Calculate the volume occupied by a particle in $fcc$ unit cell if the volume of the unit cell is $1.6 \times 10^{-23} \ cm^3$.

The number of octahedral and tetrahedral holes respectively present in a hexagonal close packed $(hcp)$ crystal of $X$ atoms are

$A$ face-centred cubic $(FCC)$ unit cell is made up of two types of atoms $A$ and $B$,in which $A$ occupies the corner positions and $B$ occupies the face centres. If atoms along one of the body diagonals are removed,what is the empirical formula of the remaining solid?

What fraction of one edge-centred octahedral void lies in one unit cell of $fcc$?

In a $ccp$ structure of $X$ atoms,$Y$ atoms occupy half of the octahedral voids. If one $X$ atom and one $Y$ atom in each unit cell are replaced by a $Z$ atom,the molecular formula of the solid will be:

The packing efficiency of the face-centered cubic $(fcc)$,body-centered cubic $(bcc)$,and simple/primitive cubic $(pc)$ lattices follows the order: