Ch3Sol2 - Si 32 33 SKETCH For{Ill shown soLU'rtoN In rec...

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Unformatted text preview: Si. 32. 33. SKETCH: For {Ill} shown: soLU'rtoN: In rec: body diagonal =anc’i Fir my]! : Plumb-arming}? = _ “fl = a: F'mu mites-tenets = (to + to) r a: J3 = {1.58mi COMMENTS: The linear densityr of [Will] us greater than that of [l l l] in FCC“. FIND: Volume fraction of'rnoousize fibers {cylinders} in a close-packed structure SKETCH: SDLUTIDN: Consider one equilateral triangle, as shown. The packing in this triangle represents that ofthe entire structure. From stereology the area fraction of this structure equals the volume fraction. Show is 1/2 a unit cell. t' 1" l' . l H" _ 3 2; _ he _ 22..» _ 2 2 3——,_=fl.3?l=91 tss V’ [fix ll" [Elm [a] r M3 W CUMMTS: This represents the maximum voltune section of fiber in a composite or fiber in a yarn, assuming fiber cross—section remains circular. FIND: Coordination ofrod in problem 31 GIVEN: Hexagonal close packing ASSUMFFIDNS: Cylinders remain round SKETCH: See problem So SOLUTIDN: fi rods surround the central one FIND: Why some structures crystallize to But]:~ yet FCC is more dense and has higher 31 34. 35. 3ft. 3?. 33. coordination number. SDLUTIGN: Not all metals can he represented by ideally packed spheres that neglect interactions between the atoms. Atoms may he not spherical, lionds may be partially covalent, or both. FIND: Calculate the packing factor ofa material with a stacking sequence of ABAC DATA: The packing factor ol‘HCF and FCC is t]. H. SGLUTTUN: The packing factor of' the new material is 0. 74 This is true regardless of the sequence, so long as [I] each plane is close packed and {2) there are no that, ER or CE blocks. CUMMENTS: The material would be unique in its stacking sequence. There are presently no known materials that stack ARMS. FIND: Efi‘ect of ] atomic vacancy every mil atoms in Na GIVEN: BCC stnicture ASSUMFHDNS: Atoms are spherical, 1 atoms per unit cell SDLUTtfll-t't in EEC, their} '- 4r _ a, — Jr: dig ltitz'tlnmelatoni = {4B ll'rtr'1 occupies volumeieell = H4fllm'3 volumetric = as, are at? f APFJ 99% occupancy = {occupied Vicell - v of liSt] of a vacancyr’eelllif'tlr’cell) = {stasis H - {rtt.-- J} fit as} weird} i". where the so comes fi‘Dm tl’te fact that we subtract the volume el‘onc vacancy from 513' unit cells, each containing 1 atoms. = H.456 CDMMENTS: Recall that the APF for BCC is 3.158, so the vacancy has reduced the APF by 394: FIND: Planes and directions of" highest density in FCC, HEP, and EDS SflLUTIDhi: These planes and directions are identified in the text. FCC {lit} stifle HCP {not} also not: [llfl] stillb- F [N 1}: Calculate the packing factor ofthe primitive cell shown in Fig. 3.34s. GWEN: The packing factor for the big cell shown in Fig. 13-h is Ill-l. SULUTIDN: A primitive cell is the smallest representation of the whole crystal. The big cell also has all the same properties of' the entire crystal. It just consists of' three primitive cells. Hence, the packing factor of the primitive cell is the same as that of the big eel], {lift-l FIND: Preferred stacking sequence GWEN: Balls are laid down and jiggled to insure low energy state .-tl I-_l- 3"}. 4f}. 41. ASSUMPTIONS: Balls have equal probability oi'equal energyr states SDLUTIDN: ABC and ABA occur with equal frequency because they are equal energy SEES. COMMENTS: In HEP and FCC structures there are thermodynamic forces that drive atoms into their position. but errors like these, stacking errors. can be expected. FIND: How many neighbors does the largest atom that can fit into an octahedral site touch‘? ASSUMPTIONS: Atoms behave as hard spheres SKETCH: Theoctahedral site in FCC is shown in Fig. 3.64s and that in the BCC is shown in 3.64s. SOLUTION: Esamine the :5 atoms surrounding the central octahedral site in Fig. 3 .6-1a. Each of