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))/(E_(e^(C)))=10^(-1)`
D. `(P_(e))/(E_(e^(C)))=10^(-4)`

Explanation

To solve the problem, we need to find the relationship between the energies 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 1: Write the de Broglie wavelength for the electron
The de Broglie wavelength (\( \lambda_e \)) for an electron is given by:
\[ \lambda_e = \frac{h}{m_e v_e} \]
where:
- \( h \) is Planck's constant,
- \( m_e \) is the mass of the electron,
- \( v_e \) is the speed of the electron.

### Step 2: Substitute the speed of the electron
Given \( v_e = \frac{c}{100} \), we substitute this into the equation:
\[ \lambda_e = \frac{h}{m_e \left(\frac{c}{100}\right)} = \frac{100h}{m_e c} \]

### Step 3: Write the de Broglie wavelength for the photon
The problem states that the de Broglie wavelength of the photon (\( \lambda_p \)) is twice that of the electron:
\[ \lambda_p = 2 \lambda_e = 2 \left(\frac{100h}{m_e c}\right) = \frac{200h}{m_e c} \]

### Step 4: Write the energy expressions
The energy of the electron (\( E_e \)) can be expressed as:
\[ E_e = \frac{1}{2} m_e v_e^2 \]
Substituting \( v_e = \frac{c}{100} \):
\[ E_e = \frac{1}{2} m_e \left(\frac{c}{100}\right)^2 = \frac{1}{2} m_e \frac{c^2}{10000} = \frac{m_e c^2}{20000} \]

The energy of the photon (\( E_p \)) is given by:
\[ E_p = \frac{hc}{\lambda_p} \]
Substituting \( \lambda_p = \frac{200h}{m_e c} \):
\[ E_p = \frac{hc}{\frac{200h}{m_e c}} = \frac{m_e c^2}{200} \]

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

### Conclusion
Thus, the relationship between the energies of the electron and the photon is:
\[ E_e = \frac{1}{100} E_p \]