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Hany E.Z.Farag

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CHAPTER 3
THE STRUCTURE OF CRYSTALLINE SOLIDS
PROBLEM SOLUTIONS
Fundamental Concepts
3.1 What is the difference between atomic structure and crystal structure?
Solution
Atomic structure relates to the number of protons and neutrons inthenucleusofanatom,aswellasthe
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.
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Metallic Crystal Structures
3.2 If the atomic radius of aluminum is 0.143 nm, calculate the volume of its unit cell in cubic meters.
Solution
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 -9 3 -29 3
VC = 16R 2 = (16) (0.143 10 m ) ( 2 )= 6.62 10 m
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3 . 3 Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic
radius R are related through a =4R/ 3.
Solution
Consider the BCC unit cell shown below
Using the triangle NOP
2 2 2 2
(NP) = a + a =2 a
And then for triangle NPQ,
(NQ) 2=( QP) 2 +( NP) 2
But NQ = 4R, R being the atomic radius. AlsQP = a. Therefore,
(4R)2 = a +2 a 2
or
4R
a = 3
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.4 For the HCP crystal structure, show that the ideal c/a ratio is 1.633.
Solution
A sketch of one-third of an HCP unit cell is shown below.
Consider the tetrahedron labeled as JKLM, which is reconstructed as
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 =2 R = a
where R is the atomic radius. Furthermore, from triangle JHM,
2 2 2
(JM ) =( JH) (MH)
or
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2 2 c
a =( JH ) + 2
Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle,
a/2 3
cos 30 = =
JH 2
and
JH = a
3
Substituting this value for JH in the above expression yields
a c a 2 c2
a2 = + = +
3 2 3 4
and, solving for c/a
c 8
a = 3 = 1.633
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.5 Show that the atomic packing factor for BCC is 0.68.
Solution
The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or
VS
APF = V
C
Since there are two spheres associated with each unit cell for BCC
4 R3 8R 3
VS = 2(sphere volume) = 2 =
3 3
Also, the unit cell has cubic symmetry, that is V. But a depends on R according to Equation 3.3, and
C
4R 64R 3
V = =
C 3 33
Thus,
V 8R /3
APF = S = = 0.68
VC 64R /3 3
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.6 Show that the atomic packing factor for HCP is 0.74.
Solution
The APF is just the total sphere volume-unit cell volume ratio. For HCP, there are the equivalent of six
spheres per unit cell, and thus
3
R 3
VS =6 3 =8 R
Now, the unit cell volume is just the product of the base area times the cell height, c. This basearea isjust three
times the area of the parallelepiped ACDE shown below.
The area of ACDE is just the length of CD times the heightBC . But CD is just a or 2R, and
BC =2 R cos (30) = 2R 3
2
Thus, the base area is just
2R 3
AREA = (3)(CD)(BC) = (3)(2R) 6 R 2 3
2
and since c = 1.633a = 2R(1.633)
) 1 S . V = (A3REA)(c) = 6R c 3 (
C
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. = (6R 2 3) (2)(1.633)R = 12 3 (1.633)R3
Thus,
V 8 R3
APF = S = = 0.74
VC 12 3 (1.633)R 3
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Density Computations
7 . 3 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol.
Compute and compare its theoretical density with the experimental value found inside the front cover.
Solution
This problem calls for a computation of the density of iron. According to Equation 3.5
nAFe
=
VCN A
For BCC, n = 2 atoms/unit cell, and
3
4R
VC= 3
Thus,
nAFe
= 3
4R
3 N A
= (2 atoms/unit cell)(55.85 g/mol)
(4)(0.124 10 -7 cm) /3 3 /(unit cel(6.022 10 23atoms/mol )
3
= 7.90 g/cm
The value given inside the front cover is 7.87 g/cm .
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 8 . 3 Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4
3
g/cm , and an atomic weight of 192.2 g/mol.
Solution
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
= nAIr
16R 3 2 )N
A
And solving for R from the above expression yields
1/3
nA Ir
R =
6N A 2
/3
= (4 atoms/unit cell 92.2 g/mol
(16)(22.4 g/cm3)(6.022 10 23 atoms/mol )( 2 )
-8
= 1.36 10 cm = 0.136 nm
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 9 . 3 Calculate the radius of a vanadium atom, given that V has a BCC crystal structure, a density of 5.96
3
g/cm , and an atomic weight of 50.9 g/mol.
