Lithium crystallizes as body centered cubic crystals. If the length of the side of unit cell is 350=pm, the atomic radius of lithium is:

Options

A. 303.1pm
B. 606.2pm
C. 151.5pm
D. 123.7pm

Answer

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) unit cell, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube.

### Step 2: Identify the Relationship Between the Atomic Radius and the Unit Cell Length
In a BCC structure, the relationship between the atomic radius (R) and the edge length of the unit cell (A) can be described using the body diagonal of the cube. The body diagonal of the cube can be calculated as:
\[ \text{Body Diagonal} = \sqrt{3} \times A \]

### Step 3: Determine the Contact Condition
In the BCC structure, the body diagonal is equal to the sum of the diameters of the atoms along that diagonal. The body diagonal consists of one atom at the center and two corner atoms. Therefore, the relationship can be expressed as:
\[ \text{Body Diagonal} = 4R \]

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

### Step 5: Solve for the Atomic Radius (R)
Rearranging the equation to solve for R gives us:
\[ R = \frac{\sqrt{3} \times A}{4} \]

### Step 6: Substitute the Given Value of A
Given that the length of the side of the unit cell (A) is 350 pm, we can substitute this value into the equation:
\[ R = \frac{\sqrt{3} \times 350 \, \text{pm}}{4} \]

### Step 7: Calculate the Value
Now, we can calculate R:
1. Calculate \(\sqrt{3} \approx 1.732\).
2. Multiply by 350 pm:
\[ 1.732 \times 350 \approx 605.2 \, \text{pm} \]
3. Divide by 4:
\[ R \approx \frac{605.2}{4} \approx 151.3 \, \text{pm} \]

### Final Answer
Thus, the atomic radius of lithium is approximately **151.3 pm**.

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