# Scanning Tunnelling Microscope

Tags:
1. Dec 17, 2012

### ZedCar

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

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. Dec 17, 2012

### TSny

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.

3. Dec 18, 2012

### ZedCar

Thanks very much TSny!

I see where I was going wrong now!