Solution
This problem asks for us to calculate the radius of a vanadium atom. For BCC, n = 2 atoms/unit cell, and
3 3
V = 4R = 64R
C 3 33
Since, from Equation 3.5
nA
= V
VCN A
nA
= V
64 R 3
33 A
and solving for R the previous equation
1/3
33 nA V
R = 64 N
A
and incorporating values of parameters given in the problem statement
/3
R = ( 33 )(2 atoms/unit cell)(50.9 g/mol)
64)(5.96 g/cm 3)(6.022 10 23 atoms/mol )
= 1.32 10 -8cm = 0.132 nm
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.10 Some hypothetical metal has the simple cubic crystal structure shown in Figure 3.24. If its atomic
weight is 70.4 g/mol and the atomic radius is 0.126 nm, compute its density.
Solution
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 =2 R (Figure 3.24). Therefore,
employment of Equation 3.5 yields
= nA = nA
VCN A (2R) N A
and incorporating values of the other parameters provided in the problem statement leads to
= (1 atom/unit cell)(70.4 g/mol)
-8
2)(1.26 10 cm )/(unit cel(6.022 10 23 atoms/mol )
7.31 g/cm 3
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3.11 Zirconium has an HCP crystal structure and a density of 6.51 g/cm .
(a) What is the volume of its unit cell in cubic meters?
(b) If the c/a ratio is 1.593, compute the values of c and a.
Solution
(a) The volume of the Zr unit cell may be computed using Equation 3.5 as
nA Zr
V C
N A
Now, for HCP, n = 6 atoms/unit cell, and for Zr, A = 91.22 g/mol. Thus,
Zr
(6 atoms/unit cell)(91.22 g/mol)
VC 3 23
(6.51 g/cm )( 6.022 10 atoms/mol )
-22 3 -28 3
= 1.396 10 cm /unit cell = 1.396 10 m /unit cell
(b) From Equation 3.S1 of the solution to Problem 3.6, for HCP
V = 6R c 3
C
But, since a = 2R, (i.e., R = a/2) then
33 a c
V =6 c 3
C 2 2
but, since c = 1.593a
3
V = 33 (1.593)a = 1.396 10 -22cm /unit cell
C 2
Now, solving for a
1/3
(2)1.396 10 -22cm 3)
a =
(3)( 3)(1.593)
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= 3.23 10 cm = 0.323 nm
And finally
c = 1.593a = (1.593)(0.323 nm) = 0.515 nm
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.12 Using atomic weight, crystal structure, and atomic radius data tabulated inside the front cover,
compute the theoretical densities of lead, chromium, coppe r, and cobalt, and then compar e these values with the
measured densities listed in this same table. The c/a ratio for cobalt is 1.623.
Solution
Since Pb has an FCC crystal structure, n = 4, and V = 16R 3 2 (Equation 3.4). Also, R = 0.175 nm (1.75
C
10 cm) and A = 207.2 g/mol. Employment of Equation 3.5 yields
Pb
nA
= Pb
VCN A
(4 atoms/unit cell)(207.2 g/mol)
-8 3 23
6)(1.75 10 cm ) ( 2 )/(unit cell)(6.022 10 atoms/mol )
3
= 11.35 g/cm
3
The value given in the table inside the front cover is 11.35 g/cm .
3
3 4 R
Chromium has a BCC crystal structure for which n = 2 and V C = a = (Equation 3.3); also ACr =
3
52.00g/mol and R = 0.125 nm. Therefore, employment of Equation 3.5 leads to
(2 atoms/unit cell)(52.00 g/mol)
-8
4)(1.25 10 cm )/(unit cel(6.022 10 23 atoms/mol )
3
3
= 7.18 g/cm
The value given in the table is 7.19 g/cm .
