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

We are given that an element crystallises in a bcc lattice with unit cell edge length $$a_{bcc} = 27$$ angstrom. We need to find the edge length when the same element crystallises in an fcc lattice.

In a bcc lattice, atoms touch along the body diagonal. The relationship between the atomic radius $$r$$ and edge length is $$4r = \sqrt{3} \times a_{bcc}$$, so $$r = \frac{\sqrt{3} \times a_{bcc}}{4}$$.

In an fcc lattice, atoms touch along the face diagonal. The relationship is $$4r = \sqrt{2} \times a_{fcc}$$, so $$r = \frac{\sqrt{2} \times a_{fcc}}{4}$$.

Since the radius of the atom remains the same: $$\frac{\sqrt{3} \times a_{bcc}}{4} = \frac{\sqrt{2} \times a_{fcc}}{4}$$.

This gives $$a_{fcc} = \frac{\sqrt{3}}{\sqrt{2}} \times a_{bcc} = \frac{1.73}{1.41} \times 27 = 1.2270 \times 27 = 33.13$$ angstrom.

Rounding off to the nearest integer, the edge length of the fcc unit cell is $$\mathbf{33}$$ angstrom.