Tunnelling Probability of 2.1eV Electron on 3.1eV Barrier

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SUMMARY

The discussion focuses on calculating the tunneling probability of a 2.1 eV electron encountering a 3.1 eV barrier with a width of 0.67 nm. The key equations used include K=2π√((E-Uo)/1.505 eV·nm²) and G=16*(E/Uo)*(1-(E/Uo)). The calculated values are K=5.12 nm-1 and G=3.5, leading to an initial tunneling probability T of 0.35%. However, the correct probability to consider is |T|², which accounts for the absolute magnitude of the wave function.

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Homework Statement



An electron of energy E = 2.10 eV is incident on a barrier of width L = 0.67 nm and height Vo = 3.10 eV

What is the probability that the electron will pass through the barrier?

Homework Equations



I know that T=Ge^-2KL.
K=2pi*sqrt((E-Uo)/1.505eVnm^2))
G=16*(E/Uo)*(1-(E/Uo))

The Attempt at a Solution



K=5.12nm^-1
G=3.5
T=3.5*exp(-2*(5.12nm^-1)*(.67nm))
nm cancels in the exponent so:
T=3.5*exp(-2*5.12*.67)=3.5*.001=.35%
This isn't the right answer though. I've tried 3.5,
35, .035, and .0035 and none of these answers are right. Any help? Thanks!
 
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If you're looking for the probability that it tunnels, that should be |T|^2, not just T (remember, all of the physical significance of the wave function has to do with it's absolute magnitude)
 

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