Iron exhibits bcc structure at room temperatu | vedclass

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(C) For $bcc$ structure,the number of atoms per unit cell $Z_{bcc} = 2$ and the relation between edge length $a$ and atomic radius $r$ is $4r = \sqrt{

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

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(C) For $bcc$ structure,the number of atoms per unit cell $Z_{bcc} = 2$ and the relation between edge length $a$ and atomic radius $r$ is $4r = \sqrt{3}a$,so $a_{bcc} = \frac{4r}{\sqrt{3}}$.For $fcc$ structure,the number of atoms per unit cell $Z_{fcc} = 4$ and the relation between edge length $a$ and atomic radius $r$ is $4r = \sqrt{2}a$,so $a_{fcc} = \frac{4r}{\sqrt{2}}$.The density $d$ is given by $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Taking the ratio: $\frac{d_{bcc}}{d_{fcc}} = \frac{Z_{bcc}}{Z_{fcc}} 	imes (\frac{a_{fcc}}{a_{bcc}})^3$.Substituting the values: $\frac{d_{bcc}}{d_{fcc}} = \frac{2}{4} 	imes (\frac{4r/\sqrt{2}}{4r/\sqrt{3}})^3 = \frac{1}{2} 	imes (\frac{\sqrt{3}}{\sqrt{2}})^3 = \frac{1}{2} 	imes \frac{3\sqrt{3}}{2\sqrt{2}} = \frac{3\sqrt{3}}{4\sqrt{2}}$.

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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