A metal crystallizes in two phases, one as fcc and another as bcc with unit cell edge lengths of 3 . 5 &#160…

We know that,

Density of the cell is given as, $\mathrm{\rho }=\frac{\mathrm{Z}\times {\mathrm{M}}_{\mathrm{W}}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}}$

Here, 

$\mathrm{Z}=$ Rank of cubic lattice

${\mathrm{M}}_{\mathrm{w}}=$ Molecular weight of lattice

${\mathrm{a}}^3 ={\left(\mathrm{edge} \mathrm{length}\right)}^3=$ Volume

${\mathrm{N}}_{\mathrm{A}}=$ Avogadro's number

For $\mathrm{fcc}, \mathrm{Z}=4$ and given edge length $\mathrm{a}=3.5 \stackrel{°}{\mathrm{A}}$

${\mathrm{\rho }}_{\mathrm{fcc}}=\frac{4\times {\mathrm{M}}_{\mathrm{w}}}{{\left(3.5\times {10}^{-8}\right)}^3\times {\mathrm{N}}_{\mathrm{A}}}$

For $\mathrm{bcc}, \mathrm{Z}=2$ and given edge length $\mathrm{a}=3 \stackrel{°}{\mathrm{A}}$

${\mathrm{\rho }}_{\mathrm{bcc}}=\frac{2\times {\mathrm{M}}_{\mathrm{w}}}{{\left(3\times {10}^{-8}\right)}^3\times {\mathrm{N}}_{\mathrm{A}}}$

$\frac{{\mathrm{\rho }}_{\mathrm{fcc}}}{{\mathrm{\rho }}_{\mathrm{bcc}}}=\frac{4\times {\left(3\right)}^3}{{\displaystyle {\left(3.5\right)}^3\times 2}}=1.26$

Hence, the ratio of density of $\mathrm{fcc}$ and $\mathrm{bcc}$ phases approximately is $1.26:1$.