Question

Metallic gold crystallises in fcc lattice with edge-length $4.070\mathring{A}$. Closest distance between gold atoms is:

• A. $2.035\mathring{A}$
• B. $8.140\mathring{A}$
• C. $2.878\mathring{A}$
• D. $1.357\mathring{A}$

Answer 1

Answer: (C)

$2.878\mathring{A}$

In $FCC$, $r=\frac{a}{2\sqrt{2}}$

$\therefore r=\frac{4.070}{2\times 1.414}\stackrel{\circ }{A}=1.439\stackrel{\circ }{A}$

Closest distance between gold atoms $=r+r$

$=2.878\stackrel{\circ }{A}$