Question about de broglie wavelengths

In summary, the conversation discusses finding the energy of a photon with a wavelength equal to the de Broglie wavelength of an electron when the electron's total energy greatly exceeds its rest energy. It also includes a repeat of the problem with a different value for the electron's total energy. The summary concludes that the unique solution for both parts is that the energy of the photon is equal to the energy of the electron.
  • #1
daselocution
23
0

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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  • #2
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.
 
  • #3
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.
 

1. What is the de Broglie wavelength theory?

The de Broglie wavelength theory, also known as wave-particle duality, states that all matter exhibits both wave-like and particle-like behavior. This means that all particles, including those with mass, have a wavelength associated with them.

2. How is the de Broglie wavelength calculated?

The de Broglie wavelength is calculated using the equation λ = h/mv, where λ is the de Broglie wavelength, h is Planck's constant, m is the mass of the particle, and v is the velocity of the particle.

3. What is the significance of the de Broglie wavelength?

The de Broglie wavelength is significant because it helps explain the wave-like behavior of particles at the quantum level. It also allows us to better understand the behavior of particles in various physical phenomena, such as diffraction and interference.

4. How does the de Broglie wavelength relate to the uncertainty principle?

The de Broglie wavelength is related to the uncertainty principle in that it is one of the fundamental principles of quantum mechanics. The uncertainty principle states that it is impossible to know both the position and momentum of a particle with absolute certainty. The de Broglie wavelength is a manifestation of this principle, as it describes the uncertainty in the position of a particle due to its wave-like nature.

5. Can the de Broglie wavelength be observed experimentally?

Yes, the de Broglie wavelength has been observed experimentally in various experiments, such as the double-slit experiment. This experiment showed that particles, such as electrons, exhibit wave-like behavior and have a corresponding de Broglie wavelength. Other experiments, such as electron diffraction, also support the existence of the de Broglie wavelength.

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