If 'a' be the edge length of the unit cell and r be the radius of an atom, then for face centred cubic lattice, the correct relation is

Options

A. `4r = sqrt(3)a`
B. `4r = sqrt(2)a`
C. `4a = sqrt(3)r`
D. None of the above

Answer

To find the relationship between the edge length 'a' of a face-centered cubic (FCC) unit cell and the radius 'r' of an atom, we can follow these steps:

### Step 1: Understand the FCC Structure
In a face-centered cubic (FCC) lattice, atoms are located at each of the corners of the cube and at the center of each face.

### Step 2: Identify the Face Diagonal
In an FCC unit cell, the atoms touch along the face diagonal. To find the relationship between 'a' and 'r', we need to calculate the length of the face diagonal.

### Step 3: Calculate the Length of the Face Diagonal
For a square face of the cube, 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 4: Relate the Face Diagonal to Atomic Radii
In the FCC structure, the face diagonal consists of the radius of one atom at the face center and the radii of two corner atoms. Therefore, the face diagonal can be expressed in terms of the atomic radius 'r':

\[ \text{Face diagonal} = 4r \]

### Step 5: Set the Two Expressions Equal
Now we can set the two expressions for the face diagonal equal to each other:

\[ a\sqrt{2} = 4r \]

### Step 6: Solve for the Relationship
To find the relationship between 'a' and 'r', we can rearrange the equation:

\[ a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \]

### Final Relation
Thus, the correct relation between the edge length 'a' of the unit cell and the radius 'r' of an atom in a face-centered cubic lattice is:

\[ a = 2\sqrt{2}r \]

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