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Unformatted text preview: Chapter 3, Problem 11 Niobium at 20°C is BCC and has an atomic radius 0110.143 11111. Calculate a value for its lattice constant a in
nanometers. Chapter 3, Solution 11 For a BCC unit cell having an edge length a and containing niobium atoms. —i(0.143 11111) =0.330 11m 4
:_R_
” 13 15 ﬁn 2 4R or Chapter 3, Problem 12 Lithium at 20°C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a
lithium atom in nanometers. Chapter 3, Solution 12 For the lithium BCC st111ct111‘e.11=11icl1 has a lattice constant of a = 0.35092 nm. the atomic radius is. Ji
2—(1
4 R 273(0.3509211111)=0.152nm Chapter 3, Problem 16 Gold is FCC and has a lattice constant of 0.40788 11m. Calculate a value for the atomic radius of a gold atom in
nanometers. Chapter 3, Solution 16 For the gold FCC structure. which has a lattice constant of (1 = 0.40?88 nm. the atomic radius is. \Ex/E 232—0: 4 T(0.40?88111n)= 0.144 nm Chapter 3, Problem 18
Palladium is FCC and has an atomic radius of 0.13? 11111. Calculate a Value for its lattice constant a in nanometers. Chapter 3, Solution 18
Letting (1 represent the FCC unit cell edge length and R the palladium atomic radius. ﬁa=4a or a=ia=im131nm)=0.337nm 2 \5 Chapter 3I Problem 23 What is the ideal ("Err ratio for HCP metals? Chapter 3, Solution 23 The ideal C/(l ratio for HCP metals is 1.633: however. the actual ratios may deviate signiﬁcantly from this value. Chapter 3, Problem 31 Draw the following directions in a BCC unit cell and list the position coordinates of the atoms whose centers are intersected by the direction 1sector:
( (a) [1001(5) [1101(6) [111]
1.1.0) Chapter 3, Solution 31 (3) Position Coordinates: (b) Position Coordinates: (c) Position Coordinates:
(0, 0, 0), (1, 0, 0) (0,0,0),(1,1,0) (0,0,0),(1,1,1) Chapter 3, Problem 32 Draw direction vectors in unit cells for The following cubic directions: (a) [ITT] (b) [1T0] (c) [ET] (0’) [113] Chapter 3, Solution 32 5 (a) (b)
: p .1:
x x = +1 .T = +1
.1’ =  J" = 
[1 I I] [l T 0]
(C) (d) Dividing by 2, _ 1 / Dividing by 3 ;
1 x — .:2 Y _ ,x
. — _ f3
1? = I _ .a
h _ 1 I .1: _ —1’;3
_ — '3’2
: = l Chapter 3, Problem 34 TWhat are the indices of the directions shown in the unit cubes of Fig. P3 .3 4? Figure 133.34 Chapter 3, Solution 34 A
I... V2 ——— New 0
1/ 7"
:4
New 0 a. Vector components: 13. Moving direction vector c. Moving direction vector for
. = 1. v = 1. : = 0 down 1/4 . vector components ward vector components
. _ 1.
. . . t " : =:_ :1 " :_:' r: 4:
Direction indices: [110] 316' x 1' J 1' ‘ _’/“ a“: x "6' ‘ 1;
Direction indices: [441] Direction indices: [166] _ 1/3
Bew 0 
I 2/3
d MOWHE dHCCtlon “3301' e. Vector components are: f. Moving direction vector up
1." v .
left "4 WW” cmnponems x I 3/4 . fr 2 1. : I 1/3. vector components are:
are: x=1. 1‘=‘/2.:= D.  d _ :14 =_1 1:1 =_,x'3
Direction indices [712] 116mm m 1035' [3' ] i ' ' ‘  — —
' ‘ Direction indices: [331] New 0 New 0 7A
/ M\ g. Moving direction vector 11. Moving direction vector
up $3. vector components up 1/4. vector components
are: x = 1. _1 = 1. : = are: x = 3/; .__1' = 1.  = Direction indices: [Ari—ﬂ] Direction indices: [3 3.3] Chapter 3, Problem 36 . . . 3 l 1 . . . . .
