Q. An alloy of Cu, Ag and Au is found to have Cu forming the 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) $Cu_{4}Ag_{4}Au$
B) $CuAg_{3}Au$
C) $CuAgCu$
D) $Cu_{4}Ag_{2}Au$
Explanation
(b)
Cu atoms are at eight corners of the cube. Therefore, the number of Cu atoms in the unit cell $= \frac{8}{8} = 1$
Ag atoms are at the face-centre of six faces. Therefore, its share in the unit cell $= \frac{6}{2} = 3$
Au atoms are at the body centre
$\therefore$ the number of Au atoms = 1
$\therefore$ The formula of the alloy is $CuAg_{3}Au$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.