Resoluções do Shackelford 6ª edição, Notas de estudo de Cultura

Baixe Resoluções do Shackelford 6ª edição e outras Notas de estudo em PDF para Cultura, somente na Docsity! Solutions to the Practice Problems from J.F. Shackelford, Introduction to Materials Science for Engineers, 6th Edition, Prentice- Hall, Upper Saddle River, NJ, 2005 Beginning with Chapter 2, a set of Practice Problems follows immediately after worked-out Sample Problems. These exercises follow directly from the preceding solutions and are intended to provide a carefully guided journey into the first calculations in each new area. The solutions provided here allow students to independently check their own approach to these problems. Section 2.1 — Atomic Structure PP 21 Calculate the number of atoms contained in à cylinder 1 um in diameter by 1 um deep of (a) magnesium and (b) lead. (See Sample Problem 2.1.) = Ebé vio atom Cu CDA Slam) My PP 2.1] 44) Ny aba bbéniola “Z93 9 fam) Ga x LESLS Cu / Ney Aloma Cx D$EL ty 7 ara = 3.38 x/0!º atima My 4 = ” Con 34.9 Lem3) PR (é) Vos bons Cgtxio a fma Cu x TR 23,9 doa) Ga x LESSA C/Mi átomos Cu TRI CN aloma Ph =259 x 40º soma Pê ES = PP 2.2 Using the density of MgO caiculated in Sample Problem 2.2, cal- culate the mass 0f an MgO refractory (temperature-resistant) brick with dimensions: SO mm x 100 mm 200 mm. PP 2.2] maoV= (3.60 9 em DE em Sam?) x (50X 00 )(a200)mm 3 = 3toxw?y = 3.60 a PP 23 Calculaie the dimensions of (a) a cube containing 1 mol of copper and (b) a cube containing 1 moi of lead. (See Sample Prob- lem 2.3.) ES 9/imel Ny PE3S] (ce cs Pu tom, f820 nm (8) espe = (2oza slea É o = 34 11.349 Leme x omog, =2bitrm In the next chapter we shall ses that MgO, Ca0, FeO, and NiQ PP 2.7 all share the NaCl crystai suuciwre. As a result, in each case he metai ions will have the same coordination number (6). The case oÉ MgO and Ca is treated in Sample Problem 2.8. Use the radius ratio calcularioa to see if it estimates the CN = 6 for Fe and NiO. PP 27) Fr Appadisa, rest = 0.087nm, Fem 0.07Enm, roa-z 0.132nm Fer Feo, Es 007 . . ivete Aº Eres 066 fr which Table 2.) q CN =b For Nio, Ds QOIinm 0.59 ama cn=b 0.132 nm Section 2.3 - The Covalent Bond In Figure 2-14 we ses the polymerization of polyethylene +4C)Huo-, illustrated, Sample Problem 2.9 illustrates poiymeriza- tion for poly(vinyl chioride) C;,HsCI5-,. Make a similar sketch to filustrate the polymerization of polypropylene A CoHsR, where R is a CH group. H H PP 2.8 é - é PrPy leme. molecule ! ! HR PP 2.8 u P e dy) prpy leme =€- = pmalteula R R — t— pres lme mer ishtre R= CH; Use a sketch to illusrate the polymerization of polystyrene +C)H5R3-,, where R is a benzene group, CsHs. I Tr Ar e-po 2-D-z “Dos 7 4 R z-q-a PP 2.9 PP 2.9 "OR e=€ styree malecule. HH Ledo pelystyme ld molecule, Calculate the reaction energy for polymerization of (a) propylene PP 2.10 (see Practice Problem 2.8) and (b) styrene (see Practice Prob- = lem 2.9). PP 2.10 (a) The bectboa reachim is Yhe Sacus, Thetre, dhe Coltulao is Ho pame : Crso- tro JA T/ mel = CO T/nol (b) Aga, (7240-680) AI feel = bo LT fimol The length of an average polyerhylene molecule in a commer- PP 2.11 cial clear plastic wrap is 0.2 um. What is the average degree of polymerization (7) for this material? (See Sample Problem 2.11.) PP ZIi] As idusfrated mm Sample Problem 2.11, L= Ant or FA "24 o.axioTm = 79 Arotabnto Im == Section 2.4 — The Metallic Bond Discuss the low coordination number (= 4) for the diamond cubic PP 2.12 suuciure found for some elemental solids, such as silicon. (See Sample Problem 2.12.) PP 2.12 À greater degree É Coveltuty iu dáe Sj-Si bomb providea evem sbmger direcfmadity and? hwer Coordnatim number. Section 2.5 — The Secondary, or van der Waals, Bond l . PP 2.13 The bond energy and bond lengih for argon are calculated (as suming a “6-12” potential) in Sample Problem 2.13. Plot E as à function of a over the range 0.33 to 0.80 om. PP 213] From Simple Problem d.13: E =l Go37uo AZ mt) + (tento EG mt) ba: aê afã 4 -! x Otoasn io? mol Section 3.1 — Seven Systems and Fourteen Lattices The note at the end of Sample Problem 3.1 comments that am PP 31 area-centered square lattice can be resolved into a simple square tatice. Sketch this equivalence. PP 31 . . . + . . . — . . . —+— Section 3.2 — Metal Structures In Sample Problem 3.2 the relationship between Jattice param- PP 3.2 eter, a, and atomic radius, r, for an fcc metal is found to be a = (4//2)r, as given in Table 3.3, Derive the similar relation- ships in Table 3.3 for (a) a bcc metal and (b) an hcp metal. PP 3.2] (a) Tue lenha, d, of a body dlragonal da: d= 4r=43 a er «£ a far (6) És rrapectito ef Fique Bm 6, tre ate Mad: aar (uide: lia shall see im Bubleo BUT Leaf Caléxa) Calculate the density of a -Fe, which is a bec metal. (Caution: A PP 3.3 different relationship between lattice parameter, «, and atomic radius, r, applies to this different crystal structure. See Practice Problem 3.2 and Table 3.3.) PP 3.3 From PP 3.L,a=(4 NE). Foro-Fe, az (INTO nm) = DARE nm Sr EH . Dama SSIS x (lobo (Uefa SE foareamjs "a torradas, ensaia = "729 9 /0m3 (ilote Lral Comedia dm» SP3.2.) Section 3.3 - Ceramic Structures Calculate the ionic packing factor of (a) CaO, (b) FeO, and (e) PP 3.4 NiO, All of these compounds share the NaCl-type structure. (d) Is there a unique IPF value for the NaCl-type structure? Explain. (See Sample Problem 3.3.) PP 3.4] (a) Q=2r a tar = A(0406 nm) + A(O/3Inm) = A S76 am Vateudy 247% (04 Ea O l0fam3 ,3 Via E 4/52 aa * 3 fr) 47 [lotoémm)* 4 (0.8.20m 3 /m COS ES nms 3 8.542 am (1) a*= [2le.087mm) 4200.1324) Ê AoPGO nas Via E = 7, “2 lho Tum)? 4 o. Bon) Ja aos am? IPF= o osdéns = Deo pão mms * Lu PO, (e) a 3. Latoorp ma Co.t32nm)] * Co TE nm Vaao LR flor)! é 00.1324m 5] 0.046S amo? 3 IPEF = D0MSnm 00747 am Lobato () Wo. do sem dy fla calutubra abme, dl 297 is a Acham of 77 nadiua Lato. PP 3.5 Calculate the density of CaO. (Sec Sample Problem 3.4.) q = 2 +2h 2 2/0 10b im) + 2 (0.132nm 20.476 um PP 3.5 Vateell zqla (0:47 rum Es O. togum? - [a(ts007 3) + 4 (f6.009)] Wobeaz no?!) N tlm ? 