In a technique used for surface analysis called Auger electron spectroscopy

In a technique used for surface analysis called auger electron spectroscopy (AES), electrons are accelerated toward a metal surface. These electrons cause the emissions of secondary electrons-called auger electrons-from the metal surface. The kinetic energy of the auger electrons depends on the composition of the surface. The presence of oxygen atoms on the surface results in auger electrons with a kinetic energy of approximately 525 eV.

What is the de Broglie wavelength of this electron?
[KE = 1\2mv^2; 1 electron volt eV = 1.602*10^-19J]


How do I do this problem? chem is hard :(

momentum of electron = p = Sqrt ( 2* m* KE)
=
de Broglie wavelength = h/p, where h is Plancks constant = 6.63* 10^-34 Js.