Question

For a BCC structure, if a = 351 pm, find r. Lithium forms a BCC structure having an edge length of a unit cell 351 pm, then find the atomic radius of lithium.

Answer 1

For a BCC (body-centered cubic) structure, the relationship between the lattice constant (a) and the atomic radius (r) is:

a = 4√(2)r/3 Solving for r, we get: r = (3a/4√(2)) Substituting the given value of a = 351 pm (or 3.51 Å), we get: r = (3 x 3.51 Å) / (4√(2)) r ≈ 1.53 Å

Therefore, the atomic radius of lithium in its BCC structure is approximately 1.53 Å.