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Absorption and emission

Using shell model diagram to relate absorption to emission. Derives relationship between emitted photon and energy levels, the Balmer-Rydberg equation. Created by Jay.

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Video transcript

- We've been talking about the Bohr model for the hydrogen atom, and we know the hydrogen atom has one positive charge in the nucleus, so here's our positively charged nucleus of the hydrogen atom and a negatively charged electron. If you're going by the Bohr model, the negatively charged electron is orbiting the nucleus at a certain distance. So, here I put the negatively charged electron a distance of r1, and so this electron is in the lowest energy level, the ground state. This is the first energy level, e1. We saw in the previous video that if you apply the right amount of energy, you can promote that electron. The electron can jump up to a higher energy level. If we add the right amount of energy, this electron can jump up to a higher energy level. So now this electron is a distance of r3, so we're talking about the third energy level here. This is the process of absorption. The electron absorbs energy and jumps up to a higher energy level. This is only temporary though, the electron is not going to stay there forever. It's eventually going to fall back down to the ground state. Let's go ahead and put that on the diagram on the right. Here's our electron, it's at the third energy level. It's eventually going to fall back down to the ground state, the first energy level. Here's the electron going back to the first energy level here. When it does that, it's going to emit a photon. It's going to emit light. When the electron drops from a higher energy level to a lower energy level, it emits light. This is the process of emission. I could represent that photon here. This is how you usually see it in textbooks. We emit a photon, which is going to have a certain wavelength. Lambda is the symbol for wavelength. We need to figure out how to relate lambda to those different energy levels. The energy of the photon is, the energy of the emitted photon is equal to the difference in energy between those two energy levels. We have energy with the third energy level and the first energy level. The difference between those... So, the energy of the third energy level minus the energy of the first energy level. That's equal to the energy of the photon. This is equal to the energy of that photon here. We know the energy of a photon is equal to h nu. Let me go ahead and write that over here. Energy of a photon is equal to h nu. H is Planck's constant, this is Planck's constant. Nu is the frequency. We want to think about wavelength. We need to relate the frequency to the wavelength. The equation that does that is of course, C is equal to lambda nu. So, C is the speed of light, lambda is the wavelength, and nu is the frequency. So, if we solve for the frequency, the frequency would be equal to the speed of light divided by lambda. Then, we're going to take all of that and plug this in to here. We get the energy of a photon is equal to Planck's constant, h, I'll write that in here, times the frequency, and the frequency is equal to c over lambda. Now we have the energy of the photon is equal to hc over lambda. Instead of using E3 and E1, let's think about a generic high energy level. Let's call this Ej now, so this is just a higher energy level, Ej. The electron falls back down to a lower energy level, which we'll call Ei. Instead of using E3 and E1, let's make it more generic, let's do Ej and Ei. Let's go ahead and plug that in now. The energy of the photon would be equal to the higher energy level, Ej minus the lower energy which is Ei. So now we have this equation, let me go and highlight it here. We have hc over lambda is equal to Ej minus Ei. Let's get some more room, and let's see if we can solve that a little bit better here. So let me write this down here. We have hc over lambda is equal to the energy of the higher energy level, minus the energy of the lower energy level, like that. Alright, so in an earlier video, I showed you how you can calculate the energy at any energy level. We derived this equation. The energy at any energy level, n, is equal to E1 divided by n squared. So, if we wanted to know the energy when n is equal to j, that would be just E1 over j squared. We could take that and we could plug it in to here. Alright, if I wanted to know the energy for the lower energy level, that was Ei, and that's equal to E1 divided by i squared. I could take all of this, I could take this and I could plug it into here. Let's, once again, get some more room. Let's write what we have so far. We have hc over lambda is equal to Ej was E1 over j squared and Ei was E1 over i squared. Okay, we could pull out an E1 on the right. So we have hc over lambda is equal to E1, and so that would give us one over j squared minus one over i squared, like that. We could divide both sides by hc, so let's do that. So on the left, we could have one over the wavelength is equal to E1 divided by hc, one over j squared minus one over i squared. Again, from an earlier video, we calculated what that E1 is equal to. So, one over lambda is equal to E1 was negative 2.17 times 10 to the negative 18 joules. So once again, you can see that calculation in an earlier video. It took us a while to get there. We are going to divide by hc, and this is one over j squared minus one over i squared. Well let's look a little bit more closely at what we have right here. So, if I, for right now, not worry about the negative sign, and just think about what we have. This is all equal to a constant. H is Planck's constant, and c is the speed of light, so we have all these constants here. We could rewrite all of these as just R. I'm going to rewrite this as R, so this would be one over lambda is equal to negative R times one over j squared minus one over i squared. So, R is called the Rydberg constant so let's see if we can solve for that. Over here R would be equal to 2.17, times 10 to the negative 18, over h is Planck's constant, that's 6.626 times 10 to the negative 34, and then c is the speed of light. So we could use 2.9979 times 10 to the eighth meters per second as the speed of light. If you do all that math, I won't do it here just to save time, but if you do all that, you'll get 1.09 times 10 to the seventh and this is one over meters. I think I might have, it's possible I had a rounding issue in here, because if you use 2.18, you get a better number. It's 1.097 times 10 to the seventh, which is the Rydberg constant. This is just, again I'm just trying to show you the idea behind the Rydberg constant here so we get this value for the Rydberg constant. So you could plug that in for R if you needed to, and we're going to be doing that in the next video. You could stop right there, and you have related the wavelength, right to your different energy levels. You could go a little bit further, let's just go a little bit further here. So, one over lambda is equal to negative R. If we pull out a negative one, that would give us one over i squared minus one over j squared. Now we have these two negative signs, right? These two negative signs out here which gives up a positive. This is now equal to one over lambda, one over the wavelength is equal to positive R, the Rydberg constant, times one over i squared minus one over j squared. Remember what i and j represented. I represented the lower energy level, and j represented the higher energy level. This is an extremely useful equation, so usually you see this called the Balmer-Rydberg equation. We've derived this equation using the assumptions of the Bohr model, and this equation is extremely useful because it explains the entire emission spectrum of hydrogen. This is again, this is why we were exploring the Bohr model in the first place. We got this equation, and in the next video we're going to see how this explains the emissions spectrum of hydrogen. Think about lambda or the wavelength, right as the light is emitted when the electron falls back down to a lower energy state.