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

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:

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

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.