.5a0 = 0.496 nm Linear density = 1/(20.5a0) = 2.015 nm-1 Linear packing fraction = (2)(0.1519 nm)(2.015 nm-1) = 0.612 For [111]: Repeat distance 30.5a0/2 = 0.3039 nm Linear density = 2/(30.5a0) ? 3.291 nm-1 Linear packing fraction = (2)(0.1519 nm)(3.291 nm-1) = 1 65 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 4 0 4 0 The (111) is close packed. 3-77 Determine the planar density and packing fraction for BCC lithium in the (100), (110), and (111) planes. Which, if any, of these planes is close packed? Solution: a0 = 3.5089 Å For (100): Planar density = 1 (3.5089 × 10−8 cm)2 = 0.0812 × 1016 cm−2 2 Packing fraction = 𝜋 ( √3 𝑎0) 𝑎2 = 0.589 For (110): Planar density = 2 (√2)(3.5089 × 10−8 cm)2 2𝜋 ( √3 𝑎0) = 0.1149 × 1016 cm−2 2 Packing fraction = (√2)(𝑎2) = 0.833 66 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 0 0 For (111): There are only (3)(⅙) = ½ points in the plane, which has an area of 0.866a 2 . 1 Planar density = ( )( 2 −8 cm)2 = 0.0469 × 1016 cm−2 0.866 3.5089 × 10 2 𝜋 ( √3 𝑎0) Packing fraction = 2 4 0.866𝑎2 = 0.34 There is no close-packed plane in BCC structures. 3-78 Suppose that FCC rhodium is produced as a 1-mm thick sheet, with the (111) plane parallel to the surface of the sheet. How many (111) interplanar spacings d111 thick is the sheet? See Appendix A for necessary data. Solution: 𝑑 = 𝑎0 = 3.796 Å = 2.1916 Å 111 √12 + 12 + 12 √3 thickness = 0.1 mm = 2.1916 × 10−8 cm 3-79 In an FCC unit cell, how many d111 are present between the (0,0,0) point and the (1,1,1) point? 4.563 × 106 × 𝑑111 67 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 𝑟 = 0.2798 Å Solution: The distance between the 0,0,0 and 1,1,1 points is √3𝑎0. The interplanar spacing is 𝑑 = 𝑎0 = 𝑎0 111 √12 + 12 + 12 √3 Therefore the number of interplanar spacings is 𝑁𝑑111 = √3𝑎0 = ( 𝑎0 ) √3 3-80 What are the stacking sequences in the FCC and HCP structures? Solution: For FCC, ABCABCABC… For HCP, ABABAB… 3-81 Determine the minimum radius of an atom that will just fit into (a) the tetrahedral interstitial site in FCC nickel; and (b) the octahedral interstitial site in BCC lithium. Solution: For the tetrahedral site in FCC nickel (a0 = 3.5167 Å): 𝑟Ni = √2 (3.5167 Å) = 1.2433 Å 4 ( 𝑟 ) = 0.225 𝑟Ni tetrahedral site 𝑟 = (1.2433 Å)(0.225) For the octahedral site in BCC lithium (a0 = 3.5089 Å): 3 70 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 𝑟K+1 𝑟Br−1 = 1.33 Å = 0.68; 1.96 Å 3-86 A particular unit cell is cubic with ions of type A located at the corners and face-centers of the unit cell and ions of type B located at the midpoint of each edge of the cube and at the body-centered position. The ions contribute to the unit cell in the usual way (1/8 ion contribution for each ion at the corners, etc.). (a) How many ions of each type are there per unit cell? (b) Describe this structure as a lattice and a basis. Check to be sure that the number of ions per unit cell given by your description of the structure as a lattice and a basis is consistent with your answer to part (a). (c) What is the coordination number of each ion? (d) What is the name commonly given to this crystal structure? Solution: For type A: 1 ion × 8 corner + 1 ion × 6 face center = 8 corner For type B: 2 face center 1 ion × 12 edge center + 1 body center = 4 edge center The FCC lattice and basis A (0, 0, 0) and one of the following four B bases: (½, 0, 0), (0, ½, 0), (0, 0, ½), (½, ½, ½). There are four lattice points per unit cell and two ions in the basis per lattice point for a total of eight ions per unit cell – which is consistent with part (a). CN = 6 for both A and B. Common name is sodium chloride. 