3
THE CHAPTER STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS 3.1 Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material. 3.2 A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell, whereas a crystal system is described only in terms of the unit cell geometry. For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic crystal system. 3.3 For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation (3.4) as
3
VC = 16R
2 = (16) 0.143 x 10
(
-9
m
)
3
2 = 6.62 x 10
-29
m
3
3.4 This problem calls for a demonstration of the relationship a = 4R
unit cell shown below
3 for BCC. Consider the BCC
Using the triangle NOP
1
(NP)2 = a 2 + a 2 = 2a 2
And then for triangle NPQ,
(NQ)2 = (QP )2 + (NP)2
But NQ = 4R, R being the atomic radius. Also, QP = a. Therefore,
(4R)2 = a 2 + 2a2 , or
4R 3
a =
3.5 We are asked to show that the ideal c/a ratio for HCP is 1.633. A sketch of one-third of an HCP unit cell is shown below.
Consider the tetrahedron labeled as JKLM, which is reconstructed as
2
The atom at point M is midway between the top and bottom faces of the unit cell--that is MH = c/2. And, since atoms at points J, K, and M, all touch one another,
JM = JK = 2R = a
where R is the atomic radius. Furthermore, from triangle JHM,
(JM) 2 = (JH)2 + (MH)2, or
a
2
= (JH)
2
c + 2
2
Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral
triangle,
cos 30 =
a/ 2 = JH a 3
3 , and 2
JH =
Substituting this value for JH in the above expression yields
2 a a2 c2 c a = + + = 2 3 4 3 2 2
and, solving for c/a
3
c = a
8 = 1.633 3
3.6 We are asked to show that the atomic packing factor for BCC is 0.68. The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or
VS V
C
APF =
Since there are two spheres associated with each unit cell for BCC
3 4R 3 8R VS = 2(sphere volume) = 2 = 3 3
3 Also, the unit cell has cubic symmetry, that is V = a . But a depends on R according to Equation C (3.3), and
4R VC = 3
3
=
64R 3 3 3
Thus,
8R 3 / 3 64R 3 / 3 3
APF =
= 0.68
3.7 This problem calls for a demonstration that the APF for HCP is 0.74. Again, the APF is just the total sphere-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus
4 R3 3 VS = 6 3 = 8R
Now, the unit cell volume is just the product of the base area times the cell height, c. This base area is just three times the area of the parallelepiped ACDE shown below.
4
The area of ACDE is just the length of CD times the height BC . But CD is just a or 2R, and
2R 3 2
BC = 2R cos (30) =
Thus, the base area is just
2R 3 2 AREA = (3)(CD)(BC) = (3)(2 R) = 6R 2
3
and since c = 1.633a = 2R(1.633)
VC = (AREA)(c) = 6R c 3 = 6 R
2
(
2
3 (2)(1.633)R = 12 3 (1.633 )R
)
3
Thus,
VS VC 8 R3 12 3 (1.633 )R 3
APF =
=
= 0.74
3.8 This problem calls for a computation of the density of iron. According to Equation (3.5)
nAFe V N
C A
=
For BCC, n = 2 atoms/unit cell, and
5
4R VC = 3
3
Thus,
=
(2 atoms/unit cell)(55.9 g/mol)
3 -7 cm 3 / 3 /(unit cell) 6.023 x 1023 atoms/mol (4 ) 0.124 x 10
(
)
(
)
= 7.90 g/cm3 3 The value given inside the front cover is 7.87 g/cm . 3.9 We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal structure. For FCC, n = 4 atoms/unit cell, and V = 16R 3 2 [Equation (3.4)]. Now, C
nAIr V N
C A
=
And solving for R from the above two expressions yields
nA Ir R = 16N 2 A = 1/3
(
(4 atoms/unit cell)(192.2 g/mol) 2 )16) 22.4 g/cm3 6.023 x 10 23 atoms/mol (
(
) (
)
1/3
= 1.36 x 10-8 cm = 0.136 nm 3.10 This problem asks for us to calculate the radius of a vanadium atom. For BCC, n = 2 atoms/unit cell, and
4R VC = 3
3
=
64R 3 3 3
6
Since,
nAV V N
C A
=
and solving for R
3 n 3A V R = 64N A
1/3
3 3 (2 atoms/unit cell)(50.9 g/mol) = 3 23 atoms/mol (64) 5.96 g/cm 6.023 x 10
( )
(
)(
)
1/3
= 1.32 x 10-8 cm = 0.132 nm 3.11 For the simple cubic crystal structure, the value of n in Equation (3.5) is unity since there is only a single atom associated with each unit cell. Furthermore, for the unit cell edge length, a = 2R. Therefore, employment of Equation (3.5) yields
= nA nA = VC NA (2R )3N
A
=
(1 atom/unit cell)(74.5 g/mol) 3 -8 23 atoms/mol (2) 1.45 x 10 cm /(unit cell) 6.023 x 10
(
)
(
)
5.07 g/cm3 3.12. (a) The volume of the Ti unit cell may be computed using Equation (3.5) as
nATi N
A
VC =
Now, for HCP, n = 6 atoms/unit cell, and for Ti, A
Ti
= 47.9 g/mol. Thus,
7
VC =
(
(6 atoms/unit cell)(47.9 g/mol) 4.51 g/cm3 6.023 x 1023 atoms/mol
) (
)
= 1.058 x 10-22 cm3/unit cell = 1.058 x 10-28 m3/unit cell (b) From the solution to Problem 3.7, since a = 2R, then, for HCP
3 3 a 2c 2
VC =
but, since c = 1.58a
3 3 (1.58)a 3 = 1.058 x 10 -22 cm3 /unit cell 2
VC =
Now, solving for a
2 1.058 x 10-22 cm3 ( ) a = (3 ) 3 (1.58)
(
( )
)
1/3
= 2.96 x 10-8 cm = 0.296 nm And finally c = 1.58a = (1.58)(0.296 nm) = 0.468 nm 3.13 This problem asks that we calculate the theoretical densities of Al, Ni, Mg, and W. Since Al has an FCC crystal structure, n = 4, and VC = 2R 2
(
)3 .
