Question

Metallic gold crystallises in fcc lattice. The length of the cubic unit cell is a which is $4.07$ $\mathring{A}$. How many nearest neighbours does each gold atom have at the distance of $2.9\mathring{A}$.

• A. 8
• B. 14
• C. 10
• D. 12

Answer 1

The correct option is D 12

As we know, in fcc, closest distance $2r=\frac{a}{\sqrt{2}}=2.9\mathring{A}$
As we know, fcc have co-ordination number $=12$
So, the number of nearest neighbours that each gold atom have $=12$