Iron crystallizes in different structures. At | vedclass

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(A) For $bcc$,the distance between nearest neighbors is $d = \frac{\sqrt{3}}{2} a_1$,so $a_1 = \frac{2d}{\sqrt{3}}$.For $fcc$,the distance between nea

Vedclass · Accepted Answer

$0.91$

Vedclass · Answer

(A) For $bcc$,the distance between nearest neighbors is $d = \frac{\sqrt{3}}{2} a_1$,so $a_1 = \frac{2d}{\sqrt{3}}$.For $fcc$,the distance between nearest neighbors is $d = \frac{a_2}{\sqrt{2}}$,so $a_2 = d\sqrt{2}$.Density ratio $\frac{ho_{bcc}}{ho_{fcc}} = \frac{Z_1 	imes M / (N_A 	imes a_1^3)}{Z_2 	imes M / (N_A 	imes a_2^3)} = \frac{Z_1}{Z_2} 	imes (\frac{a_2}{a_1})^3$.Substituting $Z_1 = 2$ $(bcc)$ and $Z_2 = 4$ $(fcc)$:$\frac{ho_{bcc}}{ho_{fcc}} = \frac{2}{4} 	imes (\frac{d\sqrt{2}}{2d/\sqrt{3}})^3 = \frac{1}{2} 	imes (\frac{\sqrt{6}}{2})^3 = \frac{1}{2} 	imes \frac{6\sqrt{6}}{8} = \frac{3\sqrt{6}}{8} \approx \frac{3 	imes 2.449}{8} = \frac{7.347}{8} \approx 0.918$.Thus,the ratio is approximately $0.91$.

Iron crystallizes in different structures. At $911 \, ^oC$,it undergoes a transition from $bcc$ to $fcc$ structure. If the distance between the nearest neighbors is the same in both structures at the transition temperature,the ratio of the densities of $bcc$ and $fcc$ structures will be ..........

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