Electronic transitions and energy | AP Chemistry | Khan Academy

In this video we're going to be talking about exciting electrons. We can interpret that both ways: that electrons can be exciting and that we're going to excite them into higher energy levels, or we're going to think about what happens when they get unexcited when they go back into lower energy levels.

To help us understand this, I'll start with a simple atom. Hydrogen is the simplest I know, and we're going to think about the version of a hydrogen that we typically see, the isotope that only has one proton in its nucleus. It typically has one or will have one electron if it's a neutral hydrogen atom, and it would normally be in its ground state if it isn't excited yet. So it's going to be in that first shell, but it can be excited to other shells. It could be excited to the second shell, or the third shell, or the fourth shell. This is obviously hand drawn, and not hand drawn that well, and this is really just to help us for visualization purposes.

We know that electrons don't orbit nuclei the way planets orbit stars. They really have both particle and wave-like properties, and they're more of a probability density function of where you might find them. These energy levels are associated with different probability densities of various energies. But this is what an electron will typically look like if we're thinking about just a neutral hydrogen atom where the electron is in its ground state.

Now let's say we're dealing with a hydrogen atom where the electron has already been excited a little bit. So instead of being in the first shell, it's already in the second shell right over here. What we're going to do is we're going to hit it with a photon that excites it even more. We know that light has both particle and wave-like properties. When we think of it as a particle, we think of it as a photon, but I will depict it like this.

So this light has a wavelength of 486 nanometers, and we know that that photon that hits it with a wavelength of 486 nanometers has sufficient energy to excite this electron. In this case, actually from n equals two, from the second shell to the fourth shell. So it'll go all the way over there; it will absorb that photon. After some time, it can come back down.

When it comes back down, I could do it over here. So after some time, that electron right over here, that excited electron, can go back from the fourth shell to the second shell. When it does that, it will emit a photon of that same wavelength. So as that does this, it will emit a photon of 486. Just like that, we already are starting to understand that photons of the right energy can excite an electron by a shell or more than one shell.

When we talk about quantum mechanics, it's this notion that photons need a certain amount of energy in order to be able to excite the electron to the next energy level or the energy level after that. Things in between don't work, and the same thing is true when you're emitting the energy. The electron is not going to go from the fourth energy level to some place in between the fourth and the third. It can't do that. It has these quantum states; it's going to be either in the fourth, or the third, or the second, or the first. There's no such thing as a three-and-a-half shell.

We can actually answer based on this. We can think about what is the energy difference between these shells. The energy difference between these shells is essentially the energy of the photon that we emit when we go from the fourth energy shell, from the fourth shell to the second shell. To figure out the energy of that photon, we just have to think about some useful formulas in quantum mechanics.

The first, and I'm going to just look at it right over here, is that the energy is equal to Planck's constant times the frequency. This thing that looks like a "v" is actually the Greek letter, the lowercase Greek letter nu, and this is what we typically use for frequency, especially when we're talking about frequencies of things like light. We also know how to go between frequency and wavelength because we see that the speed of light is equal to whatever the wavelength of that light is times the frequency of that light.

So how would we figure out the energy of one photon of 486 nanometer light? We could think about it this way: we can first figure out its frequency using c is equal to lambda times nu. Let me write this down. So we know that c, the speed of light, is equal to the wavelength of the light times the frequency of that light. If we know the wavelength, we can figure out the frequency by dividing both sides by lambda.

So let's do that. If we divide both sides by lambda, we get that the frequency of the light is going to be equal to the speed of light divided by the wavelength of the light. Remember they've given us the wavelength of the light here, 486 nanometers or at least I have given it to you. Then you could take this and plug it back into Planck's equation up here: that energy is equal to Planck's constant times the frequency to figure out the energy.

So let me write that down. The energy is going to be equal to Planck's constant times the frequency. Well, we know the frequency right over here, so it's going to be equal to Planck's constant times the speed of light divided by the wavelength of our light, which we know is 486 nanometers.

So we could say, just scroll down a little bit, that the energy is going to be equal to Planck's constant times the speed of light divided by, instead of writing the wavelength as 486 nanometers, I can write it as 486 times 10 to the negative ninth meters. A nanometer is just one billionth of a meter. Then we can just get our calculator out.

We know what Planck's constant is; they give it right over here. We know what the speed of light is right over here, and we know that we have a maximum over here. They're giving us four significant figures in each of these, and then we have three significant figures here. Our answer is going to be in terms of three significant figures.

I'm going to get Planck's constant, which is 6.626 times 10 to the negative 34th joule seconds. So let me write that down. So times 10 to the negative 34th. Then I'm going to multiply that times the speed of light, so times 2.998 times 10 to the 8th meters per second. That gets me this business. Then I'm going to divide that by, divide that by 486 times 10 to the negative ninth.

Gives me—I think we deserve a little bit of a drumroll—gives me this. If we were to look at three significant figures, this would be 4.09 times 10 to the negative 19. Planck's constant here is given in terms of joules: 4.0 times 10 to the negative 19 joules.

So what that tells us is that the difference in these energy levels is this many joules, or the energy of that photon that has a wavelength of 486 nanometers. That energy is 4.09 times 10 to the negative 19 joules.