Question

Through what potential difference must an electron be accelerated from rest to have a de Broglie wavelength of $500\mathrm{nm}$?

Short answer

The potential difference$\Delta V=6.02·{10}^{-6}\mathrm{V}$

Step-by-step solution

Given information 

To find the potential difference of electron from rest to de Broglie wave length of$500nm$

Calculation 

We can start the solution with expression for de-Broglie wavelength

$$\begin{gathered} \lambda =\frac{h}{mv} \\ v=\frac{h}{m\lambda } \\ {E}_k=U \\ \frac{m{v}^2}{2}=e\Delta V \\ \Delta V=\frac{m{v}^2}{2e} \\ \Delta V=\frac{m{h}^2}{2e{m}^2{\lambda }^2} \\ \Delta V=\frac{h^2}{2em{\lambda }^2} \\ \Delta V=\frac{2·1.6·{10}^{-19}·9.11·{10}^{-31}·{\left(500·{10}^{-9}\right)}^2}{2V} \\ \Delta V=6.02·{10}^{-6}\mathrm{V} \end{gathered}$$