In the closest packing of atoms,there are

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(B) In a closest packing of atoms (e.g.,$fcc$ or $ccp$),the number of atoms per unit cell is $n = 4$.The number of tetrahedral voids $(TV)$ is $2n = 2

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Two tetrahedral voids and one octahedral void per atom

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(B) In a closest packing of atoms (e.g.,$fcc$ or $ccp$),the number of atoms per unit cell is $n = 4$.The number of tetrahedral voids $(TV)$ is $2n = 2 	imes 4 = 8$.The number of octahedral voids $(OV)$ is $n = 4$.Therefore,the number of $TV$s per atom is $\frac{8}{4} = 2$.The number of $OV$s per atom is $\frac{4}{4} = 1$.Thus,there are $2$ tetrahedral voids and $1$ octahedral void per atom.

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