The silver metal crystallizes with a face-centred cubic lattice. The edge length of the unit cell is $4.08\stackrel{o}{A}$. Calculate the percentage of empty space in the FCC structure?

The correct option is C $26\%$
For Face centered cubic (fcc) :
The atoms along the face diagonals are in contact with each other.
Here,
$\sqrt{2}\times a=4ra=2\sqrt{2}\times r$
where, r is the radius of sphere.

$Z_{eff}\text{for fcc}=4$
So,

$\text{The total volume occupied by sphere}V=4\times \frac{4}{3}\times \pi r^3$
$\text{Volume of the cube}=a^3=(2\sqrt{2}r{)}^3$
$\text{Packing efficiency (P.F)}=\frac{4\times \frac{4}{3}\times \pi r^3}{(2\sqrt{2}r{)}^3}\times 100\%P.F=\frac{\pi \times 100}{3\sqrt{2}}\%=74.1\approx 74\%$

The percentage of empty space in the unit cell is,
$\%\text{of Empty space}=100-74=26$