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

Options

A. 144 pm
B. 353 pm
C. 288 pm
D. 408 pm

Explanation

To find the diameter of a metal atom that crystallizes in a body-centered cubic (BCC) lattice with a given edge length, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship in BCC lattice**:
In a body-centered cubic lattice, the relationship between the radius of the atom (r) and the edge length of the unit cell (a) is given by the formula:
\[ r = \frac{a \sqrt{3}}{4} \]

2. **Substitute the given edge length**:
We are given that the edge length \( a = 408 \) pm (picometers). We can substitute this value into the formula:
\[ r = \frac{408 \, \text{pm} \times \sqrt{3}}{4} \]

3. **Calculate the radius**:
First, we need to calculate \( \sqrt{3} \), which is approximately \( 1.732 \):
\[ r = \frac{408 \times 1.732}{4} \]
Now, calculate the numerator:
\[ 408 \times 1.732 \approx 707.856 \]
Now divide by 4:
\[ r \approx \frac{707.856}{4} \approx 176.964 \, \text{pm} \]

4. **Calculate the diameter**:
The diameter (D) of the atom is twice the radius:
\[ D = 2r = 2 \times 176.964 \, \text{pm} \approx 353.928 \, \text{pm} \]

5. **Round off the final answer**:
Rounding to three significant figures, we get:
\[ D \approx 354 \, \text{pm} \]

### Final Answer:
The diameter of the metal atom is approximately **354 pm**.