A metal crystallizes into two forms i.e., bcc and fcc whose edge lengths are 3.0 Å and 3.6 Å respectively. What is the ratio of densities of the two forms?
Answer & explanation
Correct answer: option 4
The correct answer is option 4. \(\frac{108}{125}\).
To find the ratio of densities of the two forms of the metal (bcc and fcc) with given edge lengths, we need to use the formula for density:
\(\text{Density} = \frac{Z \cdot M}{a^3 \cdot N_A}\)
where:
\( Z \) is the number of atoms per unit cell,
\( M \) is the molar mass,
\( a \) is the edge length of the unit cell,
\( N_A \) is Avogadro's number.
For bcc (Body-Centered Cubic):
\( Z = 2 \) (2 atoms per unit cell),
Edge length \( a_{\text{bcc}} = 3.0 \, \text{Å} = 3.0 \times 10^{-8} \, \text{cm} \).
For fcc (Face-Centered Cubic):
\( Z = 4 \) (4 atoms per unit cell),
Edge length \( a_{\text{fcc}} = 3.6 \, \text{Å} = 3.6 \times 10^{-8} \, \text{cm} \).
Let us calculate the densities:
Density of bcc:
\(\rho_{\text{bcc}} = \frac{Z_{\text{bcc}} \cdot M}{a_{\text{bcc}}^3 \cdot N_A} = \frac{2 \cdot M}{(3.0 \times 10^{-8})^3 \cdot N_A}\)
Density of fcc:
\(\rho_{\text{fcc}} = \frac{Z_{\text{fcc}} \cdot M}{a_{\text{fcc}}^3 \cdot N_A} = \frac{4 \cdot M}{(3.6 \times 10^{-8})^3 \cdot N_A}\)
Ratio of densities:
\(\frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{\frac{2 \cdot M}{(3.0 \times 10^{-8})^3 \cdot N_A}}{\frac{4 \cdot M}{(3.6 \times 10^{-8})^3 \cdot N_A}}\)
\(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{2 \cdot (3.6 \times 10^{-8})^3}{4 \cdot (3.0 \times 10^{-8})^3}\)
\(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{2}{4} \times \frac{(3.6)^3}{(3.0)^3}\)
\(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{1}{2} \times \frac{(36)^3}{(30)^3}\)
\(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{1}{2} \times \frac{(6)^3}{(5)^3}\)
\(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{1}{2} \times \frac{216}{125}\)
\(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{108}{125}\)
Thus, the ratio of the densities of the two forms is 108:125