Lithium forms body centred cube structure .The length of the side of its unirt cell is 310 pm Atomic radius of the lithium will be

Options

A. `151.8 "pm"`
B. `75.6 "pm"`
C. `134.50 "pm"`
D. `240.8 "pm"`

Explanation

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

### Step 1: Understand the BCC Structure
In a body-centered cubic (BCC) structure, atoms are located at each of the eight corners of a cube and one atom is located at the center of the cube.

### Step 2: Identify the Given Information
We are given:
- The edge length of the unit cell (A) = 310 pm (picometers).

### Step 3: Relate the Atomic Radius to the Body Diagonal
In a BCC unit cell, the body diagonal can be expressed in terms of the edge length (A):
- The body diagonal (d) = \( A \sqrt{3} \)

The body diagonal also consists of the atomic radii:
- The body diagonal can be expressed as: \( d = 4R \)
Where R is the atomic radius.

### Step 4: Set Up the Equation
From the above relationships, we can set up the equation:
\[ 4R = A \sqrt{3} \]

### Step 5: Solve for the Atomic Radius (R)
Substituting the value of A into the equation:
\[ R = \frac{A \sqrt{3}}{4} \]
\[ R = \frac{310 \, \text{pm} \times \sqrt{3}}{4} \]

### Step 6: Calculate the Value
Now, calculate the value of R:
- First, calculate \( \sqrt{3} \approx 1.732 \).
- Then substitute:
\[ R = \frac{310 \times 1.732}{4} \]
\[ R = \frac{537.92}{4} \]
\[ R \approx 134.48 \, \text{pm} \]

### Step 7: Conclusion
The atomic radius of lithium in a body-centered cubic structure is approximately 134.48 pm.

### Final Answer
The atomic radius of lithium is approximately **134.50 pm**.

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