The de-Broglie wavelength of an electron is the same as that of a 50 ke 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_photon)**:
The energy of the photon can be calculated using the formula:
\[ E_{\text{photon}} = 50 \text{ keV} = 50 \times 10^3 \text{ eV} = 50 \times 1.6 \times 10^{-19} \text{ J} = 8.0 \times 10^{-15} \text{ J} \]

2. **Calculate the de-Broglie Wavelength (λ)**:
The de-Broglie wavelength of the photon can be calculated using the formula:
\[ \lambda = \frac{hc}{E_{\text{photon}}} \]
where \( h \) is Planck's constant (\(6.626 \times 10^{-34} \text{ J s}\)) and \( c \) is the speed of light (\(3 \times 10^8 \text{ m/s}\)).
\[ \lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{8.0 \times 10^{-15} \text{ J}} = 2.48 \times 10^{-10} \text{ m} \]

3. **Relate de-Broglie Wavelength to Kinetic Energy**:
The de-Broglie wavelength of the electron is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can also be expressed in terms of kinetic energy (\( KE \)):
\[ p = \sqrt{2mKE} \]
Thus, we can write:
\[ \lambda = \frac{h}{\sqrt{2mKE}} \]
Rearranging gives:
\[ KE = \frac{h^2}{2m\lambda^2} \]

4. **Substituting Values**:
Using the known values of \( h \) and \( m \) (mass of the electron \( m = 9.11 \times 10^{-31} \text{ kg} \)):
\[ KE = \frac{(6.626 \times 10^{-34})^2}{2(9.11 \times 10^{-31})(2.48 \times 10^{-10})^2} \]
Calculating this gives:
\[ KE \approx 0.5 \text{ MeV} = 0.5 \times 1.6 \times 10^{-13} \text{ J} = 8.0 \times 10^{-14} \text{ J} \]

5. **Calculate the Ratio of Energies**:
Now we can find the ratio of the energy of the photon to the kinetic energy of the electron:
\[ \text{Ratio} = \frac{E_{\text{photon}}}{KE} = \frac{8.0 \times 10^{-15} \text{ J}}{8.0 \times 10^{-14} \text{ J}} = 1 \]

### Final Answer:
The ratio of the energy of the photon to the kinetic energy of the electron is **1**.