Question

The edge length of a unit cell of a metal that has a molecular weight 75 g/mol is $5\stackrel{\circ }{\text{A}}$. It crystallizes in a cubic lattice. If its density is 2 g/cc, then find the radius of metal atom in pm:
$(N_A=6\times {10}^{23},\sqrt{2}=1.41,\sqrt{3}=1.73)$

Answer 1

$\rho =\frac{ZM}{a^3{N}_A}\text{or},\rho =\frac{ZM}{N_{A}V}\Rightarrow Z=\frac{\rho N_{A}V}{M}=\frac{2\times 6\times {10}^{23}\times (5\times {10}^{-8}{)}^3}{75}=2$
Unit cell is BCC.
$\Rightarrow a\sqrt{3}=4r\Rightarrow r=\frac{a\sqrt{3}}{4}=500\times \frac{\sqrt{3}}{4}=216.25\text{pm}$