A photon has same wavelength as the de Broglie wavelength of electrons. Given `C=speed of light`, `v=`speed of electron. Which of the followitg relation is correct? [Here `E_e=` kinetic energy of electron, `E_(ph)=` energy of photon, `P_e=` momentum of electron and `P_(ph)=` momentum of photon]

Options

A. `(E_e)/(E_(ph))=(2C)/(v)`
B. `(E_e)/(E_(ph))=(v)/(2C)`
C. `(P_e)/(P_(ph))=(2C)/(v)`
D. `(P_e)/(P_(ph))=(C)/(v)`

Explanation

To solve the problem, we need to analyze the relationships between the photon and the electron based on their wavelengths, energies, and momenta.

### Step-by-Step Solution:

1. **Understanding the Wavelengths**:
We are given that the wavelength of the photon (\( \lambda_{ph} \)) is equal to the de Broglie wavelength of the electron (\( \lambda_{e} \)).
\[ \lambda_{ph} = \lambda_{e} \]

2. **De Broglie Wavelength of the Electron**:
The de Broglie wavelength of an electron can be expressed as:
\[ \lambda_{e} = \frac{h}{p_e} \]
where \( p_e \) is the momentum of the electron.

3. **Energy of the Photon**:
The energy of the photon can be expressed in terms of its wavelength:
\[ E_{ph} = \frac{hc}{\lambda_{ph}} \]
Since \( \lambda_{ph} = \lambda_{e} \), we can write:
\[ E_{ph} = \frac{hc}{\lambda_{e}} \]

4. **Kinetic Energy of the Electron**:
The kinetic energy of the electron can be expressed as:
\[ E_{e} = \frac{p_e^2}{2m} \]
We can also express momentum \( p_e \) in terms of the de Broglie wavelength:
\[ p_e = \frac{h}{\lambda_{e}} \]
Substituting this into the kinetic energy equation gives:
\[ E_{e} = \frac{(h/\lambda_{e})^2}{2m} = \frac{h^2}{2m\lambda_{e}^2} \]

5. **Finding the Ratio of Energies**:
Now, we can find the ratio of the kinetic energy of the electron to the energy of the photon:
\[ \frac{E_{e}}{E_{ph}} = \frac{\frac{h^2}{2m\lambda_{e}^2}}{\frac{hc}{\lambda_{e}}} \]
Simplifying this expression:
\[ \frac{E_{e}}{E_{ph}} = \frac{h^2}{2m\lambda_{e}^2} \cdot \frac{\lambda_{e}}{hc} = \frac{h}{2mc} \]

6. **Momentum of the Photon**:
The momentum of the photon can be expressed as:
\[ p_{ph} = \frac{E_{ph}}{c} = \frac{hc/\lambda_{e}}{c} = \frac{h}{\lambda_{e}} \]

7. **Finding the Ratio of Momenta**:
The momentum of the electron can be expressed as:
\[ p_{e} = \frac{h}{\lambda_{e}} \]
Thus, the ratio of the momenta is:
\[ \frac{p_{e}}{p_{ph}} = \frac{h/\lambda_{e}}{h/\lambda_{e}} = 1 \]

### Final Relations:
- The ratio of kinetic energy of the electron to the energy of the photon is:
\[ \frac{E_{e}}{E_{ph}} = \frac{h}{2mc} \]
- The ratio of momentum of the electron to the momentum of the photon is:
\[ \frac{p_{e}}{p_{ph}} = 1 \]