The relation between atomic radius ( r ) and edge length ( a ) a face - centred cubic cell is

Options

A. `r = a/2`
B. `r = a/(2sqrt2)`
C. `r = (a sqrt3)/4`
D. `(sqrt2a)/2`

Answer

To derive the relation between atomic radius (r) and edge length (a) of a face-centered cubic (FCC) cell, we can follow these steps:

### Step 1: Understand the Structure of FCC
In a face-centered cubic cell, atoms are located at each corner of the cube and at the center of each face. This means there are 8 corner atoms and 6 face atoms.

### Step 2: Identify the Face Diagonal
The face diagonal of the cube connects two opposite corners of a face. If we denote the edge length of the cube as 'a', the length of the face diagonal can be calculated using the Pythagorean theorem:
\[ \text{Face Diagonal} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

### Step 3: Relate the Face Diagonal to Atomic Radius
In an FCC structure, the face diagonal is also equal to the sum of the diameters of the atoms that touch along that diagonal. Since there are two atomic radii (r) from each of the two atoms at the corners of the face, the relationship can be expressed as:
\[ \text{Face Diagonal} = 4r \]
Thus, we can set up the equation:
\[ a\sqrt{2} = 4r \]

### Step 4: Solve for Atomic Radius (r)
Now, we can solve for the atomic radius (r):
\[ 4r = a\sqrt{2} \]
Dividing both sides by 4 gives:
\[ r = \frac{a\sqrt{2}}{4} \]

### Step 5: Simplify the Expression
To simplify further, we can express 4 as \(2 \times 2\):
\[ r = \frac{a\sqrt{2}}{2 \times 2} = \frac{a}{2\sqrt{2}} \]

### Final Relation
Thus, the final relation between atomic radius (r) and edge length (a) for a face-centered cubic cell is:
\[ r = \frac{a}{2\sqrt{2}} \]