X-rays are incident normally on a crystal of lattice constant `0.6nm`. The first order reflection on diftion from the crystal occurs at an angle of `30^(circ) .` The wavelength of the X-ray used is `alpha` nano-metre. Find `10 alpha`.

Answer

To solve the problem, we will use Bragg's Law, which relates the wavelength of X-rays to the angle of diffraction and the lattice spacing of the crystal. The formula is given by:

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

where:
- \( d \) is the lattice constant (spacing between the crystal planes),
- \( \theta \) is the angle of diffraction,
- \( n \) is the order of reflection (in this case, first order, so \( n = 1 \)),
- \( \lambda \) is the wavelength of the X-rays.

### Step-by-Step Solution:

1. **Identify the given values:**
- Lattice constant, \( d = 0.6 \, \text{nm} \)
- Angle of diffraction, \( \theta = 30^\circ \)
- Order of reflection, \( n = 1 \)

2. **Write down Bragg's Law:**
\[ 2d \sin \theta = n \lambda \]

3. **Substitute the known values into Bragg's Law:**
\[ 2 \times 0.6 \, \text{nm} \times \sin(30^\circ) = 1 \times \lambda \]

4. **Calculate \( \sin(30^\circ) \):**
\[ \sin(30^\circ) = \frac{1}{2} \]

5. **Substitute \( \sin(30^\circ) \) into the equation:**
\[ 2 \times 0.6 \, \text{nm} \times \frac{1}{2} = \lambda \]

6. **Simplify the equation:**
\[ 0.6 \, \text{nm} = \lambda \]

7. **Since \( \lambda = \alpha \, \text{nm} \), we have:**
\[ \alpha = 0.6 \, \text{nm} \]

8. **Now, calculate \( 10\alpha \):**
\[ 10\alpha = 10 \times 0.6 \, \text{nm} = 6 \, \text{nm} \]

### Final Answer:
\[ 10\alpha = 6 \, \text{nm} \]