- #1
ZedCar
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Homework Statement
The tip of a scanning tunnelling microscope is placed 1.0 nm away from a conducting surface and a potential of dV = 0.03 V is applied to the surface relative to the tip. When the tip is moved laterally to a new position on the surface the tunnelling current increases by 50%. What is the change in the tip to surface distance (in nm)? The work function of the tip and the surface are both 4.0 eV.
(a) –0.15 (b) 0.0 (c) +0.15 (d) –0.02 (e) +0.02
Homework Equations
Tunnelling probability P
P ~ exp(-2αx)
and
α = { (2me(ϕ-dV))^0.5 } / (h-bar)
I calculate α to be 1.025 x 10^10 which is correct.
Then I use :
|Below is x sub 2, and x sub 1|
P2 / P1 = exp(-2αx2) / exp(-2αx1)
1.5 below is due to 50% in question.
1.5 = exp(2αdx)
log(1.5) = 2αdx
dx = 2.0 x 10^-11
dx = 0.02nm
So answer (e).
However, the solution is given as answer (d).
So either there is a problem in the signs I have attributed to the powers in the exp terms when calculating P2/P1. Or the given solution is incorrect and I have the correct answer.
Was wondering if anyone could clarify?
Thank you.