If $M$ is the atomic mass of an element and $a$ is the edge length of the unit cell,then the formula to calculate the density $\rho$ is:

If the radius of an atom of an element which forms a body centered cubic unit cell is $173.2 \ pm$, the volume of the unit cell in $cm^3$ is:

The unit cell of copper corresponds to a face-centered cubic $(FCC)$ lattice with an edge length of $3.596 \, \mathring{A}$. The calculated density of copper in $kg / m^{3}$ is ....... .
[Molar mass of $Cu = 63.54 \, g/mol$; Avogadro Number $= 6.022 \times 10^{23} \, mol^{-1}$]

Calculate the number of unit cells present in $1 \text{ g}$ of a metal if the product of the density of the metal and the volume of the unit cell is $2.5 \times 10^{-22} \text{ g}$.

Gold has a face-centered cubic $(fcc)$ lattice with an edge length of the unit cube of $407 \, pm$. The diameter of the gold atom is .............. $pm$.

The edge length of the unit cell of a metal $(M_W = 24 \, g \, mol^{-1})$ having a cubic structure is $4.53 \, \mathring{A}$. If the density of the metal is $1.74 \, g \, cm^{-3}$,then the effective number of atoms in the unit cell is :- $(N_A = 6 \times 10^{23} \, mol^{-1})$