The radius of an atom is 100 pm. If this element crystallizes in FCC lattice, the edge length of unit cell is

Options

A. 280 pm
B. 150 pm
C. 141.4 pm
D. none of the above

Answer

To find the edge length of the unit cell for an atom with a radius of 100 pm that crystallizes in a Face-Centered Cubic (FCC) lattice, we can follow these steps:

### Step 1: Understand the relationship in FCC
In an FCC lattice, the relationship between the atomic radius (r) and the edge length (a) of the unit cell is given by the formula:
\[ 4r = \sqrt{2} \cdot a \]

### Step 2: Substitute the known values
Given that the radius \( r = 100 \) pm, we can substitute this value into the formula:
\[ 4 \cdot 100 \, \text{pm} = \sqrt{2} \cdot a \]

### Step 3: Calculate \( 4r \)
Calculating \( 4r \):
\[ 4 \cdot 100 \, \text{pm} = 400 \, \text{pm} \]

### Step 4: Rearrange the formula to solve for \( a \)
Now, we can rearrange the equation to solve for \( a \):
\[ a = \frac{4r}{\sqrt{2}} \]

### Step 5: Substitute \( 4r \) into the equation
Substituting \( 400 \, \text{pm} \) into the equation:
\[ a = \frac{400 \, \text{pm}}{\sqrt{2}} \]

### Step 6: Calculate \( a \)
To calculate \( a \), we need to divide \( 400 \, \text{pm} \) by \( \sqrt{2} \):
\[ a = \frac{400}{1.414} \, \text{pm} \]
\[ a \approx 282.84 \, \text{pm} \]

### Conclusion
Thus, the edge length of the unit cell is approximately:
\[ a \approx 282.84 \, \text{pm} \]

### Final Answer
The edge length of the unit cell is approximately **282 pm**.

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