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NEET Medium

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

A

3342\frac{3\sqrt{3}}{4\sqrt{2}}

B

4332\frac{4\sqrt{3}}{3\sqrt{2}}

C

32\frac{\sqrt{3}}{\sqrt{2}}

D

12\frac{1}{2}

Step-by-Step Solution

For BCC lattice : Z=2Z = 2, a=4r3a = \frac{4r}{\sqrt{3}}

For FCC lattice : Z=4Z = 4, a=22ra = 2\sqrt{2} r

∴d25∘Cd900∘C=(ZMNAa3)BCC(ZMNAa3)FCC\therefore \frac{d_{25^\circ\text{C}}}{d_{900^\circ\text{C}}} = \frac{\left(\frac{ZM}{N_A a^3}\right)_{\text{BCC}}}{\left(\frac{ZM}{N_A a^3}\right)_{\text{FCC}}}

=24(22r4r3)3= \frac{2}{4} \left( \frac{2\sqrt{2}r}{\frac{4r}{\sqrt{3}}} \right)^3

=3342= \frac{3\sqrt{3}}{4\sqrt{2}}

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