3-87 Would you expect NiO to have the cesium chloride, sodium chloride, or zincblende structure? Based on your answer, determine (a) the lattice parameter; (b) the density; and (c) the packing factor. Solution: 𝑟Ni+2 = 0.69 Å = 0.52; CN = 6 𝑟O−2 1.32 Å A coordination number of 8 is expected for the CsCl structure, and a coordination number of 4 is expected for ZnS. But a coordination number of 6 is consistent with the NaCl structure. 𝑎0 = 2(0.69 Å) + 2(1.32 Å) = 4.02 Å 4 ion 4 ion CN = 6 71 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 mol mol (4 ion ) (58.17 g + 16 g ) 𝜌 = cell mol mol (4.02 × 10−8 cm)3 (6.022 × 1023 atom ) 𝜌 = 7.64 g cm3 4𝜋 ion 3 3 PF = ( 3 ) (4 cell) ([0.69 Å] (4.02 Å) 3 + [1.32 Å] ) 3-88 Would you expect UO2 to have the sodium chloride, zincblende, or fluorite structure? Based on your answer, determine (a) the lattice parameter; (b) the density; and (c) the packing factor. Solution: 𝑟U+4 = 0.97 Å = 0.735; CN = 8 𝑟O−2 1.32 Å The radius ratio predicts a coordination number of 8; however, there must be twice as many oxygen ions as uranium ions in order to balance the charge. The fluorite structure will satisfy these requirements, with U = FCC position (4) and O = tetrahedral position (8). √3𝑎0 = 4𝑟U+4 + 4𝑟O−2 = 4(0.97 Å + 1.32 Å) = 9.16 Å → (4) (238.03 g ) + (8) (16 g ) 𝜌 = mol mol = (5.2885 × 10−8 cm)3 (6.022 × 1023 atom ) 4𝜋 3 3 PF = ( 3 ) (4[0.97 Å] + 8[1.32 Å] ) = (5.2885 Å) 3 3-89 Would you expect BeO to have the sodium chloride, zincblende, or fluorite structure? Based on your answer, determine (a) the lattice parameter; (b) the density; and (c) the packing factor. Solution: 𝑟Be+2 = 0.35 Å = 0.265; CN = 4 𝑟O−2 1.32 Å This coordination number signifies a zincblende structure. √3𝑎0 = 4𝑟Be+2 + 4𝑟O−2 = 4(0.35 Å + 1.32 Å) = 6.68 Å → 𝑎0 = 3.8567 Å 0.624 𝑎0 = 5.2885 Å PF = 0.678 12.13 g cm3 72 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 mol mol Zn S (4) (9.01 g + 16 g ) 𝜌 = mol mol = (3.8567 × 10−8 cm)3 (6.022 × 1023 atom ) 4𝜋 3 PF = ( 3 ) 4 ([0.35 Å] + 8[1.32 Å] 3 ) = (3.8567 Å) 3 3-90 Would you expect CsBr to have the sodium chloride, zincblende, fluorite, or cesium chloride structure? Based on your answer, determine (a) the lattice parameter; (b) the density; and (c) the packing factor. Solution: 𝑟Cs+1 = 1.67 Å = 0.852; CN = 8 𝑟Br−1 1.96 Å This coordination number signifies a cesium chloride structure. √3𝑎0 = 2𝑟Cs+1 + 2𝑟Br−1 = 2(1.67 Å + 1.96 Å) = 7.26 Å → (79.904 g + 132.91 g ) 𝜌 = mol mol = (4.1916 × 10−8 cm)3 (6.022 × 1023 atom ) 4𝜋 3 3 PF = ( 3 ) ([1.96 Å] + [1.67 Å] ) = (4.1916 Å) 3 3-91 Sketch the ion arrangement on the (110) plane of ZnS (with the zincblende structure) and compare this arrangement to that on the (110) plane of CaF2 (with the flourite structure). Compare the planar packing fraction on the (110) planes for these two materials. Solution: ZnS: √3𝑎0 = 4𝑟Zn+2 + 4𝑟S−2 = 4(0.074 nm + 0.184 nm) → 𝑎0 = 0.596 nm (2)(𝜋𝑟2 ) + (2)(𝜋𝑟2) PPF = = (√2𝑎0)𝑎0 2𝜋(0.074 nm)2 + 2𝜋(0.184 nm)2 √2(0.596 nm)2 = 0.492 0.693 𝑎0 = 4.1916 Å 0.684 2.897 g cm3 4.8 g cm3 75 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 3-98 Figure 3-45 shows the results of an x-ray diffraction experiment in the form of the intensity of the diffracted peak versus the 2θ diffraction angle. If x-rays with a wavelength of 0.15418 nm are used, determine (a) the crystal structure of the metal; (b) the indices of the planes that produce each of the peaks; and (c) the lattice parameter of the metal. Solution: The 2θ values can be estimated from the figure: 2θ sin2 θ sin2 θ/0.0077 Planar indicies d = λ/(2 sin θ) a0 = (h2 + k2 +l2)0.5 2 20.5 0.032 4 (200) 0.4332 0.8665 4 33.5 0.083 11 (311) 0.2675 0.8872 6 41.5 0.123 16 (400) 0.2201 0.8805 8 46.5 0.156 20 (420) 0.1953 0.8734 The sin2 θ values must be divided by 0.077 (one third the first sin2 θ value) in order to produce a possible sequence of numbers). The 3, 4, 8, 11, … sequence means that the material is FCC The average a0 = 0.8781 nm. 3-99 Figure 3–46 shows the results of an x-ray diffraction experiment in the form of the intensity of the diffracted peak versus the 2θ diffraction angle. If x-rays with a wavelength of 0.07107 nm are used, determine (a) the crystal structure of the metal; (b) the indices of the planes that produce each of the peaks; and (c) the lattice parameter of the metal. 