# Video: Emission and Absorption Spectra

In this video, we will learn how to determine the composition of a material from the features that appear in the spectrum of light coming from it.

16:23

### Video Transcript

In this video, we’re talking about emission and absorption spectra. We can see examples of these spectra on screen with the top image representing an absorption spectrum and the bottom image representing an emission spectrum. Note that when we have one of these, it’s called a spectrum, while more than one are referred to as spectra.

Before we look more closely at these, though, let’s consider this. It’s what’s called a continuous spectrum. The idea is that the word spectrum refers to a range of wavelengths. Here we see all the different colors that our eyes are sensitive to. And the word continuous tells us that all of those wavelengths within our given spectrum are represented. In other words, this spectrum isn’t missing any colors. Every single color, every single wavelength of visible light, is present. It’s possible to create a spectrum like this using two objects.

One is a white light, that is, a light that emits all visible wavelengths. And the other object is called a prism. This is typically a triangular object, often made of glass, that bends the different colors of light differently. This means that if we send white light into the prism, then the prism will separate out this light into all the various colors our eyes can see. And it will refract or bend these different colors by different amounts. This has the effect of spreading the different colors of light out as they move through the prism. And if we were to project all these colors onto a screen or surface, we would have a continuous spectrum.

But now let’s imagine that we change our setup a bit. Let’s say that we put something in between the white light and the prism. In particular, let’s say that we put a cold gas of a specific atomic element in front of the white light. Now, we’ll explain why this gas should be cold later on. But for now, all we need to know is that cold atoms of some element are forming a gas in front of our white light. Then, like before, we put a prism in the path of our beam of light and say that we once again project all of the light that goes through the prism, and it spread out by color onto a screen. If we did that, we might see something that looks like this.

Notice that this spectrum looks a lot like the continuous spectrum we saw earlier except for one thing. There are these dark bands, these black lines, that appear at various wavelengths in this spectrum. These dark bands represent specific wavelengths of light that are absent from the light that passes through our prism. But here’s the thing. We knew that those wavelengths of light, as well as all other visible wavelengths, were present in the white light emitted by our source. So, for them to be absent from our spectrum, like we see here, something must have happened to those particular wavelengths of light when they passed into the cold gas. Indeed, what happened is that those wavelengths were absorbed by the gas.

But this raises a question. Why is it that these particular wavelengths represented by the gaps in our spectrum were absorbed while the other ones made it through? In other words, apparently, the gas absorbed light at this wavelength and this wavelength and this one and this one and this one and so on. But it’s simply allowed to pass through light of these wavelengths here that show up in our spectrum. To answer this question of why sunlight makes it through the gas while some is absorbed, we’ll need to understand this cold gas at an atomic level.

Let’s say that this sketch here represents a single atom in our gas. In this sketch, we see the nucleus of the atom. That’s this core here at the center. And we also see these three concentric rings. These rings, according to an earlier model of the atom, represent different energy levels for electrons to inhabit. So, say we had an electron at this energy level, which is the lowest energy level, but then that that electron is capable of transitioning up to a higher energy level within the atom. For this to happen, though, it would take energy. And this energy would need to come from somewhere outside the atom.

Now, here’s something interesting. In order for this electron to transition to a higher energy level, it needs to absorb a very specific amount of energy to make that happen. Let’s say our electron were to transition from its current energy level to this outermost orbital in the atom. The energy this electron would need to absorb to make that happen, as we said, would have to come from outside the atom somewhere. This often happens through the electron absorbing a photon which has just the right amount of energy for the electron to make this transition.

Now, let’s recall for a moment the equation for the energy of a photon in terms of its wavelength. We can refer to a photon’s energy as 𝐸 sub p, and it’s equal to Planck’s constant ℎ times the speed of light in vacuum, that’s the speed of the photon, divided by its wavelength 𝜆. Now, here’s the thing. An individual photon could have any amount of energy depending on its wavelength. But in order for our atomic electron to transition up to a higher energy level, it needs a very precise amount of energy. If we call the energy required for the specific transition we’re talking about Δ𝐸, then in terms of photon energy, we need to have a photon with a specific wavelength 𝜆, so that 𝑐 over 𝜆 times ℎ is equal to that energy amount.

So, of all the possible wavelengths of photons that are incident on this atom, we could say that the electron selects only those photons that have the right wavelength to give the right amount of energy to allow the electron to transition to a higher energy level. When a photon with the right wavelength and therefore the right amount of energy comes along, the electron absorbs it and jumps up to that higher energy state. Now, that word “absorbs” is important. The electron really does absorb the incident photon in order to take its energy. This means, of course, that that photon is now gone.

