Question

Chromium metal crystallizes with a body-centered cubic lattice. The length of the unit cell edge is found to be $287pm$. Calculate the atomic radius. What would be the density of chromium in $g/c{m}^3$?

Answer 1

$a=287pm=2.87\mathring{A}$,

$Z=2\left(BCC\right)$, $M_A=52$

In body-centred cubic unit cell,

$a\sqrt{3}=4r$

where, $a=$ edge length, $r=$ radius of atom

$r=\frac{a\sqrt{3}}{4}=\frac{\sqrt{3}\times 2.87}{4}=1.2427\stackrel{\circ }{A}$

Now,

$\rho =\frac{Z\times M_A}{N_A\times a^3}$

$\rho =\frac{2\times 52}{6.023\times {10}^{23}\times 23.64\times {10}^{-24}}$

$=7.3g/c{m}^3$