What is the wavelength of a beam of `X` -rays (in `A^o` ) which is difted by a crystal in a direction making an angle of `60^(circ)` with the incident beam and corresponding to first order reflection from crystal lattice planes separated by `3xx 10^(-10) m ?`

Answer

To find the wavelength of the X-rays diffracted by a crystal, we can use Bragg's Law, which is given by the equation:

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

Where:
- \( D \) is the distance between the crystal lattice planes,
- \( \theta \) is the angle of diffraction,
- \( n \) is the order of reflection (in this case, \( n = 1 \)),
- \( \lambda \) is the wavelength of the X-rays.

### Step-by-step Solution:

1. **Identify the Given Values:**
- The distance between the lattice planes, \( D = 3 \times 10^{-10} \) m.
- The angle of diffraction, \( \theta = 60^\circ \).
- The order of reflection, \( n = 1 \).

2. **Substitute the Values into Bragg's Law:**
\[ 2D \sin \theta = n \lambda \]
Substituting the known values:
\[ 2 \times (3 \times 10^{-10}) \times \sin(60^\circ) = 1 \times \lambda \]

3. **Calculate \( \sin(60^\circ) \):**
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

4. **Substitute \( \sin(60^\circ) \) into the Equation:**
\[ 2 \times (3 \times 10^{-10}) \times \frac{\sqrt{3}}{2} = \lambda \]

5. **Simplify the Equation:**
\[ 3 \times 10^{-10} \times \sqrt{3} = \lambda \]

6. **Calculate the Value of \( \lambda \):**
- We know \( \sqrt{3} \approx 1.732 \), so:
\[ \lambda = 3 \times 10^{-10} \times 1.732 \]
\[ \lambda \approx 5.196 \times 10^{-10} \text{ m} \]

7. **Convert the Wavelength to Angstroms:**
- Since \( 1 \text{ A}^o = 10^{-10} \text{ m} \):
\[ \lambda \approx 5.196 \times 10^{-10} \text{ m} = 5.196 \times 10^{1} \text{ A}^o \approx 51.96 \text{ A}^o \]

### Final Answer:
The wavelength of the X-rays is approximately \( 51.96 \text{ A}^o \).

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