DeBroglie Wavelength with Relativistic Electron

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Discussion Overview

The discussion revolves around calculating the DeBroglie wavelength of an electron moving at 0.8 times the speed of light (0.8c), with a focus on the application of relativistic momentum. Participants explore the implications of special relativity on momentum and wavelength calculations, as well as the appropriate use of units in these calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant attempts to calculate the DeBroglie wavelength using the formula for relativistic momentum, expressing uncertainty about the correctness of their application.
  • Another participant questions the units used in the calculations and clarifies that the mass in the momentum formula should be the rest mass, not the relativistic mass.
  • A third participant reiterates the importance of using SI units and provides the correct formula for relativistic momentum, suggesting that the use of SI units may not be intuitive in high-energy physics.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the initial calculations, and there is disagreement regarding the application of the mass term in the momentum formula. The discussion remains unresolved with multiple viewpoints presented.

Contextual Notes

There are limitations regarding the assumptions made about the mass used in the momentum formula and the clarity of units applied in the calculations. The discussion does not resolve these issues.

Oaxaca
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I am trying to find the DeBroglie wavelength of an electron moving at .8c. I have never learned special relativity but I believe the momentum is affected (mass change). I used the formula p= (mv)/(1-v^2/c^2) and got a momentum of p = 2.733 E-22 and a wavelength of lamda = 2.4149 E-12. Did I apply the formula correctly for the relativistic momentum?

Thanks for any help
 
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Neither ##p=2.733\times{10}^{-22}## grain-furlongs per fortnight nor ##\lambda=2.4149\times{10}^{-12}## parsecs would be correct, but I'm pretty sure you didn't use those units. What units did you use?

The ##m## in the formula you cite is the rest mass, not the relativistic mass (and we have a FAQ on why relativistic mass is seldom used - look for a link to it in the sticky thread at the top of this forum). As long as you get that right and pay attention to the units, you should come to the right answer.
 
Nugatory said:
Neither ##p=2.733\times{10}^{-22}## grain-furlongs per fortnight nor ##\lambda=2.4149\times{10}^{-12}## parsecs would be correct, but I'm pretty sure you didn't use those units. What units did you use?

The ##m## in the formula you cite is the rest mass, not the relativistic mass (and we have a FAQ on why relativistic mass is seldom used - look for a link to it in the sticky thread at the top of this forum). As long as you get that right and pay attention to the units, you should come to the right answer.
I used kg and m/s for the electron, with momentum therefor being kg*m/s - and meters for my wavelength. Thanks for the response!
 
The correct formula is
$$\vec{p}=\frac{m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}.$$
I'm to lazy to check your formula with quantities given in SI units which are not very intuitive to use in high-energy physics and thus why I've never seen anybody using them there in scientific work in this field ;-)).
 

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