Question

As was found for the square well potential, the wave function amplitude decays exponentially in the "wall". This fact affects the electron tunnelling current between the tip and the sample in scanning tunnelling microscopy. In the following questions, you may assume that the tunneling current through a wall of width L is proportional to Ψ(L). (a) If vo E - 4.50eV, by what factor would the tunnelling current change for a wall width of 0.20 nm than for a wall width of 0.30 nm. (This problem sheds light on the sensitivity of a scanning tunnelling microscope.) (b) Someone suggests to you that a proton tunnelling microscope would work as effectively as an electron tunnelling microscope. For a 0.2 nm wall width and Vo - E -4.50eV, by what factor would the tunnelling current change if protons were used instead of electrons?

Answer 1

The tunneling current change for a potential barrier is accounted by transmission coefficient T given by, T = 16 E (v_0 - E) e^- 2 alpha L/v^2_0 - (1) where, v_0 is the potential barrier alpha^2 = 2 m(v_0 - E)/h^2 L = width of the wall a) Given, v_0 - E = 4.50 eV L = 0.20 nm = 0.2 times 10^- 9 m m = mass of electron (Lets say me) rightarrow T_1 = 16E(v_0 - E) e^- 2 alpha L/v^2_0 = 16 E(4.50) e^- 0.4 alpha times 106- 9/v^2_0 - (i) Now For L = 0.30 nm T_2 = 16 E(4.50)e^- 0.6 alpha times 10^- 9/v^2_0 - (2) rightarrow T_1/T_2 = 16E(4.50) e^- 0.4 alpha times 106- 9/v^2_0 v^2_0/16 E(4.50) e^- 0.6 alpha rightarrow T_1/T_2 = e^- 0.4 alpha times 10^- 9/e^- 0.6 alpha times 10^- 9 = e^- alpha (0.4 - 0.6) 10^- 9 rightarrow T_1/T_2 = e^0.2 alpha times 10^- 9 - (3) alpha = squareroot 2m_e (v_0 - E)/h = squareroot 2 times 9.1 times 10^- 31 times 4.50 eV/6.5821 times 10^16 e Vs