the 6 neighbors is afl r2 away. Hence. the largest sphere in the hole will just touch each of the 6 neighbors. Now examine Fig. Sonic. The 2 atoms above and below the site indicated are an I 2 away. Those at the corners oFthe unit cell plane in which the hole lies are further away (an| r“ all]. Hence, the largest atoms in the octahedral site touches otdy two neighbors - one above and one below. FIND: APP for FCC 1hath all octahedral and tetrahedral sites filled GWEN: in the text are given. _4 octahedra] sitesi’cell Err = lit-lid S tetrahedral siteslcell Err = {1.225 ASSUMPTIONS: Spherical atoms - all sites filled with “perfectly” fitting spherical atoms SflLUTIflN: 1n rec is... 4’3 = 4a _ a. —- «Iowa The volume of the cell atoms is [t-llillttr3 The VGllII'I'lE. of the octahedral atoms is 4 it {4r3}rt(tl.4l4r)" The volume ofthe tetrahedral sites id s e rersiercassrr‘ Thus, the APF with all tetrahedral and octahedral sites filled = {cell atorn volume + 4 4r 1: 3 e{ H + (a. are r + 2:11.225: )3] p' T? HE.» tetrahedral atom volume + octahedral atom volume] l' {cell volume) = FIND: Win.r is the tetrahedral site in BCC larger than octahedral site? GWEN: In the text it is shown that the octahedral holes are Ill fir and the tetrahedral holes are fill-lit. Fig. 3 6-1 shows the shes. ASSUMI’HDNS: Atoms are spherical SKETCH: See Fig. 3.6—] c and d SOLUTION: It is not obiious whg.r the tetrahedral sites are larger. It has to do with the separations of atoms. Recall in SEC that corner atoms do not touch. The tetrahedral site —- £13th 42 4c is equally spaced among 4 neighbors - iii the center of the tetrahedra. The location oftlte octahedral site is along one of the edges of the tetrahedra, directly on line between two atoms. It is squeezed by the two atoms. Hence, it is a smaller site. The tetrahedral site is irregularly shaped rather than spherical as in the octahedral site. CDMMENTS: Note that you cannot fill both tetrahedral and octahedral sites in ECC FIND: Number and size of tetrahedral and octahedral sites in FCC and BCC ASSUMPTICIHS: Atoms are spherical SKETCH: See Fig. 3.6-] a-d SOLUTICIN: These values are given in the tear Err or to'r BCC ti Der. 13.155 24 Tel. H.291 FCC 4 Del. 04H 3 Tet. 0.225 FIND: How many impurities [l I]? nm and 0114 nm in diameter can fit into crystalline Ca? ASSUMPTIONS: Atoms behave as hard spheres. DATA: According to Appendix A, Ca crystallizes into the FCC structure. The atomic radius of Ca is 1.9? A, according to Appendix C .. SKETCH: The octahedral and tetrahedral sites in FCC are shown in Fig. fie-la and h. SDLU‘I’IDN: Table 3.5-1 provides the solution. For an impurity llfl? I'll'l'l in radius, 1dr = {ll Ar“ 1.91“ A = {1.355 Similarly, for an impurity {1.94 nm in radius, it r" r = [tilAr’ 1.9? A = 0.203. Thus, the larger impurity will fit into the larger sites, the octahedral sites. There is l octahedral site per Ca atom The smaller impurity can fit into either the octahedral or tetrahedral sites. There are l octahedral and 2 tetrahedral sites = 3 available sites per Ca atom FIND: Which form of iron, FCC or ECC, is capable of dissolving more (3'? DATA: Atomic radius of Fe depends on the extent of covalent bonding It is 1.241A in ECC and 1.269A in FCC. The atomic radius of carbon is UTTA. ASSUMPTIDNS: Spherical atoms, C goes into holes SDLUTIDN: The radius of the octahedral sites, the larger ones in FCC, is {i did at l 269 = DSZSA The radius ofthe tetrahedral sites, the larger ones in ECC, is {1191 it lldlA = lleIA Therefore, carbon fits easier into FCC iron; however, there are more tetrahedral sites in FCC {24) than octahedral sites in FCC {4}. CCMMENTS: We expect the FCC crystal structure to have greater C solubility than the low temperature BCC crystal structure. The maximum solubility of C is oozes in ECC iron and 2.l% in FCC iJ'oo. FIND: Lattice parameter of Si DATA: Si assumes the diamond cubic structure, which has atoms in the FCC positions and l”: the tetrahedral sites filled. The atomic radius of Si is l.ll'ErA. ASSUMPTIGHS: Spherical atoms. .