Question

In the face-centred cubic lattice structure of gold the closest distance between gold atoms is: ('a' being the edge length of the cubic unit cell)

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

Answer 1

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

For, FCC, the closet distance between two atoms is 2R.
Taking 'a' is the edge length
$(4R{)}^2=a^2+a^2$
$\Rightarrow 16R^2=2a^2$
$\Rightarrow R=\frac{a}{2\sqrt{2}}$
$\Rightarrow 2R=\frac{a}{\sqrt{2}}$