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

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

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

$\frac{1}{2}$

answer is C.

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Detailed Solution

For bcc, z= 2 and $a = \frac{4 r}{\sqrt{2}}$
For fcc, z= 4 and $r = 2 \sqrt{2} r$
Density $(d) = \frac{z M}{N_{A} a^{3}}$
$\begin{array}{l} \frac{d_{25^{\circ} C}}{d_{900^{\circ} C}} = \frac{d_{b c c}}{d_{f c c}} = \frac{{(\frac{z}{a^{3}})}_{b c c}}{{(\frac{z}{a^{3}})}_{t c c}} \\ = \frac{\frac{2}{{(\frac{4 r}{\sqrt{3}})}^{3}}}{\frac{4}{(2 \sqrt{2} r)^{3}}} = \frac{2 \times (\sqrt{3})}{(4 r)^{3}} \times \frac{(2 \sqrt{2} r)^{3}}{4} = \frac{3 \sqrt{3}}{4 \sqrt{2}} \end{array}$