In a metal M having bcc arrangement edge length of the unit cell is 400 pm . The atomic radius of the metal is

Options

A. `173` pm
B. `100` pm
C. `141` pm
D. `200` pm

Answer

To find the atomic radius of a metal M that has a body-centered cubic (BCC) arrangement with a given edge length of the unit cell, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the BCC Structure**:
In a body-centered cubic (BCC) structure, there is one atom at each corner of the cube and one atom at the center of the cube.

2. **Identify the Relationship**:
The relationship between the atomic radius (R) and the edge length (A) of a BCC unit cell is given by the formula:
\[ R = \frac{\sqrt{3}}{4} A \]
This formula arises from the geometry of the BCC unit cell where the body diagonal can be expressed in terms of the atomic radius.

3. **Substitute the Given Edge Length**:
We are given that the edge length (A) of the unit cell is 400 pm (picometers). Substitute this value into the formula:
\[ R = \frac{\sqrt{3}}{4} \times 400 \text{ pm} \]

4. **Calculate the Value**:
First, calculate \(\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \]
Now, substitute this value into the equation:
\[ R = \frac{1.732}{4} \times 400 \text{ pm} \]
\[ R = \frac{1.732 \times 400}{4} \text{ pm} \]
\[ R = \frac{692.8}{4} \text{ pm} \]
\[ R \approx 173.2 \text{ pm} \]

5. **Final Answer**:
Rounding to the nearest whole number, the atomic radius of the metal M is approximately:
\[ R \approx 173 \text{ pm} \]

### Conclusion:
The atomic radius of the metal M with a BCC arrangement and an edge length of 400 pm is **173 pm**.