# Question Video: Determining Electron Path Difference for Generating Adjacent Intensity Maxima Physics

Electrons in a beam all have the same wavelength, 𝜆. The beam is diffracted by a regular crystal lattice. After the electrons exit the crystal, what is the difference between the distances traveled by electrons that arrive at point one and electrons that arrive at point II? [A] 𝜆 [B] 2𝜆 [C] 𝜆/2

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### Video Transcript

Electrons in a beam all have the same wavelength, 𝜆. The beam is diffracted by a regular crystal lattice. After the electrons exit the crystal, what is the difference between the distances traveled by electrons that arrive at point one and electrons that arrive at point two? (A) 𝜆, (B) two times 𝜆, (C) 𝜆 divided by two.

In our diagram, we see a beam of electrons incident on a crystal. The electrons in the beam scatter off of atoms in the crystal and are diffracted or deflected in different directions. Thinking of these traveling electrons as waves that all have the same wavelength 𝜆, we know that when they interact, these waves will interfere with one another. The result of the interference of all these waves is shown in the intensity distribution in our diagram.

We see this distribution has certain peaks. Here is a central peak, then here is a secondary peak, here a tertiary peak, and so on. And it also has valleys or minima. For example, here is a minimum point and here is a minimum point of this intensity distribution. The same labels apply to the corresponding points above, what we can call our central maximum, this main peak here, located at what’s labeled as point two.

So, point two corresponds to the central maximum of our intensity distribution. And point one, we see, corresponds to the next maximum up above the central maximum. It’s much smaller in magnitude, but still it is a peak, a maximum point. To get to point two on our diagram, electrons follow this path. To get to point one, this is the path they follow. Each one of these paths involves a distance, and our question asks about the difference between these distances. Notice that all three of our candidate answers are given in terms of the wavelength 𝜆 of our electrons.

One thing worth pointing out about this wavelength is just as the electrons in the beam before they reach the crystal have this wavelength 𝜆, so do the diffracted electrons on the other side of the crystal. Their wavelength stays constant all throughout. Since these electrons all have some wavelength, we can imagine them as a wave. The wavelength of a wave, we know, is the linear distance from one peak to an adjacent peak, or equivalently one trough to an adjacent trough, or really any linear distance covering one complete wave cycle. What we see on the right of our diagram is an intensity distribution that results from waves interfering with one another.

Say that we have a second wave that is in phase with our first one. That means that the peaks of the second wave line up with the peaks of the first wave and the troughs of the second wave line up with the troughs of the first. If these waves were to overlap, that is, to interfere with one another, the result would be what we call constructive interference. The amplitude of the resulting wave would be greater than either of the waves that make it up. In our intensity distribution, constructive interference shows up as a maximum value. It could be this central maximum value, or a secondary maximum, or so on. Each one of the peaks of the intensity distribution corresponds to this type of interference.

In reality, there are many waves that interfere to form this intensity distribution, far more than two. But to create a simplified picture, we can imagine that there’re just two waves interfering, creating an alternating series of bright and dark spots on the screen. As we’ve seen, point two is the central bright spot, and point one is the bright spot right above that one. Because these are adjacent bright spots in the intensity distribution, we know that they correspond to the smallest possible shift of these two waves relative to one another while still interfering constructively.

Let’s say, for example, that when our two interfering waves are in the positions that we’ve drawn, they form the maximum that is displayed at point two in our diagram. In order to then create the maximum that is at point one, we know that at least one of our two waves will need to shift laterally with respect to the other. And yet after that shift takes place, the waves still need to interfere constructively because, as we see, a maximum is displayed at point one.

Let’s move the pink wave then the smallest distance possible to the left so that it still interferes constructively with the blue wave, whereas currently these two peaks of these waves line up vertically. After the shift, it’s these two peaks that line up. The two waves still interfere constructively, giving us a maximum in our intensity distribution. And this has happened by shifting or sliding one of the waves a distance 𝜆.

We needed to move this exact distance so that the constructive interference of these waves was maintained and so that we would move from one maximum of our distribution to an adjacent maximum, that is, a maximum right next to it. If we had moved farther, say, a distance of two 𝜆 as in answer option (B), our two waves would still interfere constructively. We would get two maxima. But then, the maximum that we would be talking about would be this one on our intensity distribution rather than the one at point one. For this reason, we choose answer option (A). The difference between the distances traveled by electrons that arrive at point one and electrons that arrive at point two is the wavelength of the electrons, 𝜆.