Materials for Biomedical Engineering. Mohamed N. Rahaman

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Materials for Biomedical Engineering - Mohamed N. Rahaman


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3.4 (a) Packing of triangular layers of atoms in ABAB pattern to give a close‐packed hexagonal (CPH) structure; (b) atomic positions in a hexagonal unit cell; (c) unit cell commonly used for CPH structure.

Crystal structure Simple cubic BCC FCC CPH
Atomic packing fraction 0.52 0.68 0.74 0.74

      3.2.1 Unit Cells and Crystal Systems

      A standard nomenclature is used to succinctly describe the geometrical arrangement of the atoms in a unit cell and, thus, in a crystal.

      Unit Cell Parameters

Schematic illustration of geometry and parameters of a unit cell.

      Crystal Systems and Bravais Lattices

Schematic illustration of the seven crystal systems and their parameters. Schematic illustration of the 14 Bravais lattices.

      As pure metals are composed of a single type of atom, their crystal structures are rather simple. In comparison, the crystal structures of ceramics are more complex and they depend on the type of bonding, ionic, or covalent.

      3.3.1 Crystal Structure of Metals

      Example 3.1

      1 Iron (Fe) has a BCC structure and an atomic radius of 0.1241 nm. Determine (i) the number of atoms in a unit cell of Fe, (ii) the unit cell length, and (iii) the theoretical density of Fe (that is, the density of the pore‐free material). (Atomic weight of Fe = 55.85; Avogadro number = 6.023 × 1023/mol.)Solution:In the BCC structure, each unit cell has one atom at its corners and one atom at its center (Figure 3.2). As an atom at each corner is shared by eight unit cells, only one‐eighth of this atom effectively belongs to a given unit cell. Thus, the total number of atoms within a BCC unit cell is 2 (that is, 8 × 1/8 at the corners + 1 in the center).The unit cell length can be found from the geometry of the unit cell and Pythagoras theorem. Taking the radius of an iron atom = r and the unit cell length = a, from the BCC structure in Figure 3.2c, the diagonal from one corner to the opposite corner across the center of the cell c = 4r, and the diagonal across one face b = a√2. Using Pythagoras theorem, we get c 2 = a 2 + b 2, giving a = 4r/√3. Since r = 0.1241 nm, we get that a = 0.286.The theoretical density Dt is the mass of the equivalent number of atoms in the unit cell divided by the volume of the unit cell. Thus where the mass of the atom and the unit cell length were expressed in units of grams and centimeters, respectively.


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