Iron exhibits bcc structure at room temperatu | vedclass

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(A) The density of a unit cell is given by the formula: $d = \frac{ZM}{a^3 N_0}$.For a $bcc$ unit cell: $Z = 2$ and $a = \frac{4r}{\sqrt{3}}$.Thus,$d_

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

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(A) The density of a unit cell is given by the formula: $d = \frac{ZM}{a^3 N_0}$.For a $bcc$ unit cell: $Z = 2$ and $a = \frac{4r}{\sqrt{3}}$.Thus,$d_{bcc} = \frac{2M}{(\frac{4r}{\sqrt{3}})^3 N_0} = \frac{2M 	imes 3\sqrt{3}}{64r^3 N_0} = \frac{3\sqrt{3}M}{32r^3 N_0}$.For an $fcc$ unit cell: $Z = 4$ and $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$.Thus,$d_{fcc} = \frac{4M}{(2\sqrt{2}r)^3 N_0} = \frac{4M}{16\sqrt{2}r^3 N_0} = \frac{M}{4\sqrt{2}r^3 N_0}$.The ratio of densities is $\frac{d_{bcc}}{d_{fcc}} = \frac{3\sqrt{3}M}{32r^3 N_0} 	imes \frac{4\sqrt{2}r^3 N_0}{M} = \frac{3\sqrt{3}}{8\sqrt{2}}$.Wait,re-calculating: $\frac{d_{bcc}}{d_{fcc}} = \frac{3\sqrt{3}}{32} 	imes 4\sqrt{2} = \frac{3\sqrt{3}}{8\sqrt{2}}$.Actually,$\frac{d_{bcc}}{d_{fcc}} = \frac{3\sqrt{3}}{32} 	imes 4\sqrt{2} = \frac{3\sqrt{3}}{8\sqrt{2}} = \frac{3\sqrt{3}}{4 	imes 2 	imes \sqrt{2}} = \frac{3\sqrt{3}}{4 	imes \sqrt{2} 	imes \sqrt{2} 	imes \sqrt{2}} = \frac{3\sqrt{3}}{4 	imes 2\sqrt{2}} = \frac{3\sqrt{3}}{8\sqrt{2}}$.Let's re-evaluate the ratio: $\frac{d_{bcc}}{d_{fcc}} = \frac{2}{a_{bcc}^3} 	imes \frac{a_{fcc}^3}{4} = \frac{1}{2} 	imes (\frac{a_{fcc}}{a_{bcc}})^3 = \frac{1}{2} 	imes (\frac{4r/\sqrt{2}}{4r/\sqrt{3}})^3 = \frac{1}{2} 	imes (\frac{\sqrt{3}}{\sqrt{2}})^3 = \frac{1}{2} 	imes \frac{3\sqrt{3}}{2\sqrt{2}} = \frac{3\sqrt{3}}{4\sqrt{2}}$.

Iron exhibits $bcc$ structure at room temperature. Above $500^{\circ} C$,it transforms to $fcc$ structure. Find the ratio of the density of iron at room temperature to that at $500^{\circ} C$. (Assume the atomic radii and the molar mass of iron remain constant even with variation in temperature)

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