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Chemistry
The ratio of close-packed atoms to tetrahedral holes in cubic close packing is
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Q. The ratio of close-packed atoms to tetrahedral holes in cubic close packing is
The Solid State
A
$1: 1$
B
$1: 3$
C
$1: 2$
D
$2: 1$
Solution:
Every constituent has two tetrahedral voids. In ccp lattice atoms
$=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$
$\therefore $ Tetrahedral void $=4 \times 2=8$,
Thus, ratio $=4: 8:: 1: 2$.