A metal crystallises in two phases,one as fcc | vedclass

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

Vedclass · Accepted Answer

$1.26 : 1$

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(C) Density $(d)$ is given by the formula $d = \frac{Z 	imes M}{N_A 	imes a^3}$,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 $fcc$,$Z_1 = 4$ and $a_1 = 3.5 \ \mathring{A}$. For $bcc$,$Z_2 = 2$ and $a_2 = 3.0 \ \mathring{A}$. The ratio of densities is $\frac{d_{fcc}}{d_{bcc}} = \frac{Z_1}{Z_2} 	imes \frac{a_2^3}{a_1^3}$. Substituting the values: $\frac{d_{fcc}}{d_{bcc}} = \frac{4}{2} 	imes \frac{(3.0)^3}{(3.5)^3} = 2 	imes \frac{27}{42.875} = 2 	imes 0.6297 = 1.2594 \approx 1.26 : 1$.

$A$ metal crystallises in two phases,one as $fcc$ and other as $bcc$ with unit cell edge lengths of $3.5 \ \mathring{A}$ and $3.0 \ \mathring{A}$ respectively. The ratio of density of $fcc$ and $bcc$ phases approximately is

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