Wave Particle Duality For Electrons and Photons

  • #1
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Homework Statement


Discuss the concept of the wave-particle duality for electrons and photons and include an equation which connects the wave like and particle like properties.

Homework Equations




The Attempt at a Solution


So I am having trouble with how to word this question and generally getting my self in a muddle with things. Here what I have so far on the topic

Wave-Particular duality of electron: As and electron is assumed to be a particle when a stream of electrons is aimed at two slits the electron creates a interference pattern of high and low intensities which as shining light though the two slits the same same pattern is produce thus concluding the wave nature of the electron.

Wave-Particle Duality of A Photon: As light is assume to be a wave, when light is shined upon a metal with high enough frequency, it will allow electron to escape from that metal, the only way this is possible is if the light be shone on to the metal come in form of discrete packets of energy know as a photon ( a elementary particle of light), as the photon collides in a elastic collision with the electron the energy of the photon given by ##E_{photon}=hf## is transferred to the electron to give the electron enough K.E to escape.

This is the part that really confuse me and think some of the confusion has to do with the history.

But cant wave particle duality be seen in one with Compton scattering? What I mean is that Compton used the equation ##E_{photon}=hc/ \lambda## to calculate a given wave length for his x-rays and measure the the scattered wavelength with the respect to the scattering angle and showed that by comparing the results with the ##E_{photon}## he verified the particle nature of light but surely as this x-ray is being diffracted at some angle is this not equally showing the wave nature of light as well?

I assume that equation it is referring to is the De Brogile relationship as he was the first person to think if a photon show the particle/wave maybe electron, proptons and matter dose too.

so then the given equation would be

$$\lambda=h/p=h/m_ev$$
 

Answers and Replies

  • #2
scottdave
Science Advisor
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Looks like a pretty good start. A couple of misspellings. Its not the easiest concept to grasp. You might get something from this SixtySymbols video on YouTube:
 
Last edited:
  • #3
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Ah thank you I have just watch the video, much appreciated.
 

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