TEST BANK FOR Engineering Materials Science By Milton Ohring (Solution Manual)

Exam (elaborations) TEST BANK FOR Engineering Materials Science By Milton Ohring (Solution Manual) Exam (elaborations) TEST BANK FOR Engineering Materials Science By Milton Ohring (Solution Manual) SOLUTIONS TO PROBLEMS PREFACE This section of instructors materials contains solutions and answers to all problems and questions that appear in the textbook. My penmanship leaves something to be desired; therefore, I generated these solutions/answers using computer software so that the resulting product would be "readable." Furthermore, I endeavored to provide complete and detailed solutions in order that: (1) the instructor, without having to take time to solve a problem, will understand what principles/skills are to be learned by its solution; and (2) to facilitate student understanding/learning when the solution is posted. I would recommended that the course instructor consult these solutions/answers before assigning problems and questions. In doing so, he or she ensures that the students will be drilled in the intended principles and concepts. In addition, the instructor may provide appropriate hints for some of the more difficult problems. With regard to symbols, in the text material I elected to boldface those symbols that are italicized in the textbook. Furthermore, I also endeavored to be consistent relative to symbol style. However, in several instances, symbols that appear in the textbook were not available, and it was necessary to make appropriate substitutions. These include the following: the letter a (unit cell edge length, crack length) is used in place of the cursive a. And Roman F and E replace script F (Faraday's constant in Chapter 18) and script E (electric field in Chapter 19), respectively. I have exercised extreme care in designing these problems/questions, and then in solving them. However, no matter how careful one is with the preparation of a work such as this, errors will always remain in the final product. Therefore, corrections, suggestions, and comments from instructors who use the textbook (as well as their teaching assistants) pertaining to homework problems/solutions are welcomed. These may be sent to me in care of the publisher. 1 CHAPTER 2 ATOMIC STRUCTURE AND INTERATOMIC BONDING PROBLEM SOLUTIONS 2.1 (a) When two or more atoms of an element have different atomic masses, each is termed an isotope. (b) The atomic weights of the elements ordinarily are not integers because: (1) the atomic masses of the atoms generally are not integers (except for 12 C), and (2) the atomic weight is taken as the weighted average of the atomic masses of an atom's naturally occurring isotopes. 2.2 Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally occurring isotopes. 2.3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as #g/amu = èç æ ø÷ ö 1 mol 6.023 x 1023 atoms ( 1 g/mol ) 1 amu/atom = 1.66 x 10-24 g/amu (b) Since there are 453.6 g/lbm, 1 lb-mol = (453.6 g/lbm)(6.023 x 1023 atoms/g-mol) = 2.73 x 1026 atoms/lb-mol 2.4 (a) Two important quantum-mechanical concepts associated with the Bohr model of the atom are that electrons are particles moving in discrete orbitals, and electron energy is quantized into shells. (b) Two important refinements resulting from the wave-mechanical atomic model are that electron position is described in terms of a probability distribution, and electron energy is quantized into both shells and subshells--each electron is characterized by four quantum numbers. 2 2.5 The n quantum number designates the electron shell. The l quantum number designates the electron subshell. The m l quantum number designates the number of electron states in each electron subshell. The m s quantum number designates the spin moment on each electron. 