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 a unit cell in a face-centered cubic (FCC) lattice given the atomic radius, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- The radius of the atom (r) = 100 pm (picometers).

2. **Understand the relationship in FCC lattice:**
- In a face-centered cubic (FCC) lattice, the atoms touch along the face diagonal of the cube. The relationship between the edge length (a) of the unit cell and the atomic radius (r) is given by:
\[ 4r = \sqrt{2}a \]
- This equation arises because there are 4 atomic radii along the face diagonal of the cube.

3. **Rearranging the equation to find edge length (a):**
- From the equation \(4r = \sqrt{2}a\), we can solve for \(a\):
\[ a = \frac{4r}{\sqrt{2}} \]

4. **Substituting the value of r:**
- Substitute \(r = 100\) pm into the equation:
\[ a = \frac{4 \times 100 \, \text{pm}}{\sqrt{2}} \]

5. **Calculating the edge length:**
- Simplifying the equation:
\[ a = \frac{400 \, \text{pm}}{1.414} \quad (\text{since } \sqrt{2} \approx 1.414) \]
- Performing the division:
\[ a \approx 282.8 \, \text{pm} \]

6. **Final answer:**
- The edge length of the unit cell is approximately \(282.8\) pm.

### Final Result:
The edge length of the unit cell is \(282.8\) pm.

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