What is the radius of a metal atom if it crystallizes with body-centered lattice having a unit cell edge of 333 Pico meter?

Options

A. 1538.06 pm
B. 769.03 pm
C. 288.38 pm
D. 144.19 pm

Answer

To find the radius of a metal atom that crystallizes in a body-centered cubic (BCC) lattice with a given unit cell edge length, we can use the relationship between the edge length (a) and the radius (R) of the atom.

### Step-by-Step Solution:

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

2. **Substitute the given edge length:**
The problem states that the edge length \( a \) is 333 picometers. We will substitute this value into the formula:
\[ R = \frac{\sqrt{3}}{4} \times 333 \, \text{pm} \]

3. **Calculate the value:**
First, calculate \( \sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \]
Now substitute this value into the equation:
\[ R = \frac{1.732}{4} \times 333 \, \text{pm} \]
\[ R = 0.433 \times 333 \, \text{pm} \]
\[ R \approx 144.19 \, \text{pm} \]

4. **Conclusion:**
The radius of the metal atom is approximately 144.19 picometers.

### Final Answer:
The radius of the metal atom is **144.19 picometers**.