Copper also has an FCC crystal structure and therefore
(4 atoms/unit cell)(63.55 g/mol)
-8 3 23
( 1.28 10 cm )( 2) /(unit cell6.022 10 atoms/mol )
= 8.90 g/cm 3
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. The value given in the table is 8.90 g/cm
Cobalt has an HCP crystal structure, and from the solution to Problem 3.6 (Equation 3.S1),
2
VC =6 R c 3
and, since c = 1.623a and a = 2R, c = (1.623)(2R); hence
V 6R 2(1.623)(R )3 (19.48)( 3)R 3
C
(19.48)( 3)(1.25 108 cm) 3
23 3
6.59 10 cm /unit cell
Also, there are 6 atoms/unit cell for HCP. Therefore the theoretical density is
= nACo
VCN A
(6 atoms/unit cell)(58.93 g/mol)
(6.59 10-23cm /unit cel)(6.022 10 23atoms/mol )
= 8.91 g/cm3
The value given in the table is 8.9 g/cm
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3.13 Rhodium has an atomic radius of 0.1345 nm and a density of 12.41 g/cm. Determine whether it has
an FCC or BCC crystal structure.
Solution
In order to determine whether Rh has an FCC or a BCC crystal structure, we need to compute its density
for each of the crystal structures. For FCC, n = 4, and aR= 2 (Equation 3.1). Also, from Figure 2.6, its atomic
weight is 102.91 g/mol. Thus, for FCC (employing Equation 3.5)
nA nA
Rh Rh
a N (2R 2 )3N
A A
(4 atoms/unit cell)(102.91 g/mol)
=
-8 3 23
)1.345 10 cm )( 2 ) /(unit ce6.022 10 atoms/mol )
= 12.41 g/cm3
which is the value provided in the problem statement. Therefore, Rh has the FCC crystal structure.
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4 1 . 3 Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For
each determine whether its crystal stru cture is FCC, BCC, or simple cubic and then justify your determination. A
simple cubic unit cell is shown in Figure 3.24.
Alloy Atomeiigcht Density Atomic Radius
( g/mol) (g/cm ) (nm)
A 77.4 8.22 0.125
B 107.6 13.42 0.133
C 127.3 9.23 0.142
Solution
For each of these three alloys we need, by trial and error, to calculate the density using Equation 3.5, and
compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures, the respective values of n
are 1, 2, and 4, whereas the expressions for a (since V = a ) are 2R, , and 4R .
C 3
For alloy A, let us calculate assuming a simple cubic crystal structure.
nA
= A
VCN A
nA
= A
R 3N
A
= (1 atom/unit cell)(77.4 g/mol)
8 3 23
)(1.25 10 ) /(unit cel(6.022 10 atoms/mol)
= 8.22 g/cm3
Therefore, its crystal structure is simple cubic.
For alloy B, let us calculate assuming an FCC crystal structure.
nA
= B
(2R 2) N A
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-8 3 23
2 (1.33 10 cm ) /(unit cel(6.022 10 atoms/mol )
= 13.42 g/cm 3
Therefore, its crystal structure is FCC.
For alloy C, let us calculate assuming a simple cubic crystal structure.
nA C
= 3
R N A
= (1 atom/unit cell)(127.3 g/mol)
-8 3 23
)(1.42 10 cm ) /(unit cel(6.022 10 atoms/mol )
3
= 9.23 g/cm
Therefore, its crystal structure is simple cubic.
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.15 The unit cell for tin has tetragonal symmetry, with a and b lattice parameters of 0.583 and 0.318 nm,
3
respectively. If its density, atomic weight, and atomic radius are 7.30 g/cm , 118.69 g/mol, and 0.151 nm,
respectively, compute the atomic packing factor.
Solution
In order to determine the APF for Sn, we need to compute both the unit cell volume (V C) which is just the
2
a c product, 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
VCN A
n =
ASn
(7.30 g/cm 3)(5.83) (3.18)( 10-24 cm 3)(6.022 10 23 atoms/mol )
=
118.69 g/mol
= 4.00 atoms/unit cell
Therefore
4 3
V (4)3 R
APF = S =
V C (a) (c)
4
(4) ()(1.51 10 -8 cm) 3
3
= (5.83 10 -8 cm) (3.18 10 -8 cm)
= 0.534
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students enrolled in courses for which the textbook has been adopted.roduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.16 Iodine has an orthorhombic unit cell for which th e a, b, and c lattice parameters are 0.479, 0.725,
and 0.978 nm, respectively.
