Exam (elaborations) TEST BANK FOR The Science & Engineering of Materials By Donald R. Askeland, Pradeep P. Fulay (Solution Manual)

Exam (elaborations) TEST BANK FOR The Science & Engineering of Materials By Donald R. Askeland, Pradeep P. Fulay (Solution Manual) 1 Introduction to Materials Science and Engineering 1–4 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. 1–5 We would like to produce a transparent canopy for an aircraft. If we were to use a ceramic (that is, traditional window glass) canopy, rocks or birds might cause it to shatter. Design a material that would minimize damage or at least keep the canopy from breaking into pieces. Solution: We might sandwich a thin sheet of a transparent polymer between two layers of the glass. This approach, used for windshields of automobiles, will prevent the “safety” glass from completely disintegrating when it 01 Askeland Chap 9/27/05 1:48 PM Page 1 fails, with the polymer holding the broken pieces of glass together until the canopy can be replaced. Another approach might be to use a transparent, “glassy” polymer material such as polycarbonate. Some polymers have reasonably good impact properties and may resist failure. The polymers can also be toughened to resist impact by introducing tiny globules of a rubber, or elastomer, into the polymer; these globules improve the energy-absorbing ability of the composite polymer, while being too small to interfere with the optical properties of the material. 1–6 Coiled springs ought to be very strong and stiff. 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 assure 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 to fail even for relatively low forces. The silicon nitride is NOT recommended. 1–7 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 modulus of elasticity so that no permanent deformation of the material occurs. 1–8 You would like to design an aircraft that can be flown by human power nonstop for a distance of 30 km. What types of material properties would you recommend? What materials might be appropriate? Solution: Such an aircraft must possess enough strength and stiffness to resist its own weight, the weight of the human “power source”, and any aerodynamic forces imposed on it. On the other hand, it must be as light as possible to assure that the human can generate enough work to operate the aircraft. Composite materials, particularly those based on a polymer matrix, might comprise the bulk of the aircraft. The polymers have a light weight (with densities of less than half that of aluminum) and can be strengthened by introducing strong, stiff fibers made of glass, carbon, or other polymers. Composites having the strength and stiffness 2 The Science and Engineering of Materials Instructor’s Solutions Manual 01 Askeland Chap 9/27/05 1:48 PM Page 2 of steel, but with only a fraction of the weight, can be produced in this manner. 1–9 You would like to place a three-foot diameter microsatellite into orbit. The satellite will contain delicate electronic equipment that will send and receive radio signals from earth. Design the outer shell within which the electronic equipment is contained. What properties will be required, and what kind of materials might be considered? Solution: The shell of the microsatellite must satisfy several criteria. The material should have a low density, minimizing the satellite weight so that it can be lifted economically into its orbit; the material must be strong, hard, and impact resistant in order to assure that any “space dust” that might strike the satellite does not penetrate and damage the electronic equipment; the material must be transparent to the radio signals that provide communication between the satellite and earth; and the material must provide some thermal insulation to assure that solar heating does not damage the electronics. One approach might be to use a composite shell of several materials. The outside surface might be a very thin reflective metal coating that would help reflect solar heat. 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. 1–10 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 metalworking 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. Yet these portions must also possess some impact resistance, particularly so that chips do not flake off the striking face and cause injuries. 