Understanding Electronic Transitions and Bohr Model

Electronic transitions, i.e. absorption and emission, are going to be the topic in this lesson. We're going to take a look at the Bohr model of the atom. We'll see what he got right, we'll see what he got wrong, and we'll see why he even got some important things wrong that we still give him a ton of credit for his model of the atom. We'll talk about calculating the energy of these electronic transitions and how these correspond to the energies of photons that are either absorbed or emitted by an atom. You can calculate the energy of these absorbed photons, you can also calculate their frequencies or wavelengths. based on what we learned in the last lesson. Now we'll finish this lesson off by talking about emission line spectra, which can be used as a fingerprint for an atom as it turns out, and we'll look at the Lyman series, Balmer series, Paskin series, Brackett series very briefly, and talk about how the Balmer series is by far the most important. My name is Chad, and welcome to Chad's Prep, where my goal is to take the stress out of learning science. In addition to high school and college science prep, we also do MCAT, DAT, and OAT prep as well. You can find those courses at chadsprep.com. Now this lesson is part of my new general chemistry playlist. I'm releasing several lessons a week throughout the school year. So if you want to be notified every time I post a new one, subscribe to the channel and click the bell notification. All right, so let's dive into this. And we first want to talk about the Bohr model of the atom. And it turns out his model worked very well for the hydrogen atom in certain respects and pretty much didn't work for any of the other atoms as we'll see. So but for the hydrogen atom, it explained our emission spectra and our absorption spectra perfectly. just again didn't work for any multi-electron system. Well, hydrogen is the only neutral atom that has one electron, and that's why it didn't work for any of the other neutral atoms, it turns out. Now, what Mr. Bohr said is he agreed with Mr. Rutherford. He said that at the center of an atom is a nucleus, and then what he came up with is that electrons go around this nucleus at set orbits around this nucleus. By having set orbits around the nucleus, they will have set energies. Now, we're going to talk about potential energies for just a brief moment. If you've had any sort of physics, we can talk about gravitational potential energy. And it turns out when something's above the ground, it has the potential to fall. And so we ascribe to it some measure of gravitational potential energy. But it's attracted to the Earth by gravity. So any two objects having mass are attracted to each other by gravity. It's attracted to the Earth. And if I let it go, it will fall down to the Earth. And as it falls, it loses potential energy and gains kinetic energy. But the key is, whenever two objects are attracted to each other, the closer they get together, the lower their potential energy, as is the case with this marker falling to the Earth. Well, the same thing is going to be true for an electron, which is negatively charged, being attracted to the nucleus, which is going to be positively charged, having protons. And so it turns out when an electron's in this first orbit, the one that's closest to the nucleus, that's when it'll have the lowest amount of potential energy, and we'll just simply at this point say lowest energy. And the further it gets out from the nucleus, the higher the energy it's going to have. So it turns out it follows a fairly simple equation to figure that out. And so here's this lovely energy for an electron in any one of these orbits, and it's got a constant, negative 2.18 times 10 to the negative 18 joules, divided by n squared, where n happens to correspond to these orbits. The closest orbit in, we get n equals 1 value. So the next orbit, n equals 2, and the outermost orbit, n equals 3. And it just followed these lovely integers, and they just, these aren't just three orbits, there are an infinite number of orbits. But what you find though is that the further you go out, the closer these orbits get together. So n equals four would be even closer to three than three is to two. And notice three is closer to two than two is to one, and so on and so forth. So they do reach an upper limit on how far that radius goes out. So but eventually you'll reach n equals infinity. And if you look, what would be the energy of an electron in that n equals infinity state? Well, you'd plug in infinity for n here and If infinity is infinity, then infinity squared is definitely still a bigger infinity, or just still infinity. And anything divided by infinity is zero. And so, as you get to the highest energy shell possible, n equals infinity, you're essentially not even part of this atom anymore, and at that point the electron has zero potential energy. But as you get closer to the nucleus, that potential energy drops. And so if the highest energy it can have is zero, then as it drops it's going to get more and more negative. and the most negative it can be is at n equals one here. And we can do this one in our head, right? So if n equals one, one squared is one, and this constant divided by one is still this