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

$KE=\frac{1}{2}m{\upsilon }^2=\frac{p^2}{2m}thus\frac{p^2}{2m}=eVorp=\sqrt{2meV}$

Now           $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2meV}}=\frac{6.626\times {10}^{-34}Js}{{\left[\right(2\left)\right(9.1\times {10}^{-31}kg\left)\right(1.6\times {10}^{-19}C\left)\right(500V\left)\right]}^{1/2}}$

                                   $=5.49\times {10}^{-11}m=54.9pm$