Platinum crystallizes in fcc crystal with a unit cell length `a`. The atomic radius of platinum is therefore `asqrt2//2`.

Explanation

To solve the problem, we need to determine the atomic radius of platinum (Pt) which crystallizes in a face-centered cubic (FCC) structure with a unit cell length \( a \).

### Step-by-Step Solution:

1. **Understanding the FCC Structure**:
In a face-centered cubic (FCC) lattice, atoms are located at each of the corners of the cube and at the centers of each of the faces.

2. **Identifying the Face Diagonal**:
The face diagonal of the cube can be expressed in terms of the unit cell length \( a \). The face diagonal \( d \) can be calculated using the Pythagorean theorem:
\[ d = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

3. **Relating Atomic Radius to the Face Diagonal**:
In an FCC structure, the face diagonal is composed of 4 atomic radii (since there are atoms at both ends of the diagonal and one in the center):
\[ d = 4r \]
where \( r \) is the atomic radius.

4. **Setting Up the Equation**:
From the previous steps, we can set up the equation:
\[ 4r = a\sqrt{2} \]

5. **Solving for the Atomic Radius \( r \)**:
To find \( r \), we rearrange the equation:
\[ r = \frac{a\sqrt{2}}{4} \]
Simplifying this gives:
\[ r = \frac{a}{2\sqrt{2}} \]

6. **Final Result**:
Thus, the atomic radius \( r \) of platinum in an FCC structure with unit cell length \( a \) is:
\[ r = \frac{a}{2\sqrt{2}} \]

### Conclusion:
The atomic radius of platinum is \( \frac{a}{2\sqrt{2}} \).