Main content

## Electronic structure

Current time:0:00Total duration:10:46

# Emission spectrum ofÂ hydrogen

## Video transcript

- [Voiceover] I'm sure that most of you know the famous story of Isaac Newton, where he took a narrow beam of light and he put that narrow beam of light through a prism, and the prism separated the white light into all the different colors of the rainbow. And so, if you do this experiment you might see something like this rectangle up here, so all of these different colors of the rainbow. I'm going to call this a continuous spectrum. It's continuous because you see all of these colors right next to each other, so they kind of blend together, so that's a continuos spectrum. If you did a similar thing with hydrogen, you don't see a continuos spectrum. So, if you passed a current through a tube containing hydrogen gas, the electrons and the hydrogen atoms are going to absorb energy and jump up to a higher energy level. When those electrons fall down to a lower energy level, they emit light. And so, we talked about this in the last video. This is the concept of emission. If you use something like a prism or diffraction grading to separate out the light, for hydrogen you don't get a continuos spectrum. You'd see these four lines of color. So, since you see lines, we can call this a line spectrum. So, this is the line spectrum for hydrogen. So, you see one red line, and it turns out that that red line has a wavelength-- that red light has a wavelength of 656 nanometers. You'll also see a blue-green line, and so this has a wavelength of 486 nanometers; a blue line, 434 nanometers; and a violet line at 410 nanometers. And so, this emission spectrum is unique to hydrogen and so this is one way to identify elements and so this is a pretty important thing. And, since line spectrum are unique, this is pretty important to explain where those wavelengths come from. And, we can do that by using the equation we derived in the previous video. So, I call this equation the Balmer Rydberg Equation. And, you can see that 1 over Lambda, Lambda is the wavelength of light that's emitted, is equal to R, which is the Rydberg constant, times 1 over i squared, where i is talking about the lower energy level, minus 1 over j squared, where j is referring to the higher energy level. For example, let's say we were considering an excited electron that's falling from a higher energy level. n is equal to 3. So, let me write this here. So, we have an electron that's falling from n is equal to 3 down to a lower energy level, n is equal to 2. Alright, so it's going to emit light when it undergoes that transition. So, let's look at a visual representation of this. And, now, let's see if we can calculate the wavelength of light that's emitted. Alright, so, if an electron is falling from n is equal to 3 to n is equal to 2-- Let me go ahead and draw an electron here-- So, an electron is falling from n is equal to 3 energy level down to n is equal to 2, and the difference in those two energy levels, are the difference in energy is equal to the energy of the photon. And, so, that's how we calculated the Balmer Rydberg equation in the previous video. Alright, let's go ahead and calculate the wavelength of light that's emitted when the electron falls from the 3rd energy level to the 2nd. So, we have 1 over Lambda is equal to the Rydberg constant, as we saw in the previous video, is 1.097 times 10 to the 7th. The units would be 1 over meter. Alright, 1 over i squared. So, i refers to the lower energy level. Alright, so, the lower energy level is when n is equal to 2. So, we plug in 1 over 2 squared. And then, from that, we're going to subtract 1 over the higher energy level. That's n is equal to 3. Alright, so, this'll be 1 over 3 squared. So, 1 over 2 squared minus 1 over 3 squared. 1 over 2 squared, that's 1/4, so, that's .25, minus 1 over 3 squared, so that's 1 over 9, so 1/4 minus 1/9 gives us .138 repeating. And, if we multiply that number by the Rydberg constant, alright, that's 1.097 times 10 to the 7th, we get 1523611. So, let me go ahead and write that down, so, now, we have 1 over Lambda is equal to 1523611. So, to solve for Lambda, all we need to do is take 1 over that number, So, 1 over that number gives us 6.56 times 10 to the negative 7. And, that would now be in meters. So, we have Lambda is equal to 6.56 times 10 to the negative 7 meters. So, let's convert that into-- Let's go like this. Let's go 656, that's the same thing as 656 times 10 to the negative 9 meters. And, so, that's 656 nanometers. 