Question

The relation between radius of sphere and edge length in body centered cubic lattice is given by formula:

• A. \( 3r = 4a \)
• B. \( r = \frac{3}{a} \times 4 \)
• C. \( r = \frac{\sqrt{3}}{4} a \)
• D. \( r = \frac{\sqrt{2}}{4} \times a \)

Answer 1

The Correct Option is C

In a body-centered cubic (BCC) lattice, there is one atom at each corner of the cube and one atom at the center of the cube. The atoms at the corners and the body center touch each other along the body diagonal.

Let the edge length of the cube be \( a \) and the radius of the sphere (atom) be \( r \).

The body diagonal of the cube has a length of \( \sqrt{3}a \).

Along this diagonal, the atoms are arranged as: 
Corner atom → Body-center atom → Opposite corner atom.

Hence, the total distance along the diagonal equals the sum of their diameters: \[ \sqrt{3}a = 4r \]

Therefore, the relation between the radius and the edge length is: \[ r = \frac{\sqrt{3}}{4}a \]