Question

A metal crystallises in two cubic phases, fcc and bcc with edge lengths 3.5 Å and 3 Å respectively. The ratio of densities of fcc and bcc is approximately

• A. 1.36
• B. 1.26
• C. 2.16
• D. 6.13

Hint:

Memorize the number of atoms per unit cell (Z) for common cubic structures: - Simple Cubic (sc): Z = 1 - Body-Centered Cubic (bcc): Z = 2 - Face-Centered Cubic (fcc): Z = 4 This is fundamental for any calculation involving density or packing efficiency.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The density of a crystal is determined by the mass of the atoms in its unit cell and the volume of the unit cell. Different crystal structures (like face-centered cubic, fcc, and body-centered cubic, bcc) have different numbers of atoms per unit cell and can have different unit cell volumes.
Step 2: Key Formula or Approach:
The formula for the density (\(\rho\)) of a crystal is: \[ \rho = \frac{Z \times M}{N_A \times a^3} \] where: - \(Z\) = number of atoms per unit cell - \(M\) = molar mass of the element - \(N_A\) = Avogadro's number - \(a\) = edge length of the unit cell To find the ratio of densities \(\frac{\rho_{fcc}}{\rho_{bcc}}\), we can set up the ratio of their formulas. The terms \(M\) and \(N_A\) will cancel out as we are dealing with the same metal.
\[ \frac{\rho_{fcc}}{\rho_{bcc}} = \frac{Z_{fcc} / a_{fcc}^3}{Z_{bcc} / a_{bcc}^3} = \left(\frac{Z_{fcc}}{Z_{bcc}}\right) \times \left(\frac{a_{bcc}}{a_{fcc}}\right)^3 \] Step 3: Detailed Explanation:
First, identify the values for each structure:
- For fcc structure: - Number of atoms per unit cell, \(Z_{fcc} = 4\). - Edge length, \(a_{fcc} = 3.5\) Å. - For bcc structure: - Number of atoms per unit cell, \(Z_{bcc} = 2\). - Edge length, \(a_{bcc} = 3.0\) Å. Now, substitute these values into the ratio formula:
\[ \frac{\rho_{fcc}}{\rho_{bcc}} = \left(\frac{4}{2}\right) \times \left(\frac{3.0 \text{ Å}}{3.5 \text{ Å}}\right)^3 \] \[ \frac{\rho_{fcc}}{\rho_{bcc}} = 2 \times \left(\frac{3.0}{3.5}\right)^3 = 2 \times \left(\frac{6}{7}\right)^3 \] \[ \frac{\rho_{fcc}}{\rho_{bcc}} = 2 \times \frac{6^3}{7^3} = 2 \times \frac{216}{343} = \frac{432}{343} \] Now, perform the division:
\[ \frac{432}{343} \approx 1.2594... \] This value is approximately 1.26.
Step 4: Final Answer:
The ratio of the densities of fcc and bcc is approximately 1.26. Therefore, option (B) is correct.