The de-Broglie wavelength of an electron is the same as that of a 50 keV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is ( the energy equivalent of electron mass of 0.5 MeV)

Options

A. `1: 50`
B. `1: 20`
C. `20:1`
D. `50:1`

Explanation

To solve the problem, we need to find the ratio of the energy of a 50 keV X-ray photon to the kinetic energy of an electron whose de-Broglie wavelength is the same as that of the photon.

### Step-by-Step Solution:

1. **Determine the Energy of the Photon (E_p)**:
The energy of the photon is given as 50 keV.
\[ E_p = 50 \, \text{keV} = 50 \times 10^3 \, \text{eV} \]

2. **Calculate the de-Broglie Wavelength of the Photon**:
The energy of a photon can be related to its wavelength using the equation:
\[ E_p = \frac{hc}{\lambda} \]
Rearranging this gives us:
\[ \lambda = \frac{hc}{E_p} \]

3. **Substituting Values**:
We know that \( h \) (Planck's constant) is approximately \( 4.1357 \times 10^{-15} \, \text{eV s} \) and \( c \) (speed of light) is approximately \( 3 \times 10^8 \, \text{m/s} \). Therefore:
\[ \lambda = \frac{(4.1357 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{50 \times 10^3 \, \text{eV}} \]

4. **Calculate the Kinetic Energy of the Electron (K.E.)**:
The de-Broglie wavelength for the electron is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The kinetic energy of the electron can be expressed in terms of momentum:
\[ K.E. = \frac{p^2}{2m} \]
Substituting \( p = \frac{h}{\lambda} \):
\[ K.E. = \frac{h^2}{2m\lambda^2} \]

5. **Finding the Ratio**:
We need to find the ratio of the energy of the photon to the kinetic energy of the electron:
\[ \text{Ratio} = \frac{E_p}{K.E.} \]
Substituting the expressions we derived:
\[ \text{Ratio} = \frac{E_p}{\frac{h^2}{2m\lambda^2}} = \frac{2m\lambda^2E_p}{h^2} \]

6. **Substituting the Known Values**:
We know that the energy equivalent of the electron mass \( mc^2 \) is given as 0.5 MeV, which can be converted to eV:
\[ mc^2 = 0.5 \times 10^6 \, \text{eV} \]
Therefore, \( m \) can be expressed as:
\[ m = \frac{0.5 \times 10^6}{c^2} \]

7. **Final Calculation**:
Substitute the values into the ratio:
\[ \text{Ratio} = \frac{2 \times (0.5 \times 10^6) \times \lambda^2 \times (50 \times 10^3)}{h^2} \]
After simplification, we find that the ratio comes out to be:
\[ \text{Ratio} = 20:1 \]

### Conclusion:
Thus, the ratio of the energy of the photon to the kinetic energy of the electron is \( 20:1 \).