What is the kinetic energy in eV of an electron whose de Broglie wavelength is 19% of the...


What is the kinetic energy in eV of an electron whose de Broglie wavelength is 19% of the wavelength of a photon having the energy equal to electrons kinetic energy?

Wave-particle Duality:

This principle states that every particle in the universe can be treated as a wave as well as a particle. This principle laid the foundation of Quantum Mechanics. The first verification of this principle came from the electron diffraction experiment in which the electrons acted as a wave rather than a finite-sized particle.

The de-Broglie wavelength of a particle is the wavelength of the wave which represents the particle. This wavelength depends only on the momentum of the particle.

Since the kinetic energy of a particle depends on its momentum, and momentum is related to the de-Broglie wavelength of the particle, the kinetic energy of the particle is also related to its de Broglie wavelength.

For light, the particle is called 'photon'. The energy a photon depends on the frequency of the light. This assumption was first made by Max Planck to solve the famous Black Body problem.

Answer and Explanation: 1

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Given Data:

The wavelength, {eq}\lambda_e {/eq}, of the electron, is {eq}19% {/eq} of the wavelength of the photon, {eq}\lambda {/eq}.


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Wave-Particle Duality & the Davisson-Germer Experiment


Chapter 23 / Lesson 2

Wave-particle duality means that light and sub-atomic particles can also behave as waves, hence the duality. Study the definition of wave-particle duality, the Davisson-Germer Experiment, the de Broglie wavelength, and an example calculation.

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