If `E_1, E_2 and E_3` represent respectively the kinetic energies of an electron, an `alpha - particle` and a proton each having same de-Broglie wavelength, then

Options

A. `E_1gtE_3gtE_2`
B. `E_2gtE_3gtE_1`
C. `E_1gtE_2gtE_3`
D. `E_1=E_2=E_3`

Explanation

To solve the problem, we need to analyze the relationship between the kinetic energies of an electron, an alpha particle, and a proton, given that they all have the same de-Broglie wavelength.

### Step-by-Step Solution:

1. **Understanding de-Broglie Wavelength**:
The de-Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( K \) is the kinetic energy of the particle.

2. **Rearranging the Formula**:
Rearranging the formula for kinetic energy, we get:
\[ K = \frac{h^2}{2m\lambda^2} \]
This shows that the kinetic energy \( K \) is inversely proportional to the mass \( m \) of the particle when the wavelength \( \lambda \) is constant.

3. **Comparing Masses**:
- Let \( m_e \) be the mass of the electron.
- Let \( m_p \) be the mass of the proton.
- Let \( m_{\alpha} \) be the mass of the alpha particle.

The masses are approximately:
- \( m_e \) (electron) < \( m_p \) (proton) < \( m_{\alpha} \) (alpha particle)

4. **Kinetic Energy Relations**:
Since \( K \) is inversely proportional to \( m \):
- For the electron: \( E_1 = K_e \) (highest kinetic energy since \( m_e \) is the smallest)
- For the proton: \( E_3 = K_p \)
- For the alpha particle: \( E_2 = K_{\alpha} \) (lowest kinetic energy since \( m_{\alpha} \) is the largest)

5. **Order of Kinetic Energies**:
From the above relationships, we can conclude:
\[ E_1 > E_3 > E_2 \]
Thus, the order of kinetic energies is:
- \( E_1 \) (electron) > \( E_3 \) (proton) > \( E_2 \) (alpha particle)

### Final Answer:
The correct order of kinetic energies is:
\[ E_1 > E_3 > E_2 \]