Element X crystallizes in a 12 coordination f | vedclass

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Question

(D) For $fcc$ lattice: Coordination number is $12$,number of atoms per unit cell $(Z)$ is $4$,and edge length $a_1 = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$

Vedclass · Accepted Answer

$2(\sqrt{2})^3 : (\sqrt{3})^3$

Vedclass · Answer

(D) For $fcc$ lattice: Coordination number is $12$,number of atoms per unit cell $(Z)$ is $4$,and edge length $a_1 = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$.Density $d_1 = \frac{Z_1 \cdot M}{N_A \cdot a_1^3} = \frac{4 \cdot M}{N_A \cdot (2\sqrt{2}r)^3}$.For $bcc$ lattice: Coordination number is $8$,number of atoms per unit cell $(Z)$ is $2$,and edge length $a_2 = \frac{4r}{\sqrt{3}}$.Density $d_2 = \frac{Z_2 \cdot M}{N_A \cdot a_2^3} = \frac{2 \cdot M}{N_A \cdot (\frac{4r}{\sqrt{3}})^3}$.Ratio $\frac{d_1}{d_2} = \frac{4}{ (2\sqrt{2}r)^3} \cdot \frac{(\frac{4r}{\sqrt{3}})^3}{2} = \frac{2 \cdot (\frac{64r^3}{3\sqrt{3}})}{16\sqrt{2}r^3} = \frac{128}{3\sqrt{3} \cdot 16\sqrt{2}} = \frac{8}{3\sqrt{6}} = \frac{2(\sqrt{2})^3}{(\sqrt{3})^3}$.

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