Calcium crystallizes in a face centred cubic unit cell with a=0.556nm. Calculate the density if i. It contains 0.1% Frenkel defects. ii. It contains 0.1% Schottky defects.

To calculate the density of calcium in a face-centered cubic (FCC) structure with a given lattice parameter and defects, we need to follow a systematic approach. Let's break this down step by step for both types of defects: Frenkel and Schottky.

Understanding the FCC Structure

In a face-centered cubic unit cell, there are 4 atoms per unit cell. This is because there is one atom at each of the eight corners (1/8 of an atom per corner, totaling 1 atom) and one atom at the center of each face (1/2 of an atom per face, totaling 3 atoms). Therefore, the total number of atoms in the unit cell is:

• Atoms from corners: 8 corners × 1/8 = 1 atom
• Atoms from faces: 6 faces × 1/2 = 3 atoms
• Total = 1 + 3 = 4 atoms

Calculating Molar Mass and Volume

The molar mass of calcium (Ca) is approximately 40.08 g/mol. The volume of the unit cell can be calculated using the formula for the volume of a cube:

Volume (V) = a³, where a is the edge length of the unit cell.

Given that a = 0.556 nm = 0.556 × 10^-9 m, we find:

V = (0.556 × 10^-9)³ = 1.72 × 10^-28 m³.

Density Calculation without Defects

The density (ρ) of the material can be calculated using the formula:

ρ = (mass of atoms in unit cell) / (volume of unit cell).

The mass of the atoms in the unit cell can be calculated as follows:

Mass = (number of atoms) × (molar mass) / (Avogadro's number).

Using Avogadro's number (approximately 6.022 × 10²³ mol^-1), we have:

Mass = (4 atoms) × (40.08 g/mol) / (6.022 × 10²³ mol^-1) = 2.66 × 10^-25 kg.

Now, substituting the mass and volume into the density formula gives:

ρ = (2.66 × 10^-25 kg) / (1.72 × 10^-28 m³) = 1540 kg/m³.

Incorporating Frenkel Defects

Frenkel defects involve the displacement of an atom from its lattice site to an interstitial site, which does not change the total number of atoms in the crystal. If there is 0.1% Frenkel defects, the number of atoms remains effectively the same for density calculations. Thus, the density remains:

ρ_Frenkel = 1540 kg/m³.

Considering Schottky Defects

Schottky defects involve the absence of atoms from the lattice, which reduces the number of atoms in the unit cell. For 0.1% Schottky defects, we need to calculate the effective number of atoms:

Effective number of atoms = 4 atoms × (1 - 0.001) = 4 × 0.999 = 3.996 atoms.

Now, we recalculate the mass:

Mass_Schottky = (3.996 atoms) × (40.08 g/mol) / (6.022 × 10²³ mol^-1) = 2.65 × 10^-25 kg.

Substituting this mass into the density formula gives:

ρ_Schottky = (2.65 × 10^-25 kg) / (1.72 × 10^-28 m³) = 1540 kg/m³.

Final Results

In summary, the densities of calcium under different defect conditions are:

• Density with 0.1% Frenkel defects: 1540 kg/m³
• Density with 0.1% Schottky defects: 1540 kg/m³

Interestingly, in this case, the density remains unchanged despite the presence of defects, primarily due to the small percentage of defects relative to the total number of atoms in the unit cell.