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 \[=\frac{\sqrt{2}}{6}\pi =0.7404\]You need to login to perform this action.
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