Barrier Tunneling — Question 5

Question 5

A tunneling-current slope gives an effective vacuum barrier height

In a simplified tunneling microscope, current follows $I(d)=I_0e^{-2\kappa d}$ with gap d and fixed bias. Retracting the tip by $.200\,\mathrm{nm}$ reduces current by a factor of 100. Model the barrier with $\kappa=\sqrt{2m_e\Phi}/\hbar$, where $\Phi$ is the effective barrier energy. For electrons use $\hbar^2/(2m_e)=.0381\,\mathrm{eV\,nm^2}$.

1. Infer $\kappa$ from the current ratio.

2. Infer $\Phi$ and state the role of the assumed fixed prefactor.

Question 5 - Solution

A tunneling-current slope gives an effective vacuum barrier height

(a) Method and calculation

$$\ln[I(d)/I(d+\Delta d)]=2\kappa\Delta d=\ln100.$$$$\boxed{\kappa=\frac{\ln100}{.400\,\mathrm{nm}}=11.5129\,\mathrm{nm}^{-1}}.$$

(b) Interpretation and further calculation

$$\Phi=\frac{\hbar^2\kappa^2}{2m_e}=(.0381)(11.5129255)^2\,\mathrm{eV}.$$$$\boxed{\Phi=5.05006\,\mathrm{eV}},\quad\frac{d\ln I}{dd}=-2\kappa\quad\text{if }I_0\text{ is fixed}.$$

Physical check

This is an effective barrier inferred within the supplied model. Changes of electronic structure, bias or prefactor would also affect the measured slope.