Materials science callister wiley 8th solutions

CHAPTER ATOMIC STRUCTURE AND INTERATOMIC BONDING PROBLEM SOLUTIONS Fundamental Concepts Electrons in Atoms 2.1 Cite the difference between atomic mass and atomic weight Solution 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 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2.2 Chromium has four naturally-occurring isotopes: 4.34% of 52 50 Cr, with an atomic weight of 49.9460 53 amu, 83.79% of Cr, with an atomic weight of 51.9405 amu, 9.50% of Cr, with an atomic weight of 52.9407 amu, and 2.37% of 54Cr, with an atomic weight of 53.9389 amu On the basis of these data, confirm that the average atomic weight of Cr is 51.9963 amu Solution The average atomic weight of silicon (ACr ) is computed by adding fraction-of-occurrence/atomic weight products for the three isotopes Thus ACr = f 50 A Cr 50Cr + f 52 A Cr 52Cr + f 53 A Cr 53Cr + f 54 A Cr 54Cr = (0.0434)(49.9460 amu) + (0.8379)(51.9405 amu) + (0.0950)(52.9407  amu) + (0.0237)(53.9389 amu) = 51.9963 amu Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2.3 (a) How many grams are there in one amu of a material? (b) Mole, in the context of this book, is taken in units of gram-mole On this basis, how many atoms are there in a pound-mole of a substance? Solution (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 / mol  mol # g/amu =     6.022 × 10 23 atoms   amu / atom  = 1.66 × 10-24 g/amu (b) Since there are 453.6 g/lbm, lb - mol = (453.6 g/lb m ) (6.022 × 10 23 atoms/g - mol) = 2.73 × 1026 atoms/lb-mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2.4 (a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom (b) Cite two important additional refinements that resulted from the wave-mechanical atomic model Solution (a) Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells (b) Two important refinements resulting from the wave-mechanical atomic model are (1) that electron position is described in terms of a probability distribution, and (2) electron energy is quantized into both shells and subshells each electron is characterized by four quantum numbers Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2.5 Relative to electrons and electron states, what does each of the four quantum numbers specify? Solution The n quantum number designates the electron shell The l quantum number designates the electron subshell The ml quantum number designates the number of electron states in each electron subshell The ms quantum number designates the spin moment on each electron Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2.6 Allowed values for the quantum numbers of electrons are as follows: n = 1, 2, 3, l = 0, 1, 2, 3, , n –1 m l = 0, ±1, ±2, ±3, , ±l ms = ± The relationships between n and the shell designations are noted in Table 2.1 Relative to the subshells, l = corresponds to an s subshell l = corresponds to a p subshell l = corresponds to a d subshell l = corresponds to an f subshell For the K shell, the four quantum numbers for each of the two electrons in the 1s state, in the order of nlm l m s , are 2 100( ) and 100( − ) Write the four quantum numbers for all of the electrons in the L and M shells, and note which correspond to the s, p, and d subshells Solution For the L state, n = 2, and eight electron states are possible Possible l values are and 1, while possible ml 1 values are and ±1; and possible ms values are ± Therefore, for the s states, the quantum numbers are 200 ( ) 2 1 1 1 and 200 (− ) For the p states, the quantum numbers are 210 ( ) , 210 (− ) , 211 ( ) , 211 (− ) , 21(−1)( ) , and 2 2 2 21(−1)(− ) For the M state, n = 3, and 18 states are possible Possible l values are 0, 1, and 2; possible ml values are 1 0, ±1, and ±2; and possible ms values are ± Therefore, for the s states, the quantum numbers are 300 ( ) , 2 1 1 1 300 (− ) , for the p states they are 310 ( ) , 310 (− ) , 311 ( ) , 311 (− ) , 31(−1)( ) , and 31 (−1)(− ) ; for the d 2 2 2 1 1 1 1 states they are 320 ( ) , 320 (− ) , 321 ( ) , 321 (− ) , 32 (−1)( ) , 32 (−1) (− ) , 322 ( ) , 322 (− ) , 32 (−2)( ) , 2 2 2 2 and 32 (−2) (− ) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2.7 Give the electron configurations for the following ions: Fe2+, Al3+, Cu+, Ba2+, Br-, and O2- Solution The electron configurations for the ions are determined using Table 2.2 (and Figure 2.6) Fe2+: From Table 2.2, the electron configuration for an atom of iron is 1s22s22p63s23p63d64s2 In order to become an ion with a plus two charge, it must lose two electrons—in this case the two 4s Thus, the electron configuration for an Fe2+ ion is 1s22s22p63s23p63d6 Al3+: From Table 2.2, the electron configuration for an atom of aluminum is 1s22s22p63s23p1 In order to become an ion with a plus three