A certain element crystallises in a bcc lattice of unit cell edge length 27Aᵒ. If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cel in Aᵒ will be ____________. (Round off to the Nearest Integer).
[Assume each lattice point has a single atom]
[Assume √3 = 1.73, √2 = 1.41]

Solution

Chemistry ma’am · 4 steps

Same element, same atomic radius: get r from the bcc cell, then fit the fcc cell around it.

1. Where atoms touch. In bcc the atoms touch along the body diagonal, so √3 a = 4r. In fcc they touch along the face diagonal, so √2 a = 4r.
2. Radius from bcc. r = (√3 a_(bcc))/4 = (1.73 × 27)/4 = 11.68 Å. The element and the conditions are the same, so this radius carries over to the fcc cell.

   √2 a_(fcc) = 4r = √3 a_(bcc) ⇒ a_(fcc) = √3/√2 a_(bcc)

3. Put in numbers. a_(fcc) = (1.73 × 27)/1.41 = 46.71/1.41 = 33.13 Å.
4. So. The fcc edge length is about 33.1 Å, which rounds to 33.

The common slip: 22. Swapping the two diagonal relations, using √2a = 4r for bcc and √3a = 4r for fcc, gives 1.41 × 27 / 1.73 ≈ 22, not 33.

The idea to keep. When one element switches lattice, the atomic radius is the bridge. Remember where atoms touch: body diagonal in bcc (√3a = 4r), face diagonal in fcc (√2a = 4r).

Answer from the NTA final answer key. Solution by PrepWiser.