Question

Copper has FCC structure with a lattice constant 3.61 Å. The radius of the copper atom is:

• A. 1.28 Å
• B. 1.26 Å
• C. 1.23 Å
• D. 1.29 Å

Hint:

For FCC, remember the key relation \( 4r = \sqrt{2}a \). For BCC, it's \( 4r = \sqrt{3}a \). These are very common formulas in solid-state physics questions. Being able to derive or recall them quickly is essential.

Answer 1

The Correct Option is A

Step 1: Concept Overview:
Determine the atomic radius (r) of copper, given its lattice constant (a) and Face-Centered Cubic (FCC) structure.
Step 2: Formula and Method:
In an FCC lattice, atoms touch along the face diagonal. The face diagonal's length is \( \sqrt{2}a \), derived from \( \text{diagonal}^2 = a^2 + a^2 = 2a^2 \).
The face diagonal comprises 4 atomic radii (4r).
Therefore:
\[ 4r = \sqrt{2}a \]
\[ r = \frac{\sqrt{2}a}{4} \]
Step 3: Calculation:
Given:
Lattice constant, \( a = 3.61 \) Å
Applying the formula:
\[ r = \frac{\sqrt{2} \times 3.61 \text{ Å}}{4} \]
Using \( \sqrt{2} \approx 1.414 \):
\[ r \approx \frac{1.414 \times 3.61}{4} \text{ Å} \]
\[ r \approx \frac{5.10474}{4} \text{ Å} \]
\[ r \approx 1.276 \text{ Å} \]
Step 4: Solution:
The calculated atomic radius is approximately 1.276 Å, which is closest to 1.28 Å.