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Unformatted text preview: Si. 32. 33. SKETCH: For {Ill} shown: soLU'rtoN: In rec: body diagonal =anc’i Fir my]! : Plumbarming}? = _ “ﬂ = a:
F'mu mitestenets = (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 closepacked 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
SDLUTtﬂlt'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. 13h is Illl. 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],
{liftl 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 .61a.
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 = litlid 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 [tllillttr3 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 ﬁlllit. Fig. 3 61 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] ad 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. fiela and h.
SDLU‘I’IDN: Table 3.51 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 TJla. 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. 1769. 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 SﬂLljTIHN: 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 lairl, 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 soustore {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 zincblend 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; = $.99! [.32 = {1.1'5 :s CH = E, from Table libI. The difference
in electroncgatiyity is 2.44 2: bond is TWA ionic. Thus, we predict a CsCl structure, the
only MK ionic structure with EN = 3. FIND: Calculate the density of Cth. HATA: The electroncgativity osz is I. I2 and that ofEr is 2.96: tn. ‘ I 15M“, rBl— =
[Hotit. Tcrfrm. =l ﬁi‘i] as = [135 :5 [IN = E, from Table lot The difference in
electrcrlegatiyity is LE4 :1 bond is 57% ionic. Thus, we predict a Cs Cl stmcture, the
only MK ionic sinioture with EN = 8 The atomic weights onsz and Br are [32.9 and
res gimme. ASSUMPTIUNS: ions are spherical SOLUTIDN: First determine the lattice parameter: an all} '— 21'”, " 13,, ,l _ a: 4J§A 55. 56. 5?. p ' MassTh! J“  JHMGifmiEsfcelﬂif‘ﬂlﬁ r Wﬂjt’g mole) i!” t5. {123 1' mg moirmrles ' mole) J
a r: Err‘ﬂgﬁ  .79.?! — ~—_"St.t'i w?
{4. to r of" 3’ rs. {123 x to"? g L CDMMENTS: This seems quite dense for a ceramic. but the elements are ’rieair}r FIND: Show why Th0; Tcﬁi. and U01 crystallize into the ﬂuorite structure.
DATA: Eioctronegativities: Th = 1.3 and D = 3.44. etc.
ionic radii nu. = 12$]. 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 (1.321.13,
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. 31513 Most dense plane” t h
1"in .‘ {Kiel5.35”
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 DSiD bond angle in crystohaiite. Discuss the consequences if this
bond angle, the SiﬂSi bond angle, or both are allowed to vary somewhat. BETA: The crystal structure of crystohalitc is shown in F ig. 3.16. 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 SiCLSi 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. 3.76
SﬂLllTIﬂN: We need the following information to calculate the density ofcrystohalitc'
1. atomic masses 33' ...
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 Fall '08
 Tannebaum

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