A directlon vector passes through a unit cube from the l, 0,; to the E , 1,1 posrtions. What are Its dit‘ectlon
indices? Chapter 3, Solution 36 Subtracting coordinates, the vector components are: 1 3 1 3 1
r— 1— .1r—1—0—1 :=———=——
4 4 4 4 2 Clearing fractions through multiplication by 4, gives .T = —3 , 3" = 4, 3 = —2‘ The direction indices are therefore 4 Chapter 3, Problem 38 What are the directions of the [1 famin or form for a unit cube? Chapter 3, Solution 38 [100], [010], [001], [T00], [0T0], [001] Chapter 3, Problem 39 Vﬂiat are the directions of the <1 1 1) family or form for a unit cube? Chapter 3, Solution 39 [111]. [111]. [111]. [111].
[111]. [111]. [111]. [111] Chapter 3, Problem 40 1What <1 10> type directions lie on the (111) plane of a cubic unit cell'? Chapter 3, Solution 40 [011], [011], [110], [110], [101], [101] — 0 l 1
[I 01] [ ]
[011]
[10 [1 10]
Chapter 3, Problem 41
What <1 1 1> type directions lie on the (1 10) plane of a cubic unit cell‘?
Chapter 3, Solution 41
_ _ __ _ _ [111]
[lll].[lll].[lll].[lll] __
[111] [111] Chapter 3, Problem 43 Draw in unit cubes the crystal planes tliaT have the following Miller iiidices: (a) (Ill) (a) (JET) (e) (3E1) (g) (20f) (a (532) (Jr) (3—2)
(a) (10E) (0’) (213) (f) (305) (a) (Eli) (j) (13?) (I) (331) Chapter 3, Solution 43 (o. o. 0) (a, o. 0) a. For (1 l l) reciprocals b. For (103) reciprocals c. For (lil) reciprocals
are: x=1._ ‘1‘=1._ :=1 are: r=1. .1‘=30.:=l"2 are: x: 1. _1'=¥/3. :=
(0. 0. 0) .1 (0. 0. 0) +5.; .1— (3 0 3) I
V d. For (213) reciprocals e. For (351) reciprocals f. For (305) reciprocals
are: x =16. _r = 1. : = /3 are: x = 1/3.. i== 4/2 .,  =1 are: x = 3’3 :1". : = 1/2 +15 —
" ,. (213) A H: Hi
(0. 0. 0)
+93
g. For (20le reciprocals h. For (315) reciprocals i. For (532) reciprocals are:
are: x = {—3 A 31' = CO. : = 1 are: r =¥:’2, '1‘ = 1. : = #3 x =¥/2, _1== 5'3. : = Li
(0. 0. 0) +1_,3
_ r _ + :1 k. (0. 0. 0) 0.0.0)
(133) r ’3 (€31) j. For (13;) reciprocals k. For (312) reciprocals k. POI (331)1‘6Cipl‘00315
arc: x =1 ._ _1== 1/3, : = 4"": are: .1'=£""s. y = 1. : = are: x: 4/3. _1'=E{a._ : = 1 Chapter 3, Problem 44 What are the Miller indiees of the cubic cyrsyallographic planes shown in Fig. P344? Chapter 3, Solution 44 Miller Indices for Figure P3.44(a) Plane 0 based on (I), 1, 1) as origin
Planar Intercepts Reciprocals of Intercepts Plane 1) based on (1, 1, 0) as origin
Planar Intercepts Reciprocais of Intercepts 1
x=$ —=0
x
1
1:4 _:_1
.1,
1 l
:=—— —=—4
4 : The Miller indices of plane r: are (I) T 1). 1 x=1 —=—1
X —s 1 —12
1‘27 —: ' 12 y s
r :23": —:0 The Miller indices ofplane b are (E E 0). Plane c based on (1, 1, 0) as origin
Planar Intercepts Reciprocals of Intercepts Plane 0' based on (0, 0, 0) as origin
Planar Intercepts Reciprocals of Intercepts xe i=0
x
i
:13:_1 —:—1
1r 5 The Miller indices ofplane C are (0 3)‘ .T:1 121
x
i __2 l_3
‘§ i‘a The Miller indices of plane 0’ are (2 2 3). Miller Intlices for Figure P3.44(b) Plane (1 based on (1, l], 1) as origin
Planar Intercepts Reciprocals of Intercepts Plane (3 based on (0, 1, 1) as origin
Planar Intercepts Reciprocals of Intercepts 1
TZl —=—l x 1
1:00 —=0 r 1 1 The Miller indices of plane a are (I I] 3). 1
x=1 —=1 x l
12—] —=—1 r
___3 l__i
‘ 3 : 2 The Miller indices of plane I) are (2 53). Plane 6 based on (0, l, 0) as origin
Planar Intercepts Reciprocals of Intercepts Plane (I based on (0, 1, 0) as origin
Planar Intercepts Reciprocals of Intercepts {H} Figure P344 Chapter 3, Problem 49 A cubic plane has the following axial intercepts: (I = % . b = —% . C = % . What are the Miller indices of this plane'?
