Why is deBroglie λ for electrons the same as λ for photons?

[Tommy R]

Hi, I got the following question in my textbook: [translated]"Compare the wavelength of a photon and an electron where the photon and the electron have the same momentum".
My thinking is the following:
Firstly, p_p (photon) = p_e (electron).
My textbook briefly mentions the extention of the mass-energy equivalence E² = p² c² + m² c⁴, so I go by this since the particles have different speeds. The de Broglie wavelength of the electron is given by f_e = E/h = √(p² c² + m² c⁴) / h. The wavelength of the photon is the same except that the mass is 0, so it reduces to f_p = √(p² c²) / h. Since m² c⁴ ≥ 0 it follows that f_e ≥ f_p.

My textbook says that the answer is that f_e = f_p tho. Their argument is that p = h/λ holds for both the electron and the photon. But they previously state that it only applies for massless particles. They derive it from E² = p² c² + m² c⁴ by setting m to 0. [translated]"[...]E = pc which holds for massless particles.[...] we find that p=h/λ".

Why is my argument invalid and why does λ=h/p hold for the electron? Thanks!

 

[PeterDonis]

The de Broglie wavelength of the electron is given by f_e = E/h

No, it isn't. $E/h$ is a frequency, not a wavelength.

My textbook says that the answer is that f_e = f_p tho.

Does it? With $f_e$ defined as $E / h$? Or does it have a different formula for the de Broglie wavelength?

Their argument is that p = h/λ holds for both the electron and the photon.

And that is correct. Or, rearranging the formula, $\lambda = p / h$. Not $E / h$.

 

[Tommy R]

No, it isn't. $E/h$ is a frequency, not a wavelength.
Does it? With $f_e$ defined as $E / h$? Or does it have a different formula for the de Broglie wavelength?
And that is correct. Or, rearranging the formula, $\lambda = p / h$. Not $E / h$.

The deBroglie wavelength as f_e was a typo, I meant frequency. I now notice that I without really thinking about it assumed E=hf applied for the electron. I cannot this? (The book's definition of deBroglie wavelength is h/p, yes)

 

[PeterDonis]

I now notice that I without really thinking about it assumed E=hf applied for the electron.

It does. What changes between the electron and the photon is the relationship between E and p; that should be obvious from the equations you wrote down in the OP. That in turn implies a change in the relationship between $f$ and $\lambda$. (This relationship is often called a "dispersion relation" in the literature, and the difference I've just described is called a difference in the dispersion relation between massive and massless particles.)