0.108 nm am = 3.45 adm? Section 3,4 - Polymeric Structures How many unitcells are containedin 1 kg ofcommercial polyethy- PP 3.6 lene that is 50 vol % crystailine (balance amorphous) and has an overall product density of 0.940 Mgim?? (See Sample Problem 35) PP 3.6] Fist ndo MAE umit tella am pri ind intão ereto porfim. Vij,= MES 1 * (LM foca dog) Lo6 pm? th der, o. 40 Mg dm$ ” 3,0 8 Veopibllne = 0-5 (r06n10"m3 a 0.532 410m 1.8 3 Viib qell = 0.0933 sm” x (ta = L33x07* IVAI da M — D.532x/077m, tom Ê tels P33n000 Im Vamibrel! 3 = 70 eo il tells Section 3.5 —- Semiconductor Structures In Sample Problem 3.6 we find the atomic packing factor for PP 3.7 silicon to be quite low compared to the common metal struc- tures. Comment on the relationship between this characteristic and the nature of bonding in semiconductor silicon. PP 3.7 Eonding m silica és highly comi. The associated dirtehimalidy of ida bomds dominatia even eee preta of spherea fa chancteiste oÉ non-dovechimal, metadie bondig) PP 3.13 PP 3.13 PP 3.14 PP 3.14 Calculate the angies between (a) the [100] and [110] and (b) the [100] and [111] directions in the cubic system. (See Sample Problem 3.11.) (a) Se arca (I22$S )camcmerto = (b) dz are cos (Lioe 0), zarcos qe = 547º Sketch the (371) plane and its intercepts. (See Sample Problem 3.12 and Figure 3-30.) interepf af c PP 3.15 PP 3.15 Sketch the 12 members of the (110) family determined in Sam- ple Problem 3.13. (To simplify matters, you will probably want to use more than one sketch.) Calculate the linear density of atoms along the [111] direction PP 3.16 in (a) beciron and (b) fec nickel. (See Sample Problem 3.14.) (a) Fr be Fe: PP 3.16 to ds —L—s PO Rr AlelâGmm) (bh) Far fee Ni: a= Drs fo (0/25 mm) = 0,85 4m YZ 4 = 163 ama fim 03 atra hm 4 de Los Ls Do = ga" PE (03S tum) PP 3.17 Calculate the planar density of atoms in the (111) plane of (a) o beciron and (b) fcc nickel. (See Sample Problem 3.15.) (a) Fr bu Fe: PP 3.17 ds TE Ta * Pp (Odafnm) = 0.286 4m LlvZa =yZ (0.2fbam)= 0.405 nm E AL VE z am A: Ss g (0405 m)'= 0.074 “ ctemit Seus = 05 0.5 afim A O. ng + E 70% atimalhmã (b) For fe NM: HaNZ a Voa Q = 035% em PPB) gives: UV E (035tnm)= 0.500 nm Asp AB (o.smme) ta adotam * 4 ame dis = 2 sfma = atra a A O.ios nm? = EE afrma m é PP 3.23 The diffraction angles for the first three peaks in Figure 3-39 are calculated in Sample Problem 3.21. Calculate the diffraction angles for the remainder of the peaks in Figure 3-39. PP 3.23 do = B0Stm q 0.40 %ma . = O./ãz SU ate vu nm O.G0bnea O. Sum 2 ieama o = 0.H4Pam atrasar VIA = Di $OBnm, O. GIL am Laos = = = OO! nma deem 4 d, Odo nm O. LO Lnm 334 = 0.09274m = 2: HoBhma = CSM 50903 hm + o Va*rarro vas = OMS Am . a 225º Psy a RIIEn 393º or (É E) OISLA nm . . = amsm DECO gg"! a (28)5 128 qas AO Tam =H4" a Das 8 DtES2nm = º = e dos o adota or 0 725 3 Ax0.OPATaM Orar QEfdam q gps or (28) q 113º Cup 2 incam DAS RÊna sp6t ar (26)qu, me 17º x0.otas nm Section 4.1 — The Solid Solution - Chemical Imperfection Copper and nickel (which are completely soluble in each other) PP 41 satisfy the first Hume-Rothery rule of solid solubility, as shown in Sampie Problem 4.1. Aluminum and silicon are soiuble in each other to only a limited degree. Do they satisfy the first Hume- Rothery rule? PP 4.1 PE C143nm, Po; OMThm OS om = CiPam 043 nm Fhuetre, no (% redua dê > 15%) The intersútial site tor dissolving a carbon atom in a-Fe was shown PP 4.2 in Pigure 4.4. Sample Problem 4.2 shows that à carbon atom is more than four times too large for the site and, consequentiy, car- don solubility in e-Fe is quite low. Consider now the case for interstitial solution of carbon in the high-temperamre (fec) suuc- ture of y-Fe. The largest interstitial site for a carbon atom is à toi type. (a) Sketch mis interstitial solution in a manner similar to Figure 4-4, (b) Determine by how much the C atom in v-Fe is oversize. (Note that the atomic radius for fec iron is 0.127 am.) PP 4.2 |(a) x/00% = if.2% Catm af do] (b) de dra =? a rot E ndoutihal 4 Ernst O E 2 Podastidal = VR re = re E CAIS aço O BIE(a la Tn) = 0 0Saé nm or arm À 00TTmm Pombtutihal 0.0 52 nm Tamefot, Mo Cabo ado 3 Peupbly 50% %o brge. Note: This da am impwtal problem. LH Shrios dead La individual infecta do de are larger Áher those dr W= Fe, evtm Hovah JA ormall ALE for dEFe :s greter . = /.46 Section 4.2 — Point Defects - Zero-Dimensional Imperfections Calculate the density of vacant sites (in m”?) for aluminum at PP 43 660º C (just below its melting point) where the fraction of vacant attice sites is 8.82 x 1074, (See Sample Problem 4.3.) -4 -4 as -3 8.82X/0 atm x b.02X40Cadomnar mm t PP 4.3 vac. demszty Ss. 3144075 Section 4.3 - Linear Defects, or Dislocations - One-Dimensional Imperfections Calculate the magnitude of the B urgers vector for an hcp metal, PP 4.4 Meg. (See Sample Problem 4.4.) “ug PP 4.4 Fr My, da hep metal, Again bla my = Plolooam)= 0.320 hm Section 5.1 — Thermally Activated Processes Gaven the background provided by Sample Problem 5.1, calcu- late the value of the rate constant, &, for the oxidation of the magnesium alloy at 500º C. PPS.1 Va Csuosa ke “AMET e Almmank RÚTISK STE, PP 5.1 Lao “Ás E e Also se Ao o é ) Ger shco) Assu) =/o05x17 cegos isje E34 Tfmoi- Té ) 18 = " e =( “ Ys.45x107) = 0.572 Ag /1mt.5] Section 5.2 — Thermal Production of Point Defects Calculate the fraction of aluminum lattice sites vacant at (a) 500º €, (b) 200º C, and (e) room temperature (25ºC). (See Sam- ple Problem 5.2.) PP 5.2] /a) 44 So0ºc (a 7734), n -EviéT E Late" (na)e Cano Con) Pisces. = pasmo + Za PP 5.2 [b) At aoore (= 4734), = LOT6 RV) -+ em - =900x%0" Anda a tude” Cr eno-taviz)(4I E) (o) At