Question

Gold has a face centred cubic lattice with an edge length of the unit cube of 407 pm. Assuming the closest packing, the diameter of the gold atom is

• A. 576.6 pm
• B. 287.8 pm
• C. 352.5 pm
• D. 704.9 pm

Answer 1

The correct option is B 287.8 pm

For $fcc$ lattice,
$4r=a\sqrt{2}$
and $2r=diameter$
$=\frac{a\sqrt{2}}{2}=\frac{407\times \sqrt{2}}{2}=287.8\text{pm}$