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_(1)gtE_(3)gtE_(2)`
B. `E_(2)gtE_(3)gtE_(1)`
C. `E_(1)gtE_(2)gtE_(3)`
D. `E_(1)=E_(2)=E_(3)`

Answer

To solve the problem, we need to analyze the kinetic energies of an electron, an alpha particle, and a proton, all having the same de Broglie wavelength.

### Step-by-step Solution:

1. **Understand the de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

2. **Relate Kinetic Energy to Momentum**:
The kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
We can express momentum \( p \) as:
\[ p = mv \]
Therefore, we can rewrite the kinetic energy in terms of momentum:
\[ KE = \frac{p^2}{2m} \]

3. **Substituting for Momentum**:
From the de Broglie wavelength formula, we can express momentum in terms of wavelength:
\[ p = \frac{h}{\lambda} \]
Substituting this into the kinetic energy expression gives:
\[ KE = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2} \]

4. **Analyzing the Kinetic Energy**:
Since \( \lambda \) and \( h \) are constants for all three particles (electron, proton, and alpha particle), the kinetic energy depends inversely on the mass of the particle:
\[ KE \propto \frac{1}{m} \]

5. **Comparing Masses**:
- The mass of an electron (\( m_e \)) is much smaller than that of a proton (\( m_p \)).
- The mass of an alpha particle (\( m_{\alpha} \)) is approximately four times that of a proton.
- Thus, we have the order of masses:
\[ m_e < m_p < m_{\alpha} \]

6. **Relating Kinetic Energies**:
Since kinetic energy is inversely proportional to mass:
- The electron, having the smallest mass, will have the highest kinetic energy (\( E_1 \)).
- The proton will have a higher kinetic energy than the alpha particle but lower than the electron (\( E_3 \)).
- The alpha particle, having the largest mass, will have the lowest kinetic energy (\( E_2 \)).

Therefore, we can conclude:
\[ E_1 > E_3 > E_2 \]

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