‘ll SKETCH: Shouting only a portion of the FCC atoms- SGLETIDH: 2 atoms touch along U4 body diagonal _ (fr? '— 2r _ m, —- to: re — 5.43.4 as FIND: Calculate the mass density ofdiamoud and Si. GIVEN: Diamond and Si have the same crystal stmcture. DATA.- The covalent radius of carbon is 11?? A and that offii is l.l'i' A. according to Appendix C The atomic weight off is 12.01 1 and that of Si is 25.036 gimole. SKETCH: The structure ordiantond is shown in Fig. 3 T-Jla. SULUTIDN: Atoms touch along [H4 of] the bod}f diagonal. so that enema = 2r 23,0 = 3r! ‘4']. The cell volume is an} The mass of the unit cell is determined using the atomic weights and realizing that there are S atoms per unit cell. This is just twice the canal for an FCC structure, since there are two atoms per site. 4?. FIND: Calculate the densin oi'a {ato4i4— tetrahedron. DATE The ionic radius of Si“?r is 1139 A and that offlz' is 1.32 A. according to Appendix C. The atontic weight of Si is 23.09 griddle and that of U is 16.12"} gr'mole. SKETCH: See Fig. 17-69. and the following. Remember that anion and cation touch. Anions do not touch fl II 'I- 1 - - ' e soLttTtoN: observation er the figure. after focusing on the body diagonaL leads to the conclusion: (andflr’2=r+fl=>afi=2{r+fl}ffiil3 =2{1.3sn+oss mild}: l at a. To calculate mass, M, we need to determine the number of anions and cations in the tetrahedta Number of anions = H3 x 4 2 H2. Number of cations = 1 Proceeding, lilorom x to "it; + laromx ZED? “fie [1. 9714} 48. as 5!} COMMENTS: This value is much larger than the density.r of silica glass or crystal. There is a large amount of empty space surrounding the (Si04l4' tetrahedra. FIND: Lattice parameter in M30 GWEN: In M30 one ion is FCC and the other fills all the octahedral sites, 1which are at the center of each edge and the center of the cube. This is the Nat: I stmohtre. DATA: r3432. = {1156A and tea. = 1.32A fiSSUMFflDNS: lens are spherical SKETCH: plus the other ions in the FCC positions SULUTIUN: as = liners + tug.) = Efflfio + ill) ‘- 3 96A FIND: Redraw Fig. 3.?4 with 2112* at the origin SflLlj-TIHN: The figure looks the same except the ions are labeled in the opposite fashion. You can easily,r convince yourself of this by noting that the nearest atoms to the new origin are in the lair-l, lr'4, 1f4~t3rpe positions. CDMM‘ENTS: The interchange ability of atoms or ions applies to all MK structures. FIND: Sketch {Hill and (11 l} in M30. Show tetrahedral and octahedral sites. GWEN: MED has the NBCI sou-store {11m ' ASSUMP'I'IDNS: Spherical ions SKETCH: CDMMENTS: All neighbors are not equidistant in the octahedral sites. In fact, the octahedral site is the same as the tetrahedral site! There are no octahedral sites! 5]. 53. FINE: Predict the structure of Nit} BETA: I'm. = 0.15mi; rm. = 1.32s The electronegatiyity of Ni is 1.9l and that of 0 is 3 .44 ASSUMP’I‘IDNS: Ions are spherical SULUTIDN: rinsing. = seen .32 = .532 :t- CE = b, from Table 2.6-]. The difl’erence in electronegativity is 1.53 :9 bond is 44% ionic. Thus, we predict a Hat] stnicture. the oniy MK ionic structure with Chi = s. CUMMENTS: In this case. 44% is sufliciently ionic that the NaCI structure is the correct structure FIND: Predict stnicture oFSiC. DATA: The eiectronegatiyity of Si is 1.9 and that of C is 2.55 ASSUMPTIONS: Atoms are spherical SDLUTIDN: The eiectronegatiyiry difference is ass 2} 10% ionic :> covalent crystal. Thus, we predict the zinc-blend structure, which can accommodate a covalent crystal with a valence of 4 and equal number of the two types of atoms. CUMMENTS: We did not assess ionic radii at the outset because SiC does not look like an obvious ionic structure. FIND: Predict structure