# Question about de broglie wavelengths

1. Mar 8, 2013

### daselocution

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

de Broglie wavelength λ=h/p

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

E=hf=hc/λ

3. 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?

2. Mar 8, 2013

### Staff: Mentor

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. Mar 8, 2013

### daselocution

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.