7 45.5 0.146 19 (331) 0.2014 0.8781 5 35.5 0.093 12 (222) 0.2529 0.8760 3 28.5 0.061 8 (220) 0.3132 0.8858 1 17.5 0.023 3 (111) 0.5068 0.8777 76 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: The 2θ values can be estimated from the figure: 2θ sin2 θ sin2 θ/0.0077 Planar indicies d = λ/(2 sin θ) a0 = (h2 + k2 +l2)0.5 2 36.5 0.095 2 (200) 0.116 0.2320 4 51.5 0.189 4 (220) 0.0825 0.2334 6 64.5 0.285 6 (222) 0.0672 0.2327 8 75.5 0.375 8 (400) 0.0586 0.2342 The sequence 1, 2, 3, 4, 5, 6, 7, 8 (which includes the “7”) means that the material is BCC. The average a0 = 0.2327 nm. 3-100 A sample of zirconia contains cubic and monoclinic polymorphs. What will be a good analytical technique to detect the presence of these two different polymorphs? Solution: Transmission electron microscopy would work for this, if the budget and other considerations allow for it. 7 70.5 0.329 7 (321) 0.0625 0.2339 5 58.5 0.235 5 (310) 0.0739 0.2338 3 44.5 0.143 3 (211) 0.0947 0.2319 1 25.5 0.049 1 (110) 0.1624 0.2297 77 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Chapter 4: Imperfections in the Atomic and Ionic Arrangements 4-1 Gold has 5.82 × 108 vacancies/cm3 at equilibrium at 300 K. What fraction of the atomic sites is vacant at 600 K? Solution: The number of vacancies per cm3 nv is given by  Qv   nv  n exp  RT  ,   where n is the number of atoms per cm3, Qv is the energy required to produce one mole of vacancies, R is the gas constant, and T is the absolute temperature. Gold has a density of 19.302 g/cm3 and an atomic mass of 196.97 g/mol. Thus the number of atoms per cm3 is 19.302 g     1 mol  6.0221023 atoms       22 3 cm3 196.97 g 1 mol 5.90 10 atoms/cm Solving for Qv, the energy is given by    n   5.82108   Qv RT ln  v   8.314 J/(mol  K)  (300 K) ln    n   5.901022   80,438 J/mol. The fraction of the vacant atomic sites is given by nv/n. At 600 K, nv  exp   Qv   exp   80,438 J/mol   9.93108 . n  RT   8.314 J/(mol  K)  600 K        4-2 Calculate the number of vacancies per m3 for gold at 900 °C. The energy for vacancy formation is 0.86 eV/atom. 80 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024   1 vacancy  125 cells [(1/11.335 g/cm3 )]  5.821018vacancies/g   (4.949108cm)3    4-9 Cu and Ni form a substitutional solid solution. This means that the crystal structure of a Cu-Ni alloy consists of Ni atoms substituting for Cu atoms in the regular atomic positions of the FCC structure. For a Cu–30% wt% Ni alloy, what fraction of the atomic sites does Ni occupy? Solution: In 100 g of a Cu–30% wt.% Ni alloy, Cu comprises 70 g, and Ni comprises 30 g. In 70 g of Cu, there are 1 mol 6.022 1023atoms 23 70 g Cu    6.6310 Cu atoms. 63.54 g 1 mol In 30 g of Ni, there are 1 mol 6.0221023 atoms 23 30 g Cu    3.0810 Ni atoms. 58.71 g 1 mol Therefore, the Ni atoms occupy 3.081023 Ni atoms 3.081023 Ni atoms  6.631023 Cu atoms or 32% of the atomic sites.  0.32 4-10 Au and Ag form a substitutional solid solution. This means that the crystal structure of a Au-Ag alloy consists of Ag atoms substituting for Au atoms in the regular atomic positions of the FCC structure. For a Au-50 at% Ag alloy, what is the wt% Ag in the alloy? Solution: In 1 mol of a Au–50 at.