Question

20. An electron with total energy 5 eV approaches a potential barrier of height 20 eV. If the probability that the electron tunnels across the barrier is 0.03, what is the width L of the barrier? a. 0.0116 nm b. 0.116 nm c. 1.16 nm d. 11.6 nm e. 116 nm Spring 2009 21. Define K UE

Answer 1

Answer:

The width L of the barrier is b. 0.116nm

Explanation:

This is a case of tunnel effect when the probability is less than one.

First of all you find all the information the problem gives you:

Electron's total energy E=5eV

Height of potential barrier U=20eV

Crossing Probability T=0.03

Width of the barrier L=unknown

Then you need to find the equations to use in cases in which the probability is less than 0, so you have:

$T=Ge^{-2kL}$

$G=16\frac{E}{U}(1-\frac{E}{U})$

$k=\frac{\sqrt{2m(U-E)}}{h}$ where m is the mass of the electron, $m=9.11*10^{-31}Kg$, and h is the Planck constant, $h=1.054*10^{-34}J.s$

First we find G, we have:

$G=16(\frac{5.0eV}{20eV})(1-\frac{5.0eV}{20eV})$

$G=4(1-0.25)=3$

Now, to solving k, we need to find the difference (U-E) :

U - E = 20eV - 5eV = 15eV

$15eV*\frac{1.60218*10^{-19}J}{1eV} =2.403*10^{-18}J$

Then, we can find k:

$k=\frac{\sqrt{2m(U-E)}}{h}$

$k=\frac{\sqrt{2(9.11*10^{31}kg)(2.403*10^{-18}J)}}{1.054*10^{-34}\frac{J}{s}}$

$k=1.98*10^{10}$

Finally, we solve for L the probability equation T:

$T=Ge^{-2kL}$

$\frac{T}{G}=e^{-2kL}$

$ln(\frac{T}{G})=ln(e^{-2kL})$

$ln(\frac{T}{G})=-2kL$

$L=\frac{ln(\frac{T}{G})}{-2k}$

And replacing values for T, G and k, we can find the width of the barrier:

$L=\frac{ln(\frac{0.03}{3})}{-2(1.98*10^{10})}m$

$L=1.16*10^{-10}m*\frac{1*10^{9}nm}{1m}$

L=0.116nm