The wavelength of a photon and de - Broglie wavelength an electron have the same value. Given that v is the speed of electron and c is the velocity of light. `E_(e), E_(p)` is the kinetic energy of electron and energy of photon respectively while `p_(e), p_(h)` is the momentum of electron and photon respectively. Then which of the following relation is correct?

Options

A. `(E_(e))/(E_(p))=(v)/(2c)`
B. `(E_(e))/(E_(p))=(2c)/(v)`
C. `(p_(e))/(p_(h))=(v)/(2c)`
D. `(p_(e))/(p_(h))=(2c)/(v)`

Answer

To solve the problem, we need to analyze the relationship between the wavelength, momentum, and energy of a photon and an electron, given that their wavelengths are equal.

### Step-by-step Solution:

1. **Understanding Wavelengths**:
- The de Broglie wavelength (\( \lambda_e \)) of the electron is given by:
\[ \lambda_e = \frac{h}{p_e} \]
- The wavelength of the photon (\( \lambda_p \)) is given by:
\[ \lambda_p = \frac{h}{p_h} \]
- Since it is given that \( \lambda_e = \lambda_p \), we can equate the two expressions:
\[ \frac{h}{p_e} = \frac{h}{p_h} \]

2. **Equating Momenta**:
- From the above equation, we can conclude that:
\[ p_e = p_h \]

3. **Momentum Expressions**:
- The momentum of the photon (\( p_h \)) can also be expressed in terms of its energy (\( E_p \)):
\[ p_h = \frac{E_p}{c} \]
- The momentum of the electron (\( p_e \)) can be expressed in terms of its kinetic energy (\( E_e \)):
\[ p_e = \frac{E_e}{v} \]

4. **Equating Kinetic Energies**:
- Since \( p_e = p_h \), we can write:
\[ \frac{E_e}{v} = \frac{E_p}{c} \]

5. **Energy of the Photon**:
- The energy of the photon is given by:
\[ E_p = \frac{hc}{\lambda_p} \]
- Since \( \lambda_p = \lambda_e \), we can substitute:
\[ E_p = \frac{hc}{\lambda_e} \]

6. **Kinetic Energy of the Electron**:
- The kinetic energy of the electron is given by:
\[ E_e = \frac{1}{2}mv^2 \]

7. **Substituting into the Momentum Equation**:
- From the momentum equation \( \frac{E_e}{v} = \frac{E_p}{c} \):
\[ \frac{\frac{1}{2}mv^2}{v} = \frac{\frac{hc}{\lambda_e}}{c} \]
- Simplifying gives:
\[ \frac{1}{2}mv = \frac{h}{\lambda_e} \]

8. **Relating Energies**:
- We can now relate the energies:
\[ \frac{E_e}{E_p} = \frac{\frac{1}{2}mv^2}{\frac{hc}{\lambda_e}} \]
- Substituting \( \lambda_e \) gives:
\[ \frac{E_e}{E_p} = \frac{mv^2 \lambda_e}{2hc} \]

9. **Final Relation**:
- Since we know \( \lambda_e = \frac{h}{p_e} \) and \( p_e = mv \), we can express \( \lambda_e \) in terms of \( v \):
\[ \frac{E_e}{E_p} = \frac{mv^2 \cdot \frac{h}{p_e}}{2hc} \]
- This simplifies to:
\[ \frac{E_e}{E_p} = \frac{v}{2c} \]

### Conclusion:
The correct relation is:
\[ \frac{E_e}{E_p} = \frac{v}{2c} \]