For a crystal, the angle of diffraction (20) is `90^@` and the second order line has a d value of `2.28 A^@`. The wavelength (in `A^@`) of X-rays used for Bragg's diffraction is

Options

A. `1.61 A^(@)`
B. `1.14A^(@)`
C. `2.28^(@)`
D. `2.0^(@)`

Answer

To solve the problem, we will use Bragg's Law, which is given by the equation:

\[ n \lambda = 2d \sin \theta \]

Where:
- \( n \) is the order of diffraction,
- \( \lambda \) is the wavelength of the X-rays,
- \( d \) is the interplanar spacing,
- \( \theta \) is the angle of diffraction.

### Step-by-Step Solution:

1. **Identify the given values:**
- The angle of diffraction \( 2\theta = 90^\circ \)
- The order of diffraction \( n = 2 \) (since it's the second order line)
- The d value \( d = 2.28 \, \text{Å} \)

2. **Calculate \( \theta \):**
\[ \theta = \frac{2\theta}{2} = \frac{90^\circ}{2} = 45^\circ \]

3. **Substitute the values into Bragg's equation:**
\[ n \lambda = 2d \sin \theta \]
Substituting the known values:
\[ 2 \lambda = 2 \times 2.28 \, \text{Å} \times \sin(45^\circ) \]

4. **Calculate \( \sin(45^\circ) \):**
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]

5. **Substitute \( \sin(45^\circ) \) back into the equation:**
\[ 2 \lambda = 2 \times 2.28 \, \text{Å} \times \frac{1}{\sqrt{2}} \]

6. **Simplify the equation:**
\[ 2 \lambda = \frac{2 \times 2.28 \, \text{Å}}{\sqrt{2}} \]
\[ 2 \lambda = \frac{4.56 \, \text{Å}}{\sqrt{2}} \]

7. **Calculate \( \lambda \):**
\[ \lambda = \frac{4.56 \, \text{Å}}{2} = \frac{4.56}{\sqrt{2}} \, \text{Å} \]
\[ \lambda = 4.56 \times \frac{\sqrt{2}}{2} \, \text{Å} \]
\[ \lambda = 4.56 \times 0.7071 \, \text{Å} \approx 3.22 \, \text{Å} \]

8. **Final calculation:**
\[ \lambda \approx 1.612 \, \text{Å} \]

### Final Answer:
The wavelength of X-rays used for Bragg's diffraction is approximately **1.612 Å**.