The distacne between two successive atomic planes of a calcite crystal is 0.3 nm. The minimum angle for Brag scattering of 0.3 Å X-rays will be

Options

A. `1.43^(@)`
B. `1.56^(@)`
C. `2.86^(@)`
D. `30^(@)`

Explanation

To solve the problem of finding the minimum angle for Bragg scattering of 0.3 Å X-rays in a calcite crystal with a given distance between atomic planes, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Units**:
- The distance between two successive atomic planes (d) is given as 0.3 nm. We need to convert this to meters:
\[ d = 0.3 \, \text{nm} = 0.3 \times 10^{-9} \, \text{m} = 3.0 \times 10^{-10} \, \text{m} \]

2. **Convert Wavelength**:
- The wavelength (λ) of the X-rays is given as 0.3 Å. We also need to convert this to meters:
\[ \lambda = 0.3 \, \text{Å} = 0.3 \times 10^{-10} \, \text{m} = 3.0 \times 10^{-11} \, \text{m} \]

3. **Apply Bragg's Law**:
- Bragg's Law is given by the equation:
\[ n\lambda = 2d\sin\theta \]
- For the minimum angle, we take \( n = 1 \):
\[ \lambda = 2d\sin\theta \]

4. **Substitute Values**:
- Substitute the values of \( \lambda \) and \( d \):
\[ 3.0 \times 10^{-11} = 2 \times (3.0 \times 10^{-10}) \sin\theta \]

5. **Rearrange the Equation**:
- Rearranging gives:
\[ \sin\theta = \frac{3.0 \times 10^{-11}}{2 \times 3.0 \times 10^{-10}} = \frac{3.0 \times 10^{-11}}{6.0 \times 10^{-10}} = \frac{1}{20} \]

6. **Calculate the Angle**:
- To find the angle \( \theta \):
\[ \theta = \sin^{-1}\left(\frac{1}{20}\right) \]
- Using a calculator:
\[ \theta \approx 2.6^\circ \]

### Final Answer:
The minimum angle for Bragg scattering of 0.3 Å X-rays is approximately \( 2.6^\circ \).

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