An X-ray photon of wavelength `lambda` and frequency `v` colliders with an initially stationary electron ( but free to move and bounces off. If `lambda` and `v` are respectively the wavelength and frequency of the sacattered photon, then:

Options

A. `lambda'=lambda,v'=v`
B. `lambda' lt lambda,v' gtv`
C. `lambda' gt lambda,v' gtv`
D. `lambda' gt lambda, v'ltv`

Answer

To solve the problem step by step, we will analyze the interaction between the X-ray photon and the stationary electron using the principles of conservation of energy and momentum.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let the initial wavelength of the X-ray photon be \( \lambda \) and its frequency be \( v \).
- The initial energy of the photon can be expressed as:
\[ E_{\text{initial}} = h \cdot v \]
- The electron is initially stationary, so its initial kinetic energy is zero.

2. **Understand the Collision**:
- When the photon collides with the electron, it transfers some energy to the electron, causing it to move.
- After the collision, let the wavelength of the scattered photon be \( \lambda' \) and its frequency be \( v' \).

3. **Apply Conservation of Energy**:
- The total energy before the collision must equal the total energy after the collision:
\[ E_{\text{initial}} = E_{\text{final}} \]
- Therefore, we have:
\[ h \cdot v = h \cdot v' + \frac{1}{2} m v_e^2 \]
- Here, \( v_e \) is the velocity of the electron after the collision.

4. **Relate Wavelength and Frequency**:
- The relationship between wavelength and frequency is given by:
\[ v = \frac{c}{\lambda} \]
- Thus, for the scattered photon:
\[ v' = \frac{c}{\lambda'} \]

5. **Substituting Frequencies**:
- Substitute \( v \) and \( v' \) into the energy conservation equation:
\[ h \cdot \frac{c}{\lambda} = h \cdot \frac{c}{\lambda'} + \frac{1}{2} m v_e^2 \]

6. **Analyzing the Energy Loss**:
- Since the photon transfers energy to the electron, the energy of the photon after the collision must be less than before:
\[ h \cdot v' < h \cdot v \]
- This implies:
\[ v' < v \]

7. **Relating Wavelengths**:
- Since frequency is inversely proportional to wavelength, if \( v' < v \), then:
\[ \lambda' > \lambda \]
- Thus, the final wavelength of the scattered photon is greater than the initial wavelength.

8. **Conclusion**:
- The relationship we derived is:
\[ \lambda' > \lambda \]
- Therefore, the correct option is that the wavelength of the scattered photon is greater than the initial wavelength.

### Final Answer:
The relationship is \( \lambda' > \lambda \).