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

Answer

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 lattice, the relationship between the edge length (a) of the unit cell and the radius (r) of the metal atom is given by the formula:
\[ r = \frac{a}{2\sqrt{2}} \]

### Step 2: Substitute the given edge length into the formula
Given that the edge length \( a = 408 \) pm, we can substitute this value into the formula:
\[ r = \frac{408 \, \text{pm}}{2\sqrt{2}} \]

### Step 3: Calculate the radius
Now, we will calculate the radius:
1. Calculate \( 2\sqrt{2} \):
\[ 2\sqrt{2} \approx 2 \times 1.414 = 2.828 \]
2. Now divide the edge length by this value:
\[ r \approx \frac{408 \, \text{pm}}{2.828} \approx 144.22 \, \text{pm} \]

### Step 4: Calculate the diameter
The diameter (d) of the metal atom is twice the radius:
\[ d = 2r \]
Substituting the value of \( r \):
\[ d = 2 \times 144.22 \, \text{pm} \approx 288.44 \, \text{pm} \]

### Step 5: Round the answer
Rounding \( 288.44 \, \text{pm} \) gives us approximately \( 288 \, \text{pm} \).

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

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