A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :

Options

A. `k_(p)=k_(e )` and `P_(p)=P_(e )`
B. `k_(p)gtk_(e )` and `P_(p)=P_(e )`
C. `k_(p)ltk_(e )` and `P_(p)=P_(e )`
D. `k_(p)ltk_(e )` and `P_(p)ltP_(e )`

Answer

To solve the problem, we need to analyze the relationship between the de-Broglie wavelength, momentum, and kinetic energy of a moving proton and electron that have the same de-Broglie wavelength.

### Step-by-Step Solution:

1. **Understanding de-Broglie Wavelength**:
The de-Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Setting the Wavelengths Equal**:
Since the problem states that the proton and electron have the same de-Broglie wavelength, we can write:
\[ \lambda_e = \lambda_p \]
This implies:
\[ \frac{h}{p_e} = \frac{h}{p_p} \]
Thus, we can conclude that:
\[ p_e = p_p \]
where \( p_e \) is the momentum of the electron and \( p_p \) is the momentum of the proton.

3. **Relating Kinetic Energy to Momentum**:
The kinetic energy (\( K \)) of a particle can be expressed in terms of momentum (\( p \)) as:
\[ K = \frac{p^2}{2m} \]
where \( m \) is the mass of the particle.

4. **Kinetic Energy of Electron and Proton**:
For the electron:
\[ K_e = \frac{p_e^2}{2m_e} \]
For the proton:
\[ K_p = \frac{p_p^2}{2m_p} \]
Since \( p_e = p_p \), we can substitute \( p \) in both equations:
\[ K_e = \frac{p^2}{2m_e} \]
\[ K_p = \frac{p^2}{2m_p} \]

5. **Comparing Kinetic Energies**:
Since the mass of the electron (\( m_e \)) is much less than the mass of the proton (\( m_p \)), we can analyze the kinetic energies:
\[ K_e = \frac{p^2}{2m_e} \quad \text{and} \quad K_p = \frac{p^2}{2m_p} \]
Given that \( m_e < m_p \), it follows that:
\[ K_e > K_p \]
This means the kinetic energy of the electron is greater than that of the proton.

6. **Conclusion**:
From our analysis, we conclude that:
- The momentum of the electron and proton is the same.
- The kinetic energy of the electron is greater than the kinetic energy of the proton.

### Final Answer:
- Kinetic energy of the electron \( K_e \) is greater than kinetic energy of the proton \( K_p \).
- Momentum \( p_e \) is equal to momentum \( p_p \).