Crystal and Amorphous Structure Engineering Materials Crystal and Amorphous Structure in Materials

Engineering Materials Types of Atomic Bonds Atomic Bonding Strong Primary Bonds Weak Secondary Bonds Ionic Bond Covalent Bond Metallic Bond Fluctuating Dipoles Permanent Dipoles

Ionic Bond A primary bond formed by the transfer of one or Engineering Materials Types of Atomic Bonds Ionic Bond A primary bond formed by the transfer of one or more electron from an electropositive atom to an electronegative one. The ions are bonded together in a solid crystal by electrostatic forces. Example: NaCl crystal (see animation NaCl4 & 7)

Covalent Bond A primary bond resulting from the sharing of electrons. Engineering Materials Types of Atomic Bonds Covalent Bond A primary bond resulting from the sharing of electrons. Its involves the overlapping of half-filled orbitals of two atoms. I Example: Diamond, H2, H2O, methane (animation Covalent… 7)

.. . + F .. . O + .. . N + Covalent Bond Types of Atomic Bonds Engineering Materials Types of Atomic Bonds Covalent Bond F .. . + Single bond Double bond . O .. + Triple bond . N .. +

Metallic Bond A primary bond resulting from the sharing of Engineering Materials Types of Atomic Bonds Metallic Bond A primary bond resulting from the sharing of delocalized outer electrons in the form of an electron charged cloud by an aggregate of metal atoms. Example elemental sodium

Metallic Bond Types of Atomic Bonds Engineering Materials Positive ion Valance electrons in the form of electron charge clouds +

Engineering Materials Crystal and Amorphous Structure in Materials The physical structure of solid materials mainly depends on the arrangements of the atoms, ions, or molecules that make up the solid and the bonding forces between them. IF the atoms or ions of solid are arranged in pattern that repeats itself in three dimensions. Forming solid that has long range order (LRO)  crystalline solid or crystalline materials CRYSTAL a solid composed of atoms, ions, or molecules arranged in a pattern that is repeats in three dimensions. Examples metals, alloys and some ceramic materials IF the atoms or ions of solid are NOT arranged in a long range and repeatable manner Then forming solid that has short range order (SRO) Amorphous or non crystalline materials. Example liquid water has a short range order

Engineering Materials Crystal and Amorphous Structure in Materials Space lattice A three dimensional array of points each of which has identical surroundings Unit cell Is a repeating unit of a space lattice. The axial lengths (a, b, c) and the axial angles (alfa, beta, gamma) are the lattice constants of the unit cell

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials There are 7 different crystal classes according to the values of the axial lengths and the axial Angles (lattice constants, a, b, c, alfa, beta, and gamma). Cubic Tetragonal Orthorhombic Rhombohedral Hexagonal Monoclinic Triclinic Also there are 4 types of unit cell Simple Body centered Faced centered Side centered b c a

Materials Engineering

Engineering Materials Principle metallic crystal structure 1. Simple Cubic crystal structure 2. Body- Centered Cubic (BCC) crystal structure 3. Faced- centered Cubic (FCC) crystal structure 4. Hexagonal Close-Packed (HCP) crystal structure

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Body- centered cubic (BCC) crystal structure The atoms in the BCC unit cell contact each other across the cube diagonal, so the relationship between the length of the cube side (a) and the atomic radius (R) is; The central atom in the unit cell is surrounded by 8 nearest neighbor, therefore, the BCC unit cell has a coordination number of 8.

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Body- centered cubic (BCC) crystal structure Atomic packing factor (APF); The APF for the BCC unit cell is 68%; which means that 68% of the volume of the BCC unit cell is occupied by atoms and the remaining 32% is empty space. Many metals (iron, chromium, vanadium) have the BCC crystal structure at room temperature.

