A crystal has bcc structure and its lattice constant is `3.6 A`. What is the atomic radius?

Options

A. `3.6 A`
B. `1.8 A`
C. `1.27 A`
D. `1.56 A`

Answer

To find the atomic radius of a crystal with a body-centered cubic (BCC) structure and a given lattice constant, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Lattice Structure**:
- The crystal has a body-centered cubic (BCC) structure.

2. **Understand the Relationship**:
- In a BCC structure, the relationship between the atomic radius (R) and the lattice constant (a) is given by the formula:
\[ R = \frac{a \sqrt{3}}{4} \]

3. **Substitute the Given Value**:
- The lattice constant \( a \) is given as \( 3.6 \, \text{Å} \).
- Substitute this value into the formula:
\[ R = \frac{3.6 \, \text{Å} \cdot \sqrt{3}}{4} \]

4. **Calculate \( \sqrt{3} \)**:
- The value of \( \sqrt{3} \) is approximately \( 1.732 \).

5. **Perform the Calculation**:
- First, calculate \( 3.6 \cdot \sqrt{3} \):
\[ 3.6 \cdot 1.732 \approx 6.2272 \]
- Now divide by 4:
\[ R \approx \frac{6.2272}{4} \approx 1.5568 \, \text{Å} \]

6. **Round the Result**:
- Rounding \( 1.5568 \, \text{Å} \) gives approximately \( 1.56 \, \text{Å} \).

### Final Answer:
The atomic radius \( R \) is approximately \( 1.56 \, \text{Å} \).

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