Also, R = 0.143 nm (1.43
x 10-8 cm) and AAl = 26.98 g/mol. Employment of Equation (3.5) yields
= (4 atoms/unit cell)(26.98 g/mol) 3 23 -8 atoms/mol (2)(1.43 x 10 cm) 2 /(unit cell) 6.023 x 10
[
( )]
(
)
= 2.71 g/cm3 The value given in the table inside the front cover is 2.71 g/cm3.
8
Nickel also has an FCC crystal structure and therefore
= (4 atoms/unit cell)(58.69 g/mol) 3 23 -8 atoms/mol (2)(1.25 x 10 cm) 2 /(unit cell) 6.023 x 10
[
( )]
(
)
= 8.82 g/cm3 The value given in the table is 8.90 g/cm3. Magnesium has an HCP crystal structure, and from Problem 3.7,
3 3 a 2c 2
VC =
and, since c = 1.624a and a = 2R = 2(1.60 x 10-8 cm) = 3.20 x 10-8 cm
3 3 (1.624) 3.20 x 10-8 cm 2
VC =
(
)
3 = 1.38 x 10-22 cm3 /unit cell
Also, there are 6 atoms/unit cell for HCP. Therefore the theoretical density is
nAMg V N
C A
=
=
(
(6 atoms/unit cell)(24.31 g/mol) 1.38 x 10-22 cm3 /unit cell 6.023 x 10 23 atoms/mol
) (
)
= 1.75 g/cm3 The value given in the table is 1.74 g/cm3. Tungsten has a BCC crystal structure for which n = 2 and a =
and R = 0.137 nm. Therefore, employment of Equation (3.5) leads to
4R 3
; also AW = 183.85 g/mol
9
=
(2 atoms/unit cell)(183.85 g/mol) 3 -8 (4 ) 1.37 x 10 cm /(unit cell) 6.023 x 10 23 atoms/mol 3
(
)
(
)
= 19.3 g/cm3 The value given in the table is 19.3 g/cm3. 3.14 In order to determine whether Nb has an FCC or BCC crystal structure, we need to compute its density for each of the crystal structures. For FCC, n = 4, and a = 2 R 2 . Also, from Figure 2.6, its
atomic weight is 92.91 g/mol. Thus, for FCC
nA Nb 3 2R 2 N
=
(
)
A
=
(4 atoms/unit cell)(92.91 g/mol) 3 -8 23 2 /(unitcell) 6.023 x 10 atoms / mol (2) 1.43 x 10 cm
(
)( )
(
)
= 9.33 g/cm3 For BCC, n = 2, and a =
4R 3
, thus
=
(2 atoms/unit cell)(92.91 g/mol) 3 -8 (4) 1.43 x 10 cm /(unitcell) 6.023 x 10 23 atoms/ mol 3
(
)
(
)
= 8.57 g/cm3 which is the value provided in the problem. Therefore, Nb has a BCC crystal structure. 3.15 For each of these three alloys we need to, by trial and error, calculate the density using Equation (3.5), and compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures,
10
the respective values of n are 1, 2, and 4, whereas the expressions for a (since VC = a3) are 2R,
2 R 2 , and 4R / 3 .
For alloy A, let us calculate assuming a BCC crystal structure.
=
nAA V N
C A
=
(2 atoms/unit cell)(43.1 g/mol) 3 (4) 1.22 x 10-8 cm /(unit cell) 6.023 x 1023 atoms/mol 3
(
)
(
)
= 6.40 g/cm3 Therefore, its crystal structure is BCC. For alloy B, let us calculate assuming a simple cubic crystal structure.
= (1 atom/unit cell)(184.4 g/mol) 3 23 atoms/mol -8 (2 ) 1.46 x 10 cm /(unit cell) 6.023 x 10
(
)
(
)
= 12.3 g/cm3 Therefore, its crystal structure is simple cubic. For alloy C, let us calculate assuming a BCC crystal structure.
= (2 atoms/unit cell)(91.6 g/mol) 3 (4) 1.37 x 10-8 cm /(unit cell) 6.023 x 1023 atoms/mol 3
(
)
(
)
= 9.60 g/cm3 Therefore, its crystal structure is BCC.
