The de Broglie wavelength of an electron and the wavelength of a photon are same. The ratio between the energy of the photon and the momentum of the electron is

Options

A. `h`
B. `c`
C. `1//h`
D. `1//c`

Explanation

To solve the problem, we need to find the ratio between the energy of a photon and the momentum of an electron, given that the de Broglie wavelength of the electron and the wavelength of the photon are the same.

### Step-by-Step Solution:

1. **Understanding the Wavelengths**:
- Let the common wavelength be \( \lambda \).
- The de Broglie wavelength of the electron is given by:
\[ \lambda_e = \frac{h}{p_e} \]
where \( p_e \) is the momentum of the electron and \( h \) is Planck's constant.
- The wavelength of the photon is given by:
\[ \lambda = \frac{h}{p_{\gamma}} \]
where \( p_{\gamma} \) is the momentum of the photon.

2. **Relating Energy and Momentum**:
- The energy of the photon is given by:
\[ E_{\gamma} = \frac{hc}{\lambda} \]
where \( c \) is the speed of light.
- The momentum of the photon is related to its energy by:
\[ p_{\gamma} = \frac{E_{\gamma}}{c} \]

3. **Setting the Wavelengths Equal**:
- Since the de Broglie wavelength of the electron and the wavelength of the photon are the same, we have:
\[ \lambda_e = \lambda \]
- Therefore, we can write:
\[ \frac{h}{p_e} = \frac{h}{p_{\gamma}} \]
- This implies:
\[ p_e = p_{\gamma} \]

4. **Finding the Ratio**:
- Now, we can find the ratio between the energy of the photon and the momentum of the electron:
\[ \text{Ratio} = \frac{E_{\gamma}}{p_e} \]
- Substituting the expressions for \( E_{\gamma} \) and \( p_e \):
\[ \text{Ratio} = \frac{\frac{hc}{\lambda}}{p_e} \]
- Since \( p_e = p_{\gamma} = \frac{E_{\gamma}}{c} \), we can express \( p_e \) in terms of \( E_{\gamma} \):
\[ p_e = \frac{hc}{\lambda} \]
- Thus, the ratio becomes:
\[ \text{Ratio} = \frac{\frac{hc}{\lambda}}{\frac{E_{\gamma}}{c}} = \frac{hc^2}{\lambda E_{\gamma}} \]

5. **Final Expression**:
- Since \( E_{\gamma} = \frac{hc}{\lambda} \), we can substitute this back into the ratio:
\[ \text{Ratio} = \frac{hc^2}{\lambda \cdot \frac{hc}{\lambda}} = c \]

Thus, the ratio between the energy of the photon and the momentum of the electron is \( c \).