% Ag alloy, Au comprises 0.5 mol, and Ag comprises 0.5 mol. 0.5 mol Au  196.97 g Au  98.485 g Au. 1 mol 0.5 mol Ag  107.868 g Au  53.934 g Ag. 1 mol Thus the wt% Ag is 53.934 g 53.934 g +98.485 g  35.4 wt% Ag 4-11 A niobium alloy is produced by introducing tungsten substitutional atoms into the BCC structure; eventually an alloy is produced that has a lattice parameter of 0.32554 nm and a density of 11.95 g/cm3. Calculate the fraction of the atoms in the alloy that are tungsten. Solution: 11.95 g/cm 3  (xw ) (183.85 g/mol)  (2  xw ) (92.91 g/mol) (3.2554108cm)3 (6.0221023 atoms/mol) 248.186 = 183.85xW + 185.82 – 92.91xW 81 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 90.94xW  62.366 or xW  0.687 W atoms/cell There are 2 atoms per cell in BCC metals. Thus: fw = 0.687/2 = 0.344 4-12 Tin atoms are introduced into an FCC copper crystal, producing an alloy with a lattice parameter of 3.7589 × 10–8 cm and a density of 8.772 g/cm3. Calculate the atomic percentage of tin present in the alloy. Solution: 8.772 g/cm 3  (xSn ) (118.69 g/mol)  (4  xSn ) (63.54 g/mol) (3.7589108cm)3 (6.021023 atoms/mol) 280.5  55.15xSn  254.16 or xSn  0.478 Sn atoms/cell There are 4 atoms per cell in FCC metals; therefore the at% Sn is (0.478/4) = 11.95% 4-13 We replace 7.5 atomic percent of the chromium atoms in its BCC crystal with tantalum. X-ray diffraction shows that the lattice parameter is 0.29158 nm. Calculate the density of the alloy. Solution:   (2) (0.925) (51.996 g/mol)  2(0.075) (180.95 g/mol)  8.262 g/cm 3 (2.9158108cm)3(6.0221023atoms/mol) 4-14 Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For this steel, find the density and the packing factor. Solution: There is one carbon atom per 100 iron atoms, or 1 C/50 unit cells, or 1/50 C per unit cell: (2) (55.847 g/mol)  (1/50)(12 g/mol)  (2.867 108cm)3(6.021023atoms/mol)  7.89 g/cm3 2(4 /3) (1.241)3  (1/50)(4 /3) (0.77)3 Packing Factor   0.681 (2.867)3 4-15 The density of BCC iron is 7.882 g/cm3, and the lattice parameter is 0.2866 nm when hydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms; and (b) the number of unit cells on average that contain hydrogen atoms. Solution: (a) 7.882 g/cm3  2(55.847 g/mol)  x(1.00797 g/mol) (2.866108cm)3 (6.0221023atoms/mol) x = 0.0449 H atoms/cell The total atoms per cell include 2 Fe atoms and 0.0449 H atoms. Thus, f  0.0449  0.02195 H 2.0449 (b) Since there is 0.0449 H/cell, then the number of cells containing H atoms is  82 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 cells  1 / 0.0449  22.3 or 1 H in 22.3 cells 4-16 Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate (a) the number of anion vacancies per cm3; and (b) the density of the ceramic. Solution: In 10 unit cells, we expect 40 Mg + 40 O ions, but due to the defect: 40 Mg – 1 = 39 40 O – 1 = 39 1 vacancy/(10 cells)(3.96 × 10–8 cm)3 = 1.61 × 1021 vacancies/cm3   (39/40) (4) (24.312 g/mol)  (39/40) (4) (16 g/mol)  4.205 g/cm 3 (3.96108cm)3(6.021023atoms/mol) 4-17 ZnS has the zinc blende structure. If the density is 3.02 g/cm3 and the lattice parameter is 0.59583 nm, determine the number of Schottky defects (a) per unit cell; and (b) per cubic centimeter. Solution: Let x be the number of each type of ion in the unit cell. There normally are 4 of each type. (a) 3.02 g/cm3  x(65.38 g/mol)  x(32.064 g/mol) (5.9583108cm)3(6.0221023ions/mol) x  3.9478 4 – 3.9478 = 0.0522 defects/unit cell (b) # of unit cells/cm3 = 1/(5.9583 × 10–8 cm)3 = 4.728 × 1021 Schottky defects per cm3 = (4.728 × 1021)(0.0522) = 2.466 × 1020 4-18 Suppose we introduce the following point defects. (a) Mg2+ ions substitute for yttrium atoms in Y2O3; (b) Fe3+ ions substitute for magnesium ions in MgO; (c) Li1+ ions substitute for magnesium ions in MgO; and (d) Fe2+ ions replace sodium ions in NaCl. What other changes in each structure might be necessary to maintain a charge balance? Explain. Solution: (a) Remove 2 Y3+ and add 3 Mg2+ – create cation interstitial. (b) Remove 3 Mg2+ and add 2 Fe3+ – create cation vacancy. (c) Remove 1 Mg2+ and add 2 Li+ – create cation interstitial. (d) Remove 2 Na+ and add 1 Fe2+ – create cation vacancy. 