Question

A metal crystallizes in two cubic phases, face centred cubic (fcc) and body centred cubic (bcc) whose unit cell length are $3.5$ and $3.0 \, \mathring{A}$ respectively. Calculate the ratio of density of $fcc$ and $bcc$.

• A. 2.123
• B. 1.259
• C. 5.124
• D. 3.134

Answer 1

Answer: (B)

$\rho=\frac{n \times M_{m}}{N_{A} \times a^{3}}$

For fcc centred cubic cell $n=4,\, a=3.5\, \, \mathring{A}$

$\therefore \rho_{( fcc )}=\frac{4 \times M_{m}}{N_{A} \times(3.5)^{3}}$(1)

For bcc lattice

$n=2, a=3.0\, \mathring{A}$

$\therefore \rho_{( bcc )}=\frac{2 \times M_{m}}{N_{A} \times(3.0)^{3}}$(2)

From equations (1) and (2), we get

$\therefore \frac{\rho_{( fcc )}}{\rho_{( bcc )}}=\frac{4}{2} \times \frac{3^{3}}{(3.5)^{3}}$

$=\frac{4 \times 3 \times 3 \times 3}{2 \times 3.5 \times 3.5 \times 3.5}=1.259$