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Ch3Sol1 - Chapter 3 Problem Solutions 1 DISCUSS Temperature...

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Unformatted text preview: Chapter 3 - Problem Solutions 1. DISCUSS: Temperature effect and SRO and 1.110 SflLll'l‘IflN: As temperature ineresses atomic sibratinns hicrease and atoms can be regarded as roosting about their “cages" more and more. At the melting temperature the 1.110, which establishes the metallinity of many materials, vanishes; however, SRO being associated with primary bond distanees, is largely preserved. Materials that are not erysiaflhe have no long range order. At the softening temperature or glass transition temperature there is little change in SEE). Some polymers neither melt nor shon-r a glass transition; rather, they degrade by primary bond seimion, which clauses their eompoatinn. 2. FIND: Sketch the pattern ot'a sample ofwallrpaper. Is there SRO? LED? SKETCH: SDLUTIDN: The wallpaperinmyroomhas apattern aimilartothet above. Thelattioe isa parallelogram. Therein one repeat motifpernnit cell. Thewallpaperhasboth 510mm 1.3.0: It fillstheentire 1antral]! 3. DISCUSS: SRO changes with arystalmehing SBLUTIDN: Sflflchangesverylirdewiflimetting. lnfiuidstheatomieaepsrationaare abouttlresameasthebnnd distsaeesinerystallinesnlids. 4. DISCUSS: Sketch tetragorlai and orthorhombic eells. SDLUTIDN: t: m area all arlgiesgll” a E c I flahnrhmms a :L-h a: a all angles 90" a b 5. FIND: Wallpaper design showing lattiee and lattice parameters SOLUTIDN: 23 '5. “85531111": n= 15 cm, h= 33 cm, 'r=’?fl";triciinic lattice FIND: Sketch base-centered whit structure and flow cmtfotmmiun to Fig. 3.2—3 SKETCH: ' SDLUTIDN: The flatter atoms ate amounted bynoting that the cell is simple cubic and containstwu atoms pa coma: site. FINE: RedrawtheFOC structurewithanatominthefamwnerrepodtinnedatthe origin SKETCH: CflMIbIENTS: The cell looks the same! FIND: Wharthemmerltomflinthem orFCCtnuch SKETCH: 24 H}. 11. 12. SOLUTION: ecc ecoraer ttetettre eepuetedbir the“? =e- _h,= trtfi: titr 3'21: Tlmcorneraitomadoncttfluch.Theyare2.31rapart.FCC-Cornflatomaare aeparatedhyam hut atoms touch alongthecuhefacediageml. Heme, n,J§_t-.-;. 4rIJ§= 2.33r32r. Thus, comer atoms do not touch. 'I'hcyarelfliir apart. CDMIMENTS: Comer atoms are clutter in the BCC than in the FCC {see Figs. 3.3-] and 3.3-3}. FIND: Cell that charactcaiees crystalline silica. Sifl: SULUTIflNt A hexagonal lattice. CDMMENTS: flue crystalline form of Sit); is “high temperature quaa‘tz”, which has a hexagonal lattice. FIN}: Calculate the length cfa face diagonal in Cr. GIVEN: Cr is EEC. DATA: According to Appendix C, the atomic tafius of Cr is 1.25 A. SKETCH: See Fig. 3.3-]. summon: The length ofthc t‘eee diagonal it ao'iiz. them touch along the body diagonal], which has length aDNiB = 4r. Staking for an in the latter equation: a0=4rhil Thus, e0x12= 4N2 Ni =4f1.25 AN: t 43 =4_aa h FIND: p etch BATA: Cu is FCC; atomic weight 63.55 girnole; atomic radius = LETEA. ASSUMPTIBNS: Atoms are perfect spheres. SDLU'ImN: FCC : a etemeruhit cell FCC JAE = Jr J. = 3.6153 F=WWGU Ill'tflH = (a eteett ii iii. hjyetetem'e. 023 x to” ammt'e) 23.94 .x i (3.5in theethf g m comm: This compares fitrotahly withthe meaam'cd tratue offlflfi stems. FIND: Computcthedmfityofmagneehtmandcompareittothemeasuredvnlue. GWEN: HCP; measured 1Itralnc catatonic radius =1.?4A DATE Atmnic weight = 14.3 gftnole haematite: him the spite-teal SKETCH: StocFigure 3.3-4fcra complete HCPuttit cell. Next considertlic4atcms shown, whichmfi-omasectioncftheHCPcell: tta'eedefine onepianeandtlteiiootth sitsiiii'iciatt},r WthcceMcrofthc three. '1': 13. 