The interplaner distance in a crystal is `2.8 xx 10^(–8)` m. The value of maximum wavelength which can be diffracted : -

Options

A. `2.8 xx 10^(–8)` m
B. `5.6 xx 10^(–8)` m
C. `1.4 xx 10^(–8)` m
D. `7.6 xx 10^(–8)` m

Explanation

To find the maximum wavelength that can be diffracted given the interplanar distance in a crystal, we will use Bragg's Law.

### Step-by-Step Solution:

1. **Understand Bragg's Law**:
Bragg's Law is given by the equation:
\[ 2d \sin \theta = n \lambda \]
where:
- \( d \) = interplanar distance
- \( \theta \) = angle of diffraction
- \( n \) = order of diffraction (an integer)
- \( \lambda \) = wavelength of the incident light

2. **Identify the Given Values**:
We are given:
- Interplanar distance \( d = 2.8 \times 10^{-8} \) m

3. **Determine Maximum Wavelength**:
For the maximum wavelength (\( \lambda_{\text{max}} \)), we consider the case when \( \sin \theta \) is at its maximum value, which is 1. Therefore, we can rewrite Bragg's Law as:
\[ 2d = n \lambda_{\text{max}} \]
If we take \( n = 1 \) (first order diffraction), we can simplify this to:
\[ \lambda_{\text{max}} = 2d \]

4. **Calculate \( \lambda_{\text{max}} \)**:
Substituting the value of \( d \):
\[ \lambda_{\text{max}} = 2 \times (2.8 \times 10^{-8}) = 5.6 \times 10^{-8} \text{ m} \]

5. **Final Answer**:
The maximum wavelength that can be diffracted is:
\[ \lambda_{\text{max}} = 5.6 \times 10^{-8} \text{ m} \]