CHAPTER 3 STRUCTURES OF METALS AND CERAMICS
PROBLEM SOLUTIONS
Unit Cells Metallic Crystal Structures 3.2 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
APF =
V S V C
Since there are two spheres associated with each unit cell for BCC
4 R3 = 3
2(spheree volume volume)) = 2 V S = 2(spher
8 R3 3
Also, the unit cell has cubic symmetry, that is V C = a3. But a depends on R according to Equation 3.3, and
4 R 3 V C = 3
=
64 R3 3 3
Thus,
APF =
V S V C
=
8 R3 / 3 64 R3 / 3 3
= 0.68 .68
Density Computations--Metals 3.3 Molybdenum has a BCC crystal crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute its theoretical density and compare it with the experimental value found inside the front cover. Solution This problem calls for a computation of the density of molybdenum. According to Equation 3.5
=
nA Mo V N C A
For BCC, n = 2 atoms/unit cell, and realizing that V C = a3, and using Equation 3.3
V C =
4 R 3 3
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=
nA
=
Mo 3
4 R 3
N
A
(2 atoms/unit cell)(95.94 g/mol)
(4) (0.1363 10
-7 cm) /
3
unit cell ) (6.02 10 /(unit 3
23 atoms/mol)
= 10.22 g/cm3 The value given inside the front cover is 10.22 g/cm 3.
3
3.4 Calculate the radius of a palladium atom, given that Pd has an FCC crystal structure, a density of 12.0 g/cm g/cm , and an atomic weight of 106.4 g/mol. Solution For FCC, n = 4 atoms/unit cell, and V C = 16 R3 2 (Equation 3.4). Now,
=
=
nAPd V C N A
nAPd
(16 R3 2 ) N A
And solving for R from the above expression yields
R =
=
nA 1/3 Pd 16 N 2 A
atoms/unit cell) cell)106.4 g/mol (4 atoms/unit (16)(12.0 g/cm3 )(6.02 1023 atoms/mol)(
1/3 2 )
= 1.38 10-8 cm = 0.138 nm
Ceramic Crystal Structures 3.10 Show that the minimum minimum cation-to-anion radius ratio for a coordination number of 4 is 0.225. Solution In this problem we are asked to show that the minimum cation-to-anion radius ratio for a coordination number of four is 0.225. If lines are drawn from the centers of the anions, then a tetrahedron is formed. T he tetrahedron may be inscribed within a cube as shown below.
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The spheres at the apexes of the tetrahedron are drawn at the corners of the cube, and designated as positions A, B, C, and D. (These are reduced in size for the sake of clarity.) The cation resides at the center of the cube, which is designated as point E. Let us now express the cation and anion radii in terms of the cube edge length, designated designated as a. The spheres located at positions A and B touch each other along the bottom face diagonal. Thus, AB = 2 r A But
( AB) AB)2
a2
a2 2a2
or
AB = a 2 = 2r A And a =
2 r A 2
AEF will be related to There will also be an anion located at the corner, point F (not drawn), and the cube diagonal AEF will the ionic radii as
AEF AEF = 2(r A + r C ) (The line AEF has not been drawn to avoid confusion.) From the triangle ABF
FB)2 = ( AEF AEF )2 ( AB)2 + ( FB But,
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from above. Thus,
(2 r A
)2
2 r A 2 2
+
2
= 2 (r A + r C )
Solving for the r C/r A ratio leads to r C r A
6
=
2
= 0.22 0.225 5
2
Density Computations--Ceramics 3.13 Compute the atomic packing factor for the rock salt salt crystal structure in which r C/r A = 0.414. Solution From Equation 3.2
APF =
V S V C
With regard to the sphere volume, V S, there are four cation cation and four anion spheres per unit cell. cell. Thus,
4
r 3 + 3 A
V S = (4)
(4)
4
3
r C3
But, since r C/r A = 0.414 V S =
16 3
r A3
414 ) = (17.9 (17.94) 4) r (0.414) 1 (0 3
3 A
Now, for r C/r A = 0.414 the corner anions in Table 3.3 just touch one another along the cubic unit cell edges such that
V C = a3 = 2 (r A
r C )
3
2 (r A 0.414 r A ) (22.62) r A3 3
Thus
APF =
Silicate Ceramics
V S V C
=
3 (17.94) r A
(22.62) r A3
= 0.79 .79
4 –
3.18 Determine the angle between covalent bonds in an SiO4 tetrahedron. Solution
This problem asks for us to determine the angle between covalent bonds in the SiO4 tetrahedron. Below
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Now if we extend the base diagonal from one corner to the other, it is the case that
(2 y)2 = a2 + a2 = 2a2 or y =
a 2
2
Furthermore, x = a/2, and
tan =
x y
=
a /2 a 2 /2
=
1 2
From which
= tan-1
1
35.26 35.
2
Now, solving for the angle
= 180 180 90 35.26 = 54.74 54.74 Finally, the bond angle is just 2 2 , or 2 2 = (2)(54.74 (2)(54.74) = 109.48 109.48.
Crystal Systems 3.21 Sketch a unit unit cell for the face-centered orthorhombic crystal structure. Solution A unit cell for the face-centered orthorhombic crystal structure is presented below.