2.6 For the L state, n = 2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml values are 0 and ±1. Therefore, for the s states, the quantum numbers are 200( 1 2 ) and 200(- 1 2 ). For the p states, the quantum numbers are 210( 1 2 ), 210(- 1 2 ), 211( 1 2 ), 211(- 1 2 ), 21(-1)( 1 2 ), and 21(-1)(- 1 2 ). For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2; possible ml values are 0, ±1, and ±2; and possible ms values are ±1 2 . Therefore, for the s states, the quantum numbers are 300( 1 2 ), 300(- 1 2 ), for the p states they are 310( 1 2 ), 310(- 1 2 ), 311(F(1,2)), 311(-F(1,2)), 31(-1)(F(1,2)), and 31(-1)(-F(1,2)); for the d states they are 320( 1 2 ), 320(- 1 2 ), 321( 1 2 ), 321(- 1 2 ), 32(-1)( 1 2 ), 32(-1)(- 1 2 ), 322( 1 2 ), 322(- 1 2 ), 32(-2)( 1 2 ), and 32(-2)(- 1 2 ). 2.7 The electron configurations of the ions are determined using Table 2.2. Fe 2+ - 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 Fe 3+ - 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 Cu + - 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 Ba 2+ - 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 Br - - 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 S 2- - 1s 2 2s 2 2p 6 3s 2 3p 6 2.8 The Cs + ion is just a cesium atom that has lost one electron; therefore, it has an electron configuration the same as xenon (Figure 2.6). The Br - ion is a bromine atom that has acquired one extra electron; therefore, it has an electron configuration the same as krypton. 2.9 Each of the elements in Group VIIA has five p electrons. 2.10 (a) The 1s 2 2s 2 2p 6 3s 2 3p 6 3d 7 4s 2 electron configuration is that of a transition metal because of an incomplete d subshell. 3 (b) The 1s 2 2s 2 2p 6 3s 2 3p 6 electron configuration is that of an inert gas because of filled 3s and 3p subshells. (c) The 1s 2 2s 2 2p 5 electron configuration is that of a halogen because it is one electron deficient from having a filled L shell. (d) The 1s 2 2s 2 2p 6 3s 2 electron configuration is that of an alkaline earth metal because of two s electrons. (e) The 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2 electron configuration is that of a transition metal because of an incomplete d subshell. (f) The 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 electron configuration is that of an alkali metal because of a single s electron. 2.11 (a) The 4f subshell is being filled for the rare earth series of elements. (b) The 5f subshell is being filled for the actinide series of elements. 2.12 The attractive force between two ions FA is just the derivative with respect to the interatomic separation of the attractive energy expression, Equation (2.8), which is just FA = dEA dr = d(- ) A r dr = A r2 The constant A in this expression is defined in footnote 3 on page 21. Since the valences of the K+ and O2- ions are +1 and -2, respectively, Z1 = 1 and Z2 = 2, then FA = (Z1e)(Z2e) 4peor2 = (1)(2)(1.6 x 10-19 C)2 (4)(p)(8.85 x 10-12 F/m)(1.5 x 10-9 m)2 = 2.05 x 10-10 N 2.13 (a) Differentiation of Equation (2.11) yields dEN dr = A r(1 + 1) - nB r(n + 1) = 0 4 (b) Now, solving for r (= r o) A ro 2 = nB ro (n + 1) or ro = ( A ) nB 1/(1 - n) (c) Substitution for ro into Equation (2.11) and solving for E (= Eo) Eo = - A ro + B ro n = - A ( A ) nB 1/(1 - n) + B ( A ) nB n/(1 - n) 2.14 (a) Curves of E A, E R, and E N are shown on the plot below. 0.0 0.2 0.4 0.6 0.8 1.0 -7 -6 -5 -4 -3 -2 -1 0 1 2 Interatomic Separation (nm) Bonding Energy (eV) EA ER E N ro = 0.28 nm Eo = -4.6 eV 5 (b) From this plot r o = 0.28 nm E o = -4.6 eV (c) From Equation (2.11) for E N A = 1.436 B = 5.86 x 10 -6 n = 9 Thus, ro = ( A ) nB 1/(1 - n) = ë ê é û ú ù 1.436 (9)(5.86 x 10-6) 1/(1 - 9) = 0.279 nm and Eo= - 1.436 ë ê é û ú ù 1.436 (9)(5.86 x 10-6) 1/(1 - 9) + 5.86 x 10-6 ë ê é û ú ù 1.436 (9)(5.86 x 10-6) 9/(1 - 9) = - 4.57 eV 2.15 This problem gives us, for a hypothetical X+-Y- ion pair, values for ro (0.35 nm), Eo (-6.13 eV), and n (10), and asks that we determine explicit expressions for attractive and repulsive energies of Equations 2.8 and 2.9. In essence, it is necessary to compute the values of A and B in these equations. Expressions for ro and Eo in terms of n, A, and B were determined in Problem 2.13, which are as follows: ro = ( ) A nB 1/(1 - n) Eo = - A ( A ) nB 1/(1 - n) + B ( A ) nB n/(1 - n) Thus, we have two simultaneous equations with two unknowns (viz. A and B). Upon substitution of values for ro and Eo in terms of n, these equations take the forms 6 0.35 nm = ( A ) 10B 1/(1 - 10) -6.13 eV = - A ( A ) 10B 1/(1 - 10) + B ( A ) 10B 10/(1 - 10) Simultaneous solution of these two equations leads to A = 2.38 and B = 1.88 x 10-5. Thus, Equations (2.8) and (2.9) become EA = - 2.38 r ER = 1.88 x 10-5 r10 Of course these expressions are valid for r and E in units of nanometers and electron volts, respectively. 