Question

7. an electron with initial kinetic energy of 6.0 eV encounters a barrier with height 11.0 eV. What is the probability of tunneling if he width of he barrier is a. 0.80 nm b. 0.40 nm.

Answer 1

Step-by-step explanation:

In quantum tunneling through a brier, the energy of the tunneled particle is same but the probability amplitude is decreased.

The expression for the probability of tunneling is expressed as follows:

`T=16 (E)/(U_(0))\left(1-(E)/(U_(0))\right) e^{-2 \sqrt{(2 m\left(U_(0)-E\right))/(h)} L}`

Here, T is the transmission probability; E is the kinetic energy, U_{0} is the barrier height, $m$ is the mass of the electron, and L is the width of the barrier.