Question

In BCC, the distance between two nearest atoms will be 

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

Answer 1

The Correct Option is B

To determine the shortest distance between two nearest atoms in a Body-Centered Cubic (BCC) structure, we need to understand the arrangement of atoms in this type of crystal lattice.

In a BCC structure, each unit cell consists of:

• 8 corner atoms, each shared by 8 unit cells, contributing 1/8th of each atom to the unit cell.
• 1 atom at the body center, contributing entirely to the unit cell.

Now, let's consider the geometry of a BCC unit cell:

The body diagonal of the cube connects the center atom to one of the corner atoms. If the edge length of the cube is denoted by a, then the body diagonal can be expressed using Pythagorean theorem in three dimensions:

\(\text{Body Diagonal} = \sqrt{a^2 + a^2 + a^2} = \sqrt{3}a\)

Since the body-centered atom touches the corner atoms through the body diagonal, the length of the body diagonal is equal to 4 times the atomic radius r:

\(\sqrt{3}a = 4r\)

Simplifying for the atomic radius gives:

\(r = \frac{\sqrt{3}}{4}a\)

In the BCC structure, the shortest distance between two nearest atoms is not along the edge of the cube, but rather along the face diagonal, which means it is twice the atomic radius:

d = 2r = 2 \times \frac{\sqrt{3}}{4}a = \frac{\sqrt{3}}{2}a\)

Thus, the nearest distance between two atoms in a BCC structure is \(\frac{\sqrt{3}}{2}a\).

Therefore, the correct answer is: \(\frac{\sqrt{3}}{2}a\).