A photon of energy h`upsilon` and momentum `h upsilon` /c collides with an electron at rest. After the collision, the scattered electron and the scattered photon each make an angle of `45^@` with the initial direction of motion. The ratio of frequency of scattered and incident photon is

Options

A. `sqrt(2)`
B. `sqrt(2)-2`
C. 2
D. `1/(sqrt(2))`

Explanation

To solve the problem, we will use the principles of conservation of momentum and energy in the context of the Compton effect.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- A photon with energy \( E = h\nu \) and momentum \( p = \frac{h\nu}{c} \) collides with an electron at rest. The electron's initial momentum is zero.

2. **Post-Collision Angles:**
- After the collision, both the scattered photon and the scattered electron make an angle of \( 45^\circ \) with the initial direction of the photon.

3. **Using Conservation of Momentum:**
- The momentum of the system must be conserved.
- Let \( p' \) be the momentum of the scattered photon and \( p_e \) be the momentum of the scattered electron.
- The initial momentum of the photon is \( \frac{h\nu}{c} \) in the x-direction.
- After the collision, the momentum components can be expressed as:
- For the photon:
- \( p'_{x} = p' \cos(45^\circ) \)
- \( p'_{y} = p' \sin(45^\circ) \)
- For the electron:
- \( p_{ex} = p_e \cos(45^\circ) \)
- \( p_{ey} = p_e \sin(45^\circ) \)

4. **Setting Up the Momentum Conservation Equations:**
- In the x-direction:
\[ \frac{h\nu}{c} = p' \cos(45^\circ) + p_e \cos(45^\circ) \]
- In the y-direction (initially zero):
\[ 0 = p' \sin(45^\circ) - p_e \sin(45^\circ) \]

5. **Solving the y-direction Equation:**
- From the y-direction equation:
\[ p' \sin(45^\circ) = p_e \sin(45^\circ) \implies p' = p_e \]

6. **Substituting into the x-direction Equation:**
- Substitute \( p' = p_e \) into the x-direction equation:
\[ \frac{h\nu}{c} = p' \cos(45^\circ) + p' \cos(45^\circ) = 2p' \cos(45^\circ) \]
- Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \frac{h\nu}{c} = 2p' \cdot \frac{1}{\sqrt{2}} \implies p' = \frac{h\nu}{2\sqrt{2}c} \]

7. **Relating Photon Momentum to Frequency:**
- The momentum of the scattered photon can also be expressed in terms of its frequency:
\[ p' = \frac{h\nu'}{c} \]
- Equating the two expressions for \( p' \):
\[ \frac{h\nu'}{c} = \frac{h\nu}{2\sqrt{2}c} \]
- Canceling \( \frac{h}{c} \) from both sides:
\[ \nu' = \frac{\nu}{2\sqrt{2}} \]

8. **Finding the Ratio of Frequencies:**
- The ratio of the frequency of the scattered photon to the incident photon is:
\[ \frac{\nu'}{\nu} = \frac{1}{2\sqrt{2}} \]

### Final Answer:
The ratio of the frequency of the scattered photon to the incident photon is \( \frac{1}{2\sqrt{2}} \).