aste Crasr Ke), — Lozoev) “a =(m2) e (tano devia =4SIxio Mastes Section 5.3 — Point Defects and Solid-State Diffusion PPa3 Suppose that the carbon concentration gradient described in Sample Problem 5.3 occurred at 1100ºC rather than 1000ºC. Calculate the carbon atom flux for this case. PP 5.3 The only diflusue for He tumeto caleutafim da Ya di Poa : -a e Dex. tre note = De fr Cromo tm /6) e] MU Zlmol)fl 314 hat 15734) Tamo! ms AI tea Tx -DS£ =-(G mis NM = & a (Ut axo E 6X 8.23 xo 2) x = ESaxio!? arma Alia 2s) me  PP 5.4 In Sample Problem 5.4 the time to generate a given carbon concentration profile is calcutated using the error function table, The carbon content at the surface was 1.0 wt % and at 1 mm from the surface was 0.6 wt %. For this diffusion time, what is the carbon content at a distance (a) 0.5 mm from the surface and (b) 2 mm from the surface? PP 54] f)g=-2 = Sort =0.232 + ADE à (a sino m 5) (B.bfni0 ts) 0:a887 Into bom Tales] qtas D.28-0. 2387 Caom3-ertf 0.25- 0.ã0 0.0W3. 02227 er efi= o. 264% 8 Minado Cy Co I-erês Es - Co or Ca rO. Qt (. 0-0.2)04% IP Cx OP ufa =/-042 = 20073 IE — = lh) À A vlateno mA 368 W015) 0.9549 É rlepoutog em Td 4.1 o 400-0.95497 . e g427 Cro-oas E DOMIro ET” er erta=0 8230 II Cx co 2% (1.0-0.2)0%% or Cx = 0.340 A =/-0. 8230 Section 6.1 — Stress versus Strain kn Sample Problem 6.1, the basic mechanical properties of a PP 6.1 2024-T81 aluminum are calculated based on its stress-strain curve (Figure 6-3). Given at the top of page 201 are load- elongation data for a type 304 stainless steel similar to that pre- sented in Figure 6-2. This steel is similar to alloy 3(a) in Table 6.2 except that it has a different thermomechanical history, giv- ing it slightly higher strength with lower ductility. (a) Plot these data in a manner comparable to Figure 6-2. (b) Replot these data as a stress-strain curve similar to Figure 6-3. (c) Replot the initial strain data on an expanded scale, similar to that used for Figure 6-4. (d) Using the results of parts (a)-(c), calculate (d) E, (2) Y.S., () TS. and (g) percent elongation at failure for this 304 stainless steel. For parts (d)-(f), express answers in both Pa and psi units. Load (N) Gage length (mm) Load (N) Gage length (mm) 0 50.8000 35,220 50.9778 4,890 50.8102 35,720 51.0032 9,779 50.8203 40,540 51.816 14,670 50.8305 48,390 53.340 19,560 50.8406 59,030 55.880 24,450 50.8508 65,870 58.420 27,620 50.8610 69,420 60.960 29,390 508711 69,670 (maximum) 61.468 32,680 50.9016 68,150 63.500 33,950 50.9270 60,810 (fracture) 66.040 (after fracture) 34,580 50.9524 Original, specimen diamet  PP 6.1 loso(n) Clmm) est Cum) Tim) E ze tó-do ZA M aah, e |5ogoea o o 8 4 po So.gima e.oea 3p.4 0.0008 Gm l|satãos | 0.023 Tu2 | aoco4 M670 I|sotros | 0305 use o.co06 M$60 | Sorgo 00406 1844 0.000 7 atiso |sessa | 0.008 133 0.16 2740 Soft Cobro ar 0.012 aY390 Sa ff oo7i! asa ema I3bto |satou | o./ou ast cozo 33180 |5a9%2% | 04270 au Cas Isto | 50.924] 0./524 273 0.36 ISxo |sagym | oie 278 0.00387 IS |shvosa | Gãosz asa evoso fosso SA 8H (2 Sao aoz 4% |53.34% | a.s4o ara 9.0sT Ses | 5stm | Goro Ghk e.1o bs |5r4o | Zãão Sao es” 6,400 |40.%0 | votos sa e.20 bb [oLdet | to.6es sso sa! bgiso |$3.5% | 42.700 ss 0.28” oro I|bhodo | is. aso «so e.30 2 FA Y(i27um) (o) to | Crow) | ao I o o to ao 4L Gmm)  o Vis ts Pis aa o 0.00) 002 0.003 0.004 E (d) The Comstucdim im he, grega of part le) indicados Hd Es 193MPa 193 MPa 2 G00)º E 193 uso MPa 108 Az , x ZM DA + 0.14Sxto” “pai fo = aAZoxto dpsi  PP 6.7 PP 6.7 PP 6.8 PP 6.8 PP 6.9 PP 6.9 Section 6.2 — Elastic Deformation (a) Calculate the center-to-center separation distance of two Fe atoms along the (100) direction in unstressed «-iron. (b) Calculate the separation distance along that direction un- der a tensile stress of 1000 MPa. (See Sample Problem 67.) (0) fr a bee stuctuna, de epetartima distene ud ice paremetera & = Alictam = 0.2MS nm (b) es SE = CrrcomPa Vasto py= Soor o Asnehhed = 008% O MM Mum = A QRET nm ax Section 6.3 — Plastic Deformation Repeat Sample Problem 6.8, assuming that the two directions are 45º rather than 60º and 40º. Ta Des Acsf= (obI0MP)csL4S") as(4s?) = 0.345MP (50.0 pai) Section 6.4 — Hardness Suppose that a ductile iron (100-70-03, air-quenched) has a ten- sile strength of 700 MPa. What diameter impression would you expect the 3000-kg load to produce with the 10-mm-diameter ball? (See Sample Probiem 6.9.) Fique 6-28(4) gire (her TS=T00mPa) 220 BUM. 2(3000) Rao it —— - A o)[to- Viotd* 3 or da 4,0f mm  Section 6.5 - Creep and Stress Relaxation 6.10 Using an Arrhenius equation, we are able to predict the crecp PP 6.1 rate for a given alioy at 600º C in Sample Problem 6.10. For the same system, calculate the creep rate at (a) 700ºC, (b) 800ºC, and (e) 900ºC. (d) Plot the results on an Arrhenius plot similar to Figure 6-34. . PP 6.10 (a) E = CPo.sn 108% pn hun) e Teotc = Lé700? U per hun = (2x0) (e s14ya73) (b) É s quore* (PO-Sncob% pe ham) €” Camo) fergie Ko) = 4x0"? % per hun (e) Êo = Cro S mit % perbun)e = 198x0003 4 per hun = (asno se 73) td) Teo 4 jo so = o TT T 7 co. Hot N 4 Bo) ko Li uz dxteoo (Ro) PP 6.11 [E] Im Sample Problem 6.11, we are able to estimate a maximum service temperature for Inconel 718 in order to survive a stress of 690 MPa (100,000 psi) for 10,000 h. What is the maximum service temperature for this pressure vessel design that will al- low this alloy to survive (a) 100,000 h and (b) 1000 h at the same stress? (a) Fw des fimo sf 10h: 6.11 4 hu é PP eo” Tee ss Sao 72 s9s 4s é so  Co» Tessore ===. (b) for a “Lugtha time sf s0? 