An electron is accelerated through a potential difference of 500 V. Its de Broglie wavelength would be

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2=\frac{{\mathrm{p}}^2}{2\mathrm{m}}$ 

Thus $\frac{{\mathrm{p}}^2}{2\mathrm{m}}=\mathrm{eV}$ or 

$\mathrm{p}=\sqrt{2\mathrm{meV}}$

Now

 $\begin{array}{l}\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2\mathrm{meV}}}=\frac{6.626\times {10}^{-34}\mathrm{JS}}{{\left[\left(2\right)\left(9.1\times {10}^{-31}\mathrm{kg}\right)\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(500\mathrm{V}\right)\right]}^{1/2}}\\ =5.49\times {10}^{-11}\mathrm{m}=54.9\mathrm{pm}\end{array}$