The longest wavelength that can be analysed by a sodium chloride crystal of spacing `d = 2.82 Å` in the second order is -

Options

A. `2.82overset(@)A`
B. `5.64overset(@)A`
C. `8.46overset(@)A`
D. `11.28overset(@)A`

Explanation

To find the longest wavelength that can be analyzed by a sodium chloride crystal with a spacing of \( d = 2.82 \, \text{Å} \) in the second order, we can use Bragg's Law, which is given by:

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

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

### Step 1: Rearranging Bragg's Law
We need to find the wavelength \( \lambda \). Rearranging the equation gives:

\[ \lambda = \frac{2d \sin \theta}{n} \]

### Step 2: Maximizing Wavelength
To find the longest wavelength, we need to maximize \( \lambda \). The maximum value of \( \sin \theta \) is 1 (which occurs at \( \theta = 90^\circ \)). Therefore, we can write:

\[ \lambda_{\text{max}} = \frac{2d}{n} \]

### Step 3: Substituting Values
We know:
- \( d = 2.82 \, \text{Å} = 2.82 \times 10^{-10} \, \text{m} \) (since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \))
- \( n = 2 \) (since we are looking for the second order)

Substituting these values into the equation gives:

\[ \lambda_{\text{max}} = \frac{2 \times (2.82 \times 10^{-10} \, \text{m})}{2} \]

### Step 4: Simplifying the Expression
The 2's in the numerator and denominator cancel out:

\[ \lambda_{\text{max}} = 2.82 \times 10^{-10} \, \text{m} \]

### Step 5: Converting to Angstroms
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \), we can express the wavelength in angstroms:

\[ \lambda_{\text{max}} = 2.82 \, \text{Å} \]

### Final Answer
Thus, the longest wavelength that can be analyzed by the sodium chloride crystal in the second order is:

\[ \lambda_{\text{max}} = 2.82 \, \text{Å} \]