Question

Chromium metal crystallises with a body centered cubic lattice. The edge length of the unit cell is found to be $287$ pm. Calculate the atomic radius. What would be the density of chromium in $g/{cm}^3$? (atomic mass of $Cr=52.99$)

Answer 1

For body centred cubic cell atomic radius
$r=\frac{\sqrt{3}\times edgelength}{4}=\frac{\sqrt{3}\times 287}{4}=124.47pm$
Thus, atomic radius of chromium metal is $124.27$ pm.
Density of the unit cell of the solid $=\frac{Massofunitcell}{Volumeofunitcell}$
Mass of unit cell $=$ No. of atoms in unit cell $\left(z\right)\times$ mass of each atom ($m$)
For body centered cubic lattice, the number of atoms in unit cell ($z$) is
$1+(\frac{1}{8}\times 8)=1+1=2$
Mass of each atom $\left(m\right)=\frac{Atomicmass}{Avogadr{o}^{\prime }sno.}=\frac{52.99}{6.023\times {10}^{23}}$
The volume of the cubic lattice$={\left(Edgelength\right)}^3={(287\times {10}^{-10})}^3$
$\therefore$ Density of the unit cell is $=\frac{2\times 52.99}{6.023\times {10}^{23}\times {\left(287\right)}^3\times {10}^{-30}}gm/{cm}^3$
$7.44gm/{cm}^3$
Thus, the density of chromium is $gm/{cm}^3$ is $7.44$.