question_answer

Metallic gold crystallises in the fee lattice. The length of the cubic unit cell i\[a=4242\text{ }\overset{o}{\mathop{A}}\,\]. Closest distance between two gold atoms and packing factor are, respectively.

A) \[3.0\overset{o}{\mathop{A}}\,,\,\,\frac{\sqrt{2}}{6}\pi \]

B)        \[1.5\overset{o}{\mathop{A}}\,,\,\,\frac{\sqrt{2}}{6}\pi \]

C) \[3.0\overset{o}{\mathop{A}}\,,\,\,\frac{\sqrt{3}}{8}\pi \]

D)        \[1.5\overset{o}{\mathop{A}}\,,\,\,\frac{\sqrt{3}}{8}\pi \]

Correct Answer: A

Solution :

[a] In \[fcc,\,\,{{Z}_{eff}}=4.\] For fee, atomic radius \[(r)=\frac{a}{2\sqrt{2}}\] Closest distance between two \[Au\]atoms \[=2r\] \[=2\times \frac{a}{2\sqrt{2}}=\frac{2}{\sqrt{2}}=\frac{4.242}{\sqrt{2}}=\frac{4.242}{1.414}=3.0\overset{o}{\mathop{A}}\,\] (ii) Packing factor for fcc lattice \[=\frac{\sqrt{2}}{6}\pi =0.7404\]