# Potential energy of mass m is U(x)=Eₒ,when 0≤x≤1, =0when x≻1.The total energy of the particle is 2Eₒ. The De Broglie wave length of the particle is λ₁ when 0≤x≤1 and λ₂ when x≻1. what is the ratio of λ₁ and λ₂ ?

Jun 18, 2018
• For $x \in \left[0 , 1\right]$, total energy is ${E}_{1} = 2 {E}_{o}$, and potential energy is ${U}_{1} = {E}_{o}$. So kinetic energy is ${T}_{1} = {E}_{1} - {U}_{1} = {E}_{o}$

• For $x \in \left(1 , \infty\right)$, total energy again is ${E}_{2} = 2 {E}_{o}$, but potential energy is ${U}_{2} = 0$. So kinetic energy is ${T}_{2} = 2 {E}_{o}$

• $\implies {T}_{2} / {T}_{1} = 2$

From the deBroglie relation $\lambda = \frac{h}{p}$ and because $T = {p}^{2} / \left(2 m\right)$:

• $\lambda = \frac{h}{\sqrt{2 m T}} q \quad \therefore \lambda \propto \frac{1}{\sqrt{T}}$

$\implies {\lambda}_{1} / {\lambda}_{2} = \sqrt{{T}_{2} / {T}_{1}} = \sqrt{2}$