And if we think of that photon in terms of a particular wavelength of light coming from our source, we can expect that that specific wavelength will be absorbed by our gas of atoms and won’t make it through to the prism and then onto the screen. And this is the reason we see these narrow, dark bands on our spectrum. These wavelengths correspond to wavelengths of light that are absorbable, we could say, by our cold gas of atoms. That is, these are wavelengths of light that allow atomic transitions within the atom. Now, remember that we made a point of saying that this gas is a cold gas. The reason we specified that the gas is cold is because a cold atom is a relatively low-energy atom. That means that any electrons it has — let’s say that this particular atom has two electrons — will be in their lowest possible energy states.

This is important because it’s only electrons that have somewhere to go, that is, some higher energy level to inhabit that are capable of absorbing incoming radiation in order to make that transition. Imagine that instead of a cold atom, we had a very hot atom, that is, a high-energy one. In this case, our electrons might already be at the outermost or highest energy level. And that would mean that regardless of the wavelength of incoming radiation, the electrons couldn’t absorb it. They have nowhere to go. So, in order to study how a particular atomic element absorbs radiation, it’s important to use low-energy atoms of that element, in other words, cold atoms. And so that’s why our gas is a cold gas.

So, this spectrum we’re seeing, which looks mostly like a continuous spectrum except for these few dark bands we see in it, is called an absorption spectrum, and the dark bands are called absorption lines. Each atomic element has its very own specific absorption spectrum. These spectra are created the way we’ve shown here, by passing white light through a cold gas of that particular element’s atoms and then spreading that light out using a prism so that it forms a spectrum. As the title of our lesson tells us though, an absorption spectrum isn’t the only kind we can create.

Imagine that we take this gas of a particular atomic element and we heat it up so that it becomes hot. And then, say that we put a prism in front of that hot gas. It turns out that this gas will give off light, which, when spread out by the prism and projected onto a screen, creates its own spectrum. We see, though, that this spectrum looks very different from the absorption spectrum we were looking at. Whereas the absorption spectrum showed nearly all the wavelengths of visible light punctuated by a few dark bands representing absent wavelengths, this spectrum we see at the top of our screen is the opposite. It’s mostly dark, with just a few bands of present color. This is known as an emission spectrum. And emission spectra in general are recognizable because they’re mostly black. They’re mostly absent of light.

So, just like with the absorption spectrum below, we face a similar question. Why is it that these wavelengths of light are present in our emission spectrum, whereas the great majority are not? The reason for this comes down once again to the particular structure of the atom that we’re working with. Recall that we specified that the gas that creates our emission spectrum is a hot gas. One effect of heating up the gas is energizing the atoms in it, which means that the electrons in those atoms will tend to be at fairly high energy levels. Electrons of these higher energy levels, when they do move, will tend to move back towards the nucleus of the atom. That is, they’re more likely to lose energy and descend to a lower energy level.

When this happens, the energy that an electron loses in this transition is released. It’s given off as a photon, a packet of light. And here’s something interesting. The same photon wavelength that was required to energize an electron so it would make a transition from the ground energy state up to this higher state. That is precisely the wavelength of light given off when the electron makes this transition in the opposite direction, from higher energy to lower energy. So, the reason for these colored bands in our emission spectrum, that is, there are wavelengths of light present at these colors, is because electrons in the hot atoms of our gas are transitioning back down to a lower energy level and emitting photons in the process. It’s these emitted photons and their particular wavelengths that are showing up on our spectrum.

So, we now understand the reason for these dark bands in our absorption spectrum. Those wavelengths of light are absorbed by our gas. And likewise, we know the reason for these colored bands, known as emission lines, present in our emission spectrum. Those are the wavelengths of light emitted by our hot gas of the same element. Since both the absorption and emission spectrum we’re seeing now correspond to the same particular element, we see that these bands show up at the same wavelength for each spectrum. This is a strong clue that the absorption and emission spectra we’re looking at correspond to the same element.

Now, we mentioned earlier that the particular location of these bands in an emission or absorption spectrum tells us what the element was that created these spectra. That is, the specific wavelengths at which these bands happen is unique to the given element that created them. Researchers are actually able to look at a particular spectrum and work back to figure out what element or elements helped to create it. In this case, by the way, it’s the element helium that’s represented by this emission and absorption spectrum. Now, there’s one other important thing to notice about these two spectra. It’s a little hard to see, though, with these spectra the size they are on screen. So, let’s expand our absorption spectrum to see this more clearly.

