Question

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 ?

Answer 1

• For `x in [0,1]`, 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 (1,oo)`, total energy again is `E_2 = 2 E_o`, but potential energy is `U_2 = 0`. So kinetic energy is `T_2 = 2E_o`

• `implies T_2/T_1 = 2`

From the deBroglie relation `lambda = h /p` and because `T = p^2/(2m)`:

• `lambda = h / sqrt(2mT) qquad :. lambda propto 1/sqrtT`

`implies lambda_1/lambda_2 = sqrt( T_2/T_1 ) = sqrt 2`