(a) If the atomic packing factor and atomic radius are 0.547 and 0.177 nm, respectively, determine the
number of atoms in each unit cell.
(b) The atomic weight of iodine is 126.91 g/mol; compute its theoretical density.
Solution
(a) For indium, and from the definition of the APF
n R3
VS 3
APF = V = abc
C
we may solve for the number of atoms per unit cell, n, as
n = (APF)abc
4 3
3 R
Incorporating values of the above parameters provided in the problem state leads to
(0.547)(4.79 10-8 cm)(7.25 10 -8cm) (9.78 10 -8cm )
=
4 (1.77 10 -8cm )3
3
= 8.0 atoms/unit cell
(b) In order to compute the density, we just employ Equation 3.5 as
nAI
= abc N
A
= (8 atoms/unit cell)(126.91 g/mol)
(4.79 10 -8cm)(7.25 10 -8 cm) (9.78 10-8 cm ) /unit cel(6.022 10 23 atoms/mol)
3
= 4.96 g/cm
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3. 17 Titanium has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.58. If the radius
of the Ti atom is 0.1445 nm, (a) determine the unit cell volume, and (b) calculate the density of Ti and compare it
with the literature value.
Solution
(a) We are asked to calculate the unit cell volume for Ti. For HCP, from Equation 3.S1 (found in the
solution to Problem 3.6)
VC=6 R c 32
But for Ti, c = 1.58a, and a = 2R, or c = 3.16R, and
3
VC = (6)(3.16) R 3
-8 3 23 3
= (6)(3.16)( 3) 45 10 cm = 9.91 10 cm /unit cell
(b) The theoretical density of Ti is determined, using Equation 3.5, as follows:
nA Ti
= V N
C A
For HCP, n = 6 atoms/unit cell, and for Ti, A= 47.87 g/mol (as noted inside the front cover). Thus,
Ti
= (6 atoms/unit cell)(47.87 g/mol)
( 9.91 10 -23cm /unit cel)(6.022 10 23 atoms/mol )
= 4.81 g/cm 3
The value given in the literature is 4.51 g/cm .
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3.18 Zinc has an HCP crystal structure, a c/a ratio of 1.856, and a density of 7.13 g/cm . Compute the
atomic radius for Zn.
Solution
In order to calculate the atomic radius for Zn, we must use Equation 3.5, as well as the expression which
relates the atomic radius to the unit cell volume for HCP; Equation 3.S1 (from Problem 3.6) is as follows:
V =6 R c 32
C
In this case c = 1.856a, but, for HCP, a = 2R, which means that
VC=6 R (1.856)(2R)3 (1.856)(12 3 )R 3
And from Equation 3.5, the density is equal to
nA nA
= Zn = Zn 3
V C A (1.856)(12 3 )R N A
And, solving for R from the above equation leads to the following:
/3
R = nAZn
1.856)(12 3 ) N
A
And incorporating appropriate values for the parameters in this equation leads to
1/3
(6 atoms/unit cell)(65.41 g/mol)
R =
1.856)(12 3 )(7.13 g/cm 3)(6.022 10 23 atoms/mol )
= 1.33 10-8cm = 0.133 nm
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.19 Rhenium has an HCP crystal structure, an atomic radius of 0.137 nm, and a c/a ratio of 1.615.
Compute the volume of the unit cell for Re.
Solution
In order to compute the volume of the unit cell for Re, it is necessary to use Equation 3.S1 (found in Problem 3.6),
that is
V =6 R c 32
C
The problem states that c = 1.615a, and a = 2R. Therefore
VC = (1.615)(12 3 ) R3
= (1.615)(12 3 )(1.37 10 -8cm )3 = 8.63 10 -23cm 3 = 8.63 10 -2nm 3
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3.20 Below is a unit cell for a hypothetical metal.
(a) To which crystal system does this unit cell belong?
(b) What would this crystal structure be called?
