Question

An electron with initial kinetic energy 5.1 eV encounters a square potential barrier with height 10.0 eV .What is the width of the barrier if the electron has a 0.60 % probability of tunneling through the barrier? Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

$2.9\times 10^{-10}\ m$

Explanation:

E = Initial kinetic energy = 5.1 eV

$u_0$ = Potential barrier height = 10 eV

P = Probability = $\dfrac{0.6}{100}=0.6\times 10^{-2}$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

Constant A is given by

$A=16\dfrac{E}{u_0}(1-\dfrac{E}{u_0})\\\Rightarrow A=16\times \dfrac{5.1}{10}(1-\dfrac{5.1}{10})\\\Rightarrow A=3.9984$

K is given by

$K=\dfrac{\sqrt{2m(u_0-E)}}{\hslash}\\\Rightarrow K=\dfrac{\sqrt{2\times 9.1\times 10^{-31}(10-5.1)\times 1.6\times 10^{-19}}}{1.055\times 10^{-34}}\\\Rightarrow K=11322472269.49086\ /m$

We have the relation

$P=Ae^{-2Kl}\\\Rightarrow \dfrac{P}{A}=e^{-2Kl}\\\Rightarrow \dfrac{A}{P}=e^{2Kl}\\\Rightarrow ln\dfrac{A}{P}=2Kl\\\Rightarrow l=\dfrac{ln\dfrac{A}{P}}{2K}\\\Rightarrow l=\dfrac{ln\dfrac{3.9984}{0.6\times 10^{-2}}}{2\times 11322472269.49086}\\\Rightarrow l=2.87123\times 10^{-10}\ m$

The width of the barrier is $2.9\times 10^{-10}\ m$