What is the edge length of the unit cell of a body-centred cubic $(BCC)$ crystal of an element whose atomic radius is $75 \ pm$ (in $pm$)?

Calculate the radius of an atom of metal if it forms a simple cubic unit cell with an edge length of $380 \ pm$. (in $pm$)

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:

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 $1.74 \ g \ cm^{-3}$, the radius of the metal atom is ............... $pm$ $(N_A = 6 \times 10^{23} \ mol^{-1})$

Calculate the number of unit cells in $1 \ cm^3$ of an element if the unit cell edge length is $2.0 \times 10^{-8} \ cm$.

If a metal crystallises in a face-centred cubic $(FCC)$ structure with a metallic radius of $25 \ pm$, the number of unit cells in $1.0 \ cm^3$ of the lattice is: