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Chap 3 Crystal Structure Guangzhao Mao
BE1300 Winter 2002
1
Topics of Discussion ? ?
Crystalline vs. amorphous materials Unit cells of metals: ? ?
?
?
Types: FCC, BCC, and H Relation between edge length and atomic radius Atomic packing factor, coordination number, number of atoms per unit cell density
BE1300 Winter 2002
2
CaF2
Crystalline vs. Amorphous Materials ?
?
Crystalline: atoms are arranged in a repeating array over large atomic distances. ? Single crystal: the regular arrangement extends throughout the entire sample. ? Polycrystal: the entire sample is made of small crystals with boundaries among them. Amorphous: atoms lack a regular arrangement over large atomic distances.
Steel BE1300 Winter 2002
3
Example: Silicon Oxide
Quartz
Fused silica BE1300 Winter 2002
4
Crystal Structure of Metals: Unit Cell Crystal structure: the manner in which atoms, ions, and molecules are spatially arranged. ? ? ? ? ? ?
Types Edge length a vs. atomic radius R Atomic packing factor APF Number of atoms per unit cell n Coordination number CN Density ?
BE1300 Winter 2002
5
Lattice and Unit Cell ?
?
? Lattice
Lattice: 3-D infinite array of atom positions. Unit cell: the repeat unit of lattice defined by six parameters a, b, c, ? , ?, and ??
? Unit cell
BE1300 Winter 2002
6
7 Crystal Systems
BE1300 Winter 2002
7
Unit Cell Types of Metal Crystals
Face-centered cubic Body-centered cubic FCC BCC Gold, Lead, silver, etc. Iron, Tungsten, etc.
Hexagonal close-packed H Cobalt, Zinc, etc.
BE1300 Winter 2002
8
Definitions ? ? ? ? ? ?
? ?
Edge length a Atomic radius R Unit cell volume VC Occupied volume VS Atomic Packing factor APF=VS/VC Coordination number CN: number of nearest neighboring atoms Number of atoms per unit cell n A: atomic weight Density ? ? nA NA: Avogadro number VC N A BE1300 Winter 2002
9
Face-Centered Cubic Structure Express the edge length a and unit cell volume in of atomic radius R. (Ex. 3.1)
a ? 2R 2 VC ? 16 R 3 2 BE1300 Winter 2002
10
APF of FCC Structure Ex. 3.2
• n = 8 corner atoms×1/8 + 6 face atoms× ½ = 4 atoms per unit cell
BE1300 Winter 2002
11
Summary of FCC Structure a ? 2R 2 APF = 0.74 Coordination number = 12 Number of atoms per unit cell = 4
BE1300 Winter 2002
12
Express a in of R for the BCC structure.
3.4 This problem calls for a demonstration of the relationship a = 4R 3 for BCC. Consider the BCC unit cell shown below
a Q
a P N O a
Using the triangle NOP
__
2 2 2 2 (NP ) = a + a = 2a
And then for triangle NPQ,
__
Q
__
__
a N
__
2 2 2 (NQ ) = (QP ) + (NP )
__
But NQ = 4R, R being the atomic radius. Also, QP = a. Therefore,
P
a ? 4R ? 3
2 2 2 (4R) = a + 2a , or
a=
4R 3
BE1300 Winter 2002
13
APF of BCC Structure How many atoms per unit cell?
2
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
APF =
V S V C
Since there are two spheres associated with each unit cell for BCC
?4 ?R V = 2(sphere volume) = 2 ? S ? 3
3?
? = 8 ?R 3 ?
3
3 Also, the unit cell has cubic symmetry, that is V = a . But a depends on R according to C Equation (3.3), and 3
V = C
APF = 0.68
?4R ? 3 = 64R ? 3? 3 3 ? ?
Thus,
APF =
3 8 ?R /3 = 0.68 3 64R /3 3
BE1300 Winter 2002
14
Summary of BCC Structure a ? 4R ? 3 APF = 0.68 Coordination number = 8 Number of atoms per unit cell = 2
BE1300 Winter 2002
15
Chap 3 Crystal Structure (Cont’d) ?
Homework assignment #2: 3.11, 3.14, 3.16, 3.18.
?
This week is A Week. Next week is B Week.
?
Review of last lecture: ?
?
?
Give an example of a crystal. How can you tell whether it is a single crystal or a polycrystal? Give an example of an amorphous material. How can you tell that it is not crystalline? For FCC and BCC structures, describe ? ?
Relationship between a and R. APF, number of atoms per unit cell, and coordination number.
BE1300 Winter 2002
16
Summary of FCC Structure a ? 2R 2 APF = 0.74 Coordination number = 12 Number of atoms per unit cell = 4
BE1300 Winter 2002
17
Summary of BCC Structure a ? 4R ? 3 APF = 0.68 Coordination number = 8 Number of atoms per unit cell = 2
BE1300 Winter 2002
18
H structure
• Top or bottom plane: six atoms form a hexagon surrounding a center atom; • Mid-plane: three atoms form an equilateral triangle.
