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

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(C) The density of a unit cell is given by the formula: $ho = \frac{Z 	imes M}{a^3 	imes N_A}$ For $fcc$ lattice,the number of atoms per unit cell

Vedclass · Accepted Answer

$1.26:1$

Vedclass · Answer

(C) The density of a unit cell is given by the formula: $ho = \frac{Z 	imes M}{a^3 	imes N_A}$ For $fcc$ lattice,the number of atoms per unit cell $Z = 4$ and edge length $a = 3.5 \mathring{A}$. $ho_{fcc} = \frac{4 	imes M}{(3.5)^3 	imes N_A}$ For $bcc$ lattice,the number of atoms per unit cell $Z = 2$ and edge length $a = 3.0 \mathring{A}$. $ho_{bcc} = \frac{2 	imes M}{(3.0)^3 	imes N_A}$ Taking the ratio of the two densities: $\frac{ho_{fcc}}{ho_{bcc}} = \frac{4 	imes (3.0)^3}{2 	imes (3.5)^3} = \frac{4 	imes 27}{2 	imes 42.875} = \frac{108}{85.75} \approx 1.26$ Thus,the ratio of the density of $fcc$ to $bcc$ phases is approximately $1.26:1$.

$A$ metal crystallizes in two phases,one as $fcc$ and another 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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