A metal crystallizes with a body-centred cubic lattice. The length of the unit cell edge is found to be `265`pm. Calculate the atomic radius.

Explanation

To calculate the atomic radius of a metal that crystallizes in a body-centered cubic (BCC) lattice with a unit cell edge length of 265 pm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
- In a body-centered cubic (BCC) lattice, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube.

2. **Understand the Body Diagonal**:
- The body diagonal of the cube runs from one corner of the cube to the opposite corner. In a BCC structure, this diagonal contains three atomic radii (R) from the corner atom, the center atom, and the other corner atom.

3. **Calculate the Length of the Body Diagonal**:
- The length of the body diagonal (d) can be expressed in terms of the edge length (a) of the cube:
\[ d = a\sqrt{3} \]
- Here, \( a = 265 \) pm.

4. **Relate the Body Diagonal to Atomic Radii**:
- The body diagonal is equal to the sum of the atomic radii along the diagonal:
\[ d = 4R \]
- This is because the body diagonal consists of one radius from each of the two corner atoms and one radius from the center atom, totaling four radii.

5. **Set Up the Equation**:
- Combine the two equations:
\[ 4R = a\sqrt{3} \]

6. **Solve for the Atomic Radius (R)**:
- Rearranging the equation gives:
\[ R = \frac{a\sqrt{3}}{4} \]
- Substitute the value of \( a \):
\[ R = \frac{265 \times \sqrt{3}}{4} \]

7. **Calculate the Value**:
- First, calculate \( \sqrt{3} \approx 1.732 \):
\[ R = \frac{265 \times 1.732}{4} \approx \frac{459.58}{4} \approx 114.895 \text{ pm} \]

8. **Final Result**:
- The atomic radius \( R \) is approximately \( 114.895 \) pm.

### Final Answer:
The atomic radius of the metal is approximately **114.895 pm**.