BE1300 Winter 2002
19
3.5 We are asked to show that the ideal c /a ratio for H is 1.633. A sketch of one -third of an H unit cell is shown below.
a
What is the relationship between c and a?
c
M
L J K
a
Consider the tetrahedron labeled as JKLM, which is reconstructed as
M
L H
J
M
K
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,
J
L K
Atoms J, K, L, and M touch each other and form a tetrahedron
__
__
JM = JK = 2R = a
where R is the atomic radius. Furthermore, from triangle JHM,
__
__
__
2 2 2 (JM ) = (J H ) + (MH ) , or
__
2 2 ?c ? 2 a = (JH ) + ?2 ?
BE1300 Winter 2002
20
__ Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle,
L
How does JH relate to a? ?
J
H
30
a/2 K cos 30 =
a/2 3 = , and JH 2
__ JH =
a 3
__ Substituting this value for JH in the above expression yields
2 a =
2
2
? a ? 2 + ?c ? 2 = a + c ? 3? 3 4 ?2? ? ?
and, solving for c/a c = a
8 = 1.633 3
c/a = 1.633 is an ideal number. Real systems may deviate: c/a (Magnesium) = 1.633 c/a (Cium) = 1.890 c/a (Zirconium) = 1.593
BE1300 Winter 2002
21
3.7 This problem calls for a demonstration that the APF for H is 0.74. Again, the APF is just
APF of H
the total sphere -unit cell volume ratio. For H, there are the equivalent of six spheres per unit cell, and thus
Calculation of VS ?
?4 ?R V = 6? ? 3 S
3?
? = 8 ? R3 ?
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.
D
C
How many atoms per unit cell?
Calculation of VC ?
a = 2R 30
A
60
E B a = 2R
n = 3 interior atoms + 2 face atoms×1/2 + 12 corner atoms×1/6 = 6 atoms per unit cell
a = 2R
__
__
__
The area of ACDE is just the length of CD times the height BC . But CD is just a or 2R, and
__ BC = 2R cos(30 ) =
2R 3 2
Thus, the base area is just
APF = VS/VC = 0.74
__ __
?2R 3 ? ? = 6R2 3 AREA = (3)(CD )( BC ) = (3)(2R)? ? 2 ? and since c = 1.633a = 2R(1.633) V = (AREA)(c) = 6R2 c 3 = (6R2 3 )(2)(1.633)R = 12 3(1.633) R 3 C
BE1300 Winter 2002
22
Summary of H Structure a ? 2R c ? 1.633a APF = 0.74 Coordination number = 12 Number of atoms per unit cell = 6
BE1300 Winter 2002
23
Practice: Calculate the volume of the zinc crystal structure unit cell by using the following data: pure zinc has the H crystal structure with lattice constants a = 0.2665 nm and c = 0.4947 nm.
BE1300 Winter 2002
24
Practice (cont’d)
BE1300 Winter 2002
25
Calculation of Density Density = Mass of the unit cell/Volume of the unit cell
nA ? ? VC N A ?: density n: number of atoms per unit cell A: atomic weight VC: unit cell volume NA: Avogadro’s number BE1300 Winter 2002
26
Density Calculation
BE1300 Winter 2002
27
Practice: 3.12 Zirconium has an H crystal structure and a density of 6.51 g/cm3. (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. (a) The volume of the Zr unit cell may be computed using Equation (3.5) as nA Zr VC = N ? A Now, for H, n = 6 atoms/unit cell, and for Zr, A = 91.2 g/mol. Thus, Zr
VC =
(6 atoms/unit cell)(91.2 g/mol) (6.51 g/cm 3 )(6.023 x 10 23 atoms/mol)
= 1.396 x 10-22 cm3/unit cell = 1.396 x 10-28 m3/unit cell
BE1300 Winter 2002
28
(b) From the solution to Problem 3.7, since a = 2R, then, for H
VC =
3 3 a2 c 2
but, since c = 1.593a
3 3(1.593 )a VC = 2
3
= 1.396 x 10-2 2 cm 3 /unit cell
Now, solving for a
?(2)(1.396 x 10 -22 cm3 )? a=? ? (3 )( 3 )(1.593 ) ? ?
1/3
= 3.23 x 10 -8 cm = 0.323 nm
And finally c = 1.593a = (1.593)(0.323 nm) = 0.515 nm
BE1300 Winter 2002
29
Summary ?
?
Crystalline: long-range order Amorphous: no long-range order FCC, BCC, H structures: ? ? ? ? ? ?
a vs. R VC and VS APF coordination number CN number of atoms per unit cell n Density ?
