If the lattice parameter for a crystalline structure is `3.6Å`, then the atomic radius is fcc crystals is

Options

A. `1.81 Å`
B. `2.10Å`
C. `2.92 Å`
D. `1.27 Å`

Answer

To find the atomic radius of a face-centered cubic (FCC) crystal given the lattice parameter, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the Lattice Parameter**: The lattice parameter (denoted as `a`) is the length of the edge of the cubic unit cell. In this case, the lattice parameter is given as `3.6 Å`.

3. **Relate the Atomic Radius to the Lattice Parameter**: In an FCC structure, the relationship between the atomic radius (denoted as `R`) and the lattice parameter is given by the formula:
\[ \sqrt{2}a = 4R \]
This equation arises because the diagonal of the face of the cube spans across 4 atomic radii (2 radii from each of the two atoms at the ends of the diagonal).

4. **Rearrange the Formula**: To find the atomic radius `R`, we can rearrange the formula:
\[ R = \frac{\sqrt{2}a}{4} \]
or equivalently,
\[ R = \frac{a}{2\sqrt{2}} \]

5. **Substitute the Lattice Parameter**: Now, substitute the given value of the lattice parameter `a = 3.6 Å` into the formula:
\[ R = \frac{3.6 \, \text{Å}}{2\sqrt{2}} \]

6. **Calculate the Atomic Radius**: First, calculate the denominator:
\[ 2\sqrt{2} \approx 2 \times 1.414 \approx 2.828 \]
Now, divide:
\[ R \approx \frac{3.6}{2.828} \approx 1.27 \, \text{Å} \]

7. **Conclusion**: The atomic radius of the FCC crystal is approximately `1.27 Å`.

### Final Answer:
The atomic radius of the FCC crystal is `1.27 Å`.