In $CsCl$ type structure, if the radius of the cation and anion are $80 \ pm$ and $100 \ pm$ respectively, then the closest distance between two cations is:

The formula for the determination of the density of a unit cell is:

Calculate the number of particles present per unit cell if the mass of a particle is $8.0 \times 10^{-23} \ g$ and the product of density and volume of the unit cell $(\varrho \times a^3)$ is $3.2 \times 10^{-22} \ g$.

The number of atoms in $2.4 \ g$ of body-centred cubic $(BCC)$ crystal with edge length $200 \ pm$ is (density = $10 \ g \ cm^{-3}$,$N_A = 6 \times 10^{23} \ atoms \ mol^{-1}$)

$A$ metal $(X)$ of atomic weight $M \ g \ mol^{-1}$ crystallizes in a $bcc$ lattice. Its density is $d \ g \ cm^{-3}$. What is the equation for the unit cell edge length $(a)$? ($N=$ Avogadro number)

Ammonium chloride crystallizes in a body-centered cubic lattice with an edge length of the unit cell of $390 \ pm$. If the size of the chloride ion is $180 \ pm$,the size of the ammonium ion would be ........... $pm$.