4-19 Write down the defect chemistry equation for introduction of SrTiO3 in BaTiO3 using the Kröger-Vink notation. Solution: 4-20 Do amorphous and crystalline materials plastically deform by the same mechanisms? Explain. 85 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 4-29 Determine the interplanar spacing and the length of the Burgers vector for slip on the (110)[111] slip system in BCC tantalum. Repeat, assuming that the slip system is a (111)/[1 10] system. What is the ratio between the shear stresses required for slip for the two systems? Assume that k = 2 in Equation 4-2. Solution: (a) For (110)/[1 11]: b  (½) ( 3) (3.3026 Å)  2.860 Å (b) If (111)/[1 10] , then: d110  3.3026 Å  2.335 Å b  2(3.3026 Å)  4.671 Å d111  3.3026 Å  1.907 Å (c) If we assume that k = 2 in Equation 4-2, then: (d /b)a  2.335  0.8166 (d /b) 2.86 b  1.907  0.408 4.671 a  exp(2(0.8166))  0.44 b exp(2(0.408)) 4-30 The crystal shown in Figure 4-19 contains two dislocations A and B. If a shear stress is applied to the crystal as shown, what will happen to dislocations A and B? Solution: Under the action of the applied shear stress shown, dislocation A will move to the right, and dislocation B will move to the left. When the dislocations meet, they will annihilate because the combination of a negative edge dislocation and a positive edge dislocation will form perfect crystal. 4-31 Can ceramic and polymeric materials contain dislocations? Solution: Yes. All crystalline materials can contain dislocations. 12 12  02 12 12 12 86 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 4-32 Why is it that ceramic materials are brittle? Solution: Dislocations do not move easily enough because of bonding strengths, so the materials fail due to flaws such as cracks and pores before any slip can occur. 4-33 What is meant by the terms plastic and elastic deformation? Solution: Plastic deformation describes an irreversible change to the shape of an object when a force is applied. Elastic deformation is a temporary change in shape that is recovered when the force is removed. 4-34 Why is the theoretical strength of metals much higher than that observed experimentally? Solution: Slip allows metallic bonds to be broken individually rather than requiring all bonds in a sample to be broken at once, as is predicted by simply counting metallic bond strengths. 4-35 How many grams of aluminum, with a dislocation density of 1010 cm/cm3, are required to give a total dislocation length that would stretch from New York City to Los Angeles (3000 miles)? Solution: (3000 mi)(5280 ft/mi)(12 in./ft)(2.54 cm/in.) = 4.828 × 108 cm (4.828108cm) (2.699 g/cm3)    (1010cm/cm3) 0.13 g 4-36 The distance from Earth to the Moon is 240,000 miles. If this were the total length of dislocation in a cubic centimeter of material, what would be the dislocation density? Compare your answer to typical dislocation densities for metals. Solution: (240,000 mi)(5280 ft/mi)(12 in./ft)(2.54 cm/in.) = 3.86 × 1010 cm/cm3 This is reasonable as dislocation densities range from 106 cm/cm3 to 1012 cm/cm3. 4-37 Why would metals behave as brittle materials without dislocations? Solution: The dislocations allow slip, which in turn allows Without dislocations, the metal would behave as a ceramic and fail completely when its ultimate strength was reached. 