Iron exhibits bcc structure at room temperature. Above $500^{\circ} \mathrm{C}$, it transforms to fcc…
Iron exhibits bcc structure at room temperature. Above $500^{\circ} \mathrm{C}$, it transforms to fcc structure. Find the ratio of the density of iron at room temperature to that at $500^{\circ} \mathrm{C}$. (Assume the atomic radii and the molar mass of iron remain constant even with variation in temperature)
$3 \sqrt{3}: 4 \sqrt{2}$
$\sqrt{3}: \sqrt{2}$
$\sqrt{2}: \sqrt{3}$
$10: 92$
Solution
We know,
Density, $d=Z M / a^3 N_0$
For bcc unit cell, $Z=2, a=\frac{4}{\sqrt{3}} r$
$
d_{\mathrm{bcc}}=\frac{2 M}{\left(\frac{4}{\sqrt{3}} r\right)^3 N_0}
$
For fcc unit cell, $\quad Z=4, a=\frac{4}{\sqrt{2}} r$
$
d_{\mathrm{fcc}}=\frac{4 M}{\left(\frac{4}{\sqrt{2}} r\right)^3 N_0}
$
Ratio of density of iron bcc : fcc can be calculated by dividing Eq. (i) by Eq. (ii)
$
\frac{d_{\mathrm{bcc}}}{d_{\mathrm{fcc}}}=\frac{\frac{2 M}{\left(\frac{4}{\sqrt{3}} r\right)^3 N_0}}{\frac{4 M}{\left(\frac{4}{\sqrt{2}} r\right)^3 N_0}}=\frac{1}{2} \times\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^3
$
$
=\frac{1}{2} \times \frac{3}{2} \frac{\sqrt{3}}{\sqrt{2}}=\frac{3 \sqrt{3}}{4 \sqrt{2}}
$
The ratio is $3 \sqrt{3}: 4 \sqrt{2}$