11
3.16 In order to determine the APF for U, we need to compute both the unit cell volume (VC) which is just the product of the three unit cell parameters, as well as the total sphere volume (VS) which is just the product of the volume of a single sphere and the number of spheres in the unit cell (n). The value of n may be calculated from Equation (3.5) as
VC NA A
U
n =
(19.05)(2.86)(5.87)(4.95) x10 -24 6.023 x 1023 = 283.03
(
) (
)
= 4.01 atoms/unit cell Therefore
4 3 (4) R 3 = (a)(b)(c)
APF =
VS V
C
4 3 (4) ()(0.1385) 3 = (0.286)(0.587)(0.495)
= 0.536 3.17 (a) From the definition of the APF
4 3 n R 3 a 2c
APF =
VS V
C
=
we may solve for the number of atoms per unit cell, n, as
(APF)a 2c 4 R 3 3
n =
12
(0.693)(4.59)2(4.95) 10 -24 cm3 = 4 3 1.625 x 10 -8 cm
(
(
)
3
)
= 4.0 atoms/unit cell (b) In order to compute the density, we just employ Equation (3.5) as
= nAIn
A
a2 c N
=
(4 atoms/unit cell)(114.82 g/mol) 2 23 atoms/mol -8 -8 4.59 x 10 cm 4.95 x 10 cm /unit cell 6.023 x 10
(
)(
)
(
)
= 7.31 g/cm3 3. 18 (a) We are asked to calculate the unit cell volume for Be. From the solution to Problem 3.7
VC = 6R2c 3
But, c = 1.568a, and a = 2R, or c = 3.14R, and
VC = (6)(3.14) R3 3
= (6) (3.14 ) 3 0.1143 x 10-7 cm] ( )[ 3 = 4.87 x 10-23 cm3 /unit cell
(b) The density of Be is determined as follows:
nA Be V N
C A
=
For HCP, n = 6 atoms/unit cell, and for Be, A
Be
= 9.01 g/mol. Thus,
=
(
(6 atoms/unit cell)(9.01 g/mol) -23 cm 3 /unit cell 6.023 x 1023 atoms/mol 4.87 x 10
) (
)
13
= 1.84 g/cm3 3 The value given in the literature is 1.85 g/cm . 3.19 This problem calls for us to compute the atomic radius for Mg. In order to do this we must use Equation (3.5), as well as the expression which relates the atomic radius to the unit cell volume for HCP; from Problem 3.7 it was shown that
VC = 6R2c 3
In this case c = 1.624(2R). Making this substitution into the previous equation, and then solving for R using Equation (3.5) yields
nAMg R = ( 1.624) 12 3 N A
1/3
(
)
(6 atoms/unit cell)(24.31 g/mol ) = 1.624) 12 3 1.74 g/cm3 6.023 x 1023 atoms/mol (
(
)(
) (
)
1/3
= 1.60 x 10-8 cm = 0.160 nm 3.20 This problem asks that we calculate the unit cell volume for Co which has an HCP crystal structure. In order to do this, it is necessary to use a result of Problem 3.7, that is
VC = 6R2c 3
The problem states that c = 1.623a, and a = 2R. Therefore
VC = (1.623)(12 3 ) R 3
= (1.623)(12 3) 1.253 x 10
(
-8
cm
)
3
= 6.64 x 10
-23
cm
3
= 6.64 x 10
-2
nm
3
14
3.21 (a) The unit cell shown in the problem belongs to the tetragonal crystal system since a = b = 0.35 nm, c = 0.45 nm, and = = = 90. (b) The crystal structure would be called body-centered tetragonal. (c) As with BCC, n = 2 atoms/unit cell. Also, for this unit cell
VC = 3.5 x 10
(
-8
cm
) (4.5 x 10-8 cm)
2
= 5.51 x 10
-23
cm /unit cell
3
Thus,
=
nA V N
C A
=
(
(2 atoms/unit cell)(141 g/mol) -23 cm3 /unit cell 6.023 x 1023 atoms/mol 5.51 x 10
) (
)
= 8.50 g/cm3
3.22 First of all, open `Notepad" in Windows. Now enter into "Notepad" commands to generate the AuCu3 unit cell. One set of commands that may be used is as follows: [DisplayProps] Rotatez=-30 Rotatey=-15 [AtomProps] Gold=LtRed,0.14 Copper=LtYellow,0.13 [BondProps] SingleSolid=LtGray [Atoms] Au1=1,0,0,Gold Au2=0,0,0,Gold Au3=0,1,0,Gold Au4=1,1,0,Gold Au5=1,0,1,Gold Au6=0,0,1,Gold
15
Au7=0,1,1,Gold Au8=1,1,1,Gold Cu1=0.5,0,0.5,Copper Cu2=0,0.5,0.5,Copper Cu3=0.5,1,0.5,Copper Cu4=1,0.5,0.5,Copper Cu5=0.5,0.5,1,Copper Cu6=0.5,0.5,0,Copper [Bonds] B1=Au1,Au5,SingleSolid B2=Au5,Au6,SingleSolid B3=Au6,Au2,SingleSolid B4=Au2,Au1,SingleSolid B5=Au4,Au8,SingleSolid B6=Au8,Au7,SingleSolid B7=Au7,Au3,SingleSolid B8=Au3,Au4,SingleSolid B9=Au1,Au4,SingleSolid B10=Au8,Au5,SingleSolid B11=Au2,Au3,SingleSolid B12=Au6,Au7,SingleSolid Under the "File" menu of "Note Pad," click "Save As", and then assign the file for this figure a name followed by a period and "mdf"; for example, "AuCu3.mdf". And, finally save this file in the "mdf" file inside of the "Interactive MSE" folder (which may be found in its installed location). Now, in order to view the unit cell just generated, bring up "Interactive MSE", and then open any one of the three submodules under "Crystallinity and Unit Cells" or the "Ceramic Structures" module. Next select "Open" under the "File" menu, and then open the "mdf" folder. Finally, select the name you assigned to the item in the window that appears, and hit the "OK" button. The image that you generated will now be displayed. 3.23 First of all, open `Notepad" in Windows.. Now enter into "Notepad" commands to generate the AuCu unit cell. One set of commands that may be used is as follows: [DisplayProps] Rotatez=-30 Rotatey=-15 [AtomProps] Gold=LtRed,0.14 Copper=LtYellow,0.13 [BondProps] SingleSolid=LtGray [Atoms] Au1=0,0,0,Gold Au2=1,0,0,Gold Au3=1,1,0,Gold Au4=0,1,0,Gold Au5=0,0,1.27,Gold Au6=1,0,1.27,Gold
16
Au7=1,1,1.27,Gold Au8=0,1,1.27,Gold Cu1=0.5,0.5,0.635,Copper [Bonds] B1=Au1,Au2,SingleSolid B2=Au2,Au3,SingleSolid B3=Au3,Au4,SingleSolid B4=Au1,Au4,SingleSolid B5=Au5,Au6,SingleSolid B6=Au6,Au7,SingleSolid B7=Au7,Au8,SingleSolid B8=Au5,Au8,SingleSolid B9=Au1,Au5,SingleSolid B10=Au2,Au6,SingleSolid B11=Au3,Au7,SingleSolid B12=Au4,Au8,SingleSolid Under the "File" menu of "Note Pad," click "Save As", and then assign the file for this figure a name followed by a period and "mdf"; for example, "AuCu.mdf". And, finally save this file in the "mdf" file inside of the "Interactive MSE" folder (which may be found in its installed location). Now, in order to view the unit cell just generated, bring up "Interactive MSE", and then open any one of the three submodules under "Crystallinity and Unit Cells" or the "Ceramic Structures" module. Next select "Open" under the "File" menu, and then open the "mdf" folder. Finally, select the name you assigned to the item in the window that appears, and hit the "OK" button. The image that you generated will now be displayed. 3.24 A unit cell for the face-centered orthorhombic crystal structure is presented below.
