40.10.1 The Scanning Tunneling Microscope

Diffraction limits the resolution of an optical microscope to objects no smaller than about a wavelength of light—roughly 500 nm. This is more than 1000 times the size of an atom, so there is no hope of resolving atoms or molecules via traditional optical microscopy. Electron microscopes are similarly limited by the de Broglie wavelength of the electrons. Their resolution is much better than that of an optical microscope, but still not quite at the level of resolving individual atoms.

This situation changed dramatically in 1981 with the invention of the scanning tunneling microscope, or STM as it is usually called. The STM allowed scientists, for the first time, to “see” surfaces literally atom by atom. The atomic-resolution photos at the beginning of Chapter 38 and this chapter demonstrate the power of an STM. These pictures and many others you have likely seen (but may not have known where they came from) are stupendous, but how are they made?

FIGURE 40.32a shows how the scanning tunneling microscope works. A conducting probe with a very sharp tip, just a few atoms wide, is brought to within a few tenths of a nanometer of a surface. Preparing the tips and controlling the spacing are both difficult technical challenges, but scientists have learned how to do both. Once positioned, the probe can mechanically scan back and forth across the surface.

The figure labeled (“a”) illustrates the working principle of a scanning tunneling microscope.

The figure illustrates the potential energy diagram graphically in a coordinate plane.

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When we analyzed the photoelectric effect, you learned that electrons are bound inside metals by an amount of energy called the work function $E_{0}$ . A typical work function is 4 or 5 eV. This is the energy that must be supplied—by a photon or otherwise—to remove an electron from the metal. In other words, the electron’s energy in the metal is $E_{0}$ less than its energy outside the metal.

This fact is the basis for the potential-energy diagram of FIGURE 40.32b. The small air gap between the sample and the probe tip is a potential-energy barrier. The energy of an electron in the metal of the sample or the probe tip is lower than the energy of an electron in the air by $\approx 4 eV,$ the work function. The absorption of a photon with $E_{photon} > 4 eV$ would lift the electron over the barrier, from the sample to the probe. This is just the photoelectric effect. Alternatively, electrons can tunnel through the barrier if it is sufficiently narrow. This creates a tunneling current from the sample into the probe.

In operation, the tunneling current is recorded as the probe tip scans across the surface. You saw above that the tunneling current is extremely sensitive to the barrier thickness. As the tip scans over the position of an atom, the gap decreases by $\approx 0.1 nm$ and the current increases. The gap is larger when the tip is between atoms, so the current drops. Today’s STMs can sense changes in the gap of as little as 0.001 nm, or about 1% of an atomic diameter! The images you see are computer-generated from the current measurements at each position.

The STM has revolutionized the science and engineering of microscopic objects. STMs are now used to study everything from how surfaces corrode and oxidize, a topic of great practical importance in engineering, to how biological molecules are structured. Another example of quantum mechanics working for you!

CHAPTER 40 CHALLENGE EXAMPLE Tunneling in semiconductors

Quantum-mechanical tunneling can be important in semiconductors. Consider a 1.0-nm-thick layer of GaAs sandwiched between 4.0-nm-thick layers of GaAlAs. This is the situation explored in Figure 40.16, where we learned that the electron’s potential energy is 0.300 eV lower in GaAs than in GaAlAs. An electron in the GaAs layer can tunnel through the GaAlAs to escape, but this doesn’t happen instantly. In quantum mechanics, we can’t predict exactly when tunneling will occur, only the probability of it happening. Estimate the time at which the probability of escape has reached 50%.

MODEL We can model this problem by thinking of the electron as a particle bouncing back and forth between the walls of the potential well. Each time it hits a wall, it has probability $P_{tunnel}$ of tunneling and probability $P_{reflect} = 1 - P_{tunnel}$ of reflecting. The tunneling probability depends on the height and thickness of a potential barrier.

VISUALIZE FIGURE 40.33 shows the potential well. An electron in a 0.300-eV-deep, 1.0-nm-wide well is exactly the situation of Figure 40.16, so we know that the electron has a single quantum state with $E_{1} = 0.125 eV$ . The wave function decreases exponentially with distance into the potential barriers, but a very tiny amplitude—too small to see here—still exists at the far edge of the barrier.