656 nanometers. And, that should sound familiar to you. Alright, so, let's go back up here and see where we've seen 656 nanometers before. 656 nanometers is the wavelength of this red line right here. So, that red line represents the light that's emitted when an electron falls from the 3rd energy level down to the 2nd energy level. So, let's go back down to here, and, let's go ahead and show that. So, we can say that a photon, alright, a photon of red light is given off as the electron falls from the 3rd energy level to the 2nd energy level. So, that explains the red line in the line spectrum of hydrogen. So, how can we explain these other lines that we see? Alright, so, we have these other lines over here. Alright, we have this blue-green one, this blue one, and this violet one. So, if you do the math, you can use the Balmer Rydberg equation or you can do this, and you can plug in some more numbers, and you can calculate those values. So, those are electrons falling from the higher energy levels down to the 2nd energy level. So, let's go ahead and draw them on our diagram here. So, let's say an electron fell from the 4th energy level down to the 2nd. Alright, so, that energy difference, if you do the calculations, that turns out to be the blue-green line in your line spectrum. So, I'll represent the light emitted like that. And, if an electron fell from the 5th energy level down to the 2nd energy level, that corresponds to the blue line that you see on the line spectrum. And, then, finally, the violet line must be the transition from the 6th energy level down to the 2nd. So, let's go ahead and draw that in. And, so, now, we have a way of explaining this line spectrum of hydrogen that we can observe and, since we calculated this Balmer Rydberg equation using the Bohr equation, using the Bohr model I should say, the Bohr model is what allowed us to do this. The Bohr model explains these different energy levels that we see. So, when you look at the line spectrum of hydrogen, it's kind of like you're seeing energy levels. At least, that's how I like to think about it. Because, it's the only real way you can see the difference of energy. Alright, so, energy is quantized. We call this the Balmer series. So, this is called the Balmer series for hydrogen. But, there are different transitions that you can do. For example, let's think about an electron going from the 2nd energy level to the 1st. Alright, so, let's get some more room here. And, if I drew a line in here, again, not drawn to scale, think about an electron going from the 2nd energy level down to the 1st. So, from n is equal to 2 to n is equal to 1. Let's use our equation and let's calculate that wavelength next, so to speak. 1 over Lambda is equal to the Rydberg constant, 1.097 times 10 to the 7th, that's 1 over meters. And, then, we're going from the 2nd energy level to the 1st. So, this'll be 1 over the lower energy level squared. So, n is equal to 1 squared minus 1 over 2 squared. Alright, so, let's get some more room, get out the calculator here. So, 1 over 1 squared is just 1, minus 1/4. So, that's .75. And, so, if we take .75 of the Rydberg constant, let's go ahead and do that. So, 1.097 times 10 to the 7, is our Rydberg constant, and multiply that by .75, alright, so 3/4, then, we should get that number there. So, that's 8227500. So, let's write that down. 1 over the wavelength is equal to 8227500. So, to solve for that wavelength, just take 1 divided by that number and that gives you 1.21 times 10 to the negative 7, and that'd be in meters. So, the wavelength here is equal to 1.-- Let me see what that was again, 1.215 times 10 to the negative 7 meters. And, so, if you move this over 2, that's 122 nanometers. So, this is 122 nanometers. But, this is not a wavelength that we can see, so, 122 nanometers Alright, that falls into the UV region, the Ultraviolet region. So, we can't see that. We can see the ones in the visible spectrum only. And, so, this'll represent a line in a different series and you can use the Balmer Rydberg equation to calculate all of the other possible transitions for hydrogen, and that's beyond the scope of this video. So, here, I just wanted to show you that the emissions spectrum of hydrogen can be explained using the Balmer Rydberg equation which we derived using the Bohr model of the hydrogen atom. So, even though the Bohr model of the hydrogen atom is not reality, it does allow us to figure some things out, and to realize that energy is quantized.