The Solid State - Result Question 12
####12. Iron exhibits $b c c$ structure at room temperature. Above $900^{\circ} C$, it transforms to $f c c$ structure. The ratio of density of iron at room temperature to that at $900^{\circ} C$ (assuming molar mass and atomic radii of iron remains constant with temperature) is
[2018]
(a) $\frac{\sqrt{3}}{\sqrt{2}}$
(b) $\frac{4 \sqrt{3}}{3 \sqrt{2}}$
(c) $\frac{1}{2}$
(d) $\frac{3 \sqrt{3}}{4 \sqrt{2}}$
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Solution:
- (d) For bcc lattice: $Z=2, a=\frac{4 r}{\sqrt{3}}$
For fcc lattice : $Z=4, a=2 \sqrt{2} r$
$ \begin{matrix} \therefore \frac{d _{25{ }^{\circ} C}}{d _{900^{\circ} C}}=\frac{(\frac{Z M}{N_A a^{3}}) _{b c c}}{(\frac{Z M}{N_A a^{3}}) _{f c c}} \\ \quad=\frac{2}{4}(\frac{2 \sqrt{2} r}{\frac{4 r}{\sqrt{3}}})^{3}=\frac{3 \sqrt{3}}{4 \sqrt{2}} \end{matrix} $
