Question

What is the packing efficiency of silver metal in its unit cell?

• A. 52.4%
• B. 68.0%
• C. 32.0%
• D. 74.0%

Hint:

Logic Tip: You should memorize the packing efficiencies of the three main cubic lattices for quick recall: Simple Cubic (SC) = 52.4%, Body-Centered Cubic (BCC) = 68.0%, and Face-Centered Cubic (FCC/CCP) = 74.0%. Since silver forms an FCC lattice, the answer is immediately 74.0%.

Answer 1

The Correct Option is D

Concept:
Silver (Ag) crystallizes in a face-centered cubic (fcc) or cubic close-packed (ccp) lattice structure. The packing efficiency represents the percentage of total space in the unit cell that is occupied by atoms. For any fcc/ccp lattice, this value is a standard constant derived from the geometry of the unit cell.
Step 1: Identify the properties of an fcc unit cell.
In a face-centered cubic (fcc) unit cell:

• Atoms are located at all 8 corners and at the center of all 6 faces.
• The total number of effective atoms per unit cell ($Z$) is: $$Z = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \text{ atoms}$$
• The relationship between the edge length ($a$) and the atomic radius ($r$) is given by $a = 2\sqrt{2}r$.

Step 2: Calculate the packing efficiency.
The formula for packing efficiency is: $$\text{Packing Efficiency} = \frac{\text{Volume occupied by atoms in a unit cell{\text{Total volume of the unit cell \times 100$$ Substitute the respective formulas: $$\text{Efficiency} = \frac{4 \times \frac{4}{3}\pi r^3}{a^3} \times 100$$ Substitute $a = 2\sqrt{2}r$: $$\text{Efficiency} = \frac{\frac{16}{3}\pi r^3}{(2\sqrt{2}r)^3} \times 100$$ $$\text{Efficiency} = \frac{\frac{16}{3}\pi r^3}{16\sqrt{2} r^3} \times 100 = \frac{\pi}{3\sqrt{2 \times 100$$ $$\text{Efficiency} \approx \frac{3.14159}{4.2426} \times 100 \approx 74.0%$$