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Body- centered cubic (BCC) crystal structure The number of atoms in the BCC unit cell = 2 1 (at the center) + 8 * (1/8) = 2 atoms per unit cell Example 1 Iron at 20 C is BCC with atoms of atomic radius 0.124 nm. Calculate the lattice constant (a) for the cube edge of the iron unit cell. Example 2 Calculate the atomic packing factor (APF) for the BCC unit cell, assuming the atoms to be hard spheres.

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Faced- centered cubic (FCC) crystal structure The atoms in the FCC unit cell contact each other across the cube face diagonal, so the relationship between the length of the cube side (a) and the atomic radius (R) is; X2=a2+a2 a x 90 deg a Each atom is surrounded by 12 other atoms, therefore, the FCC unit cell has a coordination number of 12.

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Faced- centered cubic (FCC) crystal structure Atomic packing factor (APF); The APF for the FCC unit cell is 78%; which means that 78% of the volume of the FCC unit cell is occupied by atoms and the remaining 22% is empty space. Many metals (copper, lead, nickel) have the FCC crystal structure at elevated temperatures (912 to 1394 C).

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Faced- centered cubic (FCC) crystal structure The number of atoms in the FCC unit cell = 4 8 * (1/8) + 6 * (1/2) = 4 atoms per unit cell Example 3 Calculate the atomic packing factor (APF) for the FCC unit cell, assuming the atoms to be hard spheres.

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Hexagonal close- packed (HCP) crystal structure 60 deg c a 120 deg Each atom is surrounded by 12 other atoms, therefore, the HCP unit cell has a coordination number of 12.

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Hexagonal close- packed (HCP) crystal structure Atomic packing factor (APF); The APF for the HCP unit cell is 78%; and equal to that of FCC.

Crystal and Amorphous Structure in Materials Engineering Materials Crystal and Amorphous Structure in Materials Hexagonal close- packed (HCP) crystal structure The number of atoms in the HCP unit cell = 2 4 * (1/6) + 4 * (1/12) + 1 (center) = 2 atoms per unit cell c a Example 4 Calculate the volume of the zinc crystal structure unit cell by using the following data: Pure zinc has the HCP crystal structure with lattice constants a=0.2665 nm and c=0.4947 nm.

Engineering Materials Volume, planar, and linear density unit-cell calculation Example 5: Copper has an FCC crystal structure and an atomic radius of 0.1278 nm. Assuming the atoms to be hard sphere that touch each other along the face diagonals of the FCC unit cell, calculate a theoretical value for the density of copper in megagrams per cubic meter. The atomic mass of copper is 63.54 g/mol. 1g = 10 -6Mg

Engineering Materials Volume, planar, and linear density unit-cell calculation Example Calculate the planar atomic density on the (110) plane of the α iron BBC lattice in atoms per square milimeters. The lattice constant of α is 0.287 nm.

Engineering Materials Volume, planar, and linear density unit-cell calculation Example Calculate the planar atomic density on the (110) plane of the α iron BBC lattice in atoms per square milimeters. The lattice constant of α is 0.287 nm. (110) a

Engineering Materials Volume, planar, and linear density unit-cell calculation Example 6: Calculate the linear atomic density in the (110) direction in the copper crystal lattice in atoms milimeters. Copper is FCC and has a lattice constant of 0.361 nm

Engineering Materials Volume, planar, and linear density unit-cell calculation No. of atomic diameters intersected by the Length of line are 0.5+1+0.5=2 atoms. a (110) Length of line= length of face diagonal= Example 6: Calculate the linear atomic density in the (110) direction in the copper crystal lattice in atoms milimeters. Copper is FCC and has a lattice constant of 0.361 nm

Crystal and Amorphous Structure Engineering Materials Crystal and Amorphous Structure in Materials H W1 Molybdenum at 20 C is BCC and has an atomic radius of 0.140 nm. Calculate a value for its lattice constant a in nanometers. Lithium at 20 C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in nanometers. What is the coordination number for the atoms in the FCC crystal structure? Gold is FCC and has a lattice constant of 0.40788 nm. Calculate a value for the atomic radius of a gold atom in nanometers. 5. Calculate the atomic packing factor for the FCC structure.