First slide

Density of unit cell

Question

Iron exhibits bcc structure at room temperature. Above $900 ° C$ , it transforms to fcc structure. The ratio of density of iron at room temperature to that at $900 ° C$ (assuming molar mass and atomic radii of iron remains constant with temperature) is

Moderate

Solution

$For bcc lattice : Z = 2, a = \frac{4 r}{\sqrt{3}}$
$For fcc lattice : Z = 4, a = 2 \sqrt{2} r ∴ \frac{d_{R . T .}}{d_{900 ° C}} = \frac{{(\frac{ZM}{N_{A} a^{3}})}_{bcc}}{{(\frac{ZM}{N_{A} a^{3}})}_{fcc}}$
Given, molar mass and atom radii are constant.
$= \frac{2}{4} {(\frac{2 \sqrt{2} r}{\frac{4 r}{\sqrt{3}}})}^{3} = \frac{3 \sqrt{3}}{4 \sqrt{2}}$

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