constant. And so the energy corresponding to an electron in that first orbit is just gonna be negative 2.18 times 10 to the negative 18 joules. Okay, and we can use this equation just to calculate the energy of an electron in any of these orbits. So if we wanna go to the second orbit, n equals two. Then we're going to divide by 2 squared, divide by 4 here. So in right off your study guide there, we can see that that's going to come out to negative 5.45 times 10 to the negative 19. So, and if you look here, so 10 to the negative 19 is smaller than 10 to the negative 18, so this right here is a less negative number, so in being less negative it's closer to zero. So this is lower, more negative, this is higher, less negative. We go to n equals 3, same thing, 3 squared is 9, so this constant divided by 9, right off the study guide again is negative 2.42. times 10 to the negative 19 joules. And so for any orbit, you can use this lovely simple equation to calculate the energy of an electron that exists in that orbit according to Mr. Bohr. And you just got to plug the appropriate number of n. So n equals 4, n equals 5, n equals 6, any possible value of n, you can plug it right in and get the energy of an electron in that orbit. Now it turns out what Mr. Bohr was really trying to explain is why atoms either absorb or emit photons. And the idea is that if you've got an electron here, let's say... So in the first orbit, it turns out it can jump out to, say, the second orbit, or it can jump out to the third orbit. But if it's going to do that, we're going to need energy. Because again, an electron in these orbits are higher energy than an electron in that first orbit, where it's lowest in energy. And so where can an electron get energy? Well, it turns out that electrons can absorb photons of light, and they absorb the energy of that photon. And so it turns out for an electron to go from n equals 1 to n equals 2 orbit, it has to absorb exactly that difference in energy. And so what's interesting here is notice we have these orbits, but there's no orbits in between the first and the second. And this is what was revolutionary for Mr. Bohr. He said there's only orbits for these electrons at certain distances away. And because it's only at certain distances, that means that there's only going to be certain energies for these electrons possible, these energies we just calculated. But there isn't an energy in between these two. And so what this means is that the electron's energy in an atom is quantized. It only exists at specific quantities. And because of that, though, it turns out that only certain photons could get absorbed. So like when we go from an electron goes from n equals 1 to n equals 2, or n equals 1 to n equals 3, or n equals 1 to n equals 4, so on and so forth, that's going to correspond to a specific energy photon. as we'll see. Now if we want to go from one to three here, and that's the question on your handouts, what's delta E for that? So we could, in this case, just do E3 minus E1 is going to equal our delta E. And just take the numbers we'd already calculated and just subtract the difference, final minus initial. And so we could just go negative 2.42 times 10 to the negative 19 joules minus negative 2.18 times 10 to the negative 18 joules. So we can calculate delta E here. Let me grab my handy-dandy calculator. Alright, so negative 2.42 times 10 to the negative 19 minus a negative, i.e. plus, 2.18 times 10 to the negative 18. And we're going to get 1.938. times 10 to the negative 18 joules. Now, it turns out we could have calculated this another way, but that is the energy difference between, again, n equals 1 and n equals 3. And so if an electron is going to jump from this first orbit to this third orbit, it needs to... absorb exactly that amount of energy. And so what's true then is that the photon it's going to absorb has to have exactly that much energy. And so it turns out this difference in energy between three and one here has to equal the energy of the photon. Now in this case it was convenient that we'd already calculated, you know, n equals 1, the energy level, and the n equals 3, energy value as well, and we could just take the difference. But had we not, it turns out we can just get a simple relation between these as well. And again, e3 minus e1, using this level of formula, they're both going to have this constant. What's going to be different is the n value, that squared that we're dividing by. And so it turns out you can actually calculate delta E using this lovely formula right here as well, where you've just got the two different n values, but the constant here is just factored right out front. And it is this delta E that is also, again, equal to the energy of the photon. So, but if we look here, if we calculate the same thing out here, we could say, using that equation now, delta E equals negative 2.18 times 10 to the negative 18. Joules times 1 over n1 squared and in this case We're going to make that 3 squared minus 1 over n2 squared, which is now 1 squared. Now one thing to note about this, when you're setting this equal to the energy of a photon, so n1 and n2, so some people will tell you how you define these, and some people will say n initial, n final, which is convenient as well. Now one thing to note, the