1–11 The hull of the space shuttle consists of ceramic tiles bonded to an aluminum skin. Discuss the design requirements of the shuttle hull that led to the use of this combination of materials. What problems in producing the hull might the designers and manufacturers have faced? Solution: The space shuttle experiences extreme temperatures during re-entry into earth’s atmosphere; consequently a thermal protection system must be used to prevent damage to the structure of the shuttle (not to mention its contents!). The skin must therefore be composed of a material that has an exceptionally low thermal conductivity. The material must be capable of being firmly attached to the skin of the shuttle and to be easily repaired when damage occurs. The tiles used on the space shuttle are composed of silica fibers bonded together to produce a very low density ceramic. The thermal conductivity is so low that a person can hold on to one side of the tile while the opposite surface is red hot. The tiles are attached to the shuttle CHAPTER 1 Introduction to Materials Science and Engineering 3 01 Askeland Chap 9/27/05 1:48 PM Page 3 skin using a rubbery polymer that helps assure that the forces do not break the tile loose, which would then expose the underlying skin to high temperatures. 1–12 You would like to select a material for the electrical contacts in an electrical switching device which 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 assure 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. However, 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. 1–13 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. 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. 1–14 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 4 The Science and Engineering of Materials Instructor’s Solutions Manual 01 Askeland Chap 9/27/05 1:48 PM Page 4 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). 1–15 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, aluminum has a particularly high electrical conductivity. 1–16 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? 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. However there may be problems 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 assure 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. CHAPTER 1 Introduction to Materials Science and Engineering 5 01 Askeland Chap 9/27/05 1:48 PM Page 5 01 Askeland Chap 9/27/05 1:48 PM Page 6 7 2 Atomic Structure 2–6 (a) Aluminum foil used for storing food weighs about 0.3 g per square inch. How many atoms of aluminum are contained in one square inch of foil? Solution: In a one square inch sample: number = (0.3 g)(6.02 × 1023 atoms/mol) = 6.69 × 1021 atoms 26.981 g/mol (b) Using the densities and atomic weights given in Appendix A, calculate and compare the number of atoms per cubic centimeter in (i) lead and (ii) lithium. Solution: (i) In lead: (11.36 g/cm3)(1 cm3)(6.02 × 1023 atoms/mol) = 3.3 × 1022 atoms/cm3 207.19 g/mol (ii) In lithium: (0.534 g/cm3)(1 cm3)(6.02 × 1023 atoms/mol) = 4.63 × 1022 atoms/cm3 6.94 g/mol 2–7 (a) Using data in Appendix A, calculate the number of iron atoms in one ton (2000 pounds). Solution: (2000 lb)(454 g/lb)(6.02 × 1023 atoms/mol) = 9.79 × 1027 atoms/ton 55.847 g/mol (b) Using data in Appendix A, calculate the volume in cubic centimeters occupied by one mole of boron. Solution: (1 mol)(10.81 g/mol) = 4.7 cm3 2.3 g/cm3 02 Askeland Chap 9/27/05 1:49 PM Page 7 2–8 In order to plate a steel part having a surface area of 200 in.2 with a 0.002 in. thick layer of nickel, (a) how many atoms of nickel are required and (b) how many moles of nickel are required? Solution: Volume = (200 in.2)(0.002 in.)(2.54 cm/in.)3 = 6.555 cm3 (a) (6.555 cm3)(8.902 g/cm3)(6.02 × 1023 atoms/mol) = 5.98 × 1023 atoms 58.71 g/mol (b) (6.555 cm3)(8.902 g/cm3) = 0.994 mol Ni required 58.71 g/mol 2–9 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: 1s2 2s22p63s23p63dx4s2 (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. 2–11 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 1.8. 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. 