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

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

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