3.25 This problem asks that we list the point coordinates for all of the atoms that are associated with the FCC unit cell. From Figure 3.1b, the atom located of the origin of the unit cell has the coordinates
17
000.
Coordinates for other atoms in the bottom face are 100, 110, 010, and
11 2 2
0.
(The z
coordinate for all these points is zero.) For the top unit cell face, the coordinates are 001, 101, 111, 011, and
11 2 2
1.
(These
coordinates are the same as bottom-face coordinates except that the "0" z coordinate has been replaced by a "1".) Coordinates for only those atoms that are positioned at the centers of both side faces, and centers of both front and back faces need to be specified. For the front and back-center face atoms, the coordinates are 1
11 22
and 0
1 1
the respective coordinates are
22 1 1
, respectively. While for the left and right side center-face atoms, and
1 1 2 2
2
0
2
1 .
3.26 (a) Here we are asked list point coordinates for both sodium and chlorine ions for a unit cell of the sodium chloride crystal structure, which is shown in Figure 12.2. In Figure 12.2, the chlorine ions are situated at all corners and face-centered positions. Therefore, point coordinates for these ions are the same as for FCC, as presented in the previous problem--that is, 000, 100, 110, 010, 001, 101, 111, 011,
1 1 2 2 11 2 2
0,
11
2 2
1, 1
11 22
, 0
1 1 22
,
1
2
0
1
2
, and
1 .
Furthermore, the sodium ions are situated at the centers of all unit cell edges, and, in addition, at the unit cell center. For the bottom face of the unit cell, the point coordinates are as follows:
1 2
00 , 1 0 ,
2
1
1
2
10 , 0 0 . While, for the horizontal plane that passes through the center of
2 1 2 1 111 2 22 2
1
the unit cell (which includes the ion at the unit cell center), the coordinates are 0 0 , 1 0 ,
11 , and 01 . And for the four ions on the top face
2 2 1 1 1 2
,
01 , 1 1 ,
2
1
1 2
11 , and 0 1 .
2
1
(b) This portion of the problem calls for us to list the point coordinates of both the zinc and sulfur atoms for a unit cell of the zinc blende structure, which is shown in Figure 12.4. First of all, the sulfur atoms occupy the face-centered positions in the unit cell, which from the solution to Problem 3.25, are as follows: 000, 100, 110, 010, 001, 101, 111, 011,
1
11 22 11 1 1 22 1 2 1 2 1 1 2 2 2 2
0,
11
2 2
1,
,0
,
0
, and
1 .
3 1 1
Now, using an x-y-z coordinate system oriented as in Figure 3.4, the coordinates of the zinc atom that lies toward the lower-left-front of the unit cell has the coordinates situated toward the lower-right-back of the unit cell has coordinates of that resides toward the upper-left-back of the unit cell has the
4 44 13 1
, whereas the atom
4 44 113
. Also, the zinc atom And, the
3 33 4 44
4 44
coordinates.
coordinates of the final zinc atom, located toward the upper-right-front of the unit cell, are
.
18
3.27 A tetragonal unit in which are shown the 11
1 2
and
111 2 42
point coordinates is presented below.
3.28 This portion of the problem calls for us to draw a [12 1 ] direction within an orthorhombic unit cell (a b c, = = = 90). Such a unit cell with its origin positioned at point O is shown below. We first move along the +x-axis a units (from point O to point A), then parallel to the +y-axis 2b units (from point A to point B). Finally, we proceed parallel to the z-axis -c units (from point B to point C). The [12 1 ] direction is the vector from the origin (point O) to point C as shown.
We are now asked to draw a (210) plane within an orthorhombic unit cell. First remove the three indices from the parentheses, and take their reciprocals--i.e., 1/2, 1, and . This means that the
19
plane intercepts the x-axis at a/2, the y-axis at b, and parallels the z-axis. The plane that satisfies these requirements has been drawn within the orthorhombic unit cell below.
3.29 (a) This portion of the problem asks that a [0 1 1] direction be drawn within a monoclinic unit cell (a b c, and = = 90 ). One such unit cell with its origin at point O is sketched below. For this direction, there is no projection along the x-axis since the first index is zero; thus, the direction lies in the y-z plane. We next move from the origin along the minus y-axis b units (from point O to point R). Since the final index is a one, move from point R parallel to the z-axis, c units (to point P). Thus, the
[0 1 1] direction corresponds to the vector passing from the origin to point P, as indicated in the
figure.
20
(b) A (002) plane is drawn within the monoclinic cell shown below. We first remove the parentheses and take the reciprocals of the indices; this gives , , and 1/2. Thus, the (002) plane parallels both x- and y-axes, and intercepts the z-axis at c/2, as indicated in the drawing.