The figure illustrates the potential energy diagram of an electron in a layer of G “a” “A” s sandwiched between layers of G “a” “A” l “A” s.

FIGURE 40.33 Full Alternative Text

SOLVE Each time the electron collides with a wall of the potential well, its probability of tunneling through is $P_{tunnel} = e^{- 2 w / η}$ . The penetration distance $η$ depends on $U_{0} - E$ , the “distance” from the energy level to the top of the barrier, which in this case is

U_{0} - E = 0.300 eV - 0.125 eV = 0.175 eV = 2.8 \times 10^{- 20} J

Using this value, we can calculate the penetration distance to be

η = \frac{ℏ}{\sqrt{2 m (U_{0} - E)}} = 0.465 nm

We then find the probability of tunneling through a 4.0-nm-wide barrier to be

P_{tunnel} = e^{- 2 w / η} = 3.4 \times 10^{- 8}

It’s a very small probability, as expected. The probability of not tunneling, of reflecting back into the well, is then

P_{reflect} = 1 - P_{tunnel} = 0.999999966

You’ve seen that the probability of A or B happening is $P_{A} + P_{B}$ . Similarly, the probability of A and B happening, assuming they are independent events, is $P_{A} \times P_{B}$ . The probability of a head in a coin toss is $\frac{1}{2}$ . If you toss two coins, the probability that A is a head and B is a head is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ . If you toss three coins, the probability that all three are heads is ${(\frac{1}{2})}^{3} = \frac{1}{8}$ . If the electron is still in the potential well after $N$ bounces, it had to reflect $N$ times. The probability of this happening is

P_{in well} = P_{reflect} \times P_{reflect} \times P_{reflect} \times \dots \times P_{reflect} = (P_{reflect})^{N}

Because $P_{reflect} < 1$ , the probability of still being in the potential well decreases as $N$ increases.

We’ve focused not on $P_{escape}$ but on $P_{in well} = 1 - P_{escape}$ because staying in the well requires $N$ specific events to happen. Escape, on the other hand, could have occurred on any of $N$ attempts, so a direct calculation of $P_{escape}$ is much more complicated. If the probability of escape is 50%, then it’s also 50% probable that the electron is still in the potential well. We can find the number of reflections needed to get to the 50% probability by taking the logarithm of both sides of the equation:

\begin{array}{l} log (P_{in well}) = log ({(P_{reflect})}^{N}) = N log (P_{reflect}) \\ N = \frac{log (P_{in well})}{log (P_{reflect})} = \frac{log (0.50)}{log (0.999999966)} = 2.0 \times 10^{7} \end{array}

After 20 million reflections, the electron is 50% likely to have escaped. Although that’s a large number of reflections, it doesn’t take long because the electron is moving only a very small distance between reflections at a fairly high speed. The electron’s energy inside the potential well is entirely kinetic, $K = E = 0.125 eV = 2.0 \times 10^{- 20} J,$ so its speed is

v = \sqrt{\frac{2 K}{m}} = 2.1 \times 10^{5} m / s

The time between reflections is the time needed to travel across the 1.0-nm-wide GaAs layer:

Δ t = \frac{1.0 \times 10^{- 9} m}{2.1 \times 10^{5} m / s} = 4.8 \times 10^{- 15} s

Thus the time needed for $2.0 \times 10^{7}$ reflections is

t_{50 %} = N Δ t = 9.6 \times 10^{- 8} s = 96 ns

Because we’re making only an estimate, we can say that an electron has a 50% probability of tunneling out of the GaAs layer within about 100 ns.

REVIEW Even though the tunneling probability is very tiny, tunneling takes place very rapidly on a human time scale. An increasing number of semiconductor devices make practical use of this tunneling current. Note that no energy is lost in the tunneling process; “tunneling” is a metaphor, not a process that requires work. The electron emerges with 0.125 eV of kinetic energy.

40.10.1 The Scanning Tunneling Microscope

FIGURE 40.32 A scanning tunneling microscope.

CHAPTER 40 CHALLENGE EXAMPLE Tunneling in semiconductors

FIGURE 40.33 The potential energy of an electron in a layer of GaAs sandwiched between layers of GaAlAs.