The metal M crystallizes in a body centered lattice with cell edge. 400 pm. The atomic radius of M is

Options

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

Answer

To find the atomic radius of the metal M that crystallizes in a body-centered cubic (BCC) lattice with a cell edge of 400 pm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Body-Centered Cubic Structure**:
In a body-centered cubic lattice, there is one atom located at each of the eight corners of the cube and one atom at the center of the cube.

2. **Determine the Relationship Between Atomic Radius and Edge Length**:
In a BCC lattice, the body diagonal of the cube can be expressed in terms of the atomic radius (r) and the edge length (a). The body diagonal can be calculated using the formula:
\[ \text{Body Diagonal} = \sqrt{3}a \]
In a BCC structure, the body diagonal also equals the distance across the body center atom and two corner atoms, which is:
\[ \text{Body Diagonal} = 4r \]

3. **Set Up the Equation**:
Equating the two expressions for the body diagonal gives:
\[ \sqrt{3}a = 4r \]

4. **Rearranging the Equation to Find r**:
We can rearrange this equation to find the atomic radius (r):
\[ r = \frac{\sqrt{3}a}{4} \]

5. **Substituting the Given Edge Length**:
Now, substitute the given edge length (a = 400 pm) into the equation:
\[ r = \frac{\sqrt{3} \times 400 \, \text{pm}}{4} \]

6. **Calculating the Atomic Radius**:
First, calculate \(\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \]
Now substitute this value:
\[ r = \frac{1.732 \times 400}{4} \]
\[ r = \frac{692.8}{4} = 173.2 \, \text{pm} \]

7. **Final Result**:
The atomic radius of the metal M is approximately:
\[ r \approx 173 \, \text{pm} \]

### Final Answer:
The atomic radius of metal M is approximately **173 pm**.