Question

Using the data given below, find the type of cubic lattice to which the crystal belongs.

	$Fe$	$V$	$Pd$
$a$ in pm	$286$	$301$	$388$
$\rho$ in gm $c{m}^{-3}$	$7.86$	$5.96$	$12.16$

Answer 1

The theoretical density of the crystal is given by

$\rho =\frac{nM}{N_{A}V}=\frac{nM}{N_A{a}^3}$

$n=$ number of molecules / atoms / ions in unit cell

$M=$ Molar mass of substance

$V=$ Volume of unit cell

$a=$ edge length of unit cell

For $Fe$:-

$n=\rho \frac{N_{A}V}{M}$

Here, $\rho =7.86g/c{m}^3;N_A=6.022\times {10}^{23};a=286\times {10}^{-10}cm;M=56g$

$\Rightarrow n=\frac{7.86\times 6.022\times {10}^{23}\times (286{)}^3\times ({10}^{-10}{)}^3}{56}=2$

So it has body centered cubic lattice (BCC)

For $V$:-

$\rho =5.96g/c{m}^3;N_A=6.022\times {10}^{23};a=301\times {10}^{-10}cm;M=51g$

$\Rightarrow n=\frac{5.96\times 6.022\times {10}^{23}\times (301{)}^3\times ({10}^{-10}{)}^3}{51}=2$

So unit cell of $V$ has body centered cubic lattice (BCC)

For $Pd$:-

$\rho =12.16g/c{m}^3;N_A=6.022\times {10}^{23};a=388\times {10}^{-10}cm;M=106g$

$\Rightarrow n=\frac{12.16\times 6.022\times {10}^{23}\times (388{)}^3\times ({10}^{-10}{)}^3}{106}=4$

So unit cell of $Pd$ is face centered cubic lattic (FCC)