The distance between interatomic lattice planes is 10Å. The maximum wavelength of X–rays which are diffracted by this crystal will be :–

Options

A. 10Å
B. 20Å
C. 30Å
D. 40Å

Answer

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

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

where:
- \( d \) is the distance between the interatomic lattice planes,
- \( \theta \) is the angle of diffraction,
- \( n \) is the order of diffraction (an integer),
- \( \lambda \) is the wavelength of the X-rays.

### Step-by-Step Solution:

1. **Identify the given values**:
- The distance between interatomic lattice planes, \( d = 10 \) Å (angstroms).

2. **Understand the condition for maximum wavelength**:
- For maximum wavelength, we consider the first order of diffraction, which means \( n = 1 \).
- The maximum wavelength occurs when \( \sin \theta \) is at its maximum value, which is 1 (this happens when \( \theta = 90^\circ \)).

3. **Substitute the known values into Bragg's Law**:
- Using \( n = 1 \) and \( \sin \theta = 1 \), we can rewrite Bragg's Law:
\[ 2d \sin 90^\circ = n \lambda_{\text{max}} \]
- Since \( \sin 90^\circ = 1 \), we have:
\[ 2d = \lambda_{\text{max}} \]

4. **Calculate the maximum wavelength**:
- Substitute \( d = 10 \) Å into the equation:
\[ \lambda_{\text{max}} = 2 \times 10 \text{ Å} = 20 \text{ Å} \]

5. **Final answer**:
- The maximum wavelength of X-rays that can be diffracted by this crystal is \( 20 \) Å.

### Summary of the Solution:
The maximum wavelength of X-rays diffracted by the crystal with interatomic lattice plane distance of \( 10 \) Å is \( 20 \) Å.