Lithium forms body centred cubic structure .The length of the side of its unit cells is 351 pm.A tomic radius of lithium will be :

Options

A. 75 pm
B. 300 pm
C. 240 pm
D. 152 pm

Answer

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

### Step-by-Step Solution:

1. **Understand the BCC Structure**: In a BCC unit cell, there are atoms located at the eight corners of the cube and one atom at the center of the cube.

2. **Identify the Given Data**: The length of the side of the unit cell (denoted as 'a') is given as 351 picometers (pm).

3. **Relate the Atomic Radius to the Unit Cell**: In a BCC structure, the relationship between the atomic radius (r) and the unit cell length (a) can be derived from the body diagonal of the cube. The body diagonal connects two opposite corners of the cube and passes through the center atom.

4. **Use the Formula for Body Diagonal**: The body diagonal (d) can be expressed in terms of the unit cell length:
\[ d = \sqrt{3} \cdot a \]
In a BCC structure, the body diagonal can also be expressed in terms of the atomic radius:
\[ d = 4r \]

5. **Set the Two Expressions for the Body Diagonal Equal**:
\[ 4r = \sqrt{3} \cdot a \]

6. **Solve for the Atomic Radius (r)**:
\[ r = \frac{\sqrt{3} \cdot a}{4} \]

7. **Substitute the Value of 'a'**:
\[ r = \frac{\sqrt{3} \cdot 351 \, \text{pm}}{4} \]

8. **Calculate the Value**:
- First, calculate \(\sqrt{3} \approx 1.732\).
- Then, calculate:
\[ r = \frac{1.732 \cdot 351}{4} \approx \frac{607.092}{4} \approx 151.773 \, \text{pm} \]

9. **Round the Result**: The atomic radius of lithium is approximately 152 pm.

### Final Answer:
The atomic radius of lithium is approximately **152 picometers (pm)**.