A photon has same wavelength as the de Broglie wavelength of electrons. Given `C` = speed of light, `v=`speed of electron. Which of the followitg relation is correct? [Here `E_e=` kinetic energy of electron, `E_(ph)=` energy of photon, `P_e=` momentum of electron and `P_(ph)=` momentum of photon]

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))=(C)/(v)`

Explanation

To solve the problem, we need to derive the relationship between the kinetic energy of the electron and the energy of the photon, given that they have the same wavelength. Let's break it down step by step.

### Step 1: Write the relationship for wavelength
The de Broglie wavelength (\( \lambda \)) of an electron is given by:
\[ \lambda_e = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v \) is its speed.

### Step 2: Write the wavelength of the photon
The wavelength of a photon can be expressed as:
\[ \lambda_{ph} = \frac{c}{f} \]
where \( c \) is the speed of light and \( f \) is the frequency of the photon. The energy of the photon is given by:
\[ E_{ph} = hf \]
So, we can also write:
\[ E_{ph} = \frac{hc}{\lambda_{ph}} \]

### Step 3: Set the wavelengths equal
Since the problem states that the photon has the same wavelength as the de Broglie wavelength of the electron, we can set them equal:
\[ \lambda_e = \lambda_{ph} \]
Thus,
\[ \frac{h}{mv} = \frac{c}{f} \]

### Step 4: Express the energy of the photon
From the relationship for the energy of the photon, we can write:
\[ E_{ph} = \frac{hc}{\lambda_{ph}} = \frac{hc}{\lambda_e} \]
Substituting \( \lambda_e \) from Step 1:
\[ E_{ph} = \frac{hc}{\frac{h}{mv}} = \frac{mvc}{h} \]

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

### Step 6: Relate the energies
Now, we need to relate \( E_e \) and \( E_{ph} \). We can express the ratio of the energies:
\[ \frac{E_e}{E_{ph}} = \frac{\frac{1}{2} mv^2}{\frac{mvc}{h}} = \frac{h}{2c} \]

### Step 7: Simplify the expression
This simplifies to:
\[ \frac{E_e}{E_{ph}} = \frac{v}{2c} \]

### Conclusion
Thus, the correct relationship is:
\[ E_e = \frac{v}{2c} E_{ph} \]