- #1
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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
??
and wat would the final answer be after the 11.2x10-6
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
and wat would the final answer be after the 11.2x10-6