Question

Na atom crystallises in bcc lattice with cell edge (a) = 4.29A° The radius of Na atom is

Answer 1

Radius = (√3 * 4.29 Å) / 4

Calculating the value, we get:

Radius ≈ 1.61 Å

Therefore, the radius of the sodium (Na) atom in the bcc lattice is approximately 1.61 Å.