An electron beam of energy `10 KeV` is incident on metallic foil. If the interatomic distance is `0.55Å`. Find the angle of diffraction.

Explanation

To solve the problem step by step, we will follow the procedure outlined in the video transcript.

### Step 1: Calculate the wavelength (λ) of the electron beam
The wavelength of the electron beam can be calculated using the formula:
\[ \lambda = \frac{12.27}{\sqrt{V}} \text{ Å} \]
where \( V \) is the energy of the electron beam in kilo-electron volts (KeV).

Given:
- \( V = 10 \text{ KeV} = 10 \times 10^3 \text{ eV} \)

Substituting the value of \( V \):
\[ \lambda = \frac{12.27}{\sqrt{10 \times 10^3}} = \frac{12.27}{\sqrt{10^4}} = \frac{12.27}{100} = 0.1227 \text{ Å} \]

### Step 2: Use the diffraction formula
The relationship between the wavelength (λ), interatomic distance (d), and the angle of diffraction (φ) is given by:
\[ \lambda = d \sin \phi \]

Given:
- Interatomic distance \( d = 0.55 \text{ Å} \)

Substituting the values into the equation:
\[ 0.1227 = 0.55 \sin \phi \]

### Step 3: Solve for sin φ
Rearranging the equation to find \( \sin \phi \):
\[ \sin \phi = \frac{0.1227}{0.55} \]
Calculating this gives:
\[ \sin \phi = 0.222 \]

### Step 4: Calculate the angle φ
To find the angle φ, we take the inverse sine:
\[ \phi = \sin^{-1}(0.222) \]
Using a calculator, we find:
\[ \phi \approx 12.9^\circ \]

### Final Answer
The angle of diffraction is approximately \( 12.9^\circ \).

---