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: Understanding the Concept:
The problem requires calculating the atomic radius (r) of copper from its lattice constant (a), given that it has a Face-Centered Cubic (FCC) crystal structure.
Step 2: Key Formula or Approach:
In an FCC structure, atoms are in contact along the face diagonal of the cubic unit cell.
The length of the face diagonal is related to the lattice constant 'a' by the Pythagorean theorem: \( \text{diagonal}^2 = a^2 + a^2 = 2a^2 \), so the diagonal length is \( \sqrt{2}a \).
This face diagonal contains the radius of one corner atom, the diameter (2r) of the face-centered atom, and the radius of the opposite corner atom. Thus, the total length along the diagonal is 4r.
The relationship is:
\[ 4r = \sqrt{2}a \]
\[ r = \frac{\sqrt{2}a}{4} \]
Step 3: Detailed Explanation:
Given:
Lattice constant, \( a = 3.61 \) Å
We use the formula derived in Step 2:
\[ r = \frac{\sqrt{2} \times 3.61 \text{ Å}}{4} \]
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: Final Answer:
The calculated value, 1.276 Å, is closest to the option 1.28 Å.