3.30 (a) We are asked for the indices of the two directions sketched in the figure. For direction 1, the projection on the x-axis is zero (since it lies in the y-z plane), while projections on the y- and z-axes are b/2 and c, respectively. This is an [012] direction as indicated in the summary below
x Projections Projections in terms of a, b, and c Reduction to integers Enclosure 0 0 0a
y b/2 1/2 1
[012]
z c 1 2
Direction 2 is [112 ] as summarized below. x Projections Projections in terms of a, b, and c Reduction to integers 1/2 1 1/2 1 -1 -2 a/2 y b/2 z -c
21
Enclosure
[112 ]
(b) This part of the problem calls for the indices of the two planes which are drawn in the sketch. Plane 1 is an (020) plane. The determination of its indices is summarized below. x Intercepts Intercepts in terms of a, b, and c Reciprocals of intercepts Enclosure Plane 2 is a (22 1) plane, as summarized below. x Intercepts Intercepts in terms of a, b, and c Reciprocals of intercepts Enclosure a/2 1/2 2 y -b/2 -1/2 -2
(22 1)
y b/2 1/2 2
(020)
z c 0
a 0
z c 1 1
3.31 The directions asked for are indicated in the cubic unit cells shown below.
22
3.32 Direction A is a [ 1 10] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x Projections Projections in terms of a, b, and c Reduction to integers Enclosure -1 1 not necessary
[ 1 10]
y b
z 0c 0
-a
Direction B is a [121] direction, which determination is summarized as follows. The vector passes through the origin of the coordinate system and thus no translation is necessary. Therefore, x Projections
Projections in terms of a, b,
a 2
y
b
z
c c
Reduction 2
23
and to integers Enclosure
1 2
1
2 [121]
1 2
1
1
Direction C is a [0 1 2 ] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x Projections
Projections in terms of a, b, and c
Reduction to integers Enclosure
y
-
b 2
1 2
z
-c
-1 -2
0a
0
0
-
-1
[0 1 2 ]
Direction D is a [12 1] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x Projections
Projections in terms of a, b, and c
Reduction to integers Enclosure
a 2
1 2
y
-b
-1
-2
[12 1]
z
c 2
1 2
1
1
3.33 Direction A is a [331 ] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x Projections
Projections in terms of a, b,
y b -
z
c 3
a
24
and c
Reduction to integers Enclosure
1
3
1
3
[331 ]
-
1 3
-1
Direction B is a [4 03 ] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x Projections
Projections in terms of a, b, and c
Reduction to integers Enclosure
y 0b -
z
c 2
1 2
-
2a 3
2 3
-
0
0
[4 03 ]
-
-4
-3
Direction C is a [3 61] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x Projections
Projections in terms of a, b, and c
Reduction to integers Enclosure
y
b
1
6
[3 61]
z
c 6
1 6
-
a 2
1 2
-3
1
Direction D is a [ 1 1 1 ] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x Projections
Projections in terms of a, b,
y
b 2
z
c 2
-
a 2
25
and c
Reduction to integers Enclosure
-
1 2
1 2
-
1 2
-1
1
[111]
-1
3.34 For tetragonal crystals a = b c and = = = 90; therefore, projections along the x and y axes are equivalent, which are not equivalent to projections along the z axis. (a) Therefore, for the [011] direction, equivalent directions are the following: [101], [ 1 0 1] , [ 1 01 ] ,
[10 1] , [011 ] , [0 1 1] , and [0 1 1 ] .
(b) Also, for the [100] direction, equivalent directions are the following: [ 1 00] , [010], and [0 1 0 ] .
3.35 (a) We are asked to convert [100] and [111] directions into the four- index Miller-Bravais scheme for hexagonal unit cells. For [100] u' = 1, v' = 0, w' = 0 From Equations (3.6)
u = n n 2n (2u ' - v' ) = (2 - 0) = 3 3 3 n n n (2v' - u' ) = (0 - 1) = 3 3 3
v =
n 2n n t = - (u + v) = - - = 3 3 3
w = nw' = 0 If we let n = 3, then u = 2, v = -1, t = -1, and w = 0. Thus, the direction is represented as [uvtw] =
[2 1 1 0 ] .
For [111], u' = 1, v' = 1, and w' = 1; therefore,
u = n n (2 - 1) = 3 3 n n (2 - 1) = 3 3
v =
26
2n n n t =- + = 3 3 3
w=n If we again let n = 3, then u = 1, v = 1, t = -2, and w = 3. Thus, the direction is represented as
[112 3] .
(b) This portion of the problem asks for the same conversion of the (010) and (101) planes. A plane for hexagonal is represented by (hkil) where i = - (h + k), and h, k, and l are the same for both systems. For the (010) plane, h = 0, k = 1, l = 0, and i = - (0 + 1) = -1 Thus, the plane is now represented as (hkil) = (01 1 0) . For the (101) plane, i = - (1 + 0) = -1, and (hkil) = (10 1 1) .
3.36 For plane A we will leave the origin at the unit cell as shown. If we extend this plane back into the plane of the page, then it is a (11 1 ) plane, as summarized below.
x Intercepts Intercepts in terms of a, b, and c Reciprocals of intercepts Reduction Enclosure 1 1 a
y b 1 1 not necessary
(11 1 )
z -c -1 -1
For plane B we will leave the origin of the unit cell as shown; this is a (230) plane, as summarized below. x Intercepts
Intercepts in terms of a, b,
a 2
y
b 3
z
c
27
and c
Reciprocals of intercepts Enclosure
1 2
1 3
0
2
3 (230)
3.37 For plane A we will move the origin of the coordinate system one unit cell distance to the right along the y axis; thus, this is a (1 1 0 ) plane, as summarized below.
x Intercepts
Intercepts in terms of a, b, and c
Reciprocals of intercepts Reduction Enclosure
a 2
1 2
y
b 2
1 2
z
c
-
0 0
2 1
-2 -1
(1 1 0 )
For plane B we will leave the origin of the unit cell as shown; thus, this is a (122) plane, as summarized below. x Intercepts a y
b 2
z
c 2
Intercepts in terms of a, b, and c
Reciprocals of intercepts Reduction Enclosure
1
1
1 2
1 2
2 not necessary (122)
2
3.38 For plane A since the plane passes through the origin of the coordinate system as shown, we will move the origin of the coordinate system one unit cell distance vertically along the z axis; thus, this is a (21 1 ) plane, as summarized below.