4-38 Why is it that dislocations play an important role in controlling the mechanical properties of metallic materials, however, they do not play a role in determining the mechanical properties of glasses? Solution: Glasses do not contain dislocations. Glasses are amorphous, and therefore, they do not have defects such as dislocations. 87 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 4-39 Suppose you would like to introduce an interstitial or large substitutional atom into the crystal near a dislocation. Would the atom fit more easily above or below the dislocation line shown in Figure 4-7(c)? Explain. Solution: The atom would fit more easily into the area just below the dislocation due to the atoms being pulled apart; this allows more space into which the atom can fit. 4-40 Compare the c/a ratios for the following HCP metals, determine the likely slip processes in each, and estimate the approximate critical resolved shear stress. Explain. (See data in Appendix A.) (a) Zinc; (b) Magnesium; (c) Titanium; (d) Zirconium; (e) Rhenium; and (f) Beryllium. Solution: We expect metals with c/a > 1.633 to have a low τcrss: (a) Zn: 4.9470  1.856  low  crss 2.6648 (b) Mg: 5.209 3.2087  1.62  medium  crss (c) (d) (e) (f) Ti: 4.6831  1.587  high  crss 2.9503 Zr: 5.1477  1.593  high  crss 3.2312 Rh: 4.458  1.615  medium  crss 2.760 Be: 3.5842  1.568  high  crss 2.2858 90 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 6.7  104 3.15  105 5 0 0 4-46 Explain why hexagonal close-packed metals tend to have a limited ability to be strain hardened. Solution: HCP metals with c/a ratios > 1.633 have limited ability to strain harden because of the limited number of parallel slip systems these metals possess. Since there are only three parallel slip systems, the opportunities for dislocations to interact and become tangled are low. In addition, since these metallic systems do not have the ability to cross-slip, ductility is also limited. 4-47 Why is it that cross slip in BCC and FCC metals is easier than in HCP metals? How does this influence the ductility of BCC, FCC, and HCP metals? Solution: Cross slip is easier in BCC and FCC metals than in HCP metals because there are multiple non-parallel and intersecting slip systems in the BCC and FCC structures. This makes it possible for dislocations to move from one slip plane to another. All of the favored slip systems in HCP metals are parallel to each other making cross slip difficult. 4-48 Arrange the following metals in the expected order of increasing ductility: Cu, Ti, and Fe. Explain. Solution: From least ductile to most ductile, the order is Ti, Fe, and Cu. Ti is HCP and thus will be the most brittle due to the limited number of slip systems. BCC metals (such as Fe) tend to have higher strengths and lower ductilities than FCC metals (such as Cu) due to their higher critical resolved shear strengths. 4-49 What are the imperfections in the atomic arrangements that have a significant effect on the material behavior? Give an example of each. Solution: Imperfections include point (vacancies), line (dislocations) and surface defects (grain boundaries). 4-50 The strength of titanium is found to be 65,000 psi when the grain size is 6.7 104 in. and 82,000 psi when the grain size is 3.15 10 in. Determine (a) the constants in the Hall-Petch equation; and (b) the strength of the titanium when the grain size is reduced to 8.00 10–6 in. . Solution: 65,000    K 1    38.633 K 82,000    K 1    178.17 K (a) By solving the two simultaneous equations: 0 0 91 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 o K  121.8 psi (b)   60,293121.8/   60,293 psi  103,400 psi  103.4 ksi 4-51 A copper-zinc alloy has the properties shown in the table below: Grain diameter (mm) Strength (MPa) d–½ 0.015 170 MPa 8.165 0.025 158 MPa 6.325 0.035 151 MPa 5.345 0.050 145 MPa 4.472 Determine (a) the constants in the Hall-Petch equation; and (b) the grain size required to obtain a strength of 200 MPa. Solution: The values of d–½ are included in the table; the graph shows the relationship. We can determine K and σo either from the graph or by using two of the data points. 