Okay, that’s a bit better. Let’s take a look at, say, this absorption line compared to, say, this one here. These two lines look different. The one on the left looks wider than the one on the right. We could say that this absorption band is thicker than this one. It turns out this is not just an impression we’re getting. There really is a difference in the thickness of different spectral lines. In fact, it’s such a normal thing for this line thickness to vary that it’s common to talk about something called spectral line width. The greater the thickness of a spectral line like this one here with a relatively large thickness, the greater it’s width. But this just raises the question of what is it that makes a particular spectral line wider or narrower? In other words, what leads to the difference between this relatively wide spectral line here and this relatively narrow one?

It turns out that there are a number of influences that can lead to what’s called line broadening. When a spectral line is broaded, that just means its width is increasing. Line broadening can happen when the atoms in the gas that helped to create the spectra are heated or under pressure. And also, one of the biggest impacts on spectral line width comes from, once again, the movement of electrons within these atoms. When an electron moves to an excited state, like this electron here has, as it’s moved to the highest energy level, the length of time an electron spends there before dropping down to a lower energy level has a direct impact on the width of the spectral line that’s produced. The longer the electrons stays in this high energy state before dropping back down, the narrower the spectral line it produces will be.

And then the converse is true as well. If the electron only spends a very little bit of time in this high energy state before coming back down, then the spectral line it produces will be relatively broad or wide. Using the emission or absorption spectrum of an element, it’s even possible to measure the width, say in units of nanometers, of a given spectral line. That width corresponds to the range of wavelengths expressed in nanometers that are absorbed or emitted. Now that we’ve learned a bit about emission and absorption spectra, let’s get some practice with these ideas through an example.

A scientist has a sample of an unknown gas. In order to identify the gas, she looks at the spectrum of visible light emitted from when it is heated. This is shown in the figure. Also shown in the figure are the emission spectra of three pure gaseous elements. Which of the three elements is the unknown gas?

Okay, taking a look at our figure, we see a series of emission spectra. The top one is the spectrum of an unknown gas we want to identify. Then below that, we see the emission spectra of hydrogen, helium, and oxygen, respectively. Our question asks, which of these three elements is the unknown gas? Now, the first thing we can realize is that our unknown gas is a pure sample. It’s not a mixture of some number of other elements. In other words, it’s either entirely hydrogen or entirely helium or entirely oxygen. Knowing that, what we want to find is a match between the spectrum of our unknown gas and the spectrum of one of these other gases. And here’s how we can identify a match.

Looking at our unknown gas spectrum, we can see the particular wavelengths corresponding to emission lines of this unknown gas. So, for example, this emission line of our unknown gas is at about 422 nanometers, whereas this one is at about 437 nanometers and so on and so forth for these other emission lines in our spectrum. We want to find which of these three other gases, hydrogen, helium, or oxygen, have emission lines at the same wavelengths as our unknown sample. We can go ahead and make this comparison by eye.

Let’s start with this emission line here at about 422 nanometers. Looking at our spectrum of hydrogen, we don’t see a feature at or near that wave length. This tells us that the two emission spectra do not match. And so are unknown gas can’t be hydrogen. Next, if we consider helium, we see that this gas too has no emission line at or around 422 nanometers. That is, if we were to drop this emission line wavelength down onto our spectrum of helium, we would see that helium has no emission line at that wavelength. This means our unknown gas is not comprised of helium either.

Lastly, we look at oxygen. In this case, note that we do see a corresponding emission line at this wavelength. That’s an encouraging sign. Let’s consider another emission line in our unknown gas. This one right here, at about 437 nanometers, we see also corresponds to an emission line in oxygen. And if we continue looking at other emission lines, we see there continue to be matches between these two spectra. And this is true, we can see, as we go all the way down to the end of the visible spectrum. So, we found a match for the spectrum of our unknown gas, and that match tells us that the unknown gas is oxygen.

Let’s summarize now what we’ve learned about emission and absorption spectra. In this lesson, we saw that all atomic elements exhibit unique emission and absorption spectra. We learned that emission spectra are mostly dark, where the emission lines are bands of color representing wavelengths of light given off by a particular element. Whereas absorption spectra tend to show most visible colors of light, with a few dark bands representing absorption lines.

We also saw that the particular wavelengths, at which emission and absorption lines occur, has to do with the energy levels of the atoms creating those respective emission and absorption spectra. And lastly, we learned that the width of spectral lines varies depending on how long electrons remain in excited states, among other factors. This is a summary of emission and absorption spectra.