Question

A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of 20 and \(25 \,\mathring{A}\) respectively The ratio of densities of lattices fcc to bcc for the metal M is ___(Nearest integer)

Hint:

Remember the formula for density in crystallography: \(\rho = \frac{Z \times M}{N_A \times a^3}\). For fcc, \(Z = 4\); for bcc, \(Z = 2\). Pay close attention to units.

Answer 1

Correct Answer: 4

The formula for density is:
\[\rho = \frac{Z \times M}{N_A \times a^3},\]
where:
  \(Z\) = number of atoms per unit cell,
  \(M\) = molar mass,
  \(N_A\) = Avogadro’s number,
  \(a\) = edge length.
For fcc, \(Z = 4\), \(a = 2.0 \, \text{\AA} = 2.0 \times 10^{-10} \, \text{m}\).
For bcc, \(Z = 2\), \(a = 2.5 \, \text{\AA} = 2.5 \times 10^{-10} \, \text{m}\).
Let \(M\) be the molar mass of the metal. Then the density for fcc is:
\[\rho_{\text{fcc}} = \frac{4M}{N_A(2 \times 10^{-10})^3}.\]
And for bcc:
\[\rho_{\text{bcc}} = \frac{2M}{N_A(2.5 \times 10^{-10})^3}.\]
The ratio of densities is:
\[\frac{\rho_{\text{fcc}}}{\rho_{\text{bcc}}} = \frac{\frac{4M}{N_A(2 \times 10^{-10})^3}}{\frac{2M}{N_A(2.5 \times 10^{-10})^3}} = \frac{4}{2} \times \frac{(2.5 \times 10^{-10})^3}{(2 \times 10^{-10})^3}.\]
Simplify:
\[\frac{\rho_{\text{fcc}}}{\rho_{\text{bcc}}} = 2 \times \left(\frac{2.5}{2}\right)^3 = 2 \times \left(\frac{5}{4}\right)^3.\]
\[\frac{\rho_{\text{fcc}}}{\rho_{\text{bcc}}} = 2 \times \frac{125}{64} = \frac{250}{64} = \frac{125}{32} \approx 3.9.\]
Nearest Integer:
The nearest integer is:
\[4.\]