1. Tardigrade
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  3. Chemistry
  4. 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

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Q. Iron exhibits $bcc$ structure at room temperature. Above $900^{\circ}C$, it transforms to fcc 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

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Solution:

For $BCC$ lattice $: Z =2, a =\frac{4 r }{\sqrt{3}}$

For $FCC$ lattice $: Z =4, a =2 \sqrt{2} r$

$\therefore \frac{d_{25^{\circ} C }}{d_{900^{\circ} C }}=\frac{\left(\frac{ ZM }{ N _{ A } a ^{3}}\right)_{ BCC }}{\left(\frac{ ZM }{ N _{ A } a ^{3}}\right)_{ FCC }}=\frac{2}{4}\left(\frac{2 \sqrt{2} r }{\frac{4 r}{\sqrt{3}}}\right)^{3}=\left(\frac{3 \sqrt{3}}{4 \sqrt{2}}\right)$