Neglecting the mass variation with velocity , the ratio of the wavelength `((lamda_e)/lamda_p)` associated with an electron `(lamda_e)` having a kinetic energy E and wavelength associated with a Photon `(lamda_p)` having kinetie energy 4E is

Options

A. `2:1`
B. `1:2`
C. `4:1`
D. `1:4`

Explanation

To solve the problem, we need to find the ratio of the wavelengths associated with an electron and a photon, given their kinetic energies.

### Step-by-Step Solution:

1. **Understanding the Wavelengths**:
- The wavelength associated with a particle can be calculated using the de Broglie wavelength formula for the electron:
\[ \lambda_e = \frac{h}{p_e} \]
where \( p_e \) is the momentum of the electron.
- For a photon, the wavelength is given by:
\[ \lambda_p = \frac{h}{p_p} \]
where \( p_p \) is the momentum of the photon.

2. **Relating Kinetic Energy to Momentum**:
- The kinetic energy \( E \) of the electron is given by:
\[ E = \frac{p_e^2}{2m} \]
Rearranging gives:
\[ p_e = \sqrt{2mE} \]
- For the photon, the kinetic energy is related to its momentum by:
\[ E = pc \quad \text{(where \( c \) is the speed of light)} \]
Thus, the momentum of the photon is:
\[ p_p = \frac{E}{c} \]

3. **Substituting for Wavelengths**:
- Substitute the expressions for momentum into the wavelength formulas:
\[ \lambda_e = \frac{h}{\sqrt{2mE}} \]
\[ \lambda_p = \frac{h}{\frac{4E}{c}} = \frac{hc}{4E} \]

4. **Finding the Ratio**:
- Now, we can find the ratio of the wavelengths:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{4E}} = \frac{4E}{\sqrt{2mE} \cdot c} \]
- Simplifying this gives:
\[ \frac{\lambda_e}{\lambda_p} = \frac{4\sqrt{E}}{\sqrt{2m} \cdot c} \]

5. **Substituting Kinetic Energies**:
- Since the kinetic energy of the photon is \( 4E \), we can substitute to find the ratio:
\[ \frac{\lambda_e}{\lambda_p} = 2 \]

### Final Answer:
Thus, the ratio of the wavelength associated with the electron to the wavelength associated with the photon is:
\[ \frac{\lambda_e}{\lambda_p} = 2 : 1 \]