If the angle of incidence of X-ray of wavelength 3Å which produces a second order diffracted beam from the (100) planes in a simple cubic lattice with interlayer spacing a = 6 Å is `30^(@)`, the angle of incidence that produ ces a first-order diffracted beam from the (200) planes is

Options

A. `15^(@)`
B. `45^(@)`
C. `30^(@)`
D. `60^(@)`

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-ray,
- \( d \) is the interplanar spacing,
- \( \theta \) is the angle of incidence.

### Step 1: Calculate the interplanar spacing \( d \) for the (100) planes.

For a simple cubic lattice, the interplanar spacing \( d \) can be calculated using the formula:

\[ d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]

where \( a \) is the lattice parameter and \( (h, k, l) \) are the Miller indices of the plane. For the (100) plane, \( h = 1, k = 0, l = 0 \).

Given:
- \( a = 6 \, \text{Å} \)

Calculating \( d \):

\[ d_{(100)} = \frac{6 \, \text{Å}}{\sqrt{1^2 + 0^2 + 0^2}} = \frac{6 \, \text{Å}}{1} = 6 \, \text{Å} \]

### Step 2: Use Bragg's law to verify the second-order diffraction condition.

For the second-order diffraction from the (100) planes, we have:

- \( n = 2 \)
- \( \lambda = 3 \, \text{Å} \)
- \( d = 6 \, \text{Å} \)
- \( \theta = 30^\circ \)

Substituting these values into Bragg's law:

\[ 2 \times 3 \, \text{Å} = 2 \times 6 \, \text{Å} \sin(30^\circ) \]

Calculating \( \sin(30^\circ) = \frac{1}{2} \):

\[ 6 \, \text{Å} = 6 \, \text{Å} \times 1 \]

This confirms that the second-order diffraction condition is satisfied.

### Step 3: Calculate the angle of incidence for the first-order diffraction from the (200) planes.

Now we need to find the angle of incidence for the first-order diffraction from the (200) planes.

For the (200) plane, the Miller indices are \( h = 2, k = 0, l = 0 \).

Calculating \( d \) for the (200) planes:

\[ d_{(200)} = \frac{6 \, \text{Å}}{\sqrt{2^2 + 0^2 + 0^2}} = \frac{6 \, \text{Å}}{2} = 3 \, \text{Å} \]

Now, using Bragg's law for first-order diffraction (\( n = 1 \)):

\[ 1 \times 3 \, \text{Å} = 2 \times 3 \, \text{Å} \sin \theta \]

This simplifies to:

\[ 3 \, \text{Å} = 6 \, \text{Å} \sin \theta \]

Dividing both sides by \( 6 \, \text{Å} \):

\[ \sin \theta = \frac{3}{6} = \frac{1}{2} \]

### Step 4: Determine \( \theta \).

Since \( \sin \theta = \frac{1}{2} \), we find:

\[ \theta = 30^\circ \]

### Final Answer:

The angle of incidence that produces a first-order diffracted beam from the (200) planes is \( 30^\circ \).

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