In a solid $AB$ having the $NaCl$ structure,$A$ atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed,then the resultant stoichiometry of the solid is

What is the total number of particles present in a base-centred unit cell?

Select the correct expression for determining the Packing Fraction $(P.F.)$ of an $NaCl$ unit cell (assume ideal),if ions along an edge diagonal are absent.

$A$ cubic lattice has $A$ atoms at the body centre,$B$ atoms at the corners,and $C$ atoms at half of the face centres. The formula of the lattice is:

If the side length of a face-centered cubic $(fcc)$ unit cell of a metal is $400 \ pm$, the approximate radius of the metal atom in $pm$ is $(\sqrt{2} = 1.414)$.

$A$ solid $AB$ has a $NaCl$ structure. $B$ atoms are at the corners and all face-centered positions of the unit cell,while $A$ atoms occupy all the $O.H.V.$ (Octahedral Voids). If all the atoms from the face-centered positions along one axis are removed,what will be the resulting stoichiometry of the solid?