Question

Metallic gold crystallizes in the face-centred cubic lattice. The edge length of the cubic unit cell, $a\approx 4\stackrel{\circ }{A}$. The closest distance between centres of two gold atoms is:

• A. $4\stackrel{\circ }{A}$
• B. $3\stackrel{\circ }{A}$
• C. $4\sqrt{2}\stackrel{\circ }{A}$
• D. $2\sqrt{2}\stackrel{\circ }{A}$

Answer 1

The correct option is D $2\sqrt{2}\stackrel{\circ }{A}$

In a face-centred cubic cell, along the face diagonal
$\text{Radius}\left(r\right)=\frac{\sqrt{2}a}{4}$

The closest distance between two atoms is the distance between the centre of corner atom and the centre of face centre particle.
This is equal to distance of half of face diagonal
$\text{Nearest distance(d)}=\frac{1}{2}\times \sqrt{2}a$
$d=(\frac{1}{2}\times \sqrt{2}\times 4)\stackrel{\circ }{A}$
$d=2\sqrt{2}\stackrel{\circ }{A}$