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Current time:0:00Total duration:10:50

Video transcript

we're sure that most 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 did 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 these colors right next to each other so they kind of blend together so that's a continuous spectrum if you did this similar thing with hydrogen you don't see a continuous spectrum so if you pass 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 grating to separate out the light for hydrogen you don't get a continuous spectrum you'd see these four lines of color so since you see lines we 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 in 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 Redbird constant times 1 over I squared where I is talking about the lower energy level - one 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 three so let me write this here so we have an electron that's falling from n is equal to three down to a lower energy level n is equal to two all right so it's going to emit light when it undergoes that transition so let's let's look at the at a visual representation of this and 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 three to n is equal to two let me go ahead and draw an electron here so electrons falling from n is equal to three energy level down to n is equal to two 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 all right let's go ahead and calculate the wavelength of light that's emitted when the electron falls from the third energy level to the second so we have 1 over lambda is equal to the Redbird constant as we saw on the previous video is one point zero nine seven times ten to the seventh the units would be 1 over meter all right 1 over I squared so I I refers to the lower energy level right 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 right so this would be 1 over 3 squared so 1 over 2 squared minus 1 over 3 squared let's go ahead and get out the calculator and let's do that math so 1 over 2 squared that's 1/4 so that's 0.25 minus 1 over 3 squares that's 1 over 9 so 1/4 minus 1/9 gives us 0.138 repeating and if we multiply that number by the Redbird constant alright that's one point zero nine seven times 10 to the seventh we get one five two three six one one so let me go and write that down so now we have one over lambda is equal to one five two three six one one so to solve for lambda all we need to do is take one over that number so one over that number gives us six point five six times ten to the negative seven and that would now be in meters so we have lambda is equal to six point five six times ten to the negative seven 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 ninth meters and so that's 656 nanometers 656 nanometers and that should sound familiar to you all right 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 third energy level down to the second 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 right a photon of red light is given off as the electron falls from the third energy level to the second 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 right so we have these other lines over here right we have this blue green one this blue one and this violent one so if you do the math you can use the Balmer Rydberg equation right you can do this and you can plug in some more numbers and you can calculate those values so those are those are electrons falling from higher energy levels down to the second energy level so let's uh let's go ahead and draw them on our diagram here so let's say an electron fell from the fourth energy level down to the second right so that energy difference if you the calculation 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 fifth energy level down to the second 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 sixth energy level down to the second so let's go ahead and draw that in and so now we have a way of explaining the this line spectrum of hydrogen that we 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 so 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 your your it's the only real way you can see the difference of energy right so energy is quantized we call this the Balmer series so this is called the Balmer series for a hydrogen but there are different transitions that you could do for example let's think about let's think about an electron going from the second energy level to the first 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 second energy level down to the first 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 this be 1 over lambda is equal to the Redbird constant 1.09 7 times 10 to the 7th that's 1 over meters and then we're going from the second energy level to the first so this would be 1 over the lower energy level squared so n is equal to 1 squared minus 1 over 2 squared all right so let's 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 0.75 five and so if we take if we take 0.75 of the Redbird constant let's go ahead and do that so one point zero nine seven times ten to the seventh is a Redbird constant we multiply that by 0.75 right so 3/4 then we should get that number there so that's eight two two seven five zero zero so let's write that down one over the wavelength is equal to eight two two seven five zero two so to solve for that wavelength just take one divided by that number and that gives you one point two one times 10 to the negative 7 and that be in meters so the wavelength here is equal to one point let me see what that was again one point two one five one point two one five times ten to the negative seventh meters and so if you move this over to right that's 122 nanometers so this is 122 nanometers but this is not a wavelength that we can see so 122 nanometers right that falls into the UV region the ultraviolet region region so we can't see that we can see the ones in the visible spectrum only and so this will represent a line in a different series and you can use the Balmer Rydberg equation to calculate all 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 visit the emission 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 the hydrogen atom is not is not reality it does allow us to figure some things out and to realize that energy is quantized