Question

An electron in region I with a kinetic energy E < V_o is approaching the step potential as shown in the figure below. To determine how deep the electron can tunnel into the classical forbidden region II, calculate the penetration length l of the electron, defined as the distance $\Delta$ x where the probability density | $\Psi$ |² =   $\Psi ^{*}\Psi$ of the penetrating electron has dropped to 1/e of its value at x = 0.

Use: E = 3 eV
V(x) = 0 for x < 0   (region I)
V(x) = V_o = 4 eV     for x > 0   (region II)

Answer 1

Given: E = 3 eV rightdoublearrow V_0 - E = 1 eV V_0 = 4 eV m = 5.1 times 10^-31 kg ... electron 1 eV = 1.6 times 10^-19 C middot V The wave function in region II is psi = Ce^-kx - (1) with k = squareroot 2m(V_0 - E)/hwithstroke^2 - (2) therefore The probability density is Vertical-bar psi Vertical-bar ^2 = Vertical-bar C Vertical-bar ^2 e^-2kx - (3) for x = 1/2k, Vertical-bar psi Vertical-bar ^2 = Vertical-bar C Vertical-bar ^2/e i.e. the probability density drops to 1/e of the value at x = 0 therefore The penetration length is l = 1/2k = 1/2 squareroot hwithstroke^2/2m(V_0 - E) ... hwithstroke = h/2 pi therefore l = squareroot h^2/32 pi^2 m(V_0 - E) = h/squareroot 32 pi^2 m(V_0 - E) = (6.63 times 10^-34 J middot S)/squareroot (32 pi^2)(9.1 times 10^-31 kg)(1eV times 1.6 times 10^-19 J) l = 9.777 times 10^-11 m = 0.977 A degree