Relation between electric potential and wavelength of an electron

Question: An electron that is accelerated from rest through an electric potential difference of V has a de Broglie wavelength of λ. Investigate the relationship between V and λ." I had two arguments that led to two significantly distinct results.

The first argument: λ∝1V
. From the Planck relation, we have:

E=hf=hcλ

But after acceleration through the electric potential difference, the energy of the electron is eV, where e is the charge of the electron. As such,

eV=hcλ

λ=hce⋅1V

⟹λ∝1V

The second argument: λ∝1V√
. From de Brogile's equation, we have

p=hλ

mv=hλ

m2v2=h2λ2

m2v22m=h22mλ2

12mv2=h22m⋅1λ2

But since the kinetic energy of the electron is equal to the energy gained from accelerating through the electric potential,

eV=h22m⋅1λ2

λ2=h22meV

λ=h2me−−−−√⋅1V−−√

⟹λ∝1V−−√

I suppose that I'm having some conceptual errors because both seem valid to me. Which is the correct one, and why is the other one incorrect? Any help is greatly appreciated. Thanks!

Answer

Your second strategy is the correct one. The problem with your first attempt is your assumption that: f=cλ

This is true for light, and in fact it's true for any wave as long as we replace c
by the phase velocity of the wave. The trouble is that the velocity you're calculating using the kinetic energy is the group velocity not the phase velocity, so your equation can't be applied to the electron.

The relationship between wavelength, phase and group velocity has already been discussed at some length in the question De Broglie wavelength, frequency and velocity - interpretation.

Reference:

 http://physics.stackexchange.com/questions/116413/relation-between-electric-potential-and-wavelength-of-an-electron