4: Elhui) Tee) ido =s40 dio 595 &s 6so ss 705 Tere) ts Tséesc (ln menidoina fls neualts of Sample Prédio 6.114 amd Practice Problim 6:11, ut rede dba tender la girêa assa, à Las É service dempradns of muck lua fha do0º€ lenda do à tuo order co f- mag mito chaga dm liliboma.) PP 7.4 Section 7.4 — Thermal Shock In Sample Problem 7.4 the stress in an Al, Os tube is calculated as a result of constrained heating to 1000ºC. To what temper- ature could the furnace tube be heated to be stressed to an acceptable (but not necessarily desirable) compressive stress of 2100 MPa? PP 7.4 G= Ee = EqxaT PP 7.5 PP 7.5 ppsi E o 2100 ma, sa = = = 645º or Ta (645 as) C= 67% In Sample Problem 7.5 a temperature drop of approximately 50ºC€ caused by a water spray is seen to be sufficient to fracture an AlzO; fumace tube originally at 1000ºC. Approximately what temperature drop due to a 2.5-Ib/(s - ft?) airflow would cause a fracture? A gem dean Fiquaa 7-8 Cel netioy EA Ds rango of 1h asseciadel tl fia Cmditom, the Aeeegpirafne dp would de : & aS0% fo foco'e (x TOO “C mid rango) Section 8.1 - Impact Energy Find the necessary carbon level to ensure that a plain-carbon steel will be relatively ductile down to 0ºC. (See Sample Prob- lem 8.1.) PP 81 As m Sample Publem 8.1; wa see rm Equs 8-3 40) Pat” tua neud à emb devel <oscA. - (Bro nto Smfu XP. x1076 et) PP 8.2 Section 8.2 — Fracture Toughness What crack size is needed to produce catastrophic failure in the alloy in Sample Problem 8.2 at (a) 4 Y.S. and (b) 3 Y.5.? PP 8.2 ta (a) Eres GYTa or au Fo PP 8.3 PP 83 a= Gimps=)* mEmciseomad 2 (6) qu LEMA assuma 25S tam “vv (940 mA)]* = 0,0/21m = (9.9 imp In Sample Problem 8.3, maximum service stress for two struc- tural ceramics is calculated based on the assurance of no flaws greater than 25 um in size. Repeat these calculations given that a more economical inspection program can only guarantee detection of flaws greater than 100 um in size. (a) As q = ara, Le imamimum senvite Strenç dá citl be reduad by a Aacbr of /25/100 =0.5 . Go roguer E FIPMA nas = 169 /nPa Cb) Somdeha, Cê, tese = lozomà nos » EDEMA PP 8.4 PP 8.4 PP 8.5 Section 8.3 — Fatigue In Sample Problem 8.4, a service stress is calculated with con- sideration for fatigue loading. Using the same considerations, estimate a maximum permissible service stress for an 80-55-06 as-cast ductile iron with a Brinell hardness number of 200 (see Figure 6-28). As m Somplo Probltm 8.4, Service sbess= ES. (Ts). TS. -* 20 Fra Fique 6-284), a EHN'=200 exmespoda da TS taomA. Then, Serve stres = btt 775 MP For the system discussed in Sampie Problem 8.5, what would be the time to fracture (a) at 0ºC and (b) at room temperature, 25ºC7 PP 8.5 (a) £ =. c AT 4 widl Cs Susatos add Qu LAT 0a! gt oc, (A, 600 T; e taisuare e Niue Aron? q! or t= 2/55 (b) Masc, . (75600 )hp 314218) E'= (sinto s-) € = psjnor ds! er t= Jtbs  Section 9.3 - The Lever Rule Suppose the alloy in Sample Problem 9.3 is reheated to a tem- PP 9.3 perature at which the liquid composition is 48 wt % B and the solid-solution composition is 90 wt % B. Caiculate the amount of each phase. 20-50 (IAsj= 954 PP 9.3 1= Go-49 bg) = teia Mes E = =4º (JÁ (Ay )a 449 'zs Foge PP 9.4 In Sample Problem 9.4, we found the amount of each phase in ” a eutectoid steel at room temperature. Repeat this calculation for a steel with an overall composition of 1.13 wt % €. G.69- 113 mem (lhg) = O 831 dg = 831 PP 9.4 “= Cesto UA) dy = dig » = f!3=- Fe cs Cthg)a Out69 44 lé?s PP 9.5 In Sample Problem 9.5, the phase distribution in a partially º stabilized zirconia is calculated. Repeat this calculation for a zirconia with 5 wt % CaO. PP 9.5 Nodng dad Suth GOB 40 mel % CO: i5=to 15-a mel. mnstlme = «400 2 = =38.S5 mol *h mol. | Cubic = to-2 x 100 Q = Section 9.4 — Microstructural Development During Slow Cooling li Sample Problem 9.6, we caicuiate microstructural informa- PP 9.6 uon about the £ phase for the 70 wt % B alloy in Figure 9-35. fn a similar way, calculate (a) the amount of q phase at 7; for 1kgofa50 wt % B alioy and (b) the weight fraction of this a phase at T3, which is proeutectic. (See also Figure 9-36.) 90-56 PP 96 | (2), mi ão (149) asetds = bé7s D At To mos L2nso o. - ( Ti me go iso (ldg)20.338.43 3334 W67 PP 9.7 € meulate the amount of proeutectoid cementite at the grain º boundaries in 1 kg of the 1.13 wt % C hypereutectoid steel illustrated in Figure 9-40, (See Sample Problem 9.7.) In eltict toe nssd) do Callado ds eguilébrioa amount sê Cometito af TARIC. = 1180097, ao =40.8 Pre Sh92005 CLkg) 0.0608.64 fit 7n Sample Problem 9.8, the amount ofcarbonin ikgofa 3 wt % PP 9.8 c pn gray iron is calculated at two temperatures. Plot the amount as a function of temperature over the entire temperature range of 1135º€ to room temperature, Frecb pretudcd of = Teg = 0.50 PP 9,7 PP 9.8 From SP PE iapussto, mes tos af room demperatine, mes Jog Cidia âomasnt agplita. up f esstm fada q37€) Ar 739%, . Bo0-o dy es too - 0.68 Ubg)= asia The ntaulitna porão: 3o me 2 ? e ) IC to f ns4€ o too qo go bm do do O Tec) PP 9,9 In Sample Problem 9.9, we monitor the microstructural devel- º vpaent for i kg oí a 10 wt % Si-90 wt % Al alloy. Repeat this problem for a 20 wt % Si-80 wt % Al alloy. PP 9.9 (a) mébote (b) Sotid solubos 8 midl à toegosidom a E /00 vt 45; (e) 44 de eutechio dempinatua, STE (4) At 578%, m = 20226 CA, BO Jos 126 dp) = areias = pára (e) d+ sue, = 00 -o À = = = ue dot (ig ata vi "a egrtaga 1873 4 tbebe * CT tados = 1873 - 8545 tas 4 Si ix tuiniic a = DahoXtt3s) = 13.0 Sim tubetes (7.30 Xooag )= logo a Sim prettuhe a = (100 XESs)= ss 3 In the note at the end of Sample Problem 9.12, the point is made that the results can be easily converted to weight percent. Make these conversions. PP 9.12 PP 9.12 Thenasuits were: (it) Silas + mulirde (ii) Soy : 0% 4hOs Pulido: 60 mal 4 Mh0; Gi) 445 rol % Si0, 4 55.5 mel Umulite Fr Ci), bee 5 no need dor Comvtraimm, For (ti), OmolWAhO= O vb U AL O On fla beaia 400 imolta of LO LS: as a Tepmelsço, * do / 2(26.95)+ 3(06.00) fam = É/IE- dra, ore sO, m 40L(2h09)+ AL 1.00 )] da = 2404 ama 6itê AGE Try apa FARBER MM e TUA, For (iii), em dl bacia f 0 malta of S0, + mullido é Masini so, E S4.5fi2rot + (16.00) Tama, E A6IS Rama MS Sirala muito m ESSE [CANAL ND) e 3031600) p2(2r09) 4 a (a 0/6.00) ] leemaa im ATRG rui th SO, ee de im 36.) % ut molhe = Eaiaca nt Rm 63.4 % E Mole dont flo toledo 0Ê avllote da hermaligal by a facho +P Vs beca 1 muiito dornavla Grasists “= mol lda Componsndã (34303 425:0,). The bom auto do hormalized + TE (aj o,4 Sios) = / mole. drpeititim da ca dante Section 10.1- Time - The Third Dimension In Sample: Problem 10.1, he activation energy for crystal growth PP 10,1 in a copper alloy is calculaied. Using that result, calculate the temperature at which the growth rate would have dropped three orders of magnitude relative to the rate at 900ºC. PP 10.1 Caso = [titipo00 Tfnol)/8.314 T/timol. TT ris — de] 16º —s e = eq or, 3 [ELLA VAR dulo= = (Casais T NM» T= ESSK u Sto e Section 10,2 - The TTT Diagram In Sample Problem 10.2, we use Figure 10-7 to determine the time for 50% transformation to pearlite and bainite at 600 and 300ºC, respectively. Repeat these calculations for (a) 1% trans- formation and (b) 99% transformation. PP 10.2 (1) =13 (Mtboorc) gos (at 300ec) PP 10.2 (b) K7s Catéore) 45003 =25m (af Fc) A detailed thermal history is outlined in Sample Problem 10.3. Answer all of the questionsin that problem if only one change is made in the history; namely, step (i) is an instantaneous quench to 400ºC (not 500ºC). PP 10.3 | (a) 20/0% fre penlto + OWA (b) selo bre parlde + PO % baita (0) 250% doa prnldo + 20% mandansito (relulra à So 2nnnE dA eia 2) PP 10.3 (Dil In Sample Problem 10.4, we estimate quench rates necessary to retain austenite below the pearlite “knee.” What would be the percentage of martensite formed in each of the alloys if these quenches were continued to 200º C? PP 10.4 PP 10.4 Fiqwelo-ló srdicales fla prrindago P mes ténsita frmed till be DIO Ler OE cit GC Figos 0-1] guru 220% Ar OI HO Fiqwe O 15 gives O Ar LIZ Á RC. The time necessary for austempering is calculated for three al- PP 10.5 loys in Sample Problem 10.5. In order to do martempering (Figure 10-19), it is necessary to cool the alloy before bainite formation begins. How tong can the alloy be held at 5º above Ms before bainite formation begins in (a) 0.5 wt % C steel, (b) 0.77 wt % € steel, and (c) 1.13 wt % C steel? PP 10.5 | (a) 2/55 (as denis Eigwe lo-lé ) (b) = ass ximbos= 2a mm (Pra Espuma 10-11) (Ja E hm (Ema E Fique lO= -15) Section 10.6 - The Kinetics of Phase Transformations for Nonmetals PP 10.10 Convert the 62 moi % from Sample Problem 10.10 to weight percent. PP 10.10 | Lmet Bro, = (IizatalM0])dmas 123,03 ama Ímol MO =( 4002 L/6.00) amu = 56.08 ata L mal 0Ê 15 mol Ha allog = OBS mol. Bro, é 4.15 hist CO » = LS EI xa 743% eruão Lois (56.08) 4 0.85 (128,23)] tensa Limi of 8 mal 2. = 0.9f mal Ero, 40.02 mel (AO ) ut.) 0 o.ea(sé. Era, mea(Sbbbama , Zosa (st 00) E 0.3 PC Dna Tama AS OP E Thudot, torta mmoclnic = 24:34 2 4.0.9 mo A =6 A wh% Sectionl1.1. Ferrous Alloys PP 11.1 For every 100,000 atoms of an SAE J431 (F10009) gray castiron, how many atoms of each main alloying element are present? (Use elemental compositions in the middle of the ranges given in Table 11.5.) (See Sample Problem 11.1.) From Tese l)5, bh C= 3.40, wtib Mn= 075, mb SizlAS BB P= el, LS ea Fr a too4 Alloy , Ha bill be 3409 Cefom Assemg baleece dia Pe tlsnt nl! de tora = (Ido + 0.754 LES tolas odada = Java Fe Aimber vÊ afima m toa y adey ce: Mes BM ua soarno ll arms LOU x 10 adoma PP 11.1 sss es as .* = puxo « Marti 0 erra, No, = E x , = 3a No: RE» a = 233040?! a 4 Sat .. = Assu 40% + Oil pass x1028 q Fr 100, 000 am allen, Meo E ChompunadA fhastuost )u (08 adome = E2,136 Ve =(U7osmos/ Su + = 13,85/ Ma é Boo) 0 Dur a és Ns = (Bio) cu ya = Saas Np = Cosgatoi!/ " De a = 489 Ng = (ras ne q Dr a = 183 Sectioni1.2- Nonferrous Alloys PP 11.2 PP 11.2 A common basis for selecting nonferrous alloys is: their low density as compared to structural steels. Alloy density can be approximated as a weighted average of the densities of the con- stituent elements. In this way, calculate the densities of the aluminum alloys given in Table 6.1. =0.97152 + 00125 2 +0.01 Em a atm My =Paris(ado) + coas(247)+ 51 CIT My hu? = 2.75 My lm? DIS, 041 = 01 rocota + DONA, OS, = [0:24202.70) 400080247) + 0.053 (8.98) 200507460] My Sa 241 Ma fm! Section12.2- Glasses - Nonerystalline Materials PP 12.2 PP 12.2 In Sample Problem 12.2 we caiculated a batch formula for a common soda-lime-sílica glass. To improve chemical resistance and working properties, Al, O; is often added to the glass. This can be done by adding soda feldspar (albite), Na(AISi;)Os, to the batch formula. Calculate the formula of the glass produced when 2000 kg of the batch formula is supplemented with 100 kg of this feldspar. 2000 ba of bath girta: 0.2/6 4200049 = 432 Ay MÁ, Cs J Í O !so nu aÃ. 300 és Geo, 0.633 - 2 /abé dy Sa, Note dat Ma CAIS, Jo, = dor dALo, + 354, mal esto Ma (AISO) 5/2041 4 26.99 4 3 (24.09) + 2 (lá00) Jara = 262.28 Qua molut EMa,0 = É [202249) + (6.00 deus 30.97 tua bel. uti É AO, = É /2026.15)8 3(16.00) ] Ra = 0. PB Ama mol ut. 3Si4, = 3 faros, acigue)] Qmu = 180.27amu Thén, Loo bg Ma (MIS; Og gredds : Fon x 00 bg a tg 4,0 agaas J dg ndo by 2124 bo Mo Soda tm &y = 47 As Soa aba as Usas Calentubirea Prom Sample. Ph blêm 1 2, 4324 MO, —> tits xt324p= asa tt 40 dos. 47 a Soo bg Go, — ter x Imdy = de 81 dg ão For bla Boal produt: MAO = LB tas26) As Eat A, meo * der) dg Panoo 12444 Pa E (63.7 + 1266) Ay = (334 74 = (odds Mb E REA UIIA, ds = 17Pkb bg Motel The pesada 9 lana mada ao ct uMçO a BELL rosto = (42% tt do Ca O e xt0 2 = PLA, CAT Ms cet d = 1% atm SO, = 1ESTT mom 77H  Section 12.3- Glass-Ceramies PP 12.3 What would be the mole percentage of Li;O, Al;O3, SiO», and TO, in the first commercial glass-ceramic composition of Table 12.7? (See Sample Problem 12.3.) PP 12.3 For a Loo 3 e beca crramic, Table 2.7 Girea: 49 St + EI LO IML de