Question about de broglie wavelengths

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SUMMARY

The discussion centers on calculating the energy of a photon that has a wavelength equal to the de Broglie wavelength of an electron. The key equations used include the de Broglie wavelength formula (λ=h/p) and the energy-momentum relation (E^2 = p^2c^2 + m^2c^4). The conclusion reached is that the energy of the photon is equal to the energy of the electron (Ephoton = Eelectron = E). The second part of the problem, which assumes E = 2E0, revealed a unit mismatch error in the participant's calculations, which was subsequently corrected.

PREREQUISITES
  • Understanding of de Broglie wavelength (λ=h/p)
  • Familiarity with energy-momentum relation (E^2 = p^2c^2 + m^2c^4)
  • Knowledge of photon energy formula (E=hf=hc/λ)
  • Basic concepts of relativistic energy and momentum
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  • Study the implications of relativistic effects on particle energy and momentum
  • Explore advanced applications of de Broglie wavelength in quantum mechanics
  • Learn about the relationship between energy and wavelength in electromagnetic radiation
  • Investigate common errors in unit conversions in physics problems
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Students of physics, particularly those studying quantum mechanics and particle physics, as well as educators looking for examples of energy calculations involving photons and electrons.

daselocution
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Homework Statement


Assume that the total energy E of an electron greatly exceeds its rest energy E0. If a photon has a wavelength equal to the de Broglie wavelength of the electron, what is the photon’s energy? Repeat the prob- lem assuming E = 2E0 for the electron.

I need help with the first part of the problem--I included my answer to the second part in case it is relevant to the first.

Homework Equations



de Broglie wavelength λ=h/p

E^2 = p^2c^2 + m^2c^4

E=hf=hc/λ

The Attempt at a Solution



Part one of the problem:

Knowing that E>>Erest, I can use the mass-energy relation to show that E^2 = P^2c^2, such that E=pc. From this I know that p=E/c

I used this to show that the energy of the photon must be equal to the energy of the electron as follows:

λelectron = λphoton
λelectron = h/p = h/(E/c) = hc/Eelectron
λphoton = hc/Ephoton
hc/Eelectron = hc/Ephoton ---> Eelectron = Ephoton

From here all I can see is that there is an infinite number of solutions. I don't understand how to winnow my process down so that it yields only one solution. That said, I don't even know if my process is 100% correct.

Part two of the problem:
λ=h/pelectron=hc/Ephoton
Ephoton=c/pelectron

E^2 = p^2c^2 + E0^2 = (2Eo)^2 = 4Eo^2
3Eo^2 = p^2c^c
p=√3 * (Eo)/c

such that:
Ephoton=c/pelectron = c/(√3 * (Eo)/c), all of which are constants that I know the values of and which give me a real answer.

What say you all about the first part of the problem?
 
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daselocution said:
What say you all about the first part of the problem?
You got the correct, unique solution: Ephoton=Eelectron=E.

For the second problem, the units in your answer do not match, there has to be some error.
 
Ahh thank you very much, I made a writing mistake with the second problem that I corrected in my homework. I didn't realize that the first solution was the solution--again, thanks.
 

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