Question

A metal crystallizes into two cubic phases, face centered cubic (FCC) and body centered cubic (BCC), whose unit cell length are $4.5\stackrel{\circ }{\text{A}}$ and $3\stackrel{\circ }{\text{A}}$, respectively. Calculate the ratio of the densities of the BCC to the FCC lattice.

• A. $1.69$
• B. $2.66$
• C. $2.25$
• D. $1.25$

Answer 1

The correct option is A $1.69$

We know that, total number of atoms per unit cell of FCC is $4$ and total number of atoms per unit cell of BCC is $2$
$\text{So, density in FCC}=\frac{n_1\times \text{atomic weight}}{V_1\times N_A}$
Where, $V_1=\text{volume of unit cell}=a^3=(4.5\times {10}^{-8}{)}^3$
$\text{So, density in BCC}=\frac{n_2\times \text{atomic weight}}{V_2\times N_A}$
Where, $V_2=\text{volume of unit cell}=a^3=(3\times {10}^{-8}{)}^3$
Hence, $\frac{D_{BCC}}{D_{FCC}}=\frac{n_1\times V_2}{n_2\times V_1}=\frac{2\times (4.5\times {10}^{-8}{)}^3}{4\times (3\times {10}^{-8}{)}^3}=1.69$