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

Accepted Answer

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

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

$$	herefore \frac{d_{25^\circ	ext{C}}}{d_{900^\circ	ext{C}}} = \frac{\left(\frac{ZM}{N_A a^3}ight)_{	ext{BCC}}}{\left(\frac{ZM}{N_A a^3}ight)_{	ext{FCC}}}$$

$$= \frac{2}{4} \left( \frac{2\sqrt{2}r}{\frac{4r}{\sqrt{3}}} ight)^3$$

$$= \frac{3\sqrt{3}}{4\sqrt{2}}$$

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