Crystal
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Why Study Structures of Materials?
Lecture 3 Structures of Metals and Ceramics
Basic structural unit
2D representation of a quartz crystal lattice
2D representation of the amorphous structure.
3
Some Examples to Atomic Arrangements in Crystalline Materials
Why Study Structures of Materials?
Aluminum
Diamond
Graphite
2
Sodium Chloride
Graphite
4
Fundamental Concepts
Fundamental Concepts
Crystalline solid:
Crystal lattice:
Well-ordered, definite arrangements of molecules, atoms or ions. Crystals have an ordered, repeated Long Range Order structure.
Three-dimensional stacking of unit cells is the crystal lattice. A Lattice:
A Crystal
A Crystal
A lattice is array of points, each of which is indistinguishable from other points and has identical surroundings 5
7
Fundamental Concepts
Fundamental Concepts Unit cell:
Basis
The smallest repeating unit in a crystal is a unit cell. Unit cell is the smallest unit with all the symmetry of the entire crystal.
A basis is any object or combination of objects (atoms or molecules) that can be assigned to a lattice point
The Crystal Structure The Lattice
The Basis
The Unit Cell 6
The crystal structure is a combination of lattice + basis
8
A Lattice and a Basis - Example I
Metallic Crystal Structures Three Basic cubiccubic-crystal Structure
z x
The simple cubic (sc) crystal structure Only Polonium (Po) is crystallized in the simple cubic crystal. Each corner of the cubic lattice is occupied by an atom. The coordination number for sc is 6; each corner atom has six nearest neighbors. a=2R Unit cell length Atomic Packing Factor (APF)
p p The Crystal Structure The Lattice
p The Basis p p
APF =
The crystal structure is a combination of lattice + basis
9
A Lattice and a Basis - Example II
Volume of atoms in unit cell Unit cell v olume
1 4 8 atoms × × π R3 π 8 3 APF = = = 0.52 3 6 ( 2 R) 11
1 atom at the center of the 6 cubic faces in addition to the 8 corner atoms. Atoms touch one another across the face diagonal, so unit cell length, a and the R are related by
4R = a 2 , ⇒ a = 2 R p
The Crystal Structure
p
1/8 R 1/8 10
2
Each atom has 12 nearest-neighboring atoms. A large number of elements exhibit the FCC lattice form, including aluminum, nickel, copper, silver, lead, gold, and platinum.
or
The crystal structure is a combination of lattice + basis
x
The Face-Centered Cubic (FCC) crystal structure
p
The Basis
y
Notice: 52% of the sc unit cell volume is filled with hard spheres 48% of the volume is empty
p
The Lattice
a
a
1/2
1 4 1 3 8× + 6 × π R π 2 3 8 APF = = = 0.74 3 3 2 2R 2
(
)
Notice:74% of the fcc unit cell volume is filled with hard spheres 26% of the volume is empty
12
The Body-Centered Cubic (BCC) crystal structure p
In addition to eight corner atoms, an atom is located at the center of the cube.
p
each atom has 8 nearest neighboring atoms. Chromium, Iron, Molybdenum, Sodium and Tungsten exhibit bcc structure.
p
p
Center and corner atoms touch one another along cube diagonal, so unit cell length, a and the R are related by
(
p
p p
2R
a
a
3
1 4 1 3 4 × + 4 × + 1 × π R π 6 3 12 APF = = = 0.74 3 2 ( 2R × 2 R × sin 60) × 4R 2 3
)
The top and bottom faces of the unit cell consist of six atoms that form regular hexagons and surround a single atom in the center. Another plane that provides three additional atoms to the unit cell is situated between the top and bottom planes. each atom has 12 nearest neighboring atoms. Zinc, Titanium, Cobalt, Cadmium exhibit hcp structures.
a Consider the Primitive Unit Cell
h=a 2 3 c = 2h c 2 2 = = 1.633 a 3
Notice: 74% of the hcp unit cell volume is filled with hard spheres 26% of the volume is empty
c
a
15
DENSITY COMPUTATIONS—METALS ρ=
nA VC NA
=
# of atoms / uc. × atomic weight Vol. of uc.× Avogadroconstant
Example: At 300 ºK the lattice constant (a) for silicon is 5.43 Å. Calculate the number of silicon atoms per cubic centimeter and density of silicon at room temperature. (Atomic weight of Si: 28.09 g/mole, NA = 6.02x1023 atoms/mole and Number of Atoms Per Unit cell is 8 )
Solution
a = 2R, c
3
c a
Notice: 68% of the bcc unit cell volume is filled with hard spheres 13 32% of the volume is empty
h
a = 2 R, c = 4R 2
The Hexagonal Close-Packed (HCP) crystal structure p
Atomic Packing Factor of HCP
1 4 3 8 × + 1 × π R π 3 8 3 APF = = = 0.68 3 8 4R / 3
1/8 1
a
p
h
4R = a 3, ⇒ a = 4 R R
The Hexagonal Close-Packed (HCP) crystal structure
Number of Atoms Per Unit Volume=
Density = = 14
8 a3
=
8 (5. 43 ×10 −8 )3
22
= 5 ×10 atoms / cm
# of atoms / cm3 × atomic weight Avogadro constant
5 × 10 22 ( atoms / cm3 ) × 28 .09 ( g / mole ) = 2.33 g / cm 3 6.02 × 10 23 ( atoms / mole )
16
3
Lecture 3 Structures of Metals and Ceramics
Basic structural unit
2D representation of a quartz crystal lattice
2D representation of the amorphous structure.