és na, Vsmg catulatira from Samsol PrblmolZ: 3; T49 Stu Imelbasg a = 423) mol 4 biso x Imsi loops = AIZS mal eg All, x dt fo E OST mol Similanda, habito Tr0, = [4740 4 20H08] Que = 7990 Goa és TO Ementas = 6075 mol TotÊ no. melta =[Á23/ 40.134 40.15740075)mel = LSI7mal As a negults mol Sid, = hey nto m 27) mol LO =O3sismpamn= Ed % bol th All, = OsTismrmã= 28% mol% TO, 2.075/:547 som = 47% PP 13.4 PP 13.4 Calculate the degree of polymerization for a polyacetaimolecule with a molecular weight of 25,000 amu. (See Sample Problem 134.) < mel er (CO) mat ut: CH O - 25,008 Qumas, - P33 Lraot + ati) 16.00 Jomas Sectionl3.2- Structural Features of Polymers PP 13.5 In Sample Problem 13.5, coiled and extended molecular lengths are calcutated for a polyethylene with a degree of polymeriza- tion of 750. If the degree of polymerization of this material is increased by one-third (to n = 1000), by what percentage is (a) the coiled length and (b) the extended length increased? PP 13.5 (a) As La vn, PP 13.6 PP 13.6 1 L, he 1000 T, hd x =V5eo = L155'or amina o? 18.5 (b) As Lens Tb quill emporrtmeeo Calby moitê m ) dm imcrenas 0 33,3% A fraction of cross-link sites is calculated in Sample Problem 13.6. What actual number of sites does this represent in the 100 g of isoprene? Tregctro of Figuat 6-46 mdiutea ct Cros-lmk perman. Using Me tanto of Sampa Problem 13. 6, Motas E 0.425 N, = (0.425 1004 (oboaznio?* mol”! da [stiao)+ Rma)]a Amo! =3,Uuo?s Section13.3- Thermoplastic Polymers PP 13.7 Calculate the weight fractions for an ABS copolymer that has equal mole fractions of each component. (See Sample Problem 133) PP 137 | Filluing de Cnleutatioma of Sa-pl Publie, [3-7 1 mile Az [Bliao) + 30hoor)+ I40]tmu= 53.06 mu Lota Be [4 (120))4 ECtcor) fama = 54.09 amu Late Sa LP (t20)+ E Cito Jam = 104.14 rei E3.06 - . ULLA = coa SAM ATA =100% = 25.) % VB = ES = 25€º Als ego Mt = ds é utnS todilã ita 49.3 A = G306 4 54.04 4104.14 PP 13.8 What would be the molecular weight of a PPO polymer with a degree of polymerization of 700? (See Sample Problem 13.8.) PP 13.8 molut= n (mol ut. PPOmA) = 700 [ Piz) ACLoA lbvo Tama = 841004 mu Section13.4- Thermosetting Polymers PP 13.9 The molecular weight of a product of phenol-formaldehyde , is calculated in Sample Problem 13.9. How much water by- product is produced in the polymerization of this product? PP 13.9 |4sshnm ix Soplo Publ 13.7, 15 moltlo sÊ Ho are prduad almy «ná ova Ima of phêmel- Lormaldrhade : Uia ema phimal- fmuldulado atos en td L5L Altoor)e dé.vo Jam = 2702 amu HO Theo, mm =m PTOA Quis “o praduca E MM praduete x HR. 12 duras -— 27.0 = 14 (E) = 3.374 PP 13.10 For an elastomer similar to the one in Sample Problem 13.10, calçulate the molecular fraction of each component if there are equal weight fractions of vinylidene fluoride and hexafluoro- propylene. PP 13.10 | (005 > 503 vinylidme Aomide +50 9 Ara Huaroprepolime Vsing do nal o Senda Problem 13.10; mal. uronglidome Hlmride = É malz 0.MB/meo! 64.049 mol Áca Ávom ppolére = 507 oj= 0.333 mel 150.08 — + LHE mol 0 mal Pac. ruylidane Plrida = ATA Limal, o. 701 LM Emol a = 0333 mol mol. Pac. Arcallvorprpglena THA mor 0299 PP 14,3 Section 14.2 — Aggregate Composites Calculate the weight percent of CaO + Alz03 + SiO» in type II portland cement. (Sec Sample Problem 14.3.) PP 14.3 Fi lloning de procedume o? Seade Prsblém, 14.3, PP 14.4 ivo dy Ty pel Cambe qreldo 53 Es, Nba Gs, lt 4 CA, od 74 GAR. Total mess 0a Co va7Nszda ) + Crest) Dk Cotas Ku-da ) EMI Lg )= cotés Totel maca AlO,* (O, STD da) + fo 274) = 604 Til masa Sr0, = (0.263 Ws249)4 C34 Lg ) = Rob ds, Them, deal AUTO ALE SO, = (6ALE OA 2OL) A = go A Calculate the density of a particulate composite containing 50 vol % W particles in a copper matrix. (See Sample Problem 144.) PP 14.4 Erema A ppendtiza í, 3 Ps IAS Maas ad ps 2.93 Mg dm 1m3 Composite. arelda a soniw to. Som WU or, pefostimasdro. SCE] Mg lu? = LEMA? Section 14.3 — Property Averaging PP 14.5 Calculate the composite modulus for a composite with 50 vol % E-glass in a polyester matrix. (See Sample Problem 14.5.) PP 14.5 Ev Ent % Ep to. s)C6.IxpoSmMPa)+ es Ã7a.4 xro8mPa) 32.7 x40! MPa PP 14.6 The thermal conductivity of a particular fiberglass composite is calculated in Sample Problem 14.6, Repeat this calculation for a composite with 50 vol % E-glass in a polyester matrix. PP 14.6 | Au. uk, = (0.5)[047W/6m)] + to.5)[0.97W/Cm-K)] = 0.57 W/la.k) PP 14.7 Calculate the elastic modulus and thermai conductivity perpen- dicular to continuous reinforcing fibers for a composite with 50 vol % E-glass in a polyester matrix. (See Sample Problem 14.7.) Em PP 14.7 | Ec=5 EEE = Li fnome (72% x10IMB,) Ta sa. Lupo 3 mPa) + Ch SCI nto Em Pa) = !Abxto?mPa Ade == Ud + Y ka Lot WO 6) LAT WC) “tos Lema be )] + Co SIDO! 7 W/ 4H] = 0.29 Wim.) PP 14.8 In Sample Problem 14.8 the case of a modulus equation withn = Q is treated. Estimate the composite modulus for a reciprocal case in which 50 vol % Co aggregate is dispersedin a WC matrix. For this case, the value of n can be taken as 1. i PP 148 | Duca, 4 Eça me,“ v,E, , A = (2.5)(207 xeo mB) Cos tosnto'mPa) 4 = 20.46 (10mB) “a - e Eç= E w/0 ma,  Section 15.1 - Charge Carriers and Conduction PP 15.1 (a) The wire described in Sample Problem 15.1 shows a voltage drop of 432 mV. Calculate the voltage drop to be expected in a 0.5-mm-diameter (x 1-m-long) wire of the same alloy, also carrying a current of 10 A. (b) Repeat part (a) for a 2-mm- diameter wire. V= IR To E fe fio (33. Ino! 2 (Lim) a CcBA q (225 eos mw)? V= (WAOLI3L. ) = 73V &) Rx CB 3 Pxto! zm) (Lm) E = o.orors2 TF (tvo-su) V= Coat A = ofutva bpm PP 15.2 How many free electrons would there be in a spool of high- purity copper wire (1 mm diameter x10 m long)? (See Sample Problem 15.2.) PP 15.2 A Vere = (1040077-3)x [Cos wr008m)?] Com) = PI7nã. PP 15.3 in Sample Problem 15.3, we compare the density of free