Question

Metallic gold crystallises in a face centered cubic lattice of edge length $4.070\stackrel{o}{\text{A}}$. The closest distance between the gold atoms is:

• A. $3.035\stackrel{o}{\text{A}}$
• B. $8.140\stackrel{o}{\text{A}}$
• C. $2.878\stackrel{o}{\text{A}}$
• D. $1.357\stackrel{o}{\text{A}}$

Answer 1

The correct option is C $2.878\stackrel{o}{\text{A}}$

In an FCC crystal, the nearest neighbour of the corner atom is the face centre atom and the distance between them is $\frac{\text{face diagonal length}}{2}=\frac{\sqrt{2}a}{2}$.

$\therefore$ Nearest neighbour distance $=\frac{a\sqrt{2}}{2}=\frac{4.070\times \sqrt{2}}{2}\stackrel{o}{\text{A}}=2.878\stackrel{o}{\text{A}}$