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Scanning Tunnelling Microscope

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  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data

    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


    2. Relevant 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.


    3. The attempt at a solution
     
  2. jcsd
  3. Dec 17, 2012 #2

    TSny

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    Homework Helper
    Gold Member

    Note that you've defined dx to be equal to x1 - x2 according to the expressions above. But the "change in x" would normally be defined as (xfinal-xinitial) = x2 - x1

    An increase in tunneling probability corresponds to a decrease in tip-to-surface distance x.
     
  4. Dec 18, 2012 #3
    Thanks very much TSny!

    I see where I was going wrong now!
     
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