28
x Intercepts
Intercepts in terms of a, b, and c
Reciprocals of intercepts Reduction Enclosure
a
2
y
b
z
-c
1
2
1 1 not necessary
(21 1 )
-1 -1
2
For plane B, since the plane passes through the origin of the coordinate system as shown, we will move the origin one unit cell distance vertically along the z axis; this is a (02 1 ) plane, as summarized below. x Intercepts
Intercepts in terms of a, b, and c
Reciprocals of intercepts Reduction Enclosure
y
b 2
1 2
z
-c
a
0
-1 -1
2 not necessary
(02 1 )
3.39 The (01 1 1) and (2 1 1 0 ) planes in a hexagonal unit cell are shown below.
29
3.40 (a) For this plane we will leave the origin of the coordinate system as shown; thus, this is a (12 11) plane, as summarized below. a1 Intercepts
Intercepts in terms of a's and c
Reciprocals of intercepts Reduction Enclosure
a2 a 2 1
2
a3
a
1 1
(12 11)
z
a
1
1
c
1 1
-2
not necessary
(b) For this plane we will leave the origin of the coordinate system as shown; thus, this is a
(2 1 1 2) plane, as summarized below.
a1 Intercepts Intercepts in terms of a's and c Reciprocals of intercepts Reduction Enclosure a/2 1/2 2
a2 -a -1 -1
(2 1 1 2)
a3 -a -1 -1
z
c/2 1/2 2
not necessary
3.41 The planes called for are plotted in the cubic unit cells shown below.
30
3.42 (a) The atomic packing of the (100) plane for the FCC crystal structure is called for. An FCC unit cell, its (100) plane, and the atomic packing of this plane are indicated below.
(b) For this part of the problem we are to show the atomic packing of the (111) plane for the BCC crystal structure. A BCC unit cell, its (111) plane, and the atomic packing of this plane are indicated below.
31
3.43 (a) The unit cell in Problem 3.21 is body-centered tetragonal. Only the (100) (front face) and (0 1 0) (left side face) planes are equivalent since the dimensions of these planes within the unit cell (and therefore the distances between adjacent atoms) are the same (namely 0.45 nm x 0.35 nm), which are different than the (001) (top face) plane (namely 0.35 nm x 0.35 nm). (b) The equivalent planes are (101), (011), and (1 01) ; the dimensions of these planes within the unit cell are the same--that is 0.35 nm x (0.35 nm)2 + (0.45 nm)2
[
]
1/ 2
.
(c) All of the (111), (1 1 1) , (11 1) , and (1 1 1 ) planes are equivalent. 3.44 (a) The intersection between (110) and (111) planes results in a [ 1 10] , or equivalently, a [1 1 0] direction. (b) The intersection between (110) and (1 1 0) planes results in a [001], or equivalently, a [001 ] direction. (c) The intersection between (11 1 ) and (001) planes results in a [ 1 10] , or equivalently, a [1 1 0] direction. 3.45 (a) In the figure below is shown a [100] direction within an FCC unit cell.
For this [100] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalent of 1 atom that is centered on the direction vector. The length of this direction vector is just the unit cell edge length, 2R 2 [Equation (3.1)]. Therefore, the expression for the linear density of this plane is
LD100 = number of atoms centered on [100] direction vector length of [100] direction vector 1 atom 2R 2 1 2R 2
=
=
An FCC unit cell within which is drawn a [111] direction is shown below.
32
For this [111] direction, the vector shown passes through only the centers of the single atom at each of its ends, and, thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of this direction vector is denoted by z in this figure, which is equal to
z = x2 + y2
where x is the length of the bottom face diagonal, which is equal to 4R. Furthermore, y is the unit cell edge length, which is equal to 2R 2 [Equation (3.1)]. Thus, using the above equation, the length z may be calculate as follows:
z =
(4R)2
+ 2R 2
(
)2 =
24R2 = 2R 6
Therefore, the expression for the linear density of this plane is
LD111 = number of atoms centered on [111] direction vector length of [111] direction vector 1 atom 2R 6 1 2R 6
=
=
(b) From the table inside the front cover, the atomic radius for copper is 0.128 nm. Therefore, the linear density for the [100] direction is
LD100 (Cu) = 1 2R 2
=
1 (2)(0.128 nm) 2
= 2.76 nm -1 = 2.76 x 10 -9 m -1
While for the [111] direction
33
LD111(Cu) =
1 2R 6
=
1 (2)(0.128 nm) 6
= 1.59 nm-1 = 1.59 x 10-9 m -1
3.46 (a) In the figure below is shown a [110] direction within a BCC unit cell.
For this [110] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of this direction vector is denoted by x in this figure, which is equal to
x = z2 - y2 4R 3
where y is the unit cell edge length, which, from Equation (3.3) is equal to
. Furthermore, z is
the length of the unit cell diagonal, which is equal to 4R Thus, using the above equation, the length x may be calculate as follows:
4R 2 = - 3
32R2 2 = 4R 3 3
x =
(4R)2
Therefore, the expression for the linear density of this direction is
LD110 = number of atoms centered on [110] direction vector length of [110] direction vector 1 atom 4R 2 3 3 4R 2
=
=
A BCC unit cell within which is drawn a [111] direction is shown below.
34
For although the [111] direction vector shown passes through the centers of three atoms, there is an equivalence of only two atoms associated with this unit cell--one-half of each of the two atoms at the end of the vector, in addition to the center atom belongs entirely to the unit cell. Furthermore, the length of the vector shown is equal to 4R, since all of the atoms whose centers the vector passes through touch one another. Therefore, the linear density is equal to
LD111 = number of atoms centered on [111] direction vector length of [111] direction vector 2 atoms 1 = 4R 2R
=
(b) From the table inside the front cover, the atomic radius for iron is 0.124 nm. Therefore, the linear density for the [110] direction is
3 4R 2 3 (4)(0.124 nm) 2
-1 -9 -1
LD110 (Fe) =
=
= 2.47 nm
= 2.47 x 10
m
While for the [111] direction
1 1 = = 4.03 nm-1 = 4.03 x 10-9 m-1 2R (2)(0.124 nm)
LD111(Fe) =
3.47 (a) In the figure below is shown a (100) plane for an FCC unit cell.
