Question

4. Consider the potential step shown below with a beam of particles incident from the left. V(x)

Answer 1

(a) Energy of the incident particle is less than the height of the step E < V_0 Region 1 V_0 = 0 - hstrok^2/2 m d^2 psi_1/dx^2 = E psi_1 d^2 psi_1/dx^2 + 3 m/hstrok^2 E psi_1 = 0 d^2 psi_1/dx^2 + alpha^2 psi_1 = 0 alpha = squareroot 2 m E/ hstrok^2 Region 2 - hstrok^2/2 m d^2 psi_2/dx^2 + v_0 psi_2 = E psi_2 d^1 psi_2/dx^2 - 2 m/ hstrok^2 (v_0 - f)psi_2 = 0 d^2 psi_2/dx^2 - beta^2 psi_2 = 0 beta = squareroot 2 m/hstrok^2 (v_0 - t) \r\n General solution of equations are psi_1 = A e^i alpha x + B e^-I alpha x psi_2 = C e^beta x + D e^-beta x e^beta x rightarrow infinity as x rightarrow infinity C = 0 psi_i = D e^-beta x Applying the Boundary Condition (psi_1)_x = 0 = (psi_2)_x = 0 (d psi_1/dx)_x = 0 = (d psi_2/dx)_x = 0 we get B = I alpha + beta/I alpha - beta A