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

To solve this problem, we need to analyze the properties of a proton and an electron that have the same de-Broglie wavelength. Let's break it down step-by-step: 

Understanding the de-Broglie wavelength: The de-Broglie wavelength (\(\lambda\)) is given by the equation: \(\lambda = \frac{h}{p}\) where \(h\) is Planck’s constant and \(p\) is the momentum of the particle. If the de-Broglie wavelength is the same for both particles, we have: \(\lambda_p = \lambda_e\) which implies: \(\frac{h}{p_p} = \frac{h}{p_e}\) Thus, it follows that: \(p_p = p_e\)

Relating Kinetic Energy (K.E.) and Momentum: The kinetic energy (\(K\)) for a particle is given by: \(K = \frac{p^2}{2m}\) where \(m\) is the mass of the particle, and \(p\) is momentum. For the proton and the electron, the expressions for kinetic energy become: \(K_p = \frac{p_p^2}{2m_p}\) and \(K_e = \frac{p_e^2}{2m_e}\) Since \(p_p = p_e\), we can substitute those into the kinetic energy equations: \(K_p = \frac{p^2}{2m_p}\) and \(K_e = \frac{p^2}{2m_e}\)

Comparing Kinetic Energies: Since the mass of the proton (\(m_p\)) is greater than the mass of the electron (\(m_e\)), the kinetic energy of the proton will be less than that of the electron, given the same momentum: \(K_p < K_e\)

Therefore, the correct option is:

• \( K_p < K_e \) and \( P_p = P_e \)

This choice states that the kinetic energy of the proton is less than that of the electron and that their momenta are equal, which matches our derivation. This is consistent with both de-Broglie's hypothesis and the formulas for kinetic energy and momentum in relation to mass.