Question

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 :

• A. \( K_p = K_e \) and \( P_p = P_e \)
• B. \( K_p>K_e \) and \( P_p = P_e \)
• C. \( K_p<K_e \) and \( P_p = P_e \)
• D. \( K_p<K_e \) and \( P_p<P_e \)

Hint:

For two particles with the same de-Broglie wavelength, the lighter particle will always possess more kinetic energy.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The de-Broglie wavelength \( \lambda \) of a particle is determined by its momentum \( P \).
Kinetic energy \( K \) is related to momentum \( P \) and mass \( m \).
Step 2: Key Formula or Approach:
1. de-Broglie wavelength: \( \lambda = \frac{h}{P} \implies P = \frac{h}{\lambda} \).
2. Kinetic energy: \( K = \frac{P^2}{2m} \).
Step 3: Detailed Explanation:
Given that \( \lambda_p = \lambda_e \):
From the wavelength formula, since \( h \) is a constant, equal wavelengths imply equal momenta:
\[ P_p = P_e \]
Now, looking at the Kinetic Energy:
\[ K = \frac{P^2}{2m} \]
Since \( P \) is the same for both, \( K \) is inversely proportional to the mass:
\[ K \propto \frac{1}{m} \]
We know that the mass of a proton (\( m_p \)) is much greater than the mass of an electron (\( m_e \)):
\[ m_p>m_e \]
Therefore:
\[ K_p<K_e \]
Step 4: Final Answer:
The correct option is \( K_p<K_e \) and \( P_p = P_e \).