Chapter 3, Solution 49
. . . . . , , . . 1 l 3 l _ I
Given the anal Intercepts 01 (V3. ?"3. 1/2). the 1‘ec1p1‘ocal Intercepts are: — = 3 , — = —: , — = 2. Multlplying
x 1‘ a : by 2 to clear the fraction. the Miller indices are (6 3 4). Chapter 3, Problem 56 Rodium is FCC and has a lattice constant a of 038044 11111. Calculate the following intet‘planar spacings: (a) (3111 (b) dzoo (C) dzzo Chapter 3, Solution 56 0.38044 11111 _ 0.38044 11111 (a) {27111— _ — 0.220 11m
«2 +12 +12 43
. .3 ' '
(b) 0.200 _ 0 38044 11111 _ 0 804411111 _ 0.190 nm 22+02+02 J4 0.38044 11111 _ 0.38044 11111 (‘3) “:2 = q a
30 42A+22+0 J8 = 0.135 nm Chapter 3, Problem 72 Calculate a value for the density of FCC platinum in grams per cubic centimeter from] its lattice constant a of
0392391111} and its atomic mass of 195.09 gfinol. Chapter 3, Solution 72 First calculate the mass per unit cell based on the atomic mass and the number of atoms per unit cell of the FCC
stmctul‘e. (4 atomsf'unit cell)(195.09 gf'tnol) 6 .023 x1 023 at0111s.."'1nol —21 111ass.."'1111it cell — — 1.296X10 gt'tlnit cell The density is then found as. p _ massf'unit cell _ Inassr’unit cell _ 1.296x10—21 gt’unit cell
U volu111ef'1mit cell at3 [(039239 ><10_g 1n)3]1111it cell
3
=21._445,113 gf'1n3 i =21.45 grcm3
100 cm Chapter 3, Problem ?3 Calculate the planar atomic density in atoms per square 111illi1neter for the following Ciystal planes in 3C‘C‘
chromium. which has a lattice constant of 0.28846 nm: (a) (100). ((1) (110). (C) (l l 1). Chapter 3, Solution 73 (Solution C is on the next page.) To calculate the density. the planar area and the number of atoms contained in that area must ﬁrst be dete1'mi1 (a) The area intersected by the (1 0 0) plane inside the cubic unit cell is a2 while the number of atoms contain
4 corners)x (1.22} atom per corner) = 1 atom. The density is. _ equiv. no. of atoms whose centers are intersected by selected area pp selected area
= $92 = (1.202 >~<1019 atoms"1112) L —
(0.28846x10_ m) 1000 mm = 1.202 X 1013 atoms/mm2 (b) For the more denser packed (l l 0) plane. there are: 1 atom at center — ( 4 corners) x (1/; atom per corner) 2 2 atoms And the area is given as (ﬁZﬁ'Xﬂ) = 34'20'2 . The density is thus. 2 atoms . — (I .699 X1019 ato1ns.."'m2 )(10_6 111351111112 ) p _
1’ JE(0.28846><10’9111) : 1.699 X 1013‘ atoms/mm2 (c) The triangular (1 1 1) plane contains: (3 comers) >‘ (1,"5 atom per corner) = 3/: atom.