2–12 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 energy, (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. Solution: Ti – 1668 Zr – 1852 Hf – 2227 V – 1900 Nb – 2468 Ta – 2996 Cr – 1875 Mo – 2610 W – 3410 Mn – 1244 Tc – 2200 Re – 3180 Fe – 1538 Ru – 2310 Os – 2700 Co – 1495 Rh – 1963 Ir – 2447 Ni – 1453 Pd – 1552 Pt – 1769 200 in2 0.002 in 8 The Science and Engineering of Materials Instructor’s Solutions Manual 02 Askeland Chap 9/27/05 1:49 PM Page 8 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. 2–13 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: T(oC) Li – 180.7 Na – 97.8 K – 63.2 Rb – 38.9 Cs – 28.6 As the atomic number increases, the melting temperature decreases, opposite that found in Problem 2–12. Atomic Number 20 40 60 80 100 120 140 160 180 200 Li Na K Rb Cs Melting Temperature (Celcius) Atomic Number 3500 3000 2500 2000 1500 1000 Ti – Ni Zr – Pd Hf – Pt Melting Temperature (Celcius) CHAPTER 2 Atomic Structure 9 02 Askeland Chap 9/27/05 1:49 PM Page 9 2–14 Calculate the fraction of bonding of MgO that is ionic. Solution: EMg = 1.2 EO = 3.5 fcovalent = exp[(−0.25)(3.5 − 1.2)2] = exp(−1.3225) = 0.266 fionic = 1 − 0.266 = 0.734 ∴ bonding is mostly ionic 2–18 Beryllium and magnesium, both in the 2A column of the periodic table, are lightweight metals. Which would you expect to have the higher modulus of elasticity? Explain, considering binding energy and atom radii and using appropriate sketches of force versus interatomic spacing. Solution: 4 Be 1s22s2 E = 42 × 106 psi rBe = 1.143 Å 12 Mg 1s22s22p63s2 E = 6 × 106 psi rMg = 1.604 Å The smaller Be electrons are held closer to the core ∴ held more tightly, giving a higher binding energy. 2–19 Would you expect MgO or magnesium to have the higher modulus of elasticity? Explain. Solution: MgO has ionic bonds, which are strong compared to the metallic bonds in Mg. A higher force will be required to cause the same separation between the ions in MgO compared to the atoms in Mg. Therefore, MgO should have the higher modulus of elasticity. In Mg, E ≈ 6 × 106 psi; in MgO, E = 30 × 106 psi. 2–20 Aluminum and silicon are side-by-side in the periodic table. Which would you expect to have the higher modulus of elasticity (E)? Explain. Solution: Silicon has covalent bonds; aluminum has metallic bonds. Therefore, Si should have a higher modulus of elasticity. 2–21 Steel is coated with a thin layer of ceramic to help protect against corrosion. What do you expect to happen to the coating when the temperature of the steel is increased significantly? Explain. Solution: Ceramics are expected to have a low coefficient of thermal expansion due to strong ionic/covalent bonds; steel has a high thermal expansion coefficient. When the structure heats, steel expands more than the coating, which may crack and expose the underlying steel to corrosion. Force EBe ~ Δf /Δa Be Mg EMg ~ Δf /Δa 2rBe distance “a” 2rmg 10 The Science and Engineering of Materials Instructor’s Solutions Manual 02 Askeland Chap 9/27/05 1:49 PM Page 10 11 3 Atomic and Ionic Arrangements 3–13 Calculate the atomic radius in cm for the following: (a) BCC metal with a0 = 0.3294 nm and one atom per lattice point; and (b) FCC metal with a0 = 4.0862 Å and one atom per lattice point. Solution: (a) For BCC metals, (b) For FCC metals, 3–14 Determine the crystal structure for the following: (a) a metal with a0 = 4.9489 Å, r = 1.75 Å and one atom per lattice point; and (b) a metal with a0 = 0.42906 nm, r = 0.1858 nm and one atom per lattice point. Solution: We want to determine if “x” in the calculations below equals (for FCC) or (for BCC): (a) (x)(4.9489 Å) = (4)(1.75 Å) x = , therefore FCC (b) (x)(0.42906 nm) = (4)(0.1858 nm) x = , therefore BCC 3–15 The density of potassium, which has the BCC structure and one atom per lattice point, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate (a) the lattice parameter; and (b) the atomic radius of potassium. 3 2 3 2 r a = 2 = 2 ( .0862 Å) 4 = . 447 Å = . 447 10 8 cm ( ) ( ) × − 0 4 4 1 4 1 4 r a = 3 = 3 ( . nm) 4 = . 