A metal crystallises in two cubic phases, fcc | vedclass

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

Vedclass · Accepted Answer

$1.26$

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(B) The density of a unit cell is given by the formula $ho = \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 $fcc$, $Z_{fcc} = 4$ and $a_{fcc} = 3.5 \ \mathring{A}$.For $bcc$, $Z_{bcc} = 2$ and $a_{bcc} = 3 \ \mathring{A}$.The ratio of densities is $\frac{ho_{fcc}}{ho_{bcc}} = \frac{Z_{fcc}}{Z_{bcc}} 	imes \left(\frac{a_{bcc}}{a_{fcc}}ight)^3$.Substituting the values: $\frac{ho_{fcc}}{ho_{bcc}} = \frac{4}{2} 	imes \left(\frac{3}{3.5}ight)^3 = 2 	imes \left(\frac{6}{7}ight)^3 = 2 	imes \frac{216}{343} = \frac{432}{343} \approx 1.26$.

$A$ metal crystallises in two cubic phases, $fcc$ and $bcc$ with edge lengths $3.5 \ \mathring{A}$ and $3 \ \mathring{A}$ respectively. The ratio of densities of $fcc$ and $bcc$ is approximately

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