elec- trons in copper with the density of atoms. How many copper atoms would be in the spool of wire described in Practice Prob- tem 15.2? PP 15.3 | norafms= Cabo deck) Vime (Edit we Dal Co suto'8m Tom ) = bésSwo 33 PP 15.4 The drift velocity of the free electrons in copper is calculated in Sample Problem 15.4. How long would a typical free elec- tron take to move along the entire length of the spool of wire described in Practice Problem 15.2, under the voltage gradient of O. Vim? PP 15.4 t=1/& = (tom/(h78n0 m/s) =s Timoês x lh/3g0s =159 hm Section 15.2 - Energy Levels and Energy Bands PP 15.5 — Whatis the probability of an electron's being promoted to the conduction band in diamond at 50ºC? (See Sample Problem 15.5.) PP 15.5 / Me) ceEDaT t o eme : -& ef Pev)/Creano Ce viKY323E) 2! = Qllxr- PP 15.6 What is the probability of an electron's being promoted to the conduction band in silicon at 50º C? (See Sample Problem 15.6.) PP 15.6 A fre) º e (o. SSIS er)/C Poa ero Ce vix (323 4) l = A.3ân/009 Section 15.3 - Conductors PP 15.7 Calculate the conductivity at 200ºC of (a) copper (annealed standard) and (b) tungsten. (See Sample Problem 15.7.) PP 157 | (0) 2 £, litalr To] = a4 em? 1Zem Lit 0.00393*€"!(R00-20)%€] = AGd no! ID m = yp= Bs ont ar (8) P =(ES(n601 Lim tho00 45 "cr 20o mode | =997x0 EtZim ga Vo ma Lo.0n 108 RL! PP 15.8 Estimate the resistivity of a copper-0.06 wt % phosphorus alloy at 200º€. (See Sample Problem 15.8.) PP 15.8 | Frm Fipe, 15-12, Pare, cuosP E AS 3x0" Z.m Pluma Sample Psblta 15.8, aos, acasos? = (28.30 *2.m)[ pacosas ter!fRoo ao)xe] = 43. 200071 D.m === S PP 15.9 In Sample Problem 15.9, we find the output from a type K ther- mocouple at 800ºC. What would be the output from a Pt/90 Pt-10 Rh thermocouple? PP 15.9 | TadlelS3 shma Lda dela Pe /90PE- (ORE Brimecunlo de ope SO Fgime [515 sb deu Ma Lgpe S Hirmncaplo hua des ctg 0! Tmy «S SOO'E, PP 15.10 When the 1-2--3 superconductor in Sample Problem 15.10 is fab- ricated in a bulk specimen with dimensions S mm x 5 mm x 20 mm, the current in the long dimension at which superconduc- tivity is lost is found to be 3.25 x 10? A. What is the critical current density for this configuration? PP 15.10 As mm Semalo Proflem 15:10; entrl ho 3 25x403A Corra dão Tera E a) E = 4.30 n10 BA fm * eshiçà Aa nov fer ordira É rmaguitada lotota am rala br de Vu Piho “onfepcadrm . Section 16.1 - Visible Light PP 16.1 Calculate the energy of a single photon from the long wave- length (red) end of the visible spectrum (at 700 nm). (See Sam- ple Problem 16.1.) PP 16.1 A Es Se A = Coto rs oznexio ma) caganroEV - x 7oox «o? m Tr = Ê Te V Note: This sil be de fenil Cragelho visible laght prof. Section 16.2 - Optical Properties In Sample Problem 16.2 a critical angle of incidence is calcu- lated for light refraction from silica glass to air. What would be the critical angle if the air were to be replaced by a water environment (with n = 1.333)? PP 16.2 PP 16.2 | Ti di wu, PP 16.3 Using Fresnel's formula, caleulate the reflectance, R, of (a) a º sheet of polypropylene and (b) a sheet of polytetrafluoroethy- lene (with an average refractive index of 1.35). (See Sample Problem 16.3.) PP 16.3 | (a) Usmy Table do.2, a a p= (E = nl) = 00362 n+t LGI+ (b) R= (ist Ê o 0.0222 135+1 PP 16.4 What is the reflectance ofsingle-crystal sapphire, whichis widely used as an optical and electronic material? (Sapphire is nearly pure Al03.) (See Sample Problem 16.4.) PP 16.4 | Usmy Table 6.1) P= (= a TUn+r = (nb 1 2 (as PT) ) = povse PP 16.5 The relationship between photon energy and wavelenpth is dis- cussed in Sample Problem 16.5. A useful rule of thumbs that E (in electron volts) = KA, where à is expressed in nanometers. What is the value of K ? he PP 16.5 | E= 2º A fe a toravelimadá 2% nm, E- Co-bbag no 32 7.5 (0.2998 » 09 n/a) LI Pamo lt (e nm 0 40-80) JT = faso x ev or K= faso  Section 16.3 — Optical Systems and Devices PP 16.6 As noted in Sample Problem 16.6, the band gaps of semicon- ductors are a function of composition. By adding some GaP to the GaAs, the band gap can be increased to 1,78 eV. Calculate the laser photon wavelength corresponding to this larger band gap. PP 16.6 q e Es = Cobrar PT Mo asp mo ml) E agauotev 178eV ad = 497 x 1071 m = 617 um PP 16.7 In Sample Problem 16.7, we find the critical angle of incidence for a light ray going from a glass fiber core into its cladding. Calculate the critical angie of incidence for a single-mode fiber design in which the core index of refraction is n = 1.460 and the cladding index of refraction is very slightly smaller, n = 1.458. PP 16.7 mn é arcsin — Sladding core 1.458 LHo = arcsin = 870º PP 16.8 Repeat the wavelength calculation of Sample Problem 16.8 for a ZnSe photoconductor with a band gap of 2.67 eV. PP 16.8 2 he E = CObbabxro 3 Tso. 2448x707m/5) ab adaxio ev Ab TeV F = 4b4xiolm = 464 um  PPITA | A = ass & yo temê 0. 0033 n10%É atema, $ euê us aro? 4 = Sooxç0?” etica /m* Usma dedo mault É Sample Phsbltm 17, Luz fodnio”! x So 0wt037 area fu? = 5.20n002!ufma do? (cx do x10 28 atom /m 3 1 Table 73) PP 17.5 In Sample Problem 17.5, we calculate the probability of an elec- tron being thermally promoted to the conduction band im a P- doped silicon at 25º C. What is the probability at 50ºC? ! PP 175 | fe)- EST, ! DD 24352) /0 Po ato beviR (323%) 4 1 = F4dniorf PP 17.6 The conductivity of an n-type semiconductor at 25º € and 30º C can be found in Sample Problem 17.6. (a) Make a similar cal- culation at 50º€C and (b) plot the conductivity over the range of 25 to 50ºC as an Arrhenius-type piot similar to Figure 17-8, (c) What important assumption underties the validity of your results in parts (a) and (b)? PP 17.6 | (a) come BEAT —Cotev Re, xo *eviR AS É Carç* (AUaro3 LB tm De - k * ) = 135 1240! (6) re) TeK) Et) Cri) Lu (Rim) as ais 3.36x103 100 +.bos 30 303 3.30n/70"3 7 4.673 so 323 3.0 xt03 135 