Atomic Radius Problem with 286 pm Lattice Parameter

[vijayssonule]

i have one problem on atomic radius. i read (http://en.wikipedia.org/wiki/Atomic_radius) the artical and try to understand it but i don't found any formula for it.
The problem is:
if the lattice parameter of alpha iron is 286 pm (Pico meter),what is its atomic radius?
here we have only one variable Lattice parameter = 286 is given.
please help me to solve the problem.

 

[Astronuc]

See this discussion of atom sizes.
http://www.webelements.com/iron/atom_sizes.html

The atomic radius is approximately half the lattice parameter (l), or exactly for a simple cube, which has one atom at each corner of a cube, and so each atom is 'touching' is neighbor along the side/face of the cube.

Alpha-iron has a bcc structure, with one atom in the center of the cube, which touches the eight atoms on each corner of the cube, so that the atoms on the corners of the cube are separated and the lattice parameter is slightly greater than 2R, or R is slightly less the l/2.

Pick the a corner of the cube, and let the center atom be located at (l/2, l/2, l/2) of the cube of side, l. Find the distance from the center of the corner atom to center of the central atom, in terms of l, and that distance = 2R.