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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? |
\(\frac{125}{54}\) \(\frac{54}{125}\) \(\frac{125}{108}\) \(\frac{108}{125}\) |
\(\frac{108}{125}\) |
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 \( a_{\text{fcc}} = 3.6 \, \text{Å} = 3.6 \times 10^{-8} \, \text{cm} \). 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: \(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{2 \cdot (3.6 \times 10^{-8})^3}{4 \cdot (3.0 \times 10^{-8})^3}\) \(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{2}{4} \times \frac{(3.6)^3}{(3.0)^3}\) \(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{1}{2} \times \frac{(36)^3}{(30)^3}\) \(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{1}{2} \times \frac{(6)^3}{(5)^3}\) \(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{1}{2} \times \frac{216}{125}\) \(⇒ \frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{108}{125}\) Thus, the ratio of the densities of the two forms is 108:125 |