Question

Which physical quantity will be the same for an electron and a photon of the same wavelength?

• A. Velocity
• B. Energy
• C. Momentum
• D. Angular momentum

Hint:

The de Broglie relation \(p = h/\lambda\) is universal. It's the bridge that connects the particle property (momentum, \(p\)) to the wave property (wavelength, \(\lambda\)) for any quantum entity, whether it's a photon or a particle with mass.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question deals with the wave-particle duality, specifically the de Broglie wavelength for matter particles (like electrons) and the properties of photons. We need to compare different physical quantities for an electron and a photon that share the same wavelength \(\lambda\).
Step 2: Key Formula or Approach:
The key relationship connecting wavelength and momentum is the de Broglie relation, which applies to both matter particles and photons.
\[ p = \frac{h}{\lambda} \] where \(p\) is momentum, \(h\) is Planck's constant, and \(\lambda\) is the wavelength.
Step 3: Detailed Explanation:
Let's analyze each quantity for an electron and a photon with the same wavelength \(\lambda\).
(A) Velocity: A photon always travels at the speed of light, \(c\). An electron has rest mass, so its speed \(v\) must be less than \(c\). Therefore, their velocities are not the same.
(B) Energy: - The energy of a photon is given by \(E_{\text{photon}} = hf = \frac{hc}{\lambda}\). - The kinetic energy of a non-relativistic electron is \(E_{\text{electron}} = \frac{p^2}{2m} = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2}\). Since the formulas are different, their energies will not be the same.
(C) Momentum: - The momentum of a photon is given by \(p_{\text{photon}} = \frac{h}{\lambda}\). - The de Broglie momentum of an electron is given by \(p_{\text{electron}} = \frac{h}{\lambda}\). Since both \(h\) and \(\lambda\) are the same for the electron and the photon, their momenta must be equal.
(D) Angular momentum: This quantity is not inherently defined just by wavelength and would depend on other conditions (like orbital motion), so it is not guaranteed to be the same.
Step 4: Final Answer:
For the same wavelength \(\lambda\), both the electron and the photon will have the same momentum, \(p = h/\lambda\). Thus, option (C) is correct.