charge, it must lose three electrons—in this case two 3s and the one 3p Thus, the electron configuration for an Al3+ ion is 1s22s22p6 Cu+: From Table 2.2, the electron configuration for an atom of copper is 1s22s22p63s23p63d104s1 In order to become an ion with a plus one charge, it must lose one electron—in this case the 4s Thus, the electron configuration for a Cu+ ion is 1s22s22p63s23p63d10 Ba2+: The atomic number for barium is 56 (Figure 2.6), and inasmuch as it is not a transition element the electron configuration for one of its atoms is 1s22s22p63s23p63d104s24p64d105s25p66s2 In order to become an ion with a plus two charge, it must lose two electrons—in this case two the 6s Thus, the electron configuration for a Ba2+ ion is 1s22s22p63s23p63d104s24p64d105s25p6 Br-: From Table 2.2, the electron configuration for an atom of bromine is 1s22s22p63s23p63d104s24p5 In order to become an ion with a minus one charge, it must acquire one electron—in this case another 4p Thus, the electron configuration for a Br- ion is 1s22s22p63s23p63d104s24p6 O2-: From Table 2.2, the electron configuration for an atom of oxygen is 1s22s22p4 In order to become an ion with a minus two charge, it must acquire two electrons—in this case another two 2p Thus, the electron configuration for an O2- ion is 1s22s22p6 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2.8 Sodium chloride (NaCl) exhibits predominantly ionic bonding The Na+ and Cl- ions have electron structures that are identical to which two inert gases? Solution + The Na ion is just a sodium atom that has lost one electron; therefore, it has an electron configuration the same as neon (Figure 2.6) The Cl ion is a chlorine atom that has acquired one extra electron; therefore, it has an electron configuration the same as argon Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful The Periodic Table 2.9 With regard to electron configuration, what all the elements in Group VIIA of the periodic table have in common? Solution Each of the elements in Group VIIA has five p electrons Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 2.10 To what group in the periodic table would an element with atomic number 114 belong? Solution From the periodic table (Figure 2.6) the element having atomic number 114 would belong to group IVA According to Figure 2.6, Ds, having an atomic number of 110 lies below Pt in the periodic table and in the rightmost column of group VIII Moving four columns to the right puts element 114 under Pb and in group IVA Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21.16 The fraction of nonreflected radiation that is transmitted through a 10-mm thickness of a transparent material is 0.90 If the thickness is increased to 20 mm, what fraction of light will be transmitted? Solution In this problem we are asked to calculate the fraction of nonreflected light transmitted through a 20-mm thickness of transparent material, given that the fraction transmitted through a 10-mm thickness is 0.90 From Equation 21.18, the fraction of nonreflected light transmitted is just IT' / I 0' Using this expression we must first determine the value of β; this is possible by algebraic manipulation of Equation 21.18 Dividing both sides of the equation by I 0' , and then taking natural logarithms leads to  '  I  ln  T  = − βx I '   0 Now solving for β and also incorporating values for IT' / I 0' and x provided in the problem statement gives β =−  '  I ln  T   '  x  I0    = −   ln (0.90) = 1.05 × 10 -2 mm -1  10 mm  And computation of IT' / I 0' when x = 20 mm (Equation 21.18) is as follows: IT' = exp (− βx) I 0' [ ] = exp − (1.05 × 10−2 mm−1)(20 mm) = 0.81 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Transmission 21.17 Derive Equation 21.19, starting from other expressions given in the chapter Solution The problem asks that we derive Equation 21.19, which is IT = I (1 − R) e− βl If we examine Figure 21.7, at the front (or left) interface, some of the incident beam having intensity I0 is reflected Since IR = I0R at this surface, then IT' = I − I R = I (1 − R) in which IT' is the intensity of the nonreflected beam at the front surface that is transmitted Now there will be absorption of this transmitted beam as it passes through the solid and transparent medium according to Equation 21.18 Just inside the back (or right) interface, the beam has passed through a thickness l of this material (x = l) and, therefore, the intensity of the transmitted beam at this point (I '' ) is just T IT'' = I (1 − R) e- βl Finally, a second reflection will occur at the back interface as the beam passes out of the medium The intensity of the reflected beam (I '' ) is just R I R'' = I T'' R = I R(1 − R) e- βl And the intensity of the final