The kinetic energy of the electron will be equal to the energy what the electric field gives to it.1/2mv2=VQFrom thisv=sqrt(2VQ/m)
De Broglie wavelength:lambda=h/p=h/(mv) We know that h is the planck constant, V=300V Q is the charge of an electron and m is the mass of it.
If we substitute these into the equations above we get lambda=710-11m