Pure iron crystallizes in a body-centered cubic structure, but small amounts of impurities can stabilize a face-centered cubic structure. Which form of iron has a higher density?

In a body-centered cubic structure, there is one atom at each corner of the cube and one atom in the center of the cube. Since the corner atoms are shared with 8 adjacent unit cells, the contribution of corner atoms to each unit cell is 1/8. Therefore, the total number of atoms in a BCC unit cell is 1 (center atom) + 8 * 1/8 (corner atoms) = 2 atoms. In a face-centered cubic structure, there is one atom at each corner of the cube, and an additional atom at the center of each face of the cube. The contribution of corner atoms to each unit cell is again 1/8, while the contribution of face atoms is 1/2 as each face atom is shared with two adjacent unit cells. The total number of atoms in an FCC unit cell is 8 * 1/8 (corner atoms) + 6 * 1/2 (face atoms) = 1 + 3 = 4 atoms.

The volume of a unit cell can be calculated using the formula V = a^3, where 'a' is the edge length of the cube. Both the BCC and FCC structures have a lattice constant (edge length) of 'a.' However, to get the volume in terms of the atomic radius 'r,' we need to use geometrical relationships. For a BCC unit cell, the body diagonal is equal to 4r, and using the Pythagorean theorem in 3D, we have \[a\sqrt{3} = 4r\]. This implies that the volume of a BCC unit cell is \[V_{BCC} = a^3 = (\frac{4r}{\sqrt{3}})^3 = \frac{64r^3}{3\sqrt{3}}\]. For an FCC unit cell, the face diagonal is equal to 4r, and using the Pythagorean theorem in 2D, we have \[a\sqrt{2} = 4r\]. This implies that the volume of an FCC unit cell is \[V_{FCC} = a^3 = (\frac{4r}{\sqrt{2}})^3 = 8\sqrt{2}r^3\].

The mass of a unit cell can be calculated using the formula M = n * m, where 'n' is the number of atoms in the unit cell, and 'm' is the mass of one atom. For iron, the molar mass M(Fe) = 55.845 g/mol, and since one mole contains Avogadro's number (N) of atoms, the mass of one atom can be calculated as m = M(Fe) / N. For a BCC unit cell, the mass is \[M_{BCC} = 2 * m = \frac{2M(Fe)}{N}\]. For an FCC unit cell, the mass is \[M_{FCC} = 4 * m = \frac{4M(Fe)}{N}\].

Density can be calculated using the formula ρ = M / V, where M is the mass of the unit cell, and V is the volume of the unit cell. For a BCC unit cell, the density is \[\rho_{BCC} = \frac{M_{BCC}}{V_{BCC}} = \frac{\frac{2M(Fe)}{N}}{\frac{64r^3}{3\sqrt{3}}}\]. For an FCC unit cell, the density is \[\rho_{FCC} = \frac{M_{FCC}}{V_{FCC}} = \frac{\frac{4M(Fe)}{N}}{8\sqrt{2}r^3}\].

Comparing the densities of BCC and FCC iron, we have \[\rho_{BCC} = \frac{2M(Fe)}{64r^3 / 3\sqrt{3}} N\] \[\rho_{FCC} = \frac{4M(Fe)}{8\sqrt{2}r^3} N\] We need to compare the numerical values of these densities to see which one is higher. Simplifying the ratios to include only the constants, we get \[\frac{\rho_{BCC}}{M(Fe)} = \frac{3\sqrt{3}}{32r^3} N = 0.32476 \frac{N}{r^3}\] \[\frac{\rho_{FCC}}{M(Fe)} = \frac{4}{8\sqrt{2}r^3} N = 0.35355 \frac{N}{r^3}\] Since \(\frac{\rho_{FCC}}{M(Fe)} > \frac{\rho_{BCC}}{M(Fe)}\), we can conclude that the face-centered cubic (FCC) structure of iron has a higher density than the body-centered cubic (BCC) structure.