Xenon crystallises in face - centered cubic , and the edge of the unit cell is 620 pm .The radius of a xenon atom is

Options

A. `219.20` pm
B. `438.5` pm
C. `290.3` pm
D. `318.53` pm

Explanation

To find the radius of a xenon atom that crystallizes in a face-centered cubic (FCC) structure with an edge length of 620 pm, we can follow these steps:

### Step 1: Understand the FCC structure
In a face-centered cubic (FCC) lattice, atoms are located at each corner of the cube and at the center of each face. The relationship between the edge length (a) of the cube and the radius (r) of the atom can be derived from the geometry of the cube.

### Step 2: Use the formula for FCC
For an FCC structure, the formula that relates the edge length (a) to the radius (r) of the atom is:
\[ 4r = \sqrt{2} a \]
This formula arises because the face diagonal of the cube is equal to four times the radius of the atom.

### Step 3: Rearrange the formula to find the radius
To find the radius (r), we can rearrange the formula:
\[ r = \frac{\sqrt{2}}{4} a \]

### Step 4: Substitute the given edge length
Now, substitute the given edge length (a = 620 pm) into the rearranged formula:
\[ r = \frac{\sqrt{2}}{4} \times 620 \, \text{pm} \]

### Step 5: Calculate the radius
Now, calculate the value:
1. Calculate \(\sqrt{2} \approx 1.414\).
2. Substitute this value into the equation:
\[ r = \frac{1.414}{4} \times 620 \]
\[ r = 0.3535 \times 620 \]
\[ r \approx 219.22 \, \text{pm} \]

### Step 6: Final answer
Thus, the radius of the xenon atom is approximately:
\[ r \approx 219.20 \, \text{pm} \]

### Summary
The radius of a xenon atom that crystallizes in a face-centered cubic structure with an edge length of 620 pm is approximately 219.20 pm.

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