35
For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms associated with this FCC (100) plane. The planar section represented in the above figure is a square, wherein the side lengths are equal to the unit cell edge length, 2R 2 [Equation (3.1)]; and, thus, the area of this square is just 2 R 2 for this (100) plane is just
number of atoms centered on (100) plane area of (100) plane 2 atoms 8 R2 1 4R 2
(
)2 = 8R2.
Hence, the planar density
PD100 =
=
=
That portion of an FCC (111) plane contained within a unit cell is shown below.
There are six atoms whose centers lie on this plane, which are labeled A through F. One-sixth of each of atoms A, D, and F are associated with this plane (yielding an equivalence of one-half atom), with one-half of each of atoms B, C, and E (or an equivalence of one and one-half atoms) for a total equivalence of two atoms. Now, the area of the triangle shown in the above figure is equal to onehalf of the product of the base length and the height, h. If we consider half of the triangle, then
36
(2R)2 + h 2 = (4R)2
which leads to h = 2 R 3 . Thus, the area is equal to
4R(h) (4R) 2R 3 = = 4R2 3 2 2
Area =
(
)
And, thus, the planar density is
number of atoms centered on (111) plane area of (111) plane 2 atoms 4R 2 3 1 2R2 3
PD111 =
=
=
(b) From the table inside the front cover, the atomic radius for aluminum is 0.143 nm. Therefore, the planar density for the (100) plane is
PD100 (Al) = 1 4 R2
=
1 4(0.143 nm)2
= 12.23 nm-2 = 1.223 x 10 -17 m-2
While for the (111) plane
PD111(Al) = 1 2R2 3
=
1 2 3 (0.143 nm) 2
= 14.12 nm-2 = 1.412 x 10 -17 m-2
3.48 (a) A BCC unit cell within which is drawn a [100] plane is shown below.
For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells. Thus, there is the equivalence of 1 atom associated with this BCC (100) plane. The planar section represented in the above figure is a square, wherein the side lengths are
37
equal to the unit cell edge length,
4R 3
2
4R
3
[Equation (3.3)]; and, thus, the area of this square is just
=
16 R 2 . Hence, the planar density for this (100) plane is just 3
PD100 =
number of atoms centered on (100) plane area of (100) plane 1 atom 16 R2 3 3 16 R2
=
=
A BCC unit cell within which is drawn a [110] plane is shown below.
For this (110) plane there is one atom at each of the four cube corners through which it passes, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms associated with this BCC (110) plane. The planar section represented in the above figure is a rectangle, as noted in the figure below.
38
From this figure, the area of the rectangle is the product of x and y. The length x is just the unit cell edge length, which for BCC [Equation (3.3)] is
the triangle bounded by the lengths x, y, and z
z2 - x2
4R
3
. Now, the diagonal length z is equal to 4R. For
y=
Or
4R 2 4R 2 = (4R) - 3 3
2
y=
Thus, in terms of R, the area of this (110) plane is just
4 R 4R 2 16R2 2 Area(110) = xy = = 3 3 3
And, finally, the planar density for this (110) plane is just
number of atoms centered on (110) plane area of (110) plane 2 atoms 16R2 3 2 3 8R 2 2
PD110 =
=
=
(b) From the table inside the front cover, the atomic radius for molybdenum is 0.136 nm. Therefore, the planar density for the (100) plane is
PD100 (Mo) = 3 16R2
=
3 16(0.136 nm)2
= 10.14 nm-2 = 1.014 x 10-17 m -2
While for the (110) plane
PD110 (Mo) = 3 8R 2 2
=
3 8 (0.136 nm) 2 2
= 14.34 nm-2 = 1.434 x 10-17 m -2
3.49 (a) A (0001) plane for an HCP unit cell is show below.
39
Each of the 6 perimeter atoms in this plane is shared with three other unit cells, whereas the center atom is shared with no other unit cells; this gives rise to three equivalent atoms belonging to this plane. In terms of the atomic radius R, the area of each of the 6 equilateral triangles that have been drawn is R
2
3 , or the total area of the plane shown is 6 R
2
3 . And the planar density for this
(0001) plane is equal to
PD0001 = number of atoms centered on (0001) plane area of (0001) plane 3 atoms 6R2 3 1 2R2 3
=
=
(b) From the table inside the front cover, the atomic radius for titanium is 0.145 nm. Therefore, the planar density for the (0001) plane is
1 2R2 3 1 2 3 (0.145 nm) 2
= 13.73 nm-2 = 1.373 x 10 -17 m -2
PD0001(Ti) =
=
3.50 Unit cells are constructed below from the three crystallographic planes provided in the problem.
40
(a) This unit cell belongs to the tetragonal system since a = b = 0.40 nm, c = 0.55 nm, and = = = 90. (b) This crystal structure would be called body-centered tetragonal since the unit cell has tetragonal symmetry, and an atom is located at each of the corners, as well as the cell center. 3.51 The unit cells constructed below show the three crystallographic planes that were provided in the problem.
