An element has a body centered cubic (bcc) structure with a cell edge of 288pm. The atomic radius is:

Options

A. `sqrt3/4 xx 288pm`
B. `sqrt2/4 xx 288pm`
C. `4/sqrt3 xx 288pm`
D. `4/sqrt2 xx 288pm`

Answer

To find the atomic radius of an element with a body-centered cubic (BCC) structure given the cell edge, we can follow these steps:

### Step 1: Understand the BCC Structure
In a body-centered cubic (BCC) structure, there are atoms located at each corner of the cube and one atom at the center of the cube. The body diagonal of the cube contains the center atom and two corner atoms.

### Step 2: Relate the Cell Edge to Atomic Radius
The relationship between the edge length (A) of the BCC unit cell and the atomic radius (R) is given by the formula:
\[ A\sqrt{3} = 4R \]
This equation arises because the body diagonal of the cube (which is the longest diagonal) can be expressed in terms of the atomic radius.

### Step 3: Substitute the Given Value
We are given the edge length (A) of the BCC unit cell as 288 pm (picometers). We can substitute this value into the equation:
\[ 288\sqrt{3} = 4R \]

### Step 4: Solve for the Atomic Radius (R)
To find R, we rearrange the equation:
\[ R = \frac{288\sqrt{3}}{4} \]

Now, simplify:
\[ R = 72\sqrt{3} \]

### Step 5: Calculate the Numerical Value
To find the numerical value of R, we can calculate:
\[ \sqrt{3} \approx 1.732 \]
Thus,
\[ R \approx 72 \times 1.732 \]
\[ R \approx 124.704 \, \text{pm} \]

### Final Answer
The atomic radius (R) is approximately:
\[ R \approx 124.7 \, \text{pm} \]