170   o  K(8.165) (a) 145   o  K(4.472) 25  3.693K K  6.77 MPa/  o  114.7 MPa (b) To obtain a strength of 200 MPa: 200  114.7  6.77/ 85.3  6.77/ d = 0.0063 mm 4-52 If there were 50 grains per in2 in a photograph of a metal taken at 100× magnification, calculate the ASTM grain size number (n). Solution: N = 2 (n-1) 50 = 2(n-1) in. 8.00 106 mm d d 92 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 ln 50 = (n–1)*ln 2 3.912 = (n–1) (0.6931) 4.605 = 0.6931n n = 6.64 4-53 If the area of a photograph measured 7.812 in2 and 23 grains were documented, what would the ASTM grain size number (n) be? Solution: 4-54 You have the choice to either purchase a copper alloy that has an ASTM grain size of 5 or ASTM grain size of 8. You can’t decide if there is much of a difference between these two. Determine how many grains/in2 would appear on a photograph taken at 100× for a metal given these two ASTM grain size choices. Does this seem like a significant difference? Solution: This is a very significant difference because there is are 8 times more grains per unit area with an ASTM grain size of 8 compared to an ASTM grain size of 5. 95 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 Solution: sin(0.5/2)  ½( 3)(2.866) 2D 0.004364 = 1.241/D D = 284 Å 96 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 97 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 1: Introduction to Materials Science and Engineering A Comprehensive Examination Study Guide Latest Updated 2024 5 10 9 Chapter 5: Atom and Ion Movements in Materials 5-10 Atoms are found to move from one lattice position to another at the rate of 5 × 105 jumps/s at 400°C when the activation energy for their movement is 30,000 cal/mol. Calculate the jump rate at 750°C. Solution: 5105 c exp[  30, 000/(1.987)/(673)] Rate   o  exp(  22.434 14.759) x co exp[  30, 000/(1.987)/(1023)] 5105   x exp(  7.675)  4.64104  5 x  4.6410 4  1.0810 jumps/s 5-11 The number of vacancies in a material is related to temperature by an Arrhenius equation. If the fraction of lattice points containing vacancies is 8 × 10–5 at 600°C, determine the fraction of lattice points containing vacancies at 1000°C. Solution: 8 × 10–5 = exp[–Q/(1.987)/(873)] Q = 16,364 cal/mol f = nv/n = exp[–16,364/(1.987)/(1273)] = 0.00155 5-12 The Arrhenius equation was originally developed for comparing rates of chemical reactions. Compare the rates of a chemical reaction at 20 and 100°C by calculating the ratio of the chemical reaction rates. Assume that the activation energy for liquids in which the chemical reaction is conducted is 10 kJ/mol and that the reaction is limited by diffusion. Solution: An Arrhenius relationship gives the rate of a process or reaction as a function of temperature. In general, for an Arrhenius relationship, Rate  c exp   Q   0  RT  where c0 is a constant, Q is the activation energy, R is the universal gas constant, and T is the absolute temperature. Taking the ratio of the rates of two reactions:  Q   10 103 J/mol   c0 exp R 293 K  exp 8.314 J/ mol K293 K   Rate20˚C          0.41 Rate100˚C c exp Q   10 103 J/mol  0  R 373 K xp       8.314 J/ mol  K373 K  5-14 (a) Compare interstitial and vacancy atomic mechanisms for diffusion and (b) cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion. ,