A metal crystallizes in both fcc and bcc latt | vedclass

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(C) The density $d$ of a unit cell is given by $d = \frac{Z 	imes M}{N_A 	imes a^3}$, where $Z$ is the number of atoms per unit cell, $M$ is the mol

Vedclass · Accepted Answer

$1.01$

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(C) The density $d$ of a unit cell is given by $d = \frac{Z 	imes M}{N_A 	imes a^3}$, where $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $N_A$ is Avogadro's number, and $a$ is the edge length. For $bcc$, $Z_1 = 2$ and $a_1 = 300 \ pm$. For $fcc$, $Z_2 = 4$ and $a_2 = 380 \ pm$. The ratio of densities is $\frac{d_{bcc}}{d_{fcc}} = \frac{Z_1}{Z_2} 	imes \left(\frac{a_2}{a_1}ight)^3$. Substituting the values: $\frac{d_{bcc}}{d_{fcc}} = \frac{2}{4} 	imes \left(\frac{380}{300}ight)^3 = 0.5 	imes (1.266)^3$. $\frac{d_{bcc}}{d_{fcc}} = 0.5 	imes 2.03 = 1.015 \approx 1.01$.

$A$ metal crystallizes in both $fcc$ and $bcc$ lattice structures with unit cell edge lengths of $380 \ pm$ and $300 \ pm$ respectively. The ratio of their densities $(bcc/fcc)$ is:

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