transmitted beam (IT) becomes IT = IT'' − I R'' = I (1 − R) e- βl − I R(1 − R) e- βl = I (1 − R) e- βl which is Equation 21.19, the desired expression Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21.18 The transmissivity T of a transparent material 20 mm thick to normally incident light is 0.85 If the index of refraction of this material is 1.6, compute the thickness of material that will yield a transmissivity of 0.75 All reflection losses should be considered Solution We are asked to compute the thickness of material to yield a transmissivity of 0.75 given that T is 0.85 when l = 20 mm, n = 1.6, and for normally incident radiation The first requirement is that we calculate the value of β for this material using Equations 21.13 and 21.19 The value of R is determined using Equation 21.13 as R = = (ns − 1)2 (ns + 1)2 (1.6 − 1) = 5.33 × 10 -2 (1.6 + 1) Now, it is necessary to compute the value of β using Equation 21.19 Dividing both sides of Equation 21.19 by I0(1 – R)2 leads to IT I (1 − R) = e -βl And taking the natural logarithms of both sides of this expression gives   IT  = − βl ln  I (1 − R)  and solving for β we get β =−   IT   ln l  I (1 − R)  Since the transmissivity is T is equal to IT/I0, then the above equation takes the form β=−   T  ln  l  (1 − R)  Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Using values for l and T provided in the problem statement, as well as the value of R determined above, we solve for β as     0.85 β= −  = 2.65 × 10 -3 mm -1  ln  −2  20 mm   (1 − 5.33 × 10 )  Now, solving for l when T = 0.75 using the rearranged form of Equation 21.19 above l =− = −   T ln   β  (1 − R)    0.75 ln   2.65 × 10−3 mm−1  (1 − 5.33 × 10−2 )  = 67.3 mm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Color 21.19 Briefly explain what determines the characteristic color of (a) a metal and (b) a transparent nonmetal Solution (a) The characteristic color of a metal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is reflected (b) The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is transmitted through the material Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21.20 Briefly explain why some transparent materials appear colored while others are colorless Solution For a transparent material that appears colorless, any absorption within its interior is the same for all visible wavelengths On the other hand, if there is any selective absorption of visible light (usually by electron excitations), the material will appear colored, its color being dependent on the frequency distribution of the transmitted light beam Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Opacity and Translucency in Insulators 21.21 Briefly describe the three absorption mechanisms in nonmetallic materials The three absorption mechanisms in nonmetallic materials involve electronic polarization, electron transitions, and scattering Electronic polarization is described in Section 21.4; absorption by electron transitions is discussed in Sections 21.4 and 21.7; and scattering is discussed in Section 21.10 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21.22 Briefly explain why amorphous polymers are transparent, while predominantly crystalline polymers appear opaque or, at best, translucent Solution Amorphous polymers are normally transparent because there is no scattering of a light beam within the material However, for semicrystalline polymers, visible light will be scattered at boundaries between amorphous and crystalline regions since they have different indices of refraction This leads to translucency or, for extensive scattering, opacity, except for semicrystalline polymers having very small crystallites Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Luminescence Photoconductivity Lasers 21.23 (a) In your own words describe briefly the phenomenon of luminescence (b) What is the distinction between fluorescence and phosphorescence? Solution (a) The phenomenon of luminescence is described in Section 21.11 (b) The feature that distinguishes fluorescence from phosphorescence is the magnitude of the time interval between photon absorption and reemission events Fluorescence is for delay times less than a second; phosphorescence occurs for longer times Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21.24 In your own words, briefly describe the phenomenon of photoconductivity The phenomenon of photoconductivity is explained in Section 21.12 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21.25 Briefly explain the operation of a photographic lightmeter Solution A photographic light meter is used to measure the intensity of incident light radiation Each photon of incident light induces a valence-band-to-conduction band electron transition in