Question

1. Given the potential barrier shown, find the electron energy required for the tunneling probability to first reach 50%. [V_5eV, a-2nm] Note: The energy may be either less than or greater than the barrier height. You will want to use a graphical solution to find the answer. Plot T as a function of energy and find the lowest energy that crosses 0.5

Answer 1

Probablity of an e^- that tunnel through the basics P = e^-2 integral_xy^x_2 Y(x)dx m region Y = squareroot 2m/h^2(V(x)-E);E apha V_t P = e^-2 integral_2a^za squareroot 2m/h^2(V-E)dx P = e^-2 squareroot2m/h^2(V-E)a P = 1/2 1/2 = e^-2 squareroot 2m/h^2(V-E)a = e^-2 squareroot 2 times 9.1 times 10^-31 times 4 times pi^2(5-E)/(6.626 times 10^-34)^2 Take log on both sides we get log(1/2) = -2 squareroot 2 times 9.1 times 10^-3h4 alpha pi^2(5-E) 2 alpha 10^-9/(6.626 times 10^34)^2 +0.301 = +2 squareroot 2 times 9.1 times 10^-31 times 4 alpha pi^2(5-t)/(6.626 times 10^-34)^2 (0.302/4 alpha 10^-9)^2 = 16.36 times 10^36(5-E) E = 636 times 10^37 times 1.6 times 10^-19J \r\n E = 2.61 times 10^16J T = 4E(E-V)/4E(E-V)+V^2 sinh^2(squareroot 2m(E-V_o)Q^2/h^2) [For_E > V_0] If T = 1/2 1/2 = 4E(E-V)/4E(E-V)+V^2 sin h^2 squareroot 2m(E-V)/h^2 a^2 this happens when E/V_0 = 1/2