The de-broglie wavelength of a photon is twice the de-broglie wavelength of an electron. The speed of the electron is `v_(e)=c/100`. Then

Options

A. `(E_(e))/(E_(p))=10^(-4)`
B. `(E_(e))/(E_(p))=10^(-2)`
C. `(p_(e))/(m_(e)c)=10^(-2)`
D. `(p_(e))/(m_(e)c)=10^(-4)`

Explanation

To solve the problem, we need to find the relationship between the de Broglie wavelengths of a photon and an electron, given that the de Broglie wavelength of a photon is twice that of an electron, and the speed of the electron is \( v_e = \frac{c}{100} \).

### Step-by-Step Solution:

1. **Understanding the de Broglie Wavelength:**
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Calculate the de Broglie Wavelength of the Electron:**
The momentum of the electron (\( p_e \)) can be expressed as:
\[ p_e = m_e v_e \]
where \( m_e \) is the mass of the electron and \( v_e \) is its speed. Given \( v_e = \frac{c}{100} \), we have:
\[ p_e = m_e \left(\frac{c}{100}\right) \]
Thus, the de Broglie wavelength of the electron (\( \lambda_e \)) is:
\[ \lambda_e = \frac{h}{p_e} = \frac{h}{m_e \left(\frac{c}{100}\right)} = \frac{100h}{m_e c} \]

3. **Relate the Wavelength of the Photon:**
According to the problem, the de Broglie wavelength of the photon (\( \lambda_p \)) is twice that of the electron:
\[ \lambda_p = 2 \lambda_e = 2 \left(\frac{100h}{m_e c}\right) = \frac{200h}{m_e c} \]

4. **Calculate the Energy of the Photon and Electron:**
The energy of the photon (\( E_p \)) is given by:
\[ E_p = \frac{hc}{\lambda_p} \]
Substituting for \( \lambda_p \):
\[ E_p = \frac{hc}{\frac{200h}{m_e c}} = \frac{m_e c^2}{200} \]

The kinetic energy of the electron (\( E_e \)) is given by:
\[ E_e = \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_e \left(\frac{c}{100}\right)^2 = \frac{m_e c^2}{20000} \]

5. **Find the Ratio of Energies:**
Now, we can find the ratio of the energies:
\[ \frac{E_e}{E_p} = \frac{\frac{m_e c^2}{20000}}{\frac{m_e c^2}{200}} = \frac{200}{20000} = \frac{1}{100} \]

6. **Final Result:**
Therefore, the ratio of the energy of the electron to the energy of the photon is:
\[ \frac{E_e}{E_p} = \frac{1}{100} \]

### Conclusion:
The answer to the question is that the ratio of the energy of the electron to the energy of the photon is \( \frac{1}{100} \).