Diamond has a three-dimensional structure of | vedclass

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(D) In a face-centered cubic $(FCC)$ lattice,the number of atoms at the corners is $8 	imes (1/8) = 1$ and the number of atoms at the face centers is

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) In a face-centered cubic $(FCC)$ lattice,the number of atoms at the corners is $8 	imes (1/8) = 1$ and the number of atoms at the face centers is $6 	imes (1/2) = 3$,totaling $4$ atoms.In the diamond structure,carbon atoms occupy all the $FCC$ lattice points ($4$ atoms) and $50\,\%$ of the tetrahedral voids.The total number of tetrahedral voids in an $FCC$ unit cell is $8$.Number of carbon atoms in tetrahedral voids $= 50\,\% 	ext{ of } 8 = 4$.Total number of carbon atoms per unit cell $= 4 	ext{ (lattice points)} + 4 	ext{ (tetrahedral voids)} = 8$.

Diamond has a three-dimensional structure of $C$ atoms formed by covalent bonds. The structure of diamond has a face-centered cubic $(FCC)$ lattice where $50\,\%$ of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is $......$

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