The radius of an atom of an element is 600 pm. If it crystallizes as a face centred cubic lattice, what is the length of the side of the unit cell?

Options

A. a = 1696.8 pm
B. a = 16.9 pm
C. a =164 pm
D. a =1685.3 pm

Explanation

To find the length of the side of the unit cell for an atom that crystallizes in a face-centered cubic (FCC) lattice, we can follow these steps:

### Step 1: Understand the FCC structure
In a face-centered cubic lattice, atoms are located at each corner of the cube and at the center of each face. The effective distance between the centers of the atoms that touch along the face diagonal is crucial for calculating the edge length.

### Step 2: Use the relationship between radius and edge length
For an FCC lattice, the relationship between the atomic radius \( r \) and the edge length \( a \) of the unit cell is given by the formula:
\[ 4r = \sqrt{2}a \]
This equation arises because the face diagonal of the cube is equal to four times the radius of the atom (since there are two radii from the corner atom to the face-centered atom and two from the face-centered atom to the opposite corner atom).

### Step 3: Rearrange the formula to find \( a \)
We can rearrange the formula to solve for the edge length \( a \):
\[ a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \]

### Step 4: Substitute the given radius
The radius of the atom is given as \( 600 \, \text{pm} \). Now, we substitute this value into the equation:
\[ a = 2\sqrt{2} \times 600 \, \text{pm} \]

### Step 5: Calculate \( a \)
Now we calculate \( a \):
\[ \sqrt{2} \approx 1.414 \]
So,
\[ a = 2 \times 1.414 \times 600 \, \text{pm} \]
\[ a = 2.828 \times 600 \, \text{pm} \]
\[ a \approx 1696.8 \, \text{pm} \]

### Final Answer
The length of the side of the unit cell is approximately \( 1696.8 \, \text{pm} \).

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