Question

The volume fraction for an element in an fcc lattice is ____________ . (round off to two decimal places)

Answer 1

Correct Answer: 0.73

The face-centered cubic (fcc) lattice structure is one of the most common packing arrangements in crystalline materials. To calculate the volume fraction, we must determine the fraction of space filled by atoms in the lattice. In an fcc lattice, atoms are arranged in layers with each unit cell containing 4 atoms. When analyzing the packing, it's useful to consider the geometric relationship between the lattice parameter (a) and the atomic radius (r).

In an fcc unit cell, the diagonal of the face is equal to 4r, where r is the atomic radius. The face diagonal can also be expressed in terms of the lattice parameter a using the Pythagorean theorem:

a√2=4r

From this equation, we solve for a:

a = 2√2r

The volume of the unit cell (V_cell) is:

V_cell = a³ = (2√2r)³ = 16√2r³

Next, calculate the total volume occupied by the atoms (V_atoms) within the fcc unit cell. Since there are 4 atoms each with volume (4/3)πr³, we have:

V_atoms = 4×(4/3)πr³ = (16/3)πr³

The volume fraction (VF) is the ratio of V_atoms to V_cell:

VF = V_atoms/V_cell = ((16/3)πr³) / (16√2r³)

Simplifying the fraction:

VF = π/(3√2)

Calculating this expression:

VF ≈ 0.7405

Thus, the volume fraction for an element in an fcc lattice, rounded to two decimal places, is:

0.74