The metal niobium crystallizes in a body-centered cubic lattice.If the density of niobium is 8.57 g/cm³,what is the atomic radius of niobium?

Question

Quizplus · Accepted Answer

In a body-centered cubic (BCC) lattice, the relation between the atomic radius (r) and the edge length (a) is a = 4r/√3. The density (ρ) is given by ρ = (Z * M) / (a^3 * N_A), where Z is the number of atoms per unit cell (2 for BCC), M is the molar mass of niobium (92.906 g/mol), and N_A is Avogadro's number (6.022 x 10^23 mol^-1). Solving for a using the given density and substituting into the relation for r gives the atomic radius as approximately 142 pm.

The Metal Niobium Crystallizes in a Body-Centered Cubic Lattice