(c) Calculate the density of the material, given that its atomic weight is 141 g/mol.
Solution
(a) The unit cell shown in the problem statement belongs to the tetragonal crystal system since a = b =
0.30 nm, c = 0.40 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
8 2 8
VC= (3.0 10 cm ) (4.0 10 cm )
23 3
= 3.60 10 cm /unit cell
Thus, using Equation 3.5, the density is equal to
nA
=
VCN A
(2 atoms/unit cell)(141 g/mol)
= (3.60 10 -23cm /unit cel)(6.022 10 23 atoms/mol )
= 13.0 g/cm3
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.21 Sketch a unit cell for the body-centered orthorhombic crystal structure.
Solution
A unit cell for the body-centered orthorhombic crystal structure is presented below.
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3.22 List the point coordinates for all atoms that are associated with the FCC unit cell (Figure 3.1).
Solution
From Figure 3.1b, the atom located of the origin of the unit cell has the coordinates 000. Coordinates for
1 1
other atoms in the bottom face are 100, 110, 010, and 0 . (The z coordinate for all these points is zero.)
2 2
For the top unit cell face, the coordinates are 001, 101, 111, 011, and 1 1.
2 2
Coordinates for those atoms that are positioned at the centers of both side faces, and centers of both front
1 1 1 1
and back faces need to be specified. For the front and back-center f ace atoms, the coordinates are 1 and 0 ,
2 2 2 2
respectively. While for the left and right side center-face atoms, the respective coordinates are0 1 and 1 1 .
2 2 2 2
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.23 List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite
crystal structure (Figure 12.6).
Solution
In Figure 12.6, the barium ions are situated at all corner positions. The point coordinates for these ions are
as follows: 000, 100, 110, 010, 001, 101, 111, and 011.
1 1 1 1
The oxygen ions are located at all face-centered positions; therefore, their coordinates are 0 , 1,
2 2 2 2
1 1 1 ,0 1 1 , 1 0 1, and 111 .
2 2 2 2 2 2 2 2
1 1 1
And, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates .
2 2 2
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.24 List the point coordinates of all atoms that are associated with the diamond cubic unit cell (Figure
12.15).
Solution
First of all, one set of carbon atoms occupy all corner positions of the cubic unit cell; the coordinates of
these atoms are as follows: 000, 100, 110, 010, 001, 101, 111, and 011.
1 1
Another set of atoms reside on all of the face-centered positions, with the following coordinates: 2 2 0,
1 1 1 1 1 1 1 1 1 1
1 ,1 ,0 , 0 , and 1 .
2 2 2 2 2 2 2 2 2 2
The third set of carbon atoms are positioned within the interior of the unit cell. Using an x-y-z coordinate
system oriented as in Figure 3.4, the coordinates of the atom that lies toward the lower-left-front of the unit cell has
3 1 1
the coordinates 4 4 4, whereas the atom situated toward the lower-right-back of the unit cell has coordinates of
1 3 1. Also, the carbon atom that resides toward the upper-left-back of the unit cell has the 1 1 3 coordinates.
4 4 4 4 4 4
3 3 3
And, the coordinates of the final atom, located toward the upper-right-front of the unit cell, are .
4 4 4
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.25Sketch a tetragonal unit cell , and within that cell indicate locations of t1e1 and 1 1 3 point
2 2 4 2 4
coordinates.
Solution
1 1 1 1 3
A tetragonal unit in which are shown the 1 and point coordinates is presented below.
2 2 4 2 4
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students enrolled in courses for which the textbook hasAny other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.26 Using the Molecule Definition Utility found in both “Metallic Crystal Structures and
Crystallography” and “Ceramic Crystal Structures” m odules of VMSE, located on the book’s web site
[www.wiley.com/college/Callister (Student Companion Site)], generate a three-dimensional unit cell for the
intermetallic compound AuCu 3given the following: (1) the unit cell is cubic with an edge length of 0.374 nm, (2)
gold atoms are situated at all cube corners, and (3) copper atoms are positioned at the centers of all unit cell faces.
Solution
First of all, open the “Molecular Definition Utility”; it may be found in either of “Metallic Crystal
Structures and Crystallography” or “Ceramic Crystal Structures” modules.
