Question Video: Quantum Tunneling Current through a Potential Barrier

In scanning-tunneling microscopy (STM), tunneling-electron current is in direct proportion to the tunneling probability and tunneling probability is to a good approximation expressed by the function 𝑒^βˆ’2𝛽𝐿, where 𝛽 = 10.0 nm⁻¹ and 𝐿 is the distance of the tip of the scanning-tunneling microscope from the surface being scanned. If STM is used to detect surface features with heights of 0.00200 nm, what percent change in tunneling-electron current must the STM electronics be able to detect?

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Video Transcript

In scanning-tunneling microscopy, STM, tunneling-electron current is in direct proportion to the tunneling probability and tunneling probability is to a good approximation, expressed by the function 𝑒 to the negative two 𝛽𝐿, where 𝛽 equals 10.0 inverse nanometers and 𝐿 is the distance of the tip of the scanning-tunneling microscope from the surface being scanned. If STM is used to detect surface features with heights of 0.00200 nanometers, what percent change in tunneling-electron current must the STM electronics be able to detect?

We’ll call this percent change Δ𝑃. In our statement, we’re told that the tunneling probability of an electron in this current is given by the equation 𝑒 to the negative two 𝛽𝐿, where 𝛽 is 10.0 inverse nanometers and 𝐿 is 0.00200 nanometers. When we write out this expression for tunneling probability, inserting our values for 𝛽 and 𝐿, we can see that the units involved cancel out, leaving us with a pure number.

When we calculate that number, we find it’s roughly equal to 0.9608. This number is the probability of tunneling occurring. But we want to solve for Δ𝑃, which is the percent change in electron-tunneling current that the STM electronics must be able to detect.

One way to think of it is this. If we consider a scenario where there is no barrier to electron flow, that is, the current has nothing it needs to move through, then the probability of it making it across some distance is one. There is nothing in its way.

In our case though, we do have some sort of obstacle, which is causing us to lose a little bit of transmission. So our transmission 𝑃 is less than one, 0.9608. The change we’re looking for will be the difference between perfect transmission with no barrier at all and our actually realized transmission.

To turn this value into a percent from a decimal, we’ll multiply it by 100 percent. And then we can write that Δ𝑃 is equal to one minus 0.9608 times 100 percent. To three significant figures, this is 3.92 percent. The electronics in the microscope must be able to detect a change of at least this size.

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