Question

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

Answer: The ratio of the tunneling current is approximately $e^{0.4}$.

Step-by-step solution

Write down the probability density function at each distance

We know the probability density function is given by $|\mathrm{\Psi }{|}^2$. We'll calculate the probability densities, $P1$ and $P2$, for the STM tip at distances $0.400\mathrm{nm}$ and $0.420\mathrm{nm}$ respectively, using the formula: $|{\mathrm{\Psi }}_i{|}^2=e^{-2\times (10.0{\mathrm{nm}}^{-1})a_i}$, where i={1,2}

Calculate the probability density at a distance of $0.400\mathrm{nm}$

We can substitute $a_1=0.400\mathrm{nm}$ into the formula for the probability density: $P1=|{\mathrm{\Psi }}_1{|}^2=e^{-2\times (10.0{\mathrm{nm}}^{-1})\times 0.400\mathrm{nm}}$ Now we can calculate $P1$: $P1=e^{-2\times (10.0{\mathrm{nm}}^{-1})\times 0.400\mathrm{nm}}$ $P1=e^{-8}$

Calculate the probability density at a distance of $0.420\mathrm{nm}$

We can substitute $a_2=0.420\mathrm{nm}$ into the formula for the probability density: $P2=|{\mathrm{\Psi }}_2{|}^2=e^{-2\times (10.0{\mathrm{nm}}^{-1})\times 0.420\mathrm{nm}}$ Now we can calculate $P2$: $P2=e^{-2\times (10.0{\mathrm{nm}}^{-1})\times 0.420\mathrm{nm}}$ $P2=e^{-8.4}$

Calculate the ratio of the tunneling current

We are told that the tunneling current is proportional to the tunneling probability, which means the ratio of the tunneling current is equal to the ratio of the probability densities: $\frac{I1}{I2}=\frac{P1}{P2}$ Substitute in the values for $P1$ and $P2$: $\frac{I1}{I2}=\frac{e^{-8}}{e^{-8.4}}$ Now we can simplify the expression to get the ratio: $\frac{I1}{I2}=e^{-8+8.4}$ $\frac{I1}{I2}=e^{0.4}$ So, the ratio of the tunneling current when the tip is $0.400\mathrm{nm}$ above the surface to the current when the tip is $0.420\mathrm{nm}$ above the surface is approximately $e^{0.4}$.