In the “Step 1” window, it is necessary to define the atom types, colors for the spheres (atoms), and specify
atom sizes. Let us enter “Au” as the name for the gold atoms (since “Au” the symbol for gold), and “Cu” as the
name for the copper atoms. Next it is necessary to choose a color for each atom type from the selections that appear
in the pull-down menu—for example, “Yellow” for Au and “Red” for Cu. In the “Atom Size” window, it is
necessary to enter an atom/ion size. In the instructions for this step, it is suggested that the atom/ion diameter in
nanometers be used. From the table found inside the front cover of the textbook, the atomic radii for gold and
copper are 0.144 nm and 0.128 nm, respectively, and, therefore, their ionic diameters are twice these values (i.e.,
0.288 nm and 0.256 nm); therefore, we enter the values “0.288” and “0.256” for the two atom types. Now click on
the “Register” button, followed by clicking on the “Go to Step 2” button.
In the “Step 2” window we specify positions for all of the atoms within the unit cell; their point
coordinates are specified in the problem statement. Let’s begin with gold. Click on the yellow sphere that is
located to the right of the “Molecule Definition Utility” box. Again, Au atoms are situated at all eight corners of the
cubic unit cell. One Au will be positioned at the origin of the coordinate system—i.e., its point coordinates are 000,
and, therefore, we enter a “0” (zero) in each of the “x”, “y”, and “z” atom position boxes. Next we click on the
“Register Atom Position” button. Now we enter the coordinates of another gold atom; let us arbitrarily select the
one that resides at the corner of the unit cell that is one unit-cell length along the x-axis (i.e., at the 100 point
coordinate). Inasmuch as it is located a distance of a units along the x-axis the value of “0.374” is entered in the “x”
atom position box (since this is the value of a given in the problem statement); zeros are entered in each of the “y”
and “z” position boxes. We repeat this procedure for the remaining six Au atoms.
After this step has been completed, it is necessary to specify positions for the copper atoms, which are
located at all six face-centered sites. To begin, we click on the red sphere that is located next to the “Molecule
Definition Utility” box. The point coordinates for some of the Cu atoms are fractional ones; in these instances, the a
1 1
unit cell length (i.e., 0.374) is multiplied by the fraction. For example, one Cu atom is located 1 2 2 coordinate.
1 1
Therefore, the x, y, and z atoms positions are (1)(0.374) = 0.374, (0.374) = 0.187, and (0.374) = 0.187,
2 2
respectively.
For the gold atoms, the x, y, and z atom position entries for all 8 sets of point coordinates are as follows:
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0.374, 0, and 0
0, 0.374, and 0
0, 0, and 0.374
0, 0.374, 0.374
0.374, 0, 0.374
0.374, 0.374, 0
0.374, 0.374, 0.374
Now, for the copper atoms, the x, y, and z atom position entries for all 6 sets of point coordinates are as
follows:
0.187, 0.187, 0
0.187, 0, 0.187
0, 0.187, 0.187
0.374, 0.187, 0.187
0.187, 0.374, 0.187
0.187, 0.187, 0.374
In Step3, wemayspecifywhichatomsaretoberepresentedas being bonded to one another, and which
type of bond(s) to use (single solid, single dashed, double, and triple are possibilities), or we may elect to not
represent any bonds at all (in which case we are finished). If it is decided to show bonds, probably the best thing to
do is to represent unit cell edges as bonds. This image may be rotated by using mouse click-and-drag
Your image should appear as the following screen shot. Here the gold atoms appear lighter than the copper
atoms.
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. [Note: Unfortunately, with this version of the Molecular Definition Utility, it is not possible to save either the data
or the image that you have generated. You may use screen capture (or screen shot) software to record and store
your image.]
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Crystallographic Directions
3.27 Draw an orthorhombic unit cell, and within that cell a [121 ] direction.
Solution
This problem calls for us to draw a [121 ] 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 [121 ] direction is the vector from the origin (point O)
to point C as shown.
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.28 Sketch a monoclinic unit cell, and within that cell a [01 1] direction.