4.405 Res vibra plot: Tc) Go. E as do om. Caim) 40 L J L ) 30 &/ 32 33 24 2.5 Vrxto3 (KT!) (e) Th trbrmir bebavior extede bo SOC. Otlnwisa da supe of da pt a (6) ould not de Crstrand”, PP 17,7 In Sample Problems 17.7-17.9, detailed calculations about a P-doped Ge semiconductor are made. Assume now that the upper temperature limit of extrinsic behavior for an alumintum- doped germanium is also 100º C with an extrinsic conductivity at that point again being 6082! .m”?, Calculate (a) the level of aluminum doping in parts per billion (ppb) by weight, (b) the upper temperature for the saturation range, and (e) the extrinsic conductivity at 300 K; and (d) make a plot of the results similar to that in Sample Problem 17.9 and Figure 17-13. (a) n= Ltsttmo = 1970 m3 PP 17.7 ) h (odbmotte)(o tão mt V4") e EmJ= Ian 03 x SESI AL e lemita Le o. bossio Pata 4! Si3agãe g08 cus = Mbxio! sais Ge = Voe ppb (1) This cudtatim Ja tdenticol do ML dr Sample Problem 1.80). Bgam, Te Isa. (e) Table I13 gre E= Cole Le A à Ge, or q= ce rE/bT ho fetert) et CSA Criramio benta (373 4) 8 PLEG Bm! Ar 300 K, Gr q,e7 Ea/kT =(usI tu) e CAE 2nt0 ben p ra0) SS Oda! Cnductos hype TUE) TER) e) (Rino) Le Crtm) (4) Tre ay clidta uma: Lobensic toa ditemsia 27 mirnsie 135 intrnsie a? 5 4 323 2bprio3 do 4,09 300 JIxo3 Seg 402 404 2459203 bo 4.09 300 J33xi0'3 2.04 o 3 qm Sutura Atnge, N Crtrmsic behevrm PP 1711 PP 17.11 For intrínsic InSb, calculate the fraction of the current carried by electrons and the fraction carried by electron holes (See Sample Problem 17.13.) 7.000 0.144 Section 17.4 - Amorphous Semiconductors PP 17.12 In Sample Problem 17.14, we find that 20 mol % hydrogen has a minor effect on the final density of an amorphous silicon. Sup- pose that we make an amorphous sílicon by the decomposition of silicon tetrachioride, SiCls, rather than sitane, SiHs. Using similar assumptions, calculate the effect of 20 mol % Cl on the final density of an amorphous silicon. PP 17.12 As ix Sample Problia TUM, tabs 9 ef Clad (ivo-+)5 of Si, Mou, XÍ364S (2 Give -=)/ 2.809 as o x=25, Es cr too. x= 7hada Si q = Fo02j 330503 Bala? V PE toa =ãos ada” Which ja qm inerte Oo 33,05tm* SIRI mam = ILE 2.30 Section 17.5 — Processing of Semiconductors PP 17.13 The purity of a 99 wt % Si bar after one zone refining pass is found in Sample Problem 17.15. What would be the purity after two passes? PP 17.13 Note Bom Sursplk Poblimr 1715 foda K of Fbêxio A lendo da à Composoirm abr oe pass cÊ Jota! n3.bauo b(= 2.62 10 641) cobea do 3! Pepréstnla ha mibad CmesLradion oÉ ut, Somilanty 2 a secmdl pass, Al level = 3.bâno CA n3taxoB=t3/ xo tal or 13! parte po bilica (pb) A!  Section 17.6 - Semiconductor Devices PP 17.14 In Sample Problem 17.16, we calculate, for a given transistor, the collector current produced by increasing the emitter voltage to 50 mV. Make a continuous Plot of coliector current versus emitter voltage for this device over the range of 5 to S0mvV. PP 17.IS | Dik fe Seb Publ VUGS Len) V6v) 5 5 So as 236 se A ddiboal data sam Ie (apimA e (Rm é 164) Vim) 16 15 ISP 3s 418 45 fesolig pls: doco e (4) -S06 o O lo 20 30 4a so Viv) Section 18.3 - Ferrimagnetism PP 18.4 Calculate the magnetic moment of a unit cell of copper ferrite, (See Sample Problem 18.4.) PP 18.4 | mapnetic mma Piemito oll = (no CU famit eo O rmoatadt Cu) = Prima = Es PP 18.5 Calculate the saturation magnetization for the copper ferrite described in Practice Problem 18.4. (The lattice parameter for copper ferrite is 0.838 nm.) (See Sample Problem 18.5.) PP 18.5 | IMd= E mol. 0Ê umt cell = LIMITE RrO 3 A m3) (or3exortm)3 = LM rot A/m Section 18.4 - Metallic Magnets PP 18.6 m Sample Problem 18.6, we analyze data for a hard magnet (cunife). Use the similar data for a soft magnet (armco iron) given in Problem 18.7 to calculate the energy loss. PP 18.6 4 cond reesingmad” L do anta im Pad 87 pras os 44 Secptrto: febre Cid totrga, hss = 44 T/m$ Cote do shop Contras do Sameple Prsblim, (MÓ fe ul Ho had aque?” gart a “spo boss 0 BIn16 E Tm 3) PP 18.7 For the soft magnet referred to in Practice Problem 18.6, cal- culate the power of the magnet, as done in Sample Problem 18.7. Tt de data o Prabltm 18.8 alo aba burra : PP 18.7 E Cosbua lat) Bla) 8H (eida AS é Tha) e -as e aa -ào 2.4 az =t0 2s 0.36 o o Pietra gere: 4 (em! (It) 2 z o: pr! 5 Elab /m*) or (EM ua E 33 Thom? Combasd tia coidl dio abas of 101 T/oe in Saem plt Publto 18,7.) Section 18.5 - Ceramic Magnets PP 18.8 in Sample Problem 18.8, we use a radius ratio calculation 10 confirm the octahedral coordination of Fe** in y-Fez03. Do similar calculations for Ni?* and Fe** in the inverse spinel, nickel ferrite, introduced in Sample Problem 18.4. PP 18.8 From Apptudes À: r. ar = Fe 3+ ra 0.06 7um ” = = mat Es 0.078 hm ” =R = 0/34 hm o divina vo He Ladina pados! n/R= (0,0% mo 13anm)= 2.808 aê niRE Co ota) [Lo 1B2um Both pabos comtegondl fp 6-Lud condnatim , emite cn dá dha mereo sprel sbwctue dich Seo dovalf ita Del ontodaa alt à dr octe ledial citou. Bait ore bat ta Prova” a a o ada : É Ee tt udo (0.508) Goma. ame Pr detuedrml sita. Tia corresponda ad dig fr fo de Aun ditd value (0418). )= 259 PP 18.9 PP 18.9 (a) Calculate the magnetic moment of a unit cell of MgFe,0, in an inverse spinel structure. (b) Repeat part (a) for the case of a normal spinei structure. (e) Given that the experimental value of the unit ceil moment for MgFe> O, is 8.8 11.5, estimate the fraction of the ferrite in the inverse spinel structure. (See Sample Problem 18.9.) (a) As m Sample Problima 14 ado 18.3 (0), agachi tr fot cell = (ho. Mp host velt Viemandt Mg 2+) = 8a Kg = Ms (6) As io Sample Busblem 18.900, maguehe mente femito cell ma — Crop Liitret Mmomado Ma 2+) + Cro. Fe fit get!) Comando Fe 3t) E-(INoug) + (Usa) = ou, (e) Tatay gº Pachm o? Lervide. Dr Dmytrar same: E Coha) + Chy lPoa Je ESAs or Tous Be u)as = Eus or = (Fo-2.3) É Cm = 0,89