Question

Copper has fcc structure with edge length 495 pm. What is the radius of copper atom in pm?

Hint:

Remember relations for cubic systems:
Simple Cubic: \( a = 2r \)
BCC: \( a = \frac{4r}{\sqrt{3}} \)
FCC: \( a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \)

Answer 1

Step 1: Understanding the Concept:
In a Face-Centered Cubic (fcc) unit cell, atoms touch each other along the face diagonal.
Step 2: Key Formula or Approach:
The relationship between atomic radius (\(r\)) and edge length (\(a\)) for an fcc structure is:
\[ 4r = \sqrt{2}a \implies r = \frac{\sqrt{2}a}{4} = \frac{a}{2\sqrt{2}} \]
Step 3: Detailed Explanation:
Given: \(a = 495 \text{ pm}\)
Using the formula:
\[ r = \frac{1.414 \times 495}{4} \]
\[ r = \frac{699.93}{4} \]
\[ r \approx 174.98 \text{ pm} \approx 175 \text{ pm} \]
Step 4: Final Answer:
The radius of the copper atom is \(175 \text{ pm}\).