Tip current in STM (fractional error problem)

[Ziggy12]

Homework Statement

Suppose the STM tip current is given by

i = aV e^{-A\phi^{1/2}s}

a) Derive an expression for the fractional change in tip current as function of fractional change in tip spacing s.

b) If \phi = 4V select a reasonable set of V values so that a 1Å increase in s will cause a factor of 10 decrease in i. (a and A are given)

Homework Equations

Given in problem

The Attempt at a Solution

It's probably very simple but I'm stuck. I'm simply using error formula to get that
\delta i = \frac{\partial i}{\partial s} \delta s = -A\phi^{1/2} \cdot i \cdot \delta s

And then I move out i to the left side to get the fractional change in current.
But that doesn't seem correct. Because then I don't understand how I am supposed to solve b), since the fractional change is independent of V.

I tried other ways as well, but V seems to dissaperar out of the equation anytime I do that.
If anyone has a good idea on how to proceed, it would be welcome.
Thanks
// John[/B]

 

[mfb]

If the current changes by a factor 10, your linear approximation doesn't work any more - but then there is no formula relating a fractional change in s to a fractional change in i. There is a formula relating an absolute change in s to a fractional change in i, a more general version of the formula you derived (which only works for small changes in s).

 

[Ziggy12]

Yes I could simply divide the current at two different distances to get

\frac{i(s+\delta s)}{i(s)} = \frac{aVe^{-A\psi^{1/2}(s+\delta s)}}{aVe^{-A\psi^{1/2}s}} = e^{-A\psi^{1/2}\delta s}

but then again, this is independent of the voltage V, so I don't see how I am supposed to solve the second question.

The fractional change in current is the same regardless of the applied voltage (just like for exponential decay where the half life depends only on lambda).

 

[mfb]

this is independent of the voltage V

Not if $\phi$ (now $\psi$?) depends on the voltage.

 

[Ziggy12]

Yes sorry about that, it was a typo. Phi is the work function
Well I looked it up and the work function does depend on voltage, so I guess I have to assume that
\phi = (V-4)\hspace{0.1cm}\mathrm{Volts}, and not just 4 Volts as one could think from the problem.