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

To determine the density ratio of face centred cubic (fcc) to body centred cubic (bcc) lattices of a metal M, we must use formulas for density and unit cell properties.

For a face centred cubic (fcc) lattice:

• The formula for density is \( \rho_{\text{fcc}} = \frac{Z \cdot M}{a^3 \cdot N_A} \), where \( Z = 4 \) for fcc, \( M \) is the molar mass, \( a \) is the edge length, and \( N_A \) is Avogadro's number.

 

With edge length \( a = 20 \,\mathring{A} = 20 \times 10^{-8} \) cm, the density becomes:
\(\rho_{\text{fcc}} = \frac{4 \cdot M}{(20 \times 10^{-8})^3 \cdot N_A}\)

For a body centred cubic (bcc) lattice:

• Density formula: \( \rho_{\text{bcc}} = \frac{Z \cdot M}{a^3 \cdot N_A} \), where \( Z = 2 \) for bcc.

 

With edge length \( a = 25 \,\mathring{A} = 25 \times 10^{-8} \) cm, the density becomes:
\(\rho_{\text{bcc}} = \frac{2 \cdot M}{(25 \times 10^{-8})^3 \cdot N_A}\)

The ratio of densities \( \frac{\rho_{\text{fcc}}}{\rho_{\text{bcc}}} \) is then:

• \(\frac{\rho_{\text{fcc}}}{\rho_{\text{bcc}}} = \frac{\frac{4 \cdot M}{(20 \times 10^{-8})^3 \cdot N_A}}{\frac{2 \cdot M}{(25 \times 10^{-8})^3 \cdot N_A}} = \frac{4 \cdot (25)^3}{2 \cdot (20)^3}\)

 

Calculate the values:

• \((25)^3 = 15625 \)
• \((20)^3 = 8000 \)

 

Thus, the ratio becomes:
\(\frac{4 \cdot 15625}{2 \cdot 8000} = \frac{62500}{16000} = 3.90625\)

Rounding to the nearest integer gives 4.

Verifying the result, the calculated ratio of 4 lies within the given range [4,4]. Hence, the solution meets the requirements.