The metal M crystallizes in a body cantered lattice with cell edge 40 pm . The atomic radius of M is .

Answer

To solve the problem, we need to find the atomic radius of the metal M that crystallizes in a body-centered cubic (BCC) lattice with a given cell edge length. Let's break down the solution step by step.

### Step-by-Step Solution:

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

**Step 2: Identify the Relationship Between Atomic Radius and Cell Edge**
- The body diagonal of the cube can be expressed in terms of the cell edge length (a). The formula for the body diagonal (d) in a cubic structure is given by:
\[ d = \sqrt{3} \cdot a \]

**Step 3: Relate the Body Diagonal to Atomic Radius**
- In a BCC structure, the body diagonal contains 3 atomic radii:
- 1 radius from the center atom and 2 radii from the corner atoms.
- Therefore, the relationship can be expressed as:
\[ d = 4r \]
where \( r \) is the atomic radius.

**Step 4: Set Up the Equation**
- By substituting the expression for the body diagonal into the equation, we have:
\[ \sqrt{3} \cdot a = 4r \]

**Step 5: Solve for the Atomic Radius**
- Rearranging the equation to solve for \( r \):
\[ r = \frac{\sqrt{3} \cdot a}{4} \]

**Step 6: Substitute the Given Cell Edge Length**
- Given that the cell edge length \( a = 40 \, \text{pm} \), we substitute this value into the equation:
\[ r = \frac{\sqrt{3} \cdot 40 \, \text{pm}}{4} \]

**Step 7: Simplify the Expression**
- Simplifying the expression:
\[ r = \frac{\sqrt{3} \cdot 40}{4} = \sqrt{3} \cdot 10 \, \text{pm} \]

**Step 8: Calculate the Numerical Value**
- Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\[ r \approx 1.732 \cdot 10 \, \text{pm} \approx 17.32 \, \text{pm} \]

### Final Answer:
The atomic radius \( r \) of metal M is approximately \( 17.32 \, \text{pm} \).

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