Calculate the radius of an atom of an element in $pm$ if it forms $bcc$ unit cell structure having edge length $4.3 \times 10^{-8} \ cm$.

$KF$ has a structure similar to $NaCl$. If the density of the crystal is $2.48 \, g/cm^3$, the distance between $K^+$ and $F^-$ ions will be ............. $pm$. (Atomic masses: $K = 39 \, amu, F = 19 \, amu$)

The mass of an atom present in a unit cell is $4.4 \times 10^{-23} \ g$ and the product of density and volume of the unit cell is $1.792 \times 10^{-22} \ g$. What is the type of cubic unit cell?

In a body-centred cubic $(bcc)$ lattice of potassium, the correct relation between the atomic radius $(r)$ of potassium and the edge-length $(a)$ of the cube is:

If $TlCl$ crystallizes in a $bcc$ structure,then the sum of the radii of the cation and anion will be ........ ($a =$ edge length).

Lithium crystallises into a body-centered cubic $(BCC)$ structure. What is the radius of lithium if the edge length of its unit cell is $351 \ pm$ (in $pm$)?