3
Some Examples to Atomic Arrangements in Crystalline Materials
Why Study Structures of Materials?
Aluminum
Diamond
Graphite
2
Sodium Chloride
Graphite
4
Fundamental Concepts
Fundamental Concepts
Crystalline solid:
Crystal lattice:
Well-ordered, definite arrangements of molecules, atoms or ions. Crystals have an ordered, repeated Long Range Order structure.
Three-dimensional stacking of unit cells is the crystal lattice. A Lattice:
A Crystal
A Crystal
A lattice is array of points, each of which is indistinguishable from other points and has identical surroundings 5
7
Fundamental Concepts
Fundamental Concepts Unit cell:
Basis
The smallest repeating unit in a crystal is a unit cell. Unit cell is the smallest unit with all the symmetry of the entire crystal.
A basis is any object or combination of objects (atoms or molecules) that can be assigned to a lattice point
The Crystal Structure The Lattice
The Basis
The Unit Cell 6
The crystal structure is a combination of lattice + basis
8
A Lattice and a Basis - Example I
Metallic Crystal Structures Three Basic cubiccubic-crystal Structure
z x
The simple cubic (sc) crystal structure Only Polonium (Po) is crystallized in the simple cubic crystal. Each corner of the cubic lattice is occupied by an atom. The coordination number for sc is 6; each corner atom has six nearest neighbors. a=2R Unit cell length Atomic Packing Factor (APF)
p p The Crystal Structure The Lattice
p The Basis p p
APF =
The crystal structure is a combination of lattice + basis
9
A Lattice and a Basis - Example II
Volume of atoms in unit cell Unit cell v olume
1 4 8 atoms × × π R3 π 8 3 APF = = = 0.52 3 6 ( 2 R) 11
1 atom at the center of the 6 cubic faces in addition to the 8 corner atoms. Atoms touch one another across the face diagonal, so unit cell length, a and the R are related by
4R = a 2 , ⇒ a = 2 R p
The Crystal Structure
p
1/8 R 1/8 10
2
Each atom has 12 nearest-neighboring atoms. A large number of elements exhibit the FCC lattice form, including aluminum, nickel, copper, silver, lead, gold, and platinum.
or
The crystal structure is a combination of lattice + basis
x
The Face-Centered Cubic (FCC) crystal structure
p
The Basis
y
Notice: 52% of the sc unit cell volume is filled with hard spheres 48% of the volume is empty
p
The Lattice
a
a
1/2
1 4 1 3 8× + 6 × π R π 2 3 8 APF = = = 0.74 3 3 2 2R 2
(
)
Notice:74% of the fcc unit cell volume is filled with hard spheres 26% of the volume is empty
12
The Body-Centered Cubic (BCC) crystal structure p
In addition to eight corner atoms, an atom is located at the center of the cube.
p
each atom has 8 nearest neighboring atoms. Chromium, Iron, Molybdenum, Sodium and Tungsten exhibit bcc structure.
p
p
Center and corner atoms touch one another along cube diagonal, so unit cell length, a and the R are related by
(
p
p p
2R
a
a
3
1 4 1 3 4 × + 4 × + 1 × π R π 6 3 12 APF = = = 0.74 3 2 ( 2R × 2 R × sin 60) × 4R 2 3
)
The top and bottom faces of the unit cell consist of six atoms that form regular hexagons and surround a single atom in the center. Another plane that provides three additional atoms to the unit cell is situated between the top and bottom planes. each atom has 12 nearest neighboring atoms. Zinc, Titanium, Cobalt, Cadmium exhibit hcp structures.
a Consider the Primitive Unit Cell
h=a 2 3 c = 2h c 2 2 = = 1.633 a 3
Notice: 74% of the hcp unit cell volume is filled with hard spheres 26% of the volume is empty
c
a
15
DENSITY COMPUTATIONS—METALS ρ=
nA VC NA
=
# of atoms / uc. × atomic weight Vol. of uc.× Avogadroconstant
Example: At 300 ºK the lattice constant (a) for silicon is 5.43 Å. Calculate the number of silicon atoms per cubic centimeter and density of silicon at room temperature. (Atomic weight of Si: 28.09 g/mole, NA = 6.02x1023 atoms/mole and Number of Atoms Per Unit cell is 8 )
Solution
a = 2R, c
3
c a
Notice: 68% of the bcc unit cell volume is filled with hard spheres 13 32% of the volume is empty
h
a = 2 R, c = 4R 2
The Hexagonal Close-Packed (HCP) crystal structure p
Atomic Packing Factor of HCP
1 4 3 8 × + 1 × π R π 3 8 3 APF = = = 0.68 3 8 4R / 3
1/8 1
a
p
h
4R = a 3, ⇒ a = 4 R R
The Hexagonal Close-Packed (HCP) crystal structure
Number of Atoms Per Unit Volume=
Density = = 14
8 a3
=
8 (5. 43 ×10 −8 )3
22
= 5 ×10 atoms / cm
# of atoms / cm3 × atomic weight Avogadro constant
5 × 10 22 ( atoms / cm3 ) × 28 .09 ( g / mole ) = 2.33 g / cm 3 6.02 × 10 23 ( atoms / mole )
16
3
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