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

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

Step-by-step 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{(2m)/(((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{(2*9.1*10^(-31))/((1.055*10^(-34))^2)*(12-9)*1.6*10^(-19)}`

`k=8.86*10^(9)`

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

Using the formula of tunnel barrier

`T=(16E(v_(0)-E))/(v_(0)^2)e^(-2kl)`

Where,
`G = (16E(v_(0)-E))/(v_(0)^2)`

Put the value into the formula

`T=(16*9*(12-9))/(12^2)e^{-2*8.86*10^(9)*0.4*10^(-9)}`

`T=0.002505=2.505*10^(-3)`

`T=0.25\%`

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