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The ratio of close-packed atoms to tetrahedral holes in cubic close packing is
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$1: 2$
Every constituent has two tetrahedral voids. In $c c p$ lattice atoms
$=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$
$\therefore$ Tetrahedral void $=4 \times 2=8$,
Thus ratio $=4: 8:: 1: 2$.
$=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$
$\therefore$ Tetrahedral void $=4 \times 2=8$,
Thus ratio $=4: 8:: 1: 2$.
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