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 given the lattice constant, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the BCC Structure**:
- The body-centered cubic (BCC) structure has atoms located at each corner of the cube and one atom at the center of the cube.

2. **Identify the Lattice Constant**:
- The lattice constant (a) is provided as `3.6 Å` (angstroms).

3. **Use the Formula for Atomic Radius**:
- The formula for the atomic radius (r) in a BCC structure is given by:
\[ r = \frac{a \sqrt{3}}{4} \]

4. **Substitute the Lattice Constant into the Formula**:
- Substitute `a = 3.6 Å` into the formula:
\[ r = \frac{3.6 \times \sqrt{3}}{4} \]

5. **Calculate the Value**:
- First, calculate \(\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \]
- Now substitute this value into the equation:
\[ r = \frac{3.6 \times 1.732}{4} \]
- Calculate the numerator:
\[ 3.6 \times 1.732 \approx 6.2272 \]
- Finally, divide by 4:
\[ r \approx \frac{6.2272}{4} \approx 1.5568 \]
- Rounding this to two decimal places gives:
\[ r \approx 1.56 \, Å \]

6. **Final Answer**:
- The atomic radius is approximately `1.56 Å`.