Question

A surface is examined using a scanning tunneling microscope (STM). For the range of the gap, \(L\), between the tip of the probe and the sample surface, assume that the electron wave function for the atoms under investigation falls off exponentially as \(|\psi|=e^{-\left(10.0 \mathrm{nm}^{-1}\right) a}\). The tunneling current through the tip of the probe is proportional to the tunneling probability. In this situation, what is the ratio of the current when the tip is \(0.400 \mathrm{nm}\) above a surface feature to the current when the tip is \(0.420 \mathrm{nm}\) above the surface?

Short answer

Based on the given electron wave function, the ratio of the current when the tip is 0.400 nm above a surface feature to the current when the tip is 0.420 nm above the surface is approximately 0.8187, or about 81.87% of its original value. This indicates that the current decreases as the tip is moved further from the surface feature.

Step-by-step solution

Identify the given values and the variables needed

The electron wave function is given as \(|\psi|=e^{-\left(10.0 \mathrm{nm}^{-1}\right) a}\). The gap distances are \(L_1 = 0.400\mathrm{nm}\) and \(L_2 = 0.420\mathrm{nm}\). We want to find the ratio of the current, \(R = \frac{I_1}{I_2}\), where \(I_1\) and \(I_2\) are the currents at gap distances \(L_1\) and \(L_2\), respectively.

Calculate the tunneling probability at gap distances \(L_1\) and \(L_2\)

First, calculate the tunneling probability (electron wave function) at gap distances \(L_1\) and \(L_2\): $$ |\psi_1| = e^{-\left(10.0 \mathrm{nm}^{-1}\right) L_1} $$ $$ |\psi_2| = e^{-\left(10.0 \mathrm{nm}^{-1}\right) L_2} $$ Plug in the given values for \(L_1\) and \(L_2\): $$ |\psi_1| = e^{-\left(10.0 \mathrm{nm}^{-1}\right) 0.400\mathrm{nm}} $$ $$ |\psi_2| = e^{-\left(10.0 \mathrm{nm}^{-1}\right) 0.420\mathrm{nm}} $$

Calculate the ratio of the tunneling probability (electron wave function)

Now, we need to find the ratio of the tunneling probability, which will give us the ratio of the current: $$ R = \frac{|\psi_1|}{|\psi_2|} $$ Substitute the calculated values from step 2: $$ R = \frac{e^{-\left(10.0 \mathrm{nm}^{-1}\right) 0.400\mathrm{nm}}}{e^{-\left(10.0 \mathrm{nm}^{-1}\right) 0.420\mathrm{nm}}} $$

Simplify the ratio

Simplify the expression by combining the exponentials: $$ R = \frac{e^{-\left(10.0 \mathrm{nm}^{-1}\right) 0.400\mathrm{nm}}}{e^{-\left(10.0 \mathrm{nm}^{-1}\right) 0.420\mathrm{nm}}} = e^{-(\left(10.0 \mathrm{nm}^{-1}\right) 0.420\mathrm{nm} - \left(10.0 \mathrm{nm}^{-1}\right) 0.400\mathrm{nm})} $$ Further simplify: $$ R = e^{-\left(10.0 \mathrm{nm}^{-1}\right)(0.420\mathrm{nm} - 0.400\mathrm{nm})} $$ Plug in the values: $$ R = e^{-\left(10.0 \mathrm{nm}^{-1}\right)(0.020\mathrm{nm})} $$ Calculate the final result: $$ R = e^{-0.2} $$ $$ R \approx 0.8187 $$

Interpret the result

The ratio of the current when the tip is \(0.400\mathrm{nm}\) above a surface feature to the current when the tip is \(0.420\mathrm{nm}\) above the surface is approximately \(0.8187\). This means that the current is about \(81.87\%\) of its original value when the tip is moved from \(0.400\mathrm{nm}\) to \(0.420\mathrm{nm}\).