Question

An alloy of Cu, Ag, Au is found to have a simple cubic close packed lattice. If the Ag atoms occupy the face centres and Au is present at the body centre, the formula of the alloy will be.

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

Answer 1

The correct option is B $CuA{g}_{3}Au$

Cu atoms are at eight corners of the cube. Therefore, the number of Cu atoms in the unit cell.

$=\frac{8}{8}=1$

Au atoms are at the face-centre of six faces.

Therefore, its share in the unit cell $=\frac{6}{2}=3$

At atoms are at the body centres, so the number of Au atom$=1$

$\therefore$ The formula of the alloy is $CuA{g}_{3}Au$.