Question

The de Broglie wavelengths for an electron and a photon are λ_e and λ_p, respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two?

• A. \(λ_p ∝ λ^{2}_e\)
• B. \(λ_p ∝ λ_e\)
• C. \(λ_p ∝ \sqrtλ_e\)
• D. \(λ_p ∝ \sqrt{\frac{1}{λ_e}}\)

Answer 1

The Correct Option is A

To determine the relationship between the de Broglie wavelengths of an electron and a photon when they have the same kinetic energy, we start by considering the de Broglie wavelength formulae for both particles:

The de Broglie wavelength λ for a particle is given by:

λ = \frac{h}{p}

where h is Planck's constant and p is the momentum of the particle.

In the case of an electron, the kinetic energy (KE) is related to its momentum (p_e) by:

KE = \frac{p_e^2}{2m_e}

Solving for p_e gives:

p_e = \sqrt{2m_e \cdot KE}

Substituting in the de Broglie wavelength formula for the electron:

λ_e = \frac{h}{\sqrt{2m_e \cdot KE}}

For a photon, the kinetic energy (KE) is related directly to its momentum (p_p) and the speed of light (c) as:

KE = p_p \cdot c

So, the momentum of the photon is:

p_p = \frac{KE}{c}

Substituting in the de Broglie wavelength formula for the photon:

λ_p = \frac{h}{\frac{KE}{c}} = \frac{h \cdot c}{KE}

Since the kinetic energy of both the electron and the photon is the same, we equate the expressions for kinetic energy:

KE = \frac{h^2}{2m_e \cdot λ_e^2} = h \cdot c / λ_p

Solving for λ_p in terms of λ_e, we get:

λ_p = \frac{hc \cdot λ_e^2}{h^2} = \frac{c \cdot λ_e^2}{h}

This tells us that:

λ_p \propto λ^{2}_e

Thus, the correct relationship between the de Broglie wavelengths of the electron and the photon is λ_p \propto λ^{2}_e, which corresponds to the given correct answer.