426 nm = .426 10 8 cm ( ) ( ) × − 0 4 0 3294 0 1 1 03 Askeland Chap 9/27/05 11:34 AM Page 11 Solution: (a) Using Equation 3–5: 0.855 g/cm3 = (2 atoms/cell)(39.09 g/mol) (a0)3(6.02 × 1023 atoms/mol) a0 3 = 1.5189 × 10−22 cm3 or a0 = 5.3355 × 10−8 cm (b) From the relationship between atomic radius and lattice parameter: 3–16 The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) the lattice parameter and (b) the atomic radius of thorium. Solution: (a) From Equation 3–5: 11.72 g/cm3 = (4 atoms/cell)(232 g/mol) (a0)3(6.02 × 1023 atoms/mol) a0 3 = 1. × 10−22 cm3 or a0 = 5.0856 × 10−8 cm (b) From the relationship between atomic radius and lattice parameter: 3–17 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: 2.6 g/cm3 = (x atoms/cell)(87.62 g/mol) (6.0849 × 10−8 cm)3(6.02 × 1023 atoms/mol) x = 4, therefore FCC 3–18 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: 1.892 g/cm3 = (x atoms/cell)(132.91 g/mol) (6.13 × 10−8 cm)3(6.02 × 1023 atoms/mol) x = 2, therefore BCC 3–19 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: 7.286 g/cm3 = (x atoms/cell)(114.82 g/mol) (3.2517 × 10−8 cm)2(4.9459 × 10−8 cm)(6.02 × 1023 atoms/mol) x = 2, therefore BCT (body-centered tetragonal) r = × = × − ( )( . ) − . 4 1 7980 10 8 cm 8 cm r = × = × − ( )( . ) − . 4 1 7980 10 8 cm 8 cm 12 The Science and Engineering of Materials Instructor’s Solutions Manual 2.3103 ×10−8 cm 3 3355 03 Askeland Chap 9/27/05 11:34 AM Page 12 3–20 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: (a) The volume of the unit cell is V = a0 2c0cos30. V = (0.4546 nm)2(1.186 nm)(cos30) = 0.21226 nm3 = 2.1226 × 10−22 cm3 (b) If “x” is the number of atoms per unit cell, then: 9.808 g/cm3 = (x atoms/cell)(208.98 g/mol) (2.1226 × 10−22 cm3)(6.02 × 1023 atoms/mol) x = 6 atoms/cell 3–21 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 or V = (0.45258 nm)(0.45186 nm)(0.76570 nm) = 0.1566 nm3 = 1.566 × 10−22 cm3 (a) From the density equation: 5.904 g/cm3 = (x atoms/cell)(69.72 g/mol) (1.566 × 10−22 cm3)(6.02 × 1023 atoms/mol) x = 8 atoms/cell (b) From the packing factor (PF) equation: PF = (8 atoms/cell)(4π/3)(0.1218 nm)3 = 0.387 0.1566 nm3 3–22 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. Solution: V = (0.22858 nm)2(0.35842 nm)cos 30 = 0.01622 nm3 = 16.22 × 10−24 cm3 (a) From the density equation: 1.848 g/cm3 = (x atoms/cell)(9.01 g/mol) (16.22 × 10−24 cm3)(6.02 × 1023 atoms/mol) x = 2 atoms/cell (b) The packing factor (PF) is: PF = (2 atoms/cell)(4π/3)(0.1143 nm)3 = 0.77 0.01622 nm3 CHAPTER 3 Atomic and Ionic Arrangements 13 03 Askeland Chap 9/27/05 11:34 AM Page 13 3–23 A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of unit cells and (b) the number of iron atoms in the paper clip. (See Appendix A for required data) Solution: The lattice parameter for BCC iron is 2.866 × 10−8 cm. Therefore Vunit cell = (2.866 × 10−8 cm)3 = 2.354 × 10−23 cm3 (a) The density is 7.87 g/cm3. The number of unit cells is: number = 0.59 g = 3.185 × 1021 cells (7.87 g/cm3)(2.354 × 10−23 cm3/cell) (b) There are 2 atoms/cell in BCC iron. The number of atoms is: number = (3.185 × 1021 cells)(2 atoms/cell) = 6.37 × 1021 atoms 3–24 Aluminum foil used to package food is approximately 0.001 inch thick. Assume that all of the unit cells of the aluminum are arranged so that a0 is perpendicular to the foil surface. For a 4 in. × 4 in. square of the foil, determine (a) the total number of unit cells in the foil and (b) the thickness of the foil in number of unit cells. (See Appendix A.) Solution: The lattice parameter for aluminum is 4.04958 × 10−8 cm. Therefore: Vunit cell = (4.04958 × 10−8)3 = 6.6409 × 10−23 cm3 The volume of the foil is: Vfoil = (4 in.)(4 in.)(0.001 in.) = 0.016 in.3 = 0.262 cm3 (a) The number of unit cells in the foil is: number = 0.262 cm3 = 3.945 × 1021 cells 6.6409 × 10−23 cm3/cell (b) The thickness of the foil, in number of unit cells, is: number = (0.001 in.)(2.54 cm/in.) = 6.27 × 104 cells 4.04958 × 10−8 cm 3–27 Above 882oC, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion? Solution: We can find the volume of each unit cell. Two atoms are present in both BCC and HCP titanium unit cells, so the volumes of the unit cells can be directly compared. VBCC = (0.332 nm)3 = 0.03659 nm3 VHCP = (0.2978 nm)2(0.4735 nm)cos30 = 0.03637 nm3 ΔV = VHCP − VBCC × 100 = 0.03637 nm3 − 0.03659 nm3 × 100 = −0.6% VBCC 0.03659 nm3 Therefore titanium contracts 0.6% during cooling. 14 The Science and Engineering of Materials Instructor’s Solutions Manual