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 solve the problem of finding the maximum wavelength that can be diffracted given the interplanar distance in a crystal, we can follow these steps:

### Step 1: Understand the diffraction condition
The condition for diffraction is given by the formula:
\[ N \lambda = 2D \sin \theta \]
where:
- \( N \) is the order of diffraction,
- \( \lambda \) is the wavelength,
- \( D \) is the interplanar distance, and
- \( \theta \) is the angle of diffraction.

### Step 2: Determine the maximum value of \( \sin \theta \)
The maximum value of \( \sin \theta \) is 1. Therefore, to find the maximum wavelength, we can set \( \sin \theta = 1 \).

### Step 3: Substitute values into the diffraction condition
Substituting \( \sin \theta = 1 \) into the diffraction condition gives:
\[ N \lambda = 2D \]
For maximum wavelength, we can take \( N = 1 \):
\[ \lambda_{max} = 2D \]

### Step 4: Substitute the given interplanar distance
Given that the interplanar distance \( D = 2.8 \times 10^{-8} \) m, we can substitute this value into the equation:
\[ \lambda_{max} = 2 \times (2.8 \times 10^{-8}) \]

### Step 5: Calculate the maximum wavelength
Calculating the above expression:
\[ \lambda_{max} = 5.6 \times 10^{-8} \text{ m} \]

### Conclusion
The maximum wavelength that can be diffracted is:
\[ \lambda_{max} = 5.6 \times 10^{-8} \text{ m} \]