Finding Kinetic Energy of an Electron with 100keV X-Ray Wavelength

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Homework Help Overview

The discussion revolves around finding the kinetic energy of an electron whose de Broglie wavelength matches that of a 100 keV x-ray. The subject area includes concepts from quantum mechanics and particle physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between de Broglie wavelength and momentum, questioning the applicability of certain formulas for photons versus electrons. There is discussion about converting energy units and determining momentum.

Discussion Status

The conversation is active, with participants providing insights into the use of the de Broglie formula and the differences in momentum calculations for electrons and photons. Some guidance has been offered regarding the conversion of energy and the relationship between momentum and wavelength.

Contextual Notes

Participants are navigating the complexities of applying formulas correctly, particularly in distinguishing between the properties of photons and electrons. There is an emphasis on ensuring proper unit conversions and understanding the definitions involved.

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how do you find the kinetic energy of an electron whose de Broglie wavelength is the same as that of a 100keV x-ray?
 
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Since the de Broglie wavelength is given by the momentum of the particle, I guess you would first determine the momentum p of the photon: p = E/c (I think), then find the kinetic energy of the electron via 0.5p^2/m, where m is the electron mass.
 
it's not a photon, but can you still use that formula?
 
Yes, you can use the de Broglie formula h/p to find the wavelength of an electron or a photon. The difference between the two comes in determining p. For an electron, p = mv. For a photon, p = E/c.

So if your photon has energy E = 100 keV, convert this to Joules and use p = E/c to find the momentum of the photon.

If the de Broglie wave of the electron has the same wavelength, it must have the same momentum p.
 
ok, thanks!
 

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