A photon of wavelength `lamda` is scatted from an electron , which was at rest. The wavelength shift `Deltalamda` is three times of `lamda` and the angle of scattering `theta` is `60^@` . The angle at which the electron recoiled is `phi` . The value of tan `phi` is (electron speed is much smaller than the speed of light )

Options

A. 0.16
B. 0.22
C. 0.25
D. 0.28

Explanation

To solve the problem, we need to analyze the scattering of a photon by an electron, applying the principles of conservation of momentum. Here's the step-by-step solution:

### Step 1: Understand the Problem
We have a photon of wavelength \( \lambda \) scattering off an electron at rest. The change in wavelength \( \Delta \lambda \) is given as \( 3\lambda \), and the scattering angle \( \theta \) is \( 60^\circ \). We need to find \( \tan \phi \), where \( \phi \) is the angle at which the electron recoils.

**Hint:** Identify the initial and final states of the photon and electron.

### Step 2: Calculate the Initial and Final Momentum of the Photon
The initial momentum of the photon \( p_i \) is given by:
\[ p_i = \frac{h}{\lambda} \]

After scattering, the new wavelength of the photon becomes \( \lambda' = \lambda + \Delta \lambda = \lambda + 3\lambda = 4\lambda \). Thus, the final momentum of the photon \( p_f \) is:
\[ p_f = \frac{h}{\lambda'} = \frac{h}{4\lambda} \]

**Hint:** Use the de Broglie relation to find the momentum of the photon before and after scattering.

### Step 3: Resolve the Final Momentum of the Photon
The final momentum of the photon can be resolved into components:
- The x-component:
\[ p_{fx} = p_f \cos(60^\circ) = \frac{h}{4\lambda} \cdot \frac{1}{2} = \frac{h}{8\lambda} \]
- The y-component:
\[ p_{fy} = p_f \sin(60^\circ) = \frac{h}{4\lambda} \cdot \frac{\sqrt{3}}{2} = \frac{h\sqrt{3}}{8\lambda} \]

**Hint:** Remember to use trigonometric identities for the angles involved.

### Step 4: Conservation of Momentum in the y-direction
Initially, there is no momentum in the y-direction. Therefore, the total momentum in the y-direction after scattering must also be zero:
\[ p_{fy} - p_e \sin(\phi) = 0 \]
This gives:
\[ p_e \sin(\phi) = p_{fy} = \frac{h\sqrt{3}}{8\lambda} \]

**Hint:** Set up the equation based on the conservation of momentum.

### Step 5: Conservation of Momentum in the x-direction
For the x-direction, we have:
\[ p_i = p_{fx} + p_e \cos(\phi) \]
Substituting the known values:
\[ \frac{h}{\lambda} = \frac{h}{8\lambda} + p_e \cos(\phi) \]
Rearranging gives:
\[ p_e \cos(\phi) = \frac{h}{\lambda} - \frac{h}{8\lambda} = \frac{7h}{8\lambda} \]

**Hint:** Use the conservation of momentum to relate the initial and final momenta.

### Step 6: Finding \( \tan \phi \)
Now we have two equations:
1. \( p_e \sin(\phi) = \frac{h\sqrt{3}}{8\lambda} \)
2. \( p_e \cos(\phi) = \frac{7h}{8\lambda} \)

Dividing these two equations gives:
\[ \tan(\phi) = \frac{p_e \sin(\phi)}{p_e \cos(\phi)} = \frac{\frac{h\sqrt{3}}{8\lambda}}{\frac{7h}{8\lambda}} = \frac{\sqrt{3}}{7} \]

**Hint:** Use the definition of tangent to relate sine and cosine.

### Final Answer
Thus, the value of \( \tan \phi \) is:
\[ \tan \phi = \frac{\sqrt{3}}{7} \]