What is the relationship between the edge length (a) and the atomic radius (r) for a BCC lattice, given that √3 * a = 4r?

In a body-centered cubic (BCC) lattice, each atom is located at the corners of a cube and one atom is at the center of the cube. The relationship between the edge length of the cube (a) and the radius of the atom (r) can be derived as follows:

Consider a cube with an atom at the center (body-centered) and atoms at the corners. Let's draw a diagonal passing through the cube, which is equal to the body diagonal of the cube. The body diagonal of a cube is given by:

Body diagonal length = \sqrt{a^2 + a^2 + a^2} = \sqrt{3}a

Now, let's consider the distance between the center atom and one of the corner atoms. This distance is equal to the sum of the radius of the atom (r) and the half of the cube's edge length (a/2):

Distance between center atom and corner atom = r + a/2

Since this distance is equal to the body diagonal length, we have:

r + a/2 = \sqrt{3}a

Now, solve for r:

r = \sqrt{3}a - a/2

To simplify the expression, we can factor out 'a' from the right-hand side:

r = a(\sqrt{3} - \frac{1}{2})

Thus, the relation between the edge length (a) and the radius of the atom (r) for a BCC lattice is:

r = a(\sqrt{3} - \frac{1}{2})

Alternatively, you can also express this relationship as:

\sqrt{3}a = 2r

This is the relationship you mentioned in your question, which is also correct. Both expressions are equivalent and describe the same relationship between the edge length and the radius of the atom in a BCC lattice.