Relation between electric potential and wavelength of an electron

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

The first argument: $\lambda \propto \frac{1}{V}$. From the Planck relation, we have: $$E = hf = h\frac{c}{\lambda}$$

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, $$eV = h\frac{c}{\lambda}$$ $$\lambda = \frac{hc}{e}\cdot\frac{1}{V}$$ $$\implies \lambda \propto\frac{1}{V}$$

The second argument: $\lambda \propto \frac{1}{\sqrt{V}}$. From de Brogile's equation, we have

$$p = \frac{h}{\lambda}$$ $$mv = \frac{h}{\lambda}$$ $$m^2v^2 = \frac{h^2}{\lambda^2}$$ $$\frac{m^2v^2}{2m} = \frac{h^2}{2m\lambda^2}$$ $$\frac{1}{2}mv^2 = \frac{h^2}{2m}\cdot\frac{1}{\lambda^2}$$ But since the kinetic energy of the electron is equal to the energy gained from accelerating through the electric potential, $$eV = \frac{h^2}{2m}\cdot\frac{1}{\lambda^2}$$ $$\lambda^2 = \frac{h^2}{2meV}$$ $$\lambda = \frac{h}{\sqrt{2me}}\cdot\frac{1}{\sqrt{V}}$$ $$\implies \lambda \propto \frac{1}{\sqrt{V}}$$

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!

• @JohnRennie that should be an answer Commented Jun 2, 2014 at 9:05

Your second strategy is the correct one.

$$f = \frac{c}{\lambda}$$
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.
• What, then, would be the correct relation between $f$ and $\lambda$ for an electron? Commented Jun 6, 2014 at 6:35
• For an electron, or indeed any particle, with momentum $p$ and total energy $E$: $\lambda = h/p$ and $f = E/p$. For non-relativistic particles some algebra gives $v_p \approx c^2/v$. Commented Jun 6, 2014 at 7:13