which both electrons and holes are produced, as depicted in Figure 21.5a The magnitude of the photoinduced current resulting from these transitions is registered, which is proportional to the numbers of electrons and holes, and thus, the number of incident photons, or, equivalently, the intensity of the incident light radiation Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21.26 In your own words, describe how a ruby laser operates Section 21.13 contains a description of the operation of a ruby laser Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 21.27 Compute the difference in energy between metastable and ground electron states for the ruby laser Solution This problem asks for the difference in energy between metastable and ground electron states for a ruby laser The wavelength of the radiation emitted by an electron transition from the metastable to ground state is cited as 0.6943 µm The difference in energy between these states, ∆E, may be determined from a combined form of Equations 21.6 and 21.2, as ∆E = hν = = hc λ (4.13 × 10−15 eV - s)(3 × 10 m /s) 6.943 × 10−7 m = 1.78 eV Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Optical Fibers in Communications 21.28 At the end of Section 21.14 it was noted that the intensity of light absorbed while passing through a 16-kilometer length of optical fiber glass is equivalent to the light intensity absorbed through for a 25-mm thickness of ordinary window glass Calculate the absorption coefficient β of the optical fiber glass if the value of β for the window glass is × 10–4 mm–1 Solution This problem asks for us to determine the value of the absorption coefficient for optical fiber glass given that β for window glass is × 10-4 mm-1; furthermore, the intensity of nonabsorbed light transmitted through a 25-mm thickness of window glass is equivalent to the nonabsorbed light transmitted through a 16-km length of the optical fiber material Using Equation 21.18, it is first necessary to compute the fraction of light transmitted I' through the window glass—i.e., T Thus I 0' IT' = e−βx I' −4 −1 = e− (5 × 10 mm )(25 mm) = 0.9876 Now, solving for β from Equation 21.18 leads to    IT'  β = − ln  x I'   0 And substitution into this expression the above value for IT' (0.9876) as well as parameters for the optical fiber I' glass—x = 16 km = 16 × 103 m = 16 × 106 mm—yields β=− ln (0.9876) = 7.80 × 10−10 mm−1 16 ×10 mm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful DESIGN PROBLEM Atomic and Electronic Interactions 21.D1 Gallium arsenide (GaAs) and gallium phosphide (GaP) are compound semiconductors that have room-temperature band gap energies of 1.42 and 2.25 eV, respectively, and form solid solutions in all proportions Furthermore, the band gap of the alloy increases approximately linearly with GaP additions (in mol%) Alloys of these two materials are used for light-emitting diodes wherein light is generated by conduction band-to-valence band electron transitions Determine the composition of a GaAs–GaP alloy that will emit orange light having a wavelength of 0.60 μm Solution This problem stipulates that GaAs and GaP have room-temperature band gap energies of 1.42 and 2.25 eV, respectively, that they form solid solutions in all proportions, that alloys of these two semiconductors are used for light-emitting diodes wherein light is generated by conduction band-to-valence band electron transitions, and that the band gap of a GaAs-GaP alloy increases approximately linearly with GaP additions (in mol%) We are asked to determine the composition of an alloy that will emit red light having a wavelength of 0.60 µm It first becomes necessary to compute the band-gap energy corresponding to this wavelength of light using Equation 21.3 as Eg = = hc λ (4.13 × 10−15 eV - s)(3 × 10 m /s) 0.60 × 10−6 m = 2.065 eV Realizing that at mol% GaP, Eg = 1.42 eV, while at 100 mol% GaP, Eg = 2.25 eV, it is possible to set up the relationship 2.25 eV − 2.065 eV 100 mol% − CGaP = 2.25 eV − 1.42 eV 100 mol% − mol% Solving for CGaP, the composition of GaP, we get CGaP = 77.7 mol% Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful ... bonding energy E between the two ions (c) Mathematically determine the r and E values using the solutions to Problem 2.14 and compare these with the graphical results from part (b) Solution (a)... copyright owner is unlawful 2.22 What type(s) of bonding would be expected for each of the following materials: brass (a copper-zinc alloy), rubber, barium sulfide (BaS), solid xenon, bronze, nylon,... permission of the copyright owner is unlawful CHAPTER THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS Fundamental Concepts 3.1 What is the difference between atomic structure and crystal structure?