The K.E. of electron and photon is same then relation between their De-Broglie wavelength :

Options

A. `lamda_pltlamda_e`
B. `lamda_p=lamda_e`
C. `lamda_pgtlamda_e`
D. `lamda_p=2lamda_e`

Explanation

To solve the problem, we need to find the relationship between the de Broglie wavelengths of an electron and a photon, given that their kinetic energies are the same. Let's break this down step by step.

### Step 1: Write the expressions for kinetic energy
The kinetic energy (K.E.) of an electron can be expressed as:
\[ K.E. = \frac{1}{2} m_e v_e^2 \]
where \( m_e \) is the mass of the electron and \( v_e \) is its velocity.

The energy of a photon is given by:
\[ E = \frac{hc}{\lambda_p} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda_p \) is the wavelength of the photon.

### Step 2: Set the kinetic energy of the electron equal to the energy of the photon
Since the kinetic energy of the electron and the energy of the photon are equal, we can write:
\[ \frac{1}{2} m_e v_e^2 = \frac{hc}{\lambda_p} \]

### Step 3: Express the de Broglie wavelength of the electron
The de Broglie wavelength \( \lambda_E \) of the electron can be expressed as:
\[ \lambda_E = \frac{h}{p_e} \]
where \( p_e \) is the momentum of the electron. The momentum can be expressed as:
\[ p_e = m_e v_e \]
Thus, the de Broglie wavelength becomes:
\[ \lambda_E = \frac{h}{m_e v_e} \]

### Step 4: Substitute the expression for momentum into the kinetic energy equation
Substituting \( p_e \) into the kinetic energy equation gives:
\[ \frac{1}{2} m_e v_e^2 = \frac{hc}{\lambda_p} \]
Substituting \( v_e \) from the de Broglie wavelength:
\[ \frac{1}{2} m_e v_e^2 = \frac{hc}{\lambda_p} \]
Now, substituting for \( v_e \):
\[ \frac{1}{2} m_e \left(\frac{h}{\lambda_E}\right)^2 = \frac{hc}{\lambda_p} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \frac{1}{2} \frac{m_e h^2}{\lambda_E^2} = \frac{hc}{\lambda_p} \]
Cancelling \( h \) from both sides:
\[ \frac{1}{2} \frac{m_e h}{\lambda_E^2} = \frac{c}{\lambda_p} \]

### Step 6: Relate the wavelengths
Rearranging gives:
\[ \frac{\lambda_p}{\lambda_E^2} = \frac{m_e h}{2c} \]
This shows that the ratio of the wavelengths depends on the mass of the electron and the speed of light.

### Step 7: Conclusion
Since \( c \) (the speed of light) is always greater than \( v_e \) (the speed of the electron), we can conclude that:
\[ \lambda_p > 2 \lambda_E \]
Thus, the relationship between the de Broglie wavelengths is:
\[ \lambda_p > \lambda_E \]

### Final Answer
The de Broglie wavelength of the photon is greater than that of the electron:
\[ \lambda_p > \lambda_E \]