What happens to de Broglie wavelength of electrons when it is accelerated by increasingly higher potential difference?

"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!

Text Solution

Solution : Here, `V=100` Volts.The de- Broglie wavelength `lambda " is " lambda =(1.227)/(sqrt(V))nm`. <br> `=(1.227)/(sqrt(100))=(1.227)/(10) =0.1227=0.123 nm`. <br> The value of de Broglie wavelength is associated with the wavelength of X-ray.