Question

In a face centred cubic lattice, the nearest distance between two atoms is :
Here,
$a:\text{Edge cell length}$

• A. $a$
• B. $\frac{a}{4}$
• C. $\frac{a}{\sqrt{2}}$
• D. $\sqrt{2}a$

Answer 1

The correct option is C $\frac{a}{\sqrt{2}}$

In the fcc crytsal lattice, the atoms are present at corners of the cube and at the face-centres of the cube. Here, the corner atoms and the face-centre atoms are in contact along the face diagonal.
The nearest distance is the distance between centre of these atoms. This distance is the half of the length of face diagonal which is equal to:
$\frac{1}{2}\times \sqrt{2}a\Rightarrow \text{Nearest distance:}\frac{a}{\sqrt{2}}$