of Cafl. DATA: The electronegatisity ofCa is 1.04:} and that oft] is 3.44; rm. = (199$, res. = 1.321%. ASST} MPTI'DNS: lens are spherical SDLUTIDN: mutiny; = $.99! [.32 = {1.1'5 :s CH = E, from Table lib-I. The difference in electroncgatiyity is 2.44 2: bond is TWA ionic. Thus, we predict a CsCl structure, the only MK ionic structure with EN = 3. FIND: Calculate the density of Cth. HATA: The electroncgativity osz is I. I2 and that ofEr is 2.96: tn. ‘ I 15M“, rBl— = [Hot-it. Tcrfrm. =l fii‘i] as = [135 :5 [IN = E, from Table lot The difference in electrcrlegatiyity is LE4 :1 bond is 57% ionic. Thus, we predict a Cs Cl stmcture, the only MK ionic sinioture with EN = 8 The atomic weights onsz and Br are [32.9 and res gimme. ASSUMPTIUNS: ions are spherical SOLUTIDN: First determine the lattice parameter: an all} '— 21'”, "- 1-3,, ,l _ a: 4J§A 55. 56. 5?. p ' Mass-Th! J“ - JHMGifmiEsfc-elflif‘fllfi r Wfljt’g mole) i!” t5. {123 1' mg moirmrles -' mole) J a r: Err‘flgfi - .79.?! --— ~—_"St.t'i w? {4. to r of" 3’ rs. {123 x to"? g L CDMMENTS: This seems quite dense for a ceramic. but the elements are ’rieair}r FIND: Show why Th0; Tcfii. and U01 crystallize into the fluorite structure. DATA: Eioctronegativities: Th = 1.3 and D = 3.44. etc. ionic radii nu. = 12$]. rm. = 1 32, etc. ASSUMPTI’DNS: loos are spherical SULUTIDN: Three criteria need to be satisfied for fluon'te structure: I. stoiicliiometr}r ofmeta] to oxide = 1.2 2. Electrooegativity difierenoe large '1’} ionic structures, eg TitoI 2.14 : sets 3. radius ratio needs to he (1.32-1.13, eg for Th0: I 02!] 3. = 11?"? :5 CH = 3 for To and 4 forfl FIND: Density ofUU: DATA: U0; has the fluorite structure {see problem 54] U in FCC positions 2:» dice“ :3: 3 Dilch in tetrahedral positions; at wt. [1 = to U gi’mole and U = 233.0 gtmole. rm- = I [03. £1.32 aSSUMPTlflN S: Spherical ions [10 [1.1;— _ ' _ “farms - 132,4. 4? 5.4?A_a5 F rem A’. (s x or s 4 r 23s.: rs. res r m“) -'- Masts "Vol __ = “a ’i“ m4 2: m“ —- £9.94 cm! SflLUTIDNt FIND: Density of Cali; and identify densest planes and directions. DATA: Cali: has the fluon'te structure, of course; Team = DEER and IF. = 1.339; atomic weights are 4D G3 and 9.0 gfntole. ASSUMFTlflNS: Spherical ions SKETCH: 331' SS. 59 [Irr‘lllrg _ rL‘ui'I ; rF- _fir‘- = _a: =' W -I- . .1 rite : hr“ _ 3'3! g cm ,9 - Wms'l’m’ JIM—- SDLUTIUN: Atoms touch along the cuhE diagonal 1r on}: (life he sssfi Most dense direction” try me; .- (stirs; 3.3543 W (Hill: on + rear 5. 315-13 Most dense plane” t h 1"in .‘ {Kiel-5.35” CUMMENTS: Linear and plans: densities lose meaning when two difi'erent ions or atoms are present. Alternative definitions oflirteer and planar density can he more usefiil at times, but diE‘erent ions are never equivalent. [See problem 23). F IND: Calculate the D-Si-D bond angle in crystohaiite. Discuss the consequences if this bond angle, the Si-fl-Si bond angle, or both are allowed to vary somewhat. BETA: The crystal structure of crystohalitc is shown in F ig. 3.1-6. The cell is both using {SiDJ‘ tetrahedra. ASSUMP’I'IDNS: Spherical atoms SflLUflflN: Since the D are tetrahedrally arranged about Si, the bond angle is the tetrahedral angle, which was derived in problem 19, 139.5“. As the bond angles increasingly vary, the material loses long range order. It becomes a glass. Such is the structure oi" silica glass. it is the Si-CLSi bond angle that seems to 1.rary the most. CUMMENTS: See problem 59 FIND: Idemify information needed to calculate the density of crystooalite, one crystalline form of Siflz. ASSUMPTIUNS: Spherical atoms SKETCH: See Fig. 3.7-6 SflLllTIflN: We need the following information to calculate the density ofcrystohalitc' 1. atomic masses 33' ...
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Ch3Sol2 - Si 32 33 SKETCH For{Ill shown soLU'rtoN In rec...

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