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) lattice, the atoms touch each other along the face diagonal.
Step 2: Key Formula or Approach:
For an fcc unit cell:
\[ 4r = \sqrt{2}a \]
or
\[ r = \frac{a}{2\sqrt{2}} \]
where \( r \) is the radius of the atom and \( a \) is the edge length.
Step 3: Detailed Explanation:
Given: \( a = 495 \text{ pm} \)
Using the formula:
\[ r = \frac{495}{2 \times 1.4142} \]
\[ r = \frac{495}{2.8284} \]
\[ r \approx 175.01 \text{ pm} \]
Step 4: Final Answer:
The radius of the copper atom is approximately 175 pm.