14. This is thelbasic tetrahedron OFFICE and HCP. {1115 9111 the unit cell} SOLUTION: p = Wolmne p = (16 x 24.3 mule} I {6.623 x 11212” unmarmuremnm’r 7:16“ 6:113) = 1.13 gran? com-mm: 11.: value in tat-11:5 is a my 1.14 gm; FIND: ShowthattheciamiuinHCPmaterialsslmuldbeabom1.63. Explainwhysorm materials difl‘er 13mm 1.63. ASSUMPTIONS: Mommspherical SKETCH: Useflreskerch from prublern 12. Wencedto detemfirrethednsest distance ufatom‘tfi'omtheplmcdefinedbyammsljandl Thelinefromxtudisuisectcdby PM 1,2,3 Infill-.11.: Thuathedimfiomplm 1,2,3tout6m4132136fthebudy diagonal. Tv-rioethisdistmisn SOLUTION: waflwflmfiwjfzm 1" Mada = w = ifJE== 3.633 2rd?!- COMHENTS: or“: usually is nut 1.633 because atoms are not 11611766111? spherical and band: may be part1}-r trivalent. m: Vellum 611mg: Mic-d with HCP to BBC transition GIVEN: HCP: au= 3.674516, I: = 11.65255. BCC'. a. = 4.116%. 65301191101115 Atoms an: spherical SULUTTOM Vm=rfiwagc 2mm 3.34 = 1' @213. am y (11.3525) = 133. 59141. no: = a: =(J.fl66f= 133.5%! Thacmhowu,adiflthmberufmm1neachunitoefl.Thus,mneedto calculaterhemlumeperatum: 26 15. 1s 1".-'. 18. 19. W Emmy, - (Hit) em. = 23. so ,4! I“? GIMME = (If?) VBC'C = 33L¢fi All {Vietnam - museum) Watchman: = (33 .415-23.lfl]f23.] c = 45% increase comers: This is a huge increase, l{}_3t5A3fp-er atom, which is much larger than anticipated. FIND: Cite the fire meters with the lowest densities. SDLIJ'TIUN: We begin by examining the metals that have the Iowess atomic weighs. Using Appendix A, we note that Li {0.53}, Be (1.35}, 31134), Na(fl.9?), Mg {1T4}, and Al {23(1), [(3186), Ca (1.55}, St: (3.13], Ti (4.54) are the first entries, 1Isrhere the numbers in[ ) are densities in gicm3. CUM: He istoxic, but not so in alloy form. his stirrer}.I high perfonnence metal. Bis used in fiber form for composite rehiforcement. Na rqsontcneous11}.r bursts into flames upon exocmmtoair. LLK, andCasreslsonot stablessmetsls. Mgis comntooiyused (eg. "stag" mes). Al is very common. Ti is usett in bicycle construction and aerospace applications. FIND: Brsvais Lattice GWEN: Atoms are in 1, fl, 0 - and is, is, {Hype positicm; cubic scucturc SKETCH: 2 I SOLUTIflN: All the type positions are shown, which defined .a FCC cell. FIND: IdmfifythePhnefmmedbythe hstercepts I, -l,-2and thepianenormal. GIVEN:mbicstmcture SflLUflflN:Tl1eplaneisthereciprocai ofthe intercepss=tL bar 1, be: 1.32) with fiaclionschred=(2.hsr2,bsrl]. Thcocnnattoaplsnehssthessmehdiccsin direction fonnat [2,barlbar I]. FIND: The tetrahedral angle by crossing vectors directed towards ends of face diagonals inacubcinwhichthecentaistheorigin. 2? 20. 21. 22. SKETCH: _ He.ne.112;=[111] inane-112] = {111‘} -_ ._ w'+w'+1111" . B—_A_H_EDBH___WEE—WMEWHWiJ-l summers: Thus, msfl =(-1+1- ntrfifip -11"3 :16 =109.5”,the“tetrshedrei angle”. Cfll'l'mlENTt-‘i: Remember this angle. It is impenam in mystsllngrsphy end in euvsiem bending. FIND: Determine sngleshetweenfllflhnd Hilfifllflhnd{Ill},snd(llfl)snd [111]. GIVEN: Cuhie system SDLU’I‘IDN: Furthe[HUI]+(llljsetwewillusethefacttlmmesnglebetweenthe plane nermsls equals that et‘ the plane intersection. Furthethirdsetweusethefiietthet [Dfll]liesin {110}. We selves" usingthe _ veeturdutpruduet‘. a.fiffljlflfl}_eusfl=flfx_fl=90" E E: A 3 mg i. {HE}uflflj*eus3=2f151fi_fi=3i3° e. we;- rm; _dt}sfi' =eJ§ _ e= 54.?” FIND: Indiees nfp-nints, directions and planes in Fig. HP3.1 SDLUTIDN: Fer A-F, the origin is the hettum-heek—left cube earner {as in Figure 3.44} unless indicated otherwise. his 0, D, mum-,1, VgCis [[111] uithreepecttu the fient-bottum—righi earner; I] is [30;] 1with respect to the frunt—top-rigln W; E is (flfl ] ); and F is {112}. For GL, the urigin is the bonumfianefi earner unless indieated etherwise. G is 0, 0,113; H is It}, id, 0; I is [It'll] with respect to the neuter nfthe right face; I is [111] with respect In the fi'ulit-tnplef’t eurner; K is [110]; and L is {3231. FIND: Mines ef'pnints, direetiens end pisnes in Fig. I-IP3.2 IILUTIDN: Fur A—F, the origin is the tep-hsek—lefl tube earner {its in Figure 3.4-2) unless indicated otherwise. AisflJrLfl; Bis'ri, 5’5“]; Cis[2gl]withrespeettnthe beet-right edgeeenter; Dislllfl] with respect tetheeenter efthehettem flee; E is (I gmwithrespeettu thetnp-hseit-right euheenmensnd Fisfflil). FmG—Lflnutiginis thehettum-hsek—lefi numeruuiess indiestedntherwise. Gris 'ri, '25, (1; His 1 1 l' Iis [2“ I I‘ :l g] with respect to the tun-frent-right center; I is [IUD] with respect tn the heck-bottom- 23 23. 24. 25. 26. right earner; K is {111.} with [meet to the fient-Mttem—right earner; and L is {[122}. FIND: Iduttitivplanewithimu'cepts Era-2, landgiveitanemlal GWEN: (habieeystem SDLUTIGN: The ree'rpreee] efthe intercepts is 2, -l, 'x&. Clearing {'Zrhetitrahhbyr nafltiplfingby 2 gives 4, —2, ]. Thus, the plane may be identified [431}. The plant normal is [431]. COMENTS: In euhie system; the plane and its nemtal have the same indiees FINn: Sketch {1 DO} and {mill} ina tetragena] eell. GIVEN: Hangman" a=hate suLunnN The-{Hmfieateat‘arnflyefplanee Hence {mo}mc1udes{me},{m1)u an) andtmy 'I‘ms{lflfl}and{flfll]ateintheeamefanfily. SKETCH: cum The {010} family has (me) and (n 10} and diflTers from the {we} and {001}. FIND: Planar demity omen}, (110}, and {111} in FCC SKETCH: Nmethat the atoms have how shrunk to usable fl'll'Ollgh'm timing. HIWF {m x4+ I} £a§= 2:13;? = my? =a.25IrJ prflJ ={H4 14+ m :2) firm: .1945): Efsfir’: fixfir’ = a. 13; ,5? pm!) = rm: 1: 3 + M x swarm x h) = arrraflfix'zflm firzfik {125v ,3 snLU'rteN: In FCC m5 =4r 43:: NH comm: masfimplanardensiqrofumbumpulminmc FIND: Plum densityefflflfi) and (11m inBCC SKETCH: pflflfl) = (1:: x 4).:ai = 3:16 r3 = 9.19: r1 prmy= {:34 .1: 4+ J) : {aux WE) ~= ya§fi= 3:3,: 2 = 9.2:: r2 SULU’I'IUIN: 1:: BBC, a.J3= 4: J... =4rxJ§ma§= :5,le cumnm: Time, m: plmdensity «(110) > {1:101:11 EEC IT. 13. FIND:Dete1-minethe pecking factors fartheunit eel]: inFig. 3.1-1. ASSIFMFI'IONE: All of the drawing 1:: pm eflhe motif. SDLUTIDN: Thepaekingfueteris I fl= lflfl%. 'I'Irereisnuurmeedspmafeulfur which MCEscherierenumed. 29. FEB: Separaliun efclose-packed planesinFCCaudHCP GWEN: (HueepnckedplanesinFCCare{lll)u:dthee-plmeinHCP ASSMUNS: Alumsareapherica] SKETCH: Proceedushg themeeketehasinpreblanll 9.1;le Edefineaelosepmkedplane. Numberdrestsonthehelefomwdiryflje uflrerthree. 'I'hedistamefmmplmel 2 3tuamm4isli’3thedistaneefi'umxt04 sinee],2,3isa{111}-typeplane. - SDLUTIDN: The distance between planes iszxrfr’flqafimhm 41,516 = 2r _a,= r43 Hume, the distance hetwean close packed planes =(Efljr fifi= Mir 30.. FIND:Linenrdensityof[lm]and[Ili]inFCC 3E] ...
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