$KCl$ has the same structure as $NaCl$. If ${r_{Na^+}}/{r_{Cl^-}} = 0.55$ and ${r_{Na^+}}/{r_{K^+}} = 0.74$,then the ratio of the edge lengths of $KCl$ and $NaCl$ is:

If the crystal structure of $KCl$ is the same as that of $NaCl$,and the radius ratios are $r_{Na^+}/r_{Cl^-} = 0.55$ and $r_{K^+}/r_{Cl^-} = 0.74$,calculate the ratio of the edge lengths of the unit cells of $KCl$ and $NaCl$.

The edge length of a cube is $300 \ pm$. Its body diagonal would be: (in $pm$)

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 $\left[\varrho \times N_{A} = 5.5 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}\right]$.

An element crystallizes in a $bcc$ lattice. The atomic radius of the element is $2.598 \ \mathring{A}$. What is the volume (in $cm^3$) of one unit cell?

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 $3.2 \times 10^{-22} \ g$.