Quantum-electron tunneling barrier

  • Thread starter Discostu19
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  • #1
assuming that the electron has an incident energy of 8eV and that the barrier is of height Vo=13eV and width L=5x10-10m calculate the probability that the electron will be able to tunnel through the barrier.


Relevant equations= k = (sqrt(2m(V-E)) /h


-answer

(sqrt(2x9.1x10-31(13-8)x1.6x10-19)x2pi all divided by h=(6.63x10-34)

giving 1.14x1010??

then using e-2kL = e-2x1.14x1010x5x10-10

=11.2x10-6

is this right :confused: ??
and wat would the final answer be after the 11.2x10-6
 

Answers and Replies

  • #2
turin
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assuming that the electron has an incident energy of 8eV and that the barrier is of height Vo=13eV and width L=5x10-10m calculate the probability that the electron will be able to tunnel through the barrier.


Relevant equations= k = (sqrt(2m(V-E)) /h


-answer

(sqrt(2x9.1x10-31(13-8)x1.6x10-19)x2pi all divided by h=(6.63x10-34)

giving 1.14x1010??

then using e-2kL = e-2x1.14x1010x5x10-10

=11.2x10-6

is this right :confused: ??
No. You need to compare the transmitted piece of the wavefunction to the incident piece of the wave function. Have you studied "scattering" yet? This is a 1-D scattering problem.
 
  • #3
erm i think we have yes
we've done a lot of electron tunnelling but not sure.
 

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