$A$ metal crystallises with a face-centred cubic $(fcc)$ lattice. The edge length of the unit cell is $408 \, pm$. The diameter of the metal atom is ............. $pm$.

The density of $\beta-Fe$ is $7.6 \ g \ cm^{-3}$. It crystallizes in a cubic lattice with $a = 290 \ pm$. What is the value of $Z$? $(Fe = 56 \ g \ mol^{-1}; N_{A} = 6.022 \times 10^{23} \ mol^{-1})$

Calculate the number of unit cells in $10.8 \ g$ of metal,given that $\rho a^3 = 7.2 \times 10^{-22} \ g$.

Calculate the volume of the unit cell of an element having a molar mass of $63.5 \ g \ mol^{-1}$ that forms an $fcc$ structure $\left[\varrho \times N_{A} = 5.5 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}\right]$.

The volume of a simple cubic unit cell is $x \times 10^{-23} \ cm^3$. Calculate the value of $x$ if the volume occupied by a particle in it is $2.1 \times 10^{-23} \ cm^3$.

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length of $405 \ pm$. If its density is $2.7 \times 10^3 \ kg \ m^{-3}$, the nature of the cubic unit cell is: $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$