What is the de Broglie wavelength of the wave associated with an electron that has been accelerated through a potential difference of 50.0 V?

The gain of kinetic energy by an electron is eV.

$\begin{array}{l}\frac{1}{2}m{v}^2=eV\\ v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2\left(1.60\times {10}^{-19}\right)\left(50\right)}{\left(9.11\times {10}^{-31}\right)}}\\ =4.19\times {10}^6{\mathrm{ms}}^{-1}\end{array}$

Thus, the electron's de Broglie wavelength is

$\begin{array}{c}\lambda =\frac{h}{mv}=\frac{6.63\times {10}^{-34}}{\left(9.1\times {10}^{-31}\right)\left(4.19\times {19}^6\right)}\\ =1.74\times {10}^{-10}\mathrm{m}\end{array}$