Question

An alloy of copper, silver and gold is found to have copper constituting the $\left(FCC\right)$ lattice. If silver atoms occupy the edge centres and gold is present at body centre, the alloy has a formula

• A. $C{u}_{4}A{g}_{2}Au$
• B. $C{u}_{4}A{g}_{4}Au$
• C. $C{u}_{4}A{g}_{3}Au$
• D. $CuAgAu$

Answer 1

Answer: (C)

$C{u}_{4}A{g}_{3}Au$

Number of Cu atoms in the fcc unit cell $=8\times \frac{1}{8}+6\times \frac{1}{2}=4$
Ag atoms occupying the edge centres $=12\times \frac{1}{4}=3$
Au atoms present at the body centre $=1$
Hence, the ratio of $Cu:Ag:Au=4:3:1$. Therefore, the formula is $C{u}_{4}A{g}_{3}Au$.