2.16 (a) Differentiating Equation (2.12) with respect to r yields dE dr = C r2 - De-r/r r At r = ro, dE/dr = 0, and C r2 o = De-ro/r r (2.12b) Solving for C and substitution into Equation (2.12) yields an expression for Eo as Eo = De-ro/r èç æ ø÷ ö 1 - ro r (b) Now solving for D from Equation (2.12b) above yields D = Cre ro/r ro 2 7 Substitution of this expression for D into Equation (2.12) yields an expression for Eo as Eo = C ro èç æ ø÷ ö r ro - 1 2.17 (a) The main differences between the various forms of primary bonding are: Ionic--there is electrostatic attraction between oppositely charged ions. Covalent--there is electron sharing between two adjacent atoms such that each atom assumes a stable electron configuration. Metallic--the positively charged ion cores are shielded from one another, and also "glued" together by the sea of valence electrons. (b) The Pauli exclusion principle states that each electron state can hold no more than two electrons, which must have opposite spins. 2.18 Covalently bonded materials are less dense than metallic or ionically bonded ones because covalent bonds are directional in nature whereas metallic and ionic are not; when bonds are directional, the atoms cannot pack together in as dense a manner, yielding a lower mass density. 2.19 The percent ionic character is a function of the electron negativities of the ions XA and XB according to Equation (2.10). The electronegativities of the elements are found in Figure 2.7. For TiO2, XTi = 1.5 and XO = 3.5, and therefore, %IC = [1 - e(-0.25)(3.5 - 1.5) ] 2 x 100 = 63.2% For ZnTe, XZn = 1.6 and XTe = 2.1, and therefore, %IC = [1 - e(-0.25)(2.1 - 1.6) ] 2 x 100 = 6.1% For CsCl, XCs = 0.7 and XCl = 3.0, and therefore, %IC = [1 - e(-0.25)(3.0 - 0.7) ] 2 x 100 = 73.4% For InSb, XIn = 1.7 and XSb = 1.9, and therefore, 8 %IC = [1 - e(-0.25)(1.9 - 1.7) ] 2 x 100 = 1.0% For MgCl2, XMg = 1.2 and XCl = 3.0, and therefore, %IC = [1 - e(-0.25)(3.0 - 1.2) ] 2 x 100 = 55.5% 2.20 Below is plotted the bonding energy versus melting temperature for these four metals. From this plot, the bonding energy for copper (melting temperature of 1084°C) should be approximately 3.6 eV. The experimental value is 3.5 eV. -1000 0 1000 2000 4 6 8 10 Bonding Energy (eV) W Fe Al Hg 3.6 eV Melting Temperature (C) 2.21 For germanium, having the valence electron structure 4s 2 4p 2 , N' = 4; thus, there are 8 - N' = 4 covalent bonds per atom. For phosphorus, having the valence electron structure 3s 2 3p 3 , N' = 5; thus, there are 8 - N' = 3 covalent bonds per atom. For selenium, having the valence electron structure 4s 2 4p 4 , N' = 6; thus, there are 8 - N' = 2 covalent bonds per atom. For chlorine, having the valence electron structure 3s 2 3p 5 , N' = 7; thus, there is 8 - N' = 1 covalent bond per atom. 2.22 For brass, the bonding is metallic since it is a metal alloy. For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.) 9 For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table. For solid xenon, the bonding is van der Waals since xenon is an inert gas. For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin). For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.) For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table. 2.23 The intermolecular bonding for HF is hydrogen, whereas for HCl, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature. 2.24 The geometry of the H 2 O molecules, which are hydrogen bonded to one another, is more restricted in the solid phase than for the liquid. This results in a more open molecular structure in the solid, and a less dense solid phase. 10 CHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS 3.1 Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material. 3.2 A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell, whereas a crystal system is described only in terms of the unit cell geometry. For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic crystal system. 