(a) This unit cell belongs to the orthorhombic crystal system since a = 0.25 nm, b = 0.30 nm, c = 0.20 nm, and = = = 90. (b) This crystal structure would be called face-centered orthorhombic since the unit cell has orthorhombic symmetry, and an atom is located at each of the corners, as well as at each of the face centers. (c) In order to compute its atomic weight, we employ Equation (3.5), with n = 4; thus
41
A =
VCN A
n
(18.91 g/cm3 )(2.0)(2.5)(3.0) (x 10-24 cm3 /unit cell)(6.023 x 1023 atoms/mol) =
4 atoms/unit cell
= 42.7 g/mol 3.52 Although each individual grain in a polycrystalline material may be anisotropic, if the grains have random orientations, then the solid aggregate of the many anisotropic grains will behave isotropically. 3.53W From the table, aluminum has an FCC crystal structure and an atomic radius of 0.1431 nm. Using Equation (3.1), the lattice parameter a may be computed as
a = 2 R 2 = (2)(0.1431 nm) 2 = 0.4048 nm
Now, the interplanar spacing d110 maybe determined using Equation (3.3W) as
d110 = a (1)
2
+ (1)
2
+ (0)
2
=
0.4048 nm = 0.2862 nm 2
3.54W We must first calculate the lattice parameter using Equation (3.3) and the value of R cited in Table 3.1 as
a= 4R 3 = (4)(0.1249 nm) 3 = 0.2884 nm
Next, the interplanar spacing may be determined using Equation (3.3W) according to
d 310 = a (3)
2
+ (1)
2
+ (0)
2
=
0.2884 nm = 0.0912 nm 10
And finally, employment of Equation (3.2W) yields
42
sin =
n (1)(0.0711 nm) = = 0.390 2d (2)(0.0912 nm)
= sin-1(0.390) = 22.94
And
2 = (2)(22.94) = 45.88
3.55W From the table, -iron has a BCC crystal structure and an atomic radius of 0.1241 nm. Using Equation (3.3) the lattice parameter, a, may be computed as
a= 4R 3 = (4)(0.1241 nm) 3 = 0.2866 nm
Now, the d111 interplanar spacing may be determined using Equation (3.3W) as
d111 = a (1)
2
+ (1)
2
+ (1)
2
=
0.2866 nm = 0.1655 nm 3
And, similarly for d211
d 211 = a (2)
2
+ (1)
2
+ (1)
2
=
0.2866 nm = 0.1170 nm 6
3.56W (a) From the data given in the problem, and realizing that 36.12 = 2, the interplanar spacing for the (311) set of planes may be computed using Equation (3.3W) as
d 311 = n (1)(0.0711 nm) = = 0.1147 nm 36.12 2 sin (2) sin 2
(b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation (3.3W), and then R from Equation (3.1) since Rh has an FCC crystal structure. Therefore,
a = d 311 (3)2 + (1)2 + (1)2 = (0.1147 nm)
( 11)= 0.3804
nm
43
And
R =
a 0.3804 nm = = 0.1345 nm 2 2 2 2
3.57W. (a) From the data given in the problem, and realizing that 75.99 = 2, the interplanar spacing for the (211) set of planes may be computed using Equation (3.2W) as
d 211 = n (1)(0.1659 nm) = = 0.1348 nm 75.99 2 sin (2) sin 2
(b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation (3.3W), and then R from Equation (3.3) since Nb has a BCC crystal structure. Therefore,
a = d 211 (2)2 + (1)2 + (1)2 = (0.1347 nm)
( 6 ) = 0.3300
nm
And
a 3 (0.3300 nm) 3 = = 0.1429 nm 4 4
R=
3.58W The first step to solve this problem is to compute the interplanar spacing using Equation (3.2W). Thus,
d hkl (1)(0.1542 nm) n = = 0.2035 nm 44.53 2 sin (2) sin 2
Now, employment of both Equations (3.3W) and (3.1), and the value of R for nickel from Table 3.1 (0.1246 nm) leads to
2 2 2
h
+ k
+ l =
a d
=
hkl
2R d
2
hkl
=
(2)(0.1246 nm) 2 (0.2035 nm)
= 1.732
This means that
44
h2 + k2 + l2 = (1.732)2 = 3.0
By trial and error, the only three integers which are all odd or even, the sum of the squares of which equals 3.0 are 1, 1, and 1. Therefore, the set of planes responsible for this diffraction peak is the (111) set. 3.59W For each peak, in order to compute the interplanar spacing and the lattice parameter we must employ Equations (3.3W) and (3.2W), respectively. For the first peak which occurs at 31.3
d111 = n (1)(0.1542 nm) = = 0.2858 nm 31.3 2 sin (2) sin 2
And
a = dhkl (h)2 + (k)2 + (l) 2 = d111 (1)2 + (1)2 + (1)2 = (0.2858 nm) 3 = 0.4950 nm
Similar computations are made for the other peaks which results are tabulated below: Peak Index 200 220 311 222 2 36.6 52.6 62.5 65.5 dhkl(nm) 0.2455 0.1740 0.1486 0.1425 a (nm) 0.4910 0.4921 0.4929 0.4936
3.60W The first four diffraction peaks that will occur for BCC consistent with h + k + l being even are (110), (200), (211), and (220). 3.61W (a) Since W has a BCC crystal structure, only those peaks for which h + k + l are even will appear. Therefore, the first peak results by diffraction from (110) planes. (b) For each peak, in order to calculate the interplanar spacing we must employ Equation (3.2W). For the first peak which occurs at 40.2
45
d110 =
n (1)(0.1542 nm) = = 0.2244 nm 40.2 2 sin (2) sin 2
(c) Employment of Equations (3.3W) and (3.3) is necessary for the computation of R for W as
R =
a 4
3
=
(dhkl)( 3)
3
(h)2 + (k)2 + (l) 2 4
=
(0.2244 nm)(
)
(1)2 + (1)2 + (0)2 4
= 0.1374 nm 3.62 A material in which atomic bonding is predominantly ionic in nature is less likely to form a noncrystalline solid upon solidification than a covalent material because covalent bonds are directional whereas ionic bonds are nondirectional; it is more difficult for the atoms in a covalent material to assume positions giving rise to an ordered structure.
46