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Answer:
one half as large , initial velocity is two times larger
Explanation:
Momentum is conserved.
p₁ + p₂ = p₁' + p₂'
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
m₁ = m₂ =m , v₂= 0
v₁' =v₂'
mv₁ = 2mv₁'
v₁ = 2v₁'
Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)
So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀
Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)
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:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5
