Lithium forms body centred cube structure .The length of the side of its unirt cell is 325 pm Atomic radius of the lithium will be

Options

A. 75 pm
B. 300 pm
C. 240 pm
D. 141 pm

Explanation

To find the atomic radius of lithium in a body-centered cubic (BCC) structure, we can follow these steps:

### Step-by-Step Solution:

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

2. **Understand the Body Diagonal**:
- In a BCC structure, the body diagonal of the cube contains three atomic radii (R). The formula for the length of the body diagonal (d) in terms of the edge length (a) is given by:
\[ d = a\sqrt{3} \]

3. **Relate the Body Diagonal to Atomic Radii**:
- The body diagonal can also be expressed in terms of the atomic radius (R):
\[ d = 4R \]
- This is because the body diagonal consists of one atom at the center and two corner atoms, contributing a total of 4 radii.

4. **Set the Equations Equal**:
- Since both expressions represent the body diagonal, we can set them equal to each other:
\[ a\sqrt{3} = 4R \]

5. **Solve for the Atomic Radius (R)**:
- Rearranging the equation gives us:
\[ R = \frac{a\sqrt{3}}{4} \]

6. **Substitute the Given Edge Length**:
- The problem states that the edge length (a) is 325 pm (picometers). We substitute this value into the equation:
\[ R = \frac{325 \times \sqrt{3}}{4} \]

7. **Calculate the Value**:
- First, calculate \(\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \]
- Now substitute this value into the equation:
\[ R = \frac{325 \times 1.732}{4} \]
- Calculate the numerator:
\[ 325 \times 1.732 \approx 563.9 \]
- Now divide by 4:
\[ R \approx \frac{563.9}{4} \approx 140.975 \text{ pm} \]

8. **Round the Result**:
- Rounding this value gives us approximately:
\[ R \approx 141 \text{ pm} \]

### Final Answer:
The atomic radius of lithium is approximately **141 pm**.

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