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The ratio of closed packed atoms to tetrahedral holes in cubic close packing is

$(a)\;1:1\qquad(b)\;1:2\qquad(c)\;1:3\qquad(d)\;2:1$

1 Answer

Every constituent has two tetrahedral voids.
In CCP lattice atoms = $\large\frac{1}{8}\times8+6\times\large\frac{1}{2}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=1+3$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=4$
$\therefore Tetrahedral\; Void = 4\times2$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=8$
Thus ratio = 4:8 :: 1:2
Hence answer is (b)
answered Feb 28, 2014 by sharmaaparna1
 

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