In Q.72,An electron and a photon possess the same de Broglie wavelength. If E e and E p h are, respectively, if the velocity of electron is 25% of the velocity of photon, then `(E_e)/(E_(ph))` equals

Options

A. 0.04356
B. 0.044
C. 0.0472
D. 0.125

Explanation

To solve the problem, we need to find the ratio of the energies of an electron and a photon, given that they have the same de Broglie wavelength and that the velocity of the electron is 25% of the velocity of the photon.

### Step-by-Step Solution:

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

2. **Momentum of the Electron**:
For the electron, the momentum \( p_e \) can be expressed as:
\[ p_e = m_e v_e \]
where \( m_e \) is the mass of the electron and \( v_e \) is its velocity.

3. **Momentum of the Photon**:
For the photon, the momentum \( p_{ph} \) is given by:
\[ p_{ph} = \frac{E_{ph}}{c} \]
where \( E_{ph} \) is the energy of the photon and \( c \) is the speed of light.

4. **Equating the Wavelengths**:
Since the electron and photon have the same wavelength:
\[ \lambda_e = \lambda_{ph} \]
This implies:
\[ \frac{h}{p_e} = \frac{h}{p_{ph}} \]
Therefore:
\[ p_e = p_{ph} \]

5. **Energy of the Electron**:
The energy of the electron \( E_e \) is given by:
\[ E_e = \frac{p_e^2}{2m_e} \]

6. **Energy of the Photon**:
The energy of the photon \( E_{ph} \) is given by:
\[ E_{ph} = h \cdot f = \frac{hc}{\lambda} \]

7. **Substituting for Momentum**:
From the momentum equality \( p_e = p_{ph} \), we can express the energies:
\[ E_e = \frac{(m_e v_e)^2}{2m_e} = \frac{m_e v_e^2}{2} \]
and
\[ E_{ph} = \frac{hc}{\lambda} \]

8. **Using the Velocity Relationship**:
Given that \( v_e = 0.25c \), we substitute this into the energy of the electron:
\[ E_e = \frac{m_e (0.25c)^2}{2} = \frac{m_e \cdot 0.0625c^2}{2} = \frac{0.0625 m_e c^2}{2} = \frac{0.03125 m_e c^2}{1} \]

9. **Finding the Ratio**:
Now, we need to find the ratio \( \frac{E_e}{E_{ph}} \):
\[ \frac{E_e}{E_{ph}} = \frac{0.03125 m_e c^2}{\frac{hc}{\lambda}} \]
Since \( \lambda \) is the same for both, we can simplify:
\[ \frac{E_e}{E_{ph}} = \frac{0.03125 m_e c^2 \lambda}{hc} \]
Using \( \lambda = \frac{h}{p_e} \) and substituting \( p_e = m_e v_e \):
\[ \frac{E_e}{E_{ph}} = \frac{0.03125 m_e c^2 \cdot \frac{h}{m_e v_e}}{hc} = \frac{0.03125 c^2}{v_e} \]
Substituting \( v_e = 0.25c \):
\[ \frac{E_e}{E_{ph}} = \frac{0.03125 c^2}{0.25c} = 0.125 \]

### Final Answer:
\[ \frac{E_e}{E_{ph}} = \frac{1}{8} = 0.125 \]