(a) An electron with initial kinetic energy 32 eV encounters a square barrier with height 41 eV and width 0.25 nm. What is the probability that the electron will tunnel through the barrier? (b) A proton with the same kinetic energy encounters the same barrier. What is the probability that the proton will tunnel through the barrier?

= = An electron having total energy E 4.60 eV approaches a rectangular energy barrier with U■5.10 eV and L-950 pm as shown in the figure below. Classically, the electron cannot pass through the barrier because E < U. Quantum-mechanically, however, the probability of tunneling is not zero. Energy E U 0 i (a) Calculate this probability, which is the transmission coefficient. (Use 9.11 x 10-31 kg for the mass of an electron, 1.055 x 10-34] s for h, and note that there are 1.60 x 10-19 J per eV.) (b) To what value would the width L of the potential barrier have to be increased for the chance of an incident 4.60-eV electron tunneling through the barrier to be one in one million? nm

The wavefunction for a quantum particle tunnelling through a potential barrier of thickness L has the form ψ(x) = Ae−Cx in the classically forbidden region where A is a constant and C is given by C^2 = 2m(U − E) /h_bar^2 .   (a) Show that this wavefunction is a solution to Schrodinger’s Equation.   (b) Why is the probability of tunneling through the barrier proportional to e ^−2CL?