Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?

Options

A. 157
B. 181
C. 108
D. 128

Explanation

To find the radius of a copper atom in a face-centered cubic (FCC) structure, we can follow these steps:

### Step 1: Understand the FCC Structure
In an FCC unit cell, atoms are located at each corner of the cube and at the center of each face. Each corner atom contributes 1/8th of its volume to the unit cell, and each face-centered atom contributes 1/2.

### Step 2: Identify the Relationship Between Edge Length and Atomic Radius
In an FCC unit cell, the relationship between the edge length (a) and the atomic radius (r) can be derived from the face diagonal. The face diagonal of the cube can be expressed as:
\[ \text{Face diagonal} = \sqrt{2} \cdot a \]

### Step 3: Relate the Face Diagonal to Atomic Radius
The face diagonal also equals the sum of the diameters of the atoms along that diagonal. Since there are 4 atomic radii along the face diagonal (2 radii from each of the two face-centered atoms and 2 radii from the corner atoms), we can write:
\[ \text{Face diagonal} = 4r \]

### Step 4: Set Up the Equation
Equating the two expressions for the face diagonal gives us:
\[ \sqrt{2} \cdot a = 4r \]

### Step 5: Solve for the Atomic Radius
Rearranging the equation to solve for r, we get:
\[ r = \frac{\sqrt{2} \cdot a}{4} \]

### Step 6: Substitute the Given Edge Length
Given that the edge length \( a = 361 \, \text{pm} \):
\[ r = \frac{\sqrt{2} \cdot 361}{4} \]

### Step 7: Calculate the Radius
Now, we can compute the radius:
\[ r = \frac{1.414 \cdot 361}{4} \]
\[ r = \frac{510.074}{4} \]
\[ r \approx 127.52 \, \text{pm} \]

### Step 8: Round the Result
Rounding to three significant figures, we find:
\[ r \approx 128 \, \text{pm} \]

Thus, the radius of the copper atom is approximately **128 pm**.

---