What is the number of unit cells present in $3.9 \mathrm{~g}$ of potassium if it crystallizes in BCC…
What is the number of unit cells present in $3.9 \mathrm{~g}$ of potassium if it crystallizes in BCC Structure?
$\frac{\mathrm{N}_{\mathrm{A}}}{10}$
$\mathrm{N}_{\mathrm{A}} \times 10$
$2 \mathrm{~N}_{\mathrm{A}}$
$\frac{\mathrm{N}_{\mathrm{A}}}{20}$
Solution
$\begin{aligned} & \text { Number of atoms }=\frac{\text { Mass }}{\text { Atomicmass }} \times \mathrm{N}_{\mathrm{A}}=\frac{3.9}{39} \times \mathrm{N}_{\mathrm{A}}=0.1 \mathrm{~N}_{\mathrm{A}} \\ & \text { In BCC unit cell, } \mathrm{n}=2 \\ & \therefore \text { Number of unit cells }=\frac{0.1 \mathrm{~N}_{\mathrm{A}}}{2}=\frac{\mathrm{N}_{\mathrm{A}}}{20}\end{aligned}$