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# Given an alloy of $Cu$ , $Ag$ , and $Au$ in which $Cu$ atoms constitute the CCP arrangement. If the hypothetical formula of the alloy is $C{u_4}A{g_3}Au$ . What are the probable locations of $Ag$ and $Au$ atoms?A) $Ag$ - all tetrahedral voids; $Au$ - all octahedral voidsB) $Ag$ - $\dfrac{3}{8}$ the tetrahedral voids; $Au$ - $\dfrac{1}{4}$ the octahedral voidsC) $Ag$ - $\dfrac{1}{2}$ octahedral voids; $Au$ - $\dfrac{1}{2}$ tetrahedral voidsD) $Ag$ - all octahedral voids ; $Au$ - all tetrahedral voids

Last updated date: 22nd Feb 2024
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Hint: Three layers (ABCABC...) of hexagonally organised atoms make up a CCP structure. Since they contact six atoms in their layer, plus three atoms in the layer above and three atoms in the layer below, atoms in a CCP configuration have a coordination number of $12$ .

We know that Copper atoms ( $Cu$ ) form a CCP (Cubic Close Packed) structure, which is equivalent to an FCC (Face Centered Cubic) arrangement, according to the question.
$\Rightarrow \dfrac{1}{8}\left( {Number.of.corners} \right) + \dfrac{1}{2} \times 6 \\ \Rightarrow \dfrac{1}{8}\left( 8 \right) + \dfrac{1}{2} \times 6 \\ \Rightarrow 1 + 3 \\ \Rightarrow 4copper.atoms \\$
The number of octahedral voids (spaces) equals the number of atoms in the FCC arrangement, which is $4$ .
Since Ag has three atoms and a cube has eight corners, the fraction would be $Ag$ has $\dfrac{3}{8}$ tetrahedral voids and $Au$ has $\dfrac{1}{4}$ octahedral voids.
Hence, the correct option is B) $Ag$ - $\dfrac{3}{8}$ the tetrahedral voids; $Au$ - $\dfrac{1}{4}$ the octahedral voids.