Gold has a face centered cubic unit cell and has the density ππ. π π ππβπ . Calculate the length along an edge. (π΄π’ = 197 π πππ β1 )Β
For FCC crystal structure "n=\\frac{1}{8}\\cdot8+\\frac{1}{2}\\cdot6=4" atoms per unit cell and "V=a^3" .
"\\rho=\\frac{n\\cdot M}{V\\cdot N_A}=\\frac{n\\cdot M}{a^3\\cdot N_A}\\to a=(\\frac{n\\cdot M}{\\rho\\cdot N_A})^{1\/3}="
"=(\\frac{4\\cdot 0.197}{19300\\cdot 6.023\\cdot10^{23}})^{1\/3}=4.077\\cdot10^{-10}\\ (m)" . Answer
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