BE1300 Winter 2002
30
BE1300 Winter 2002
1
Topics of Discussion ? ?
Crystalline vs. amorphous materials Unit cells of metals: ? ?
?
?
Types: FCC, BCC, and H Relation between edge length and atomic radius Atomic packing factor, coordination number, number of atoms per unit cell density
BE1300 Winter 2002
2
CaF2
Crystalline vs. Amorphous Materials ?
?
Crystalline: atoms are arranged in a repeating array over large atomic distances. ? Single crystal: the regular arrangement extends throughout the entire sample. ? Polycrystal: the entire sample is made of small crystals with boundaries among them. Amorphous: atoms lack a regular arrangement over large atomic distances.
Steel BE1300 Winter 2002
3
Example: Silicon Oxide
Quartz
Fused silica BE1300 Winter 2002
4
Crystal Structure of Metals: Unit Cell Crystal structure: the manner in which atoms, ions, and molecules are spatially arranged. ? ? ? ? ? ?
Types Edge length a vs. atomic radius R Atomic packing factor APF Number of atoms per unit cell n Coordination number CN Density ?
BE1300 Winter 2002
5
Lattice and Unit Cell ?
?
? Lattice
Lattice: 3-D infinite array of atom positions. Unit cell: the repeat unit of lattice defined by six parameters a, b, c, ? , ?, and ??
? Unit cell
BE1300 Winter 2002
6
7 Crystal Systems
BE1300 Winter 2002
7
Unit Cell Types of Metal Crystals
Face-centered cubic Body-centered cubic FCC BCC Gold, Lead, silver, etc. Iron, Tungsten, etc.
Hexagonal close-packed H Cobalt, Zinc, etc.
BE1300 Winter 2002
8
Definitions ? ? ? ? ? ?
? ?
Edge length a Atomic radius R Unit cell volume VC Occupied volume VS Atomic Packing factor APF=VS/VC Coordination number CN: number of nearest neighboring atoms Number of atoms per unit cell n A: atomic weight Density ? ? nA NA: Avogadro number VC N A BE1300 Winter 2002
9
Face-Centered Cubic Structure Express the edge length a and unit cell volume in of atomic radius R. (Ex. 3.1)
a ? 2R 2 VC ? 16 R 3 2 BE1300 Winter 2002
10
APF of FCC Structure Ex. 3.2
• n = 8 corner atoms×1/8 + 6 face atoms× ½ = 4 atoms per unit cell
BE1300 Winter 2002
11
Summary of FCC Structure a ? 2R 2 APF = 0.74 Coordination number = 12 Number of atoms per unit cell = 4
BE1300 Winter 2002
12
Express a in of R for the BCC structure.
3.4 This problem calls for a demonstration of the relationship a = 4R 3 for BCC. Consider the BCC unit cell shown below
a Q
a P N O a
Using the triangle NOP
__
2 2 2 2 (NP ) = a + a = 2a
And then for triangle NPQ,
__
Q
__
__
a N
__
2 2 2 (NQ ) = (QP ) + (NP )
__
But NQ = 4R, R being the atomic radius. Also, QP = a. Therefore,
P
a ? 4R ? 3
2 2 2 (4R) = a + 2a , or
a=
4R 3
BE1300 Winter 2002
13
APF of BCC Structure How many atoms per unit cell?
2
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
APF =
V S V C
Since there are two spheres associated with each unit cell for BCC
?4 ?R V = 2(sphere volume) = 2 ? S ? 3
3?
? = 8 ?R 3 ?
3
3 Also, the unit cell has cubic symmetry, that is V = a . But a depends on R according to C Equation (3.3), and 3
V = C
APF = 0.68
?4R ? 3 = 64R ? 3? 3 3 ? ?
Thus,
APF =
3 8 ?R /3 = 0.68 3 64R /3 3
BE1300 Winter 2002
14
Summary of BCC Structure a ? 4R ? 3 APF = 0.68 Coordination number = 8 Number of atoms per unit cell = 2
BE1300 Winter 2002
15
Chap 3 Crystal Structure (Cont’d) ?
Homework assignment #2: 3.11, 3.14, 3.16, 3.18.
?
This week is A Week. Next week is B Week.
?
Review of last lecture: ?
?
?
Give an example of a crystal. How can you tell whether it is a single crystal or a polycrystal? Give an example of an amorphous material. How can you tell that it is not crystalline? For FCC and BCC structures, describe ? ?
Relationship between a and R. APF, number of atoms per unit cell, and coordination number.
BE1300 Winter 2002
16
Summary of FCC Structure a ? 2R 2 APF = 0.74 Coordination number = 12 Number of atoms per unit cell = 4
BE1300 Winter 2002
17
Summary of BCC Structure a ? 4R ? 3 APF = 0.68 Coordination number = 8 Number of atoms per unit cell = 2
BE1300 Winter 2002
18
H structure
• Top or bottom plane: six atoms form a hexagon surrounding a center atom; • Mid-plane: three atoms form an equilateral triangle.
BE1300 Winter 2002
19
3.5 We are asked to show that the ideal c /a ratio for H is 1.633. A sketch of one -third of an H unit cell is shown below.
a
What is the relationship between c and a?
c
M
L J K
a
Consider the tetrahedron labeled as JKLM, which is reconstructed as
M
L H
J
M
K
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,
J
L K
Atoms J, K, L, and M touch each other and form a tetrahedron
__
__
JM = JK = 2R = a
where R is the atomic radius. Furthermore, from triangle JHM,
__
__
__
2 2 2 (JM ) = (J H ) + (MH ) , or
__
2 2 ?c ? 2 a = (JH ) + ?2 ?
BE1300 Winter 2002
20
__ Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle,
L
How does JH relate to a? ?
J
H
30
a/2 K cos 30 =
a/2 3 = , and JH 2
__ JH =
a 3
__ Substituting this value for JH in the above expression yields
2 a =
2
2
? a ? 2 + ?c ? 2 = a + c ? 3? 3 4 ?2? ? ?
and, solving for c/a c = a
8 = 1.633 3
c/a = 1.633 is an ideal number. Real systems may deviate: c/a (Magnesium) = 1.633 c/a (Cium) = 1.890 c/a (Zirconium) = 1.593
BE1300 Winter 2002
21
3.7 This problem calls for a demonstration that the APF for H is 0.74. Again, the APF is just
APF of H
the total sphere -unit cell volume ratio. For H, there are the equivalent of six spheres per unit cell, and thus
Calculation of VS ?
?4 ?R V = 6? ? 3 S
3?
? = 8 ? R3 ?
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.
D
C
How many atoms per unit cell?
Calculation of VC ?
a = 2R 30
A
60
E B a = 2R
n = 3 interior atoms + 2 face atoms×1/2 + 12 corner atoms×1/6 = 6 atoms per unit cell
a = 2R
__
__
__
The area of ACDE is just the length of CD times the height BC . But CD is just a or 2R, and
__ BC = 2R cos(30 ) =
2R 3 2
Thus, the base area is just
APF = VS/VC = 0.74
__ __
?2R 3 ? ? = 6R2 3 AREA = (3)(CD )( BC ) = (3)(2R)? ? 2 ? and since c = 1.633a = 2R(1.633) V = (AREA)(c) = 6R2 c 3 = (6R2 3 )(2)(1.633)R = 12 3(1.633) R 3 C
BE1300 Winter 2002
22
Summary of H Structure a ? 2R c ? 1.633a APF = 0.74 Coordination number = 12 Number of atoms per unit cell = 6
BE1300 Winter 2002
23
Practice: Calculate the volume of the zinc crystal structure unit cell by using the following data: pure zinc has the H crystal structure with lattice constants a = 0.2665 nm and c = 0.4947 nm.
BE1300 Winter 2002
24
Practice (cont’d)
BE1300 Winter 2002
25
Calculation of Density Density = Mass of the unit cell/Volume of the unit cell
nA ? ? VC N A ?: density n: number of atoms per unit cell A: atomic weight VC: unit cell volume NA: Avogadro’s number BE1300 Winter 2002
26
Density Calculation
BE1300 Winter 2002
27
Practice: 3.12 Zirconium has an H crystal structure and a density of 6.51 g/cm3. (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. (a) The volume of the Zr unit cell may be computed using Equation (3.5) as nA Zr VC = N ? A Now, for H, n = 6 atoms/unit cell, and for Zr, A = 91.2 g/mol. Thus, Zr
VC =
(6 atoms/unit cell)(91.2 g/mol) (6.51 g/cm 3 )(6.023 x 10 23 atoms/mol)
= 1.396 x 10-22 cm3/unit cell = 1.396 x 10-28 m3/unit cell
BE1300 Winter 2002
28
(b) From the solution to Problem 3.7, since a = 2R, then, for H
VC =
3 3 a2 c 2
but, since c = 1.593a
3 3(1.593 )a VC = 2
3
= 1.396 x 10-2 2 cm 3 /unit cell
Now, solving for a
?(2)(1.396 x 10 -22 cm3 )? a=? ? (3 )( 3 )(1.593 ) ? ?
1/3
= 3.23 x 10 -8 cm = 0.323 nm
And finally c = 1.593a = (1.593)(0.323 nm) = 0.515 nm
BE1300 Winter 2002
29
Summary ?
?
Crystalline: long-range order Amorphous: no long-range order FCC, BCC, H structures: ? ? ? ? ? ?
a vs. R VC and VS APF coordination number CN number of atoms per unit cell n Density ?
BE1300 Winter 2002
30
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