3.3 For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation (3.4) as VC = 16R3Ö` 2 = (16)(0.143 x 10-9 m)3Ö` 2 = 6.62 x 10-29 m3 3.4 This problem calls for a demonstration of the relationship a = 4RÖ` 3 for BCC. Consider the BCC unit cell shown below a a N O P Q a Using the triangle NOP 11 (NP __ ) 2 = a2 + a2 = 2a2 And then for triangle NPQ, (NQ __ )2 = (QP __ )2 + (NP __ )2 But NQ __ = 4R, R being the atomic radius. Also, QP __ = a. Therefore, (4R)2 = a2 + 2a2, or a = 4R Ö` 3 3.5 We are asked to show that the ideal c/a ratio for HCP is 1.633. A sketch of one-third of an HCP unit cell is shown below. c a a J M K L Consider the tetrahedron labeled as JKLM, which is reconstructed as 12 J K L M H The atom at point M is midway between the top and bottom faces of the unit cell--that is MH __ = c/2. And, since atoms at points J, K, and M, all touch one another, JM __ = JK __ = 2R = a where R is the atomic radius. Furthermore, from triangle JHM, (JM __ )2 = ( JH __ )2 + (MH __ )2, or a2 = ( JH __ )2 + ( ) c 2 2 Now, we can determine the JH __ length by consideration of triangle JKL, which is an equilateral triangle, J L K H a/2 30 cos 30° = a/2 JH = Ö` 3 2 , and JH __ = a Ö` 3 13 Substituting this value for JH __ in the above expression yields a2 = èç æ ø÷ ö a Ö` 3 2 + (c) 2 2 = a2 3 + c2 4 and, solving for c/a c a = Ö` 8 3 = 1.633 3.6 We are asked to show that the atomic packing factor for BCC is 0.68. The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = VS VC Since there are two spheres associated with each unit cell for BCC VS = 2(sphere volume) = 2 èç æ ø÷ ö 4pR3 3 = 8pR3 3 Also, the unit cell has cubic symmetry, that is V C = a3 . But a depends on R according to Equation (3.3), and VC = èç æ ø÷ ö 4R Ö` 3 3 = 64R3 3Ö` 3 Thus, APF = 8pR3/3 64R3/3Ö` 3 = 0.68 3.7 This problem calls for a demonstration that the APF for HCP is 0.74. Again, the APF is just the total sphere-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus 14 VS = 6 èç æ ø÷ ö 4pR3 3 = 8pR3 Now, the unit cell volume is just the product of the base area times the cell height, c. This base area is just three times the area of the parallelepiped ACDE shown below. A B C D E a = 2R a = 2R a = 2R 60 30 The area of ACDE is just the length of CD __ times the height BC __ . But CD __ is just a or 2R, and BC __ = 2R cos(30°) = 2RÖ` 3 2 Thus, the base area is just AREA = (3)(CD __ )(BC __ ) = (3)(2R) èç æ ø÷ ö 2RÖ` 3 2 = 6R2Ö` 3 and since c = 1.633a = 2R(1.633) VC = (AREA)(c) = 6R2cÖ` 3 = (6R2Ö` 3)(2)(1.633)R = 12Ö` 3(1.633)R3 Thus, APF = VS VC = 8pR3 12Ö` 3(1.633)R3 = 0.74 3.8 This problem calls for a computation of the density of iron. According to Equation (3.5) 15 r = nAFe VCNA For BCC, n = 2 atoms/unit cell, and VC = èç æ ø÷ ö 4R Ö` 3 3 Thus, r = (2 atoms/unit cell)(55.9 g/mol) [(4)(0.124 x 10-7 cm)3/`Ö3]3/(unit cell)(6.023 x 1023 atoms/mol) = 7.90 g/cm3 The value given inside the front cover is 7.87 g/cm 3 . 3.9 We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal structure. For FCC, n = 4 atoms/unit cell, and V C = 16R3Ö` 2 [Equation (3.4)]. Now, r = nAIr VCNA And solving for R from the above two expressions yields R = è ç æ ø ÷ ö nAIr 16rNAÖ` 2 1/3 = ë ê é û ú ù (4 atoms/unit cell)(192.2 g/mol) (Ö` 2)(16)(22.4 g/cm3)(6.023 x 1023 atoms/mol) 1/3 = 1.36 x 10-8 cm = 0.136 nm 3.10 This problem asks for us to calculate the radius of a vanadium atom. For BCC, n = 2 atoms/unit cell, and 16 VC = èç æ ø÷ ö 4R Ö` 3 3 = 64R3 3Ö` 3 Since, r = nAV VCNA and solving for R R = è ç æ ø ÷ ö 3Ö` 3nAV 64rNA 1/3 = ëê ê é ûú ú ù (3Ö` 3)(2 atoms/unit cell)(50.9 g/mol) (64)(5.96 g/cm3)(6.023 x 1023 atoms/mol) 1/3 = 1.32 x 10-8 cm = 0.132 nm 3.11 For the simple cubic crystal structure, the value of n in Equation (3.5) is unity since there is only a single atom associated with each unit cell. Furthermore, for the unit cell edge length, a = 2R. Therefore, employment of Equation (3.5) yields r = nA VCNA = nA (2R)3NA = (1 atom/unit cell)(70.4 g/mol) [(2)(1.26 x 10-8 cm)]3/unit cell(6.023 x 1023 atoms/mol) = 7.30 g/cm3 3.12. (a) The volume of the Zr unit cell may be computed using Equation (3.5) as VC = nAZr rNA Now, for HCP, n = 6 atoms/unit cell, and for Zr, A Zr = 91.2 g/mol. Thus, 17 VC