1 1 .— JE \/ 6 a
The area is equal to: :5}? = 2(1) Ta = T (F. The density is thus. DZ atom — (9.8 l 3 ><1018 titans"1112 )(10_6 111251111112 ) ply—ﬂ T(0.28846><£0_9 111)2 = 9.813 X 10:2 atoms/mm2 Chapter 3, Problem 76 Calculate the linear atomic density in atoms per millimeter for the following directions in BC‘C vanadium. which
a lattice constant of0.3039 nm: (or) [100]. (E?) [110]. (c) [111]. Chapter 3, Solution 76 [100] [11 general. the linear atomic density is derived from: _ no. of atomic diam. intersected by selected length of direction line P; .
selected length of hue
(a) For the [100] direction of BCC‘ vanadium.
no. atom dia. 1 atom 6
pg— — 9 I 3 I —3.29X1l] mm
a (U. 3039 nm)( 1 0' 111."nm)( l 0 1111mm)
(b) For the [l 10] direction of BC‘C‘ vanadium.
no. atom dia. 1 atom
p.— — 6 I —2.33 ><1o6 mm
ﬁn £03039 nm)(10' ﬁlm"11111)
(c) For the [l l 1] direction of BCTC‘ vanadium.
_ no. atom dia. 2 atoms —3.8l] x106 mm p... — .
‘ ﬁn Jim .3 03 9 11111)(10'6 numnln) Chapter 3, Problem 80 Pure iron goes through a polymorphic change from BCC to FCC upon heating through 9130C. Calculate the volume
change associated with the change in crystal structure from BCC to FCC if at 9120C the BCC unit cell has a lattice
constant a = 0.293 11111 and the FCC unit cell a = 0.363 11111. Chapter 3, Solution 80 First detennine the individual voltunes per atom for the iron 3CC and FCC crystal sti'ucttu‘es: 3 3,.  3 ‘
a 11m unit cell _ (0.293 mm) _ 0.01258 Huffman V _
ECG 2 atomst’unit cell 2 atoms 3 3...  3
a 11m unit cell _ (0.363 mm) _ 0.01196 lung'satom V. —
FCC 4 atomsfunit cell 4 atoms Thus the change in volume due to iron‘s allotropic trausfonnation is. VFCC — VBCC (100%) _ 0.01 196 nn13.."'ator11 — 0.01258 lun3..="ato1n % Volume change = a ‘
VBCC 0.0125 8 mn' r'atom (100%) : 4.94% Chapter 3, Problem 91 An Xray' diffractometer recorder chart for an element that has either the BCC or the FCC crystal structure show
diffraction peaks at the following 26 angles: 40.6630. 41314". 69.1440. and 83.448“. (Wavelength 3'. ofthe incoming radiation 1Ju'as 0.15405 nnl.)
a) Determine the clystal structure of the element. b) Determine the lattice constant of the eletnent. c) Identify the element. Chapter 3, Solution 91 (a) Comparing the sin‘6 term for the ﬁrst two angles: 26 6' sin 6 sin3 :9 40.663" 20.3315° 0.34745 0.120?2
47.314“ 23.657“ 0.40126 0.16101 $411391 _ 0.120tz 51111:}2 0.1610] (b) The lattice constant also depends upon the ﬁrst 5111269 term. as well as. the Miller indic es of the ﬁrst
of FCC principal diffracting planes. {111}. 113 +43 +23 _ 0.15405. 11m 12 +12 +12
31113 91 2 0.120?2 — 0.3839? nm (c) From Appendix I. the FCC metal whose lattice constant is closest to 0.3839? nm is
iridium (11‘) which has a lattice constant of 0.3 8389 tun. ...
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