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

Answer

To find the atomic radius of the metal M that crystallizes in a body-centered cubic (BCC) lattice with a cell edge of 40 pm, we can follow these steps:

### Step 1: Understand the BCC Structure
In a body-centered cubic lattice, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube. The arrangement can be visualized as follows:
- Corner atoms contribute 1/8 of an atom each (since each corner is shared by 8 neighboring cubes).
- The center atom contributes a full atom.

### Step 2: Relate Atomic Radius to Edge Length
For a body-centered cubic lattice, the relationship between the edge length (a) and the atomic radius (r) can be derived from the geometry of the cube. The body diagonal of the cube can be expressed in terms of the edge length and the atomic radius.

The body diagonal (d) of the cube can be calculated using the Pythagorean theorem:
\[ d = \sqrt{a^2 + a^2 + a^2} = \sqrt{3}a \]

In a BCC structure, the body diagonal is also equal to the diameter of the atom at the center and the two corner atoms:
\[ d = 4r \]

### Step 3: Set Up the Equation
From the two expressions for the body diagonal, we can set them equal to each other:
\[ \sqrt{3}a = 4r \]

### Step 4: Solve for Atomic Radius
Rearranging the equation to solve for the atomic radius (r):
\[ r = \frac{\sqrt{3}}{4}a \]

### Step 5: Substitute the Given Edge Length
Now we can substitute the given edge length (a = 40 pm) into the equation:
\[ r = \frac{\sqrt{3}}{4} \times 40 \, \text{pm} \]

### Step 6: Calculate the Value
Calculating the value:
1. First, calculate \(\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \]
2. Now substitute this value:
\[ r = \frac{1.732}{4} \times 40 \]
\[ r = 0.433 \times 40 \]
\[ r = 17.3 \, \text{pm} \]

### Final Answer
The atomic radius of metal M is approximately **17.3 pm**.

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