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: $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$

If the angle of incidence of $X$-ray of wavelength $3 \mathring{A}$ which produces a second order diffracted beam from the $(100)$ planes in a simple cubic lattice with interlayer spacing $a = 6 \mathring{A}$ is $30^{\circ}$,the angle of incidence that produces a first order diffracted beam from the $(200)$ planes is $.... \, ^{\circ}$

Calculate the number of unit cells in $3 \ g$ of a metal that crystallises in a simple cubic unit cell with an edge length of $336 \ pm$. (Density of metal $= 9.4 \ g \ cm^{-3}$)

What is the edge length of a simple cubic unit cell if the radius of the atom is $174 \ pm$ (in $pm$)?

An element has a body-centered cubic structure with a unit cell edge length of $400 \ pm$. The atomic mass of the element is $24 \ g \ mol^{-1}$. What is the density of the element (in $g \ cm^{-3}$)? $(N_{A} = 6 \times 10^{23} \ mol^{-1})$

Polonium (atomic mass = $209$) crystallises in a simple cubic structure with a density of $9.32 \ g \ cm^{-3}$. Its lattice parameter (in $pm$) is closest to: