Question

Consider iron below 912°C, where its structure is BCC. Given the density of iron as 7.86 g/cm and its atomic mass as 55.85 g/mole, calculate the lattice parameter of the unit cell and the radius of the Fe atom

Answer 1

To calculate the lattice parameter and the atomic radius of iron with a BCC structure, use the atomic mass and density to find the volume of a unit cell and then apply the relationship between the lattice parameter and the atomic radius for BCC crystals.

Step-by-step explanation:

The question asks to calculate the lattice parameter of the unit cell and the radius of an Fe (iron) atom, assuming a body-centered cubic (BCC) crystal structure for iron below 912°C. We know that in a BCC structure, there are two atoms per unit cell: one-eighth of an atom at each of the eight corners of the cube plus one atom at the center. The density (ρ) of iron is given as 7.86 g/cm³, and the atomic mass (M) is 55.85 g/mole.

To find the lattice parameter (a), we use the formula:

1. Convert the atomic mass to kilograms: M = 55.85 g/mole × (1 kg / 1000 g) = 0.05585 kg/mole.
2. Calculate the volume of one mole of iron atoms based on the given density and atomic mass: V = M / ρ.
3. Determine the number of atoms per unit cell in BCC, which is 2.
4. Calculate the volume of the unit cell (V_cell) as V_cell = V / 2 since there are two atoms per unit cell in BCC.
5. Find the lattice parameter of the unit cell (a) by taking the cube root of the volume of the unit cell: a = ∛V_cell.
6. To find the atomic radius (r), use the relationship for the BCC structure: a = 4r/√3.
7. Solve for r: r = a √3 / 4.

Through these calculations, the lattice parameter and the radius of the Fe atom can be determined.