Question

Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is $100pm$, then choose the correct option regarding densities of unit cell :
Take $\sqrt{\frac{2}{3}}\approx 0.82$

Answer 1

For bcc unit cell,
$\text{Edge length, a}=\frac{4r}{\sqrt{3}}=\frac{4\times 100}{\sqrt{3}}=\frac{400}{\sqrt{3}}pm$
$\text{Number of atoms per unit cell, Z}=2$

$\text{Density}=\frac{Z\times \text{Molar mass}}{\text{Avogadro's number}\times a^3}$
For both the unit cells, avogadro's number and molar mass is constant.
So,
we will compare the ratio $\frac{Z}{a^3}$ for both
For bcc :
${\left(\frac{Z}{a^3}\right)}_{bcc}=\frac{2\times 3\sqrt{3}}{(400{)}^3}..\left(i\right)$

For fcc unit cell,
$\text{Edge length, a}=\frac{4r}{\sqrt{2}}=\frac{4\times 100}{\sqrt{2}}=\frac{400}{\sqrt{2}}pm$
For fcc :
${\left(\frac{Z}{a^3}\right)}_{fcc}=\frac{4\times 2\sqrt{2}}{(400{)}^3}..\left(ii\right)$

$Dividing\left(ii\right)by\left(i\right)$
$\frac{{\left(\frac{Z}{a^3}\right)}_{fcc}}{{\left(\frac{Z}{a^3}\right)}_{bcc}}=\frac{4\sqrt{2}}{3\sqrt{3}}=0.82\times \frac{4}{3}=1.09$
$\frac{{\left(\frac{Z}{a^3}\right)}_{fcc}}{{\left(\frac{Z}{a^3}\right)}_{bcc}}>1{\rho }_{fcc}>{\rho }_{bcc}$