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 establish the relationship between the energy and momentum of a photon and an electron, given that their wavelengths are the same.

### Step-by-Step Solution:

1. **Photon Energy and Wavelength**:
The energy of a photon is given by the formula:
\[ E_p = \frac{hc}{\lambda_p} \]
where \(E_p\) is the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda_p\) is the wavelength of the photon.

2. **De Broglie Wavelength of Electron**:
The de Broglie wavelength of an electron is given by:
\[ \lambda_e = \frac{h}{p_e} \]
where \(p_e\) is the momentum of the electron.

3. **Kinetic Energy of the Electron**:
The kinetic energy \(E_e\) of the electron can be expressed as:
\[ E_e = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the electron and \(v\) is its velocity.

4. **Relating Wavelengths**:
Since it is given that \(\lambda_e = \lambda_p\), we can equate the two expressions for the wavelengths:
\[ \frac{h}{p_e} = \frac{hc}{E_p} \]
From this, we can express the momentum of the electron in terms of the energy of the photon:
\[ p_e = \frac{E_p}{c} \]

5. **Finding the Relationship between Energies**:
Now, substituting \(p_e\) into the kinetic energy expression for the electron:
\[ E_e = \frac{1}{2} m v^2 = \frac{1}{2} \cdot \frac{h}{\lambda_e} \cdot v^2 \]
We can express the momentum of the electron as:
\[ p_e = mv = \frac{h}{\lambda_e} \]
Therefore, we can express \(E_e\) in terms of \(E_p\):
\[ E_e = \frac{1}{2} \cdot \frac{h}{\lambda_e} \cdot \left(\frac{h}{\lambda_e}\right) = \frac{h^2}{2\lambda_e^2} \]

6. **Final Relation**:
Since we have \(\lambda_e = \lambda_p\), we can substitute \(\lambda_p\) into the equation:
\[ E_e = \frac{h^2}{2\lambda_p^2} \]
Now, substituting \(E_p\) back into the equation, we find:
\[ \frac{E_e}{E_p} = \frac{v}{2c} \]

### Conclusion:
The correct relationship derived from the above steps is:
\[ \frac{E_e}{E_p} = \frac{v}{2c} \]