Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):Â
K = ½mv²Â
v = √(2K/m)Â
λ = h/(mv)Â
= h/(m√(2K/m))Â
= h/√(2Km)Â
So λ is proportional to 1/√K.Â
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:Â
K = 6Uâ‚€Â
Outer the potential well the potential energy is Uâ‚€, soÂ
K = 5Uâ‚€Â
(because kinetic and potential energies add up to 6Uâ‚€)Â
Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:Â
1/√(5Uâ‚€) : 1/√(6Uâ‚€)Â
= √6 : √5