Question

A free electron with initial energy of 9 eV encounters a rectangular potential step of energy 12 eV and width of 0.4 nm. What is the probability that the electron will tunnel the barrier ?

Answer 1

Answer:

The  probability that the electron will tunnel the barrier is 0.25%.

Explanation:

Given that,

Initial energy = 9 eV

Potential energy = 12 eV

Width = 0.4 nm

Using formula of transmission coefficient

$T\approx e^{-2kL}$

$T=Ge^{-2kl}$.....(I)

We need to calculate the value of k

$k=\sqrt{\dfrac{2m}{(\dfrac{h}{2\pi})^2}(v_{0}-E)}$

Where, m = mass of electron

E = initial energy

v=potential energy

Put the value into the formula

$k=\sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}\times(12-9)\times1.6\times10^{-19}}$

$k=8.86\times10^{9}$

We need to calculate the probability that the electron will tunnel the barrier

Using the formula of tunnel barrier

$T=\dfrac{16E(v_{0}-E)}{v_{0}^2}e^{-2kl}$

Where, $G = \dfrac{16E(v_{0}-E)}{v_{0}^2}$

Put the value into the formula

$T=\dfrac{16\times9\times(12-9)}{12^2}e^{-2\times8.86\times10^{9}\times0.4\times10^{-9}}$

$T=0.002505=2.505\times10^{-3}$

$T=0.25\%$

Hence, The  probability that the electron will tunnel the barrier is 0.25%.