An electron with speed v and a photon with speed c have the same de-Broglic wavelength. If the kinetic energy and momentum of electron is `E_(e)` and `P_(e)` and that of photon is `E_(ph)` and `P_(ph)` respectively, then correct statement is -

Options

A. `(E_(e))/(E_(ph))=(2c)/(v)`
B. `(E_(e))/(E_(ph))=(v)/(2c)`
C. `(P_(e))/(P_(ph))=(2c)/(v)`
D. `(P_(e))/(P_(ph))=(v)/(2c)`

Explanation

To solve the problem, we need to analyze the relationship between the de Broglie wavelength, kinetic energy, and momentum of an electron and a photon. Let's break it down step by step.

### Step 1: Understand 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.

### Step 2: Write the expressions for the electron and photon
For the photon:
- The momentum (\( P_{ph} \)) of a photon 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.

For the electron:
- The momentum (\( P_{e} \)) of the electron is given by:
\[ P_{e} = m_{e} v \]
where \( m_{e} \) is the mass of the electron and \( v \) is its velocity.

### Step 3: Set the de Broglie wavelengths equal
Since the de Broglie wavelengths of the electron and the photon are equal, we have:
\[ \frac{h}{P_{ph}} = \frac{h}{P_{e}} \]
This simplifies to:
\[ P_{ph} = P_{e} \]

### Step 4: Substitute the expressions for momentum
Substituting the expressions for momentum:
\[ \frac{E_{ph}}{c} = m_{e} v \]
From this, we can express the energy of the photon in terms of the electron's properties:
\[ E_{ph} = m_{e} v c \]

### Step 5: Write the kinetic energy of the electron
The kinetic energy (\( E_{e} \)) of the electron is given by:
\[ E_{e} = \frac{1}{2} m_{e} v^2 \]

### Step 6: Find the ratio of energies
Now we can find the ratio of the energies of the electron and the photon:
\[ \frac{E_{e}}{E_{ph}} = \frac{\frac{1}{2} m_{e} v^2}{m_{e} v c} \]
This simplifies to:
\[ \frac{E_{e}}{E_{ph}} = \frac{v}{2c} \]

### Conclusion
Thus, the correct statement regarding the energies of the electron and the photon is:
\[ \frac{E_{e}}{E_{ph}} = \frac{v}{2c} \]