Solution
This problem asks that a [0 11] direction be drawn within a monoclinic unit cell (a ≠ b ≠ c, and = =
90º ≠ ). One such unit cell with its origin at point O is sketched below. For this di rection, 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 11] direction corresponds to the vector passing from the origin (point
O) to point P, as indicated in the figure.
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.29 What are the indices for the directions indicated by the two vectors in the sketch below?
Solution
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, b/2 and c, respectively. This is[012] direction as indicated in the summary below.
x y z
Projections 0a b/2 c
Projections in terms of a, b, and c 0 1/2 1
Reduction to integers 0 1 2
e r u s o l c n E [012]
Direction 2 is [112]as summarized below.
x y z
Projections a/2 b/2 -c
Projections in terms of, b, and c 1/2 1/2 -1
Reduction to integers 1 1 -2
Enclosure [112]
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.30 Within a cubic unit cell, sketch the following directions:
(a) [1 10] , (e) [11 1] ,
(b) [12 1] , (f) [1 22] ,
(c) [01 2] , (g) [123] ,
(d) [13 3] , (h) [103] .
Solution
The directions asked for are indicated in the cubic unit cells shown below.
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.31 Determine the indices for the directions shown in the following cubic unit cell:
Solution
Direction A is a [011]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 y z
Projections 0a – b – c
Projections in terms of a, b, and c 0 –1 –1
Reduction to integers not necessary
Enclosure [01 1]
Direction B is a [210] 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 y z
Projections –a b 0 c
2
1
Projections in terms of a, b, and c –1 0
2
Reduction to integers –2 1 0
Enclosure [210]
Direction C is a [112] 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
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a b
Projections c
2 2
Projections in terms of a, b, and c 1 1 1
2 2
Reduction to integers 1 1 2
Enclosure [112]
Direction D is a[112] 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 y z
Projections a b – c
2 2
1 1
Projections in terms of a, b, and c –1
2 2
Reduction to integers 1 1 –2
Enclosure [112]
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.32 Determine the indices for the directionsshown in the following cubic unit cell:
Solution
Direction A is a [30] 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 y z
Projections – 2a b 0 c
3 2
2 1
Projections in terms of a, b, and c – 0
3 2
Reduction to integers –4 3 0
Enclosure [430]
Direction B is a [232] 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 y z
2a 2c
Projections – b
3 3
Projections in terms of a, b, and c 2 –1 2
3 3
Reduction to integers 2 –3 2
Enclosure [232]
Direction C is a [133] 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
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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a
Projections – b – c
3
Projections in terms of a, b, and c 1 –1 –1
3
Reduction to integers 1 –3 –3
Enclosure [133]
Direction D is a [136] 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 y z
a b
Projections – c
6 2
Projections in terms of a, b, and c 1 1 –1
6 2
Reduction to integers 1 3 –6
Enclosure [136]
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.33 For tetragonal crystals, cite the indices of directi ons that are equivalent to each of the following
directions:
(a) [001]
(b) [110]
(c) [010]
Solution
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 [001] direction, there is only one equivalent direction: [00 1].
(b) For the [110] direction, equivalent directions are as follows: [1 10], [110], and [110]
(b) Also, for the [010] direction, equivalent directions are the following: [010], [100] , and [100].
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.34 Convert the [100] and [111] directions into the four-index Miller–Bravais scheme for hexagonal unit
cells.
Solution
For [100]
u ' = 1,
v ' = 0,
w ' = 0
From Equations 3.6
1 1 2
u = (2u' v' ) = [(2)(1) 0] =
3 3 3
v = 1(2vΥ uΥ) = 1 [(2)(0) 1] = 1
3 3 3
2 1 1
t= (u + v) =
3 3 3
w = w' = 0
It is necessary to multiply these numbers by 3 in order to reduce them to the lowest set of integers. Thus, the
direction is represented as [uvtw] = [21 10].
For [111], u' = 1, v' = 1, and w' = 1; therefore,
1 1
u = 3[(2)(1) 1] = 3
1 1
v = [(2)(1) 1] =
3 3
t = 1 2
3 3 3
w = 1
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by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. If we again multiply these numbers by 3, then u = 1, v = 1, t

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