In a close-packed arrangement,$A$ type of atoms are at corners,$B$ type of atoms are at face centers,$C$ type of atoms are at tetrahedral voids,and $D$ type of atoms are at octahedral voids. If all the atoms along any one body diagonal are removed,what will be the formula of the compound?

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

The $CORRECT$ statement$(s)$ for cubic close packed $(ccp)$ three dimensional structure is(are)
$(A)$ The number of the nearest neighbours of an atom present in the topmost layer is $12$
$(B)$ The efficiency of atom packing is $74 \%$
$(C)$ The number of octahedral and tetrahedral voids per atom are $1$ and $2$,respectively
$(D)$ The unit cell edge length is $2\sqrt{2}$ times the radius of the atom

If $z$ represents the number of atoms in a unit cell of a close-packed structure with an $ABCABC$ stacking sequence,then the number of tetrahedral voids in the unit cell is ....

In a crystal,atoms of $A$ form an $FCC$ lattice,atoms of $B$ occupy all octahedral voids,and atoms of $C$ occupy $25\%$ of the tetrahedral voids. The possible molecular formula of the compound is: