Question

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the barrier if the barrier width a is (a) 1.00 nm [5p nm 5pt]? Suggestion: use T as given in Lecture 9-10

Answer 1

The tunnel probability for 0.5 nm and 1.00 nm are
`5.45*10^(-4)` and
`7.74*10^(-8)` respectively.

Step-by-step explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of
`\beta`

Using formula of
`\beta`

`\beta=\sqrt{(2m)/((h)/(2\pi))(v_(0)-E)}`

Put the value into the formula

`\beta = \sqrt{(2*9.1*10^(-31))/((1.055*10^(-34))^2)(5.0-2)*1.6*10^(-19)}`

`\beta=8.86*10^(9)`

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

`T=(16E(V_(0)-E))/(V_(0)^2)e^(-2\beta a)`

Put the value into the formula

`T=(16* 2(5.0-2.0))/(5.0^2)e^{-2*8.86*10^(9)*0.5*10^(-9)}`

`T=5.45*10^(-4)`

(b). We need to calculate the tunnel probability for width 1.00 nm

`T=(16* 2(5.0-2.0))/(5.0^2)e^{-2*8.86*10^(9)*1.00*10^(-9)}`

`T=7.74*10^(-8)`

Hence, The tunnel probability for 0.5 nm and 1.00 nm are
`5.45*10^(-4)` and
`7.74*10^(-8)` respectively.