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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 :

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  [tex]5.45\times10^{-4}[/tex] and [tex]7.74\times10^{-8}[/tex] respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of [tex]\beta[/tex]

Using formula of [tex]\beta[/tex]

[tex]\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}[/tex]

Put the value into the formula

[tex]\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}[/tex]

[tex]\beta=8.86\times10^{9}[/tex]

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

Using formula of tunnel barrier

[tex]T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}[/tex]

Put the value into the formula

[tex]T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}[/tex]

[tex]T=5.45\times10^{-4}[/tex]

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

[tex]T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}[/tex]

[tex]T=7.74\times10^{-8}[/tex]

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  [tex]5.45\times10^{-4}[/tex] and [tex]7.74\times10^{-8}[/tex] respectively.

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