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Example 3.1

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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.

Table 3.2 Crystal structure and atomic radius of some common metals at room temperature.

Metal Crystal structurea Atomic radius (nm) Metal Crystal structurea Atomic radius (nm)
Aluminum FCC 0.1431 Molybdenum BCC 0.1363
Cadmium CPH 0.1490 Nickel FCC 0.1246
Chromium BCC 0.1249 Platinum FCC 0.1387
Cobalt CPH 0.1253 Silver FCC 0.1445
Copper FCC 0.1278 Tantalum BCC 0.1430
Gold FCC 0.1442 Titanium CPH 0.1445
Iron (α) BCC 0.1241 Tungsten BCC 0.1371
Lead FCC 0.1750 Zinc CPH 0.1332

a FCC, face‐centered cubic; CPH, close‐packed hexagonal; BCC, body‐centered cubic.

Materials for Biomedical Engineering

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