a metal crystallizes with a face-centered cubic lattice.The edge of the unit cell is `408` pm. The diameter of the metal atom is :

Options

A. 288 pm
B. 408 pm
C. 144 pm
D. 204 pm

Explanation

To find the diameter of the metal atom in a face-centered cubic (FCC) lattice, we can follow these steps:

### Step 1: Understand the relationship between edge length and atomic radius in FCC
In a face-centered cubic (FCC) unit cell, the relationship between the edge length (a) and the radius (r) of the atom is given by the formula:
\[ r = \frac{a}{2\sqrt{2}} \]

### Step 2: Given values
We are given:
- Edge length (a) = 408 pm

### Step 3: Calculate the radius of the atom
Using the formula from Step 1, we can substitute the value of a:
\[ r = \frac{408 \, \text{pm}}{2\sqrt{2}} \]

### Step 4: Calculate \( \sqrt{2} \)
We know that:
\[ \sqrt{2} \approx 1.414 \]

### Step 5: Substitute \( \sqrt{2} \) into the equation
Now we can calculate:
\[ r = \frac{408 \, \text{pm}}{2 \times 1.414} \]
\[ r = \frac{408 \, \text{pm}}{2.828} \]

### Step 6: Perform the division
Calculating the above gives:
\[ r \approx 144.22 \, \text{pm} \]

### Step 7: Calculate the diameter of the atom
The diameter (d) of the atom is twice the radius:
\[ d = 2r = 2 \times 144.22 \, \text{pm} \]
\[ d \approx 288.44 \, \text{pm} \]

### Step 8: Round the answer
Rounding to three significant figures, we get:
\[ d \approx 288 \, \text{pm} \]

### Final Answer
The diameter of the metal atom is approximately **288 pm**.

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