energy of a photon can never be a negative number. And so depending on what order you do these, you might come out with a negative number, and you just realize you need to take the absolute value in that case. But whether or not you come out with a positive or negative number actually is going to determine whether or not a photon of light is either absorbed or emitted. So what we're going to find out is that if you go from a lower energy orbit, one closer in, out to a higher energy orbit, one further away, you need to absorb a photon. But if the electron actually moves the opposite direction and goes from a higher energy orbit to a lower energy orbit, it actually emits a photon instead. And you'll find out that when you calculate delta E, you'd probably get a negative number. with the negative meaning nothing more than that energy is being emitted instead of absorbed, kind of like what we did with delta H is negative for exothermic, delta H is positive for endothermic, same kind of thing. But when you go to plug it in for the energy of the photon, always just take the absolute value because there's no such thing as a negative energy for a photon. Again, whether delta E is positive or negative just refers to whether it's absorbed or emitted. Cool, but if we calculate this out here once again, and we're going to take negative 2.18 times 10 to the negative 18, times, and I'll use parentheses, 1 9th minus 1, and we get 1.938 times 10 to the negative 18 joules, the exact same number. So two different ways to calculate this, but this equation is often given as well. And once again, it's this energy, 1.938 times 10 to the negative 18 joules, that's equal to the energy of a photon. And if you recall from the last chapter, the last lesson, I should say. The energy of a photon was also equal to h nu or hc over lambda. And so in getting this delta E, which is the energy of the photon, you could then use it to find either the frequency or the wavelength. And the question on your handout here was to find that energy and wavelength. So we want the wavelength, which means we're going to use this relation right here to pull it off. And so from here, we'll take this. Energy of the photon equals hc over lambda. We can rearrange that to be lambda equals hc over the energy over the photon. And h, Planck's constant, 6.626 times 10 to the negative 34 joule seconds. We learned that one in the last lesson. Probably not something you need to memorize. Probably something you'll be provided with. And the speed of light, c, 3.0 times 10 to the eighth. meters per second, also a constant you're often provided with, but some of you'll have to memorize that one. And then divided by the energy of the photon, we found to be 1.938 times 10 to the negative 18 joules. Cool. And we'll let our calculator do the work here for us. So 6.626 times 10 to the negative 34 times 3 times 10 to the eighth. Divided by the last answer we just had, the 1.938 times 10 to the negative 18. Oh, I just screwed that up. Let's try this again. 6.626 times 10 to the negative 34 times 3 times 10 to the eighth divided by 1.938 times 10 to the negative 18 equals 1.02 times 10 to the negative 7. Actually, 1.03. Round that up. 1.03 times 10 to the negative 7, and that's going to be in units of meters. Notice our joules are going to cancel right here, our seconds are going to cancel right here, and we're left with an answer in meters. But this is more commonly probably going to be expressed in nanometers, in which case we'd multiply by 10 to the 9th and get 103 nanometers. So notice that's a shorter wavelength than the 400 nanometer violet light, so it's definitely shorter. wavelength and the visible spectrum, so it's probably ultraviolet, maybe even x-rays, but probably ultraviolet would be my guess. Cool, but this is kind of how you calculate the energy of photons that are either absorbed or emitted, and again, you can do it if you just want to calculate the energy of whatever two levels you're given. Given this expression, you can and then take the difference, but again, there is a nice lovely plug-and-chug calculation for it just using the two different n values like so. Get that energy of the photon. and then either set it equal to h nu or hc over lambda. Now one thing to note, turns out early on, it was more common to see the wavelengths of these different emitted or absorbed photons. And so Mr. Rydberg came up with this lovely expression right here with this number right here. being what's called the Rydberg constant. And so he turns out, he figured out and said just going straight to wavelength, not worrying about the energy of photons or frequency of photons, just for the wavelength alone, that 1 over lambda here was equal to his Rydberg constant, and again times the same difference right here. So 1 over n1 squared, 1 over n2 squared. And so he kind of simplified the process. If all you want to get to is just the wavelength, you can go straight through this one equation, the Rydberg equation here. So that's kind of the deal there. Again for his equation to work so that you don't come out with a negative number, n1 has to be the smaller n value and n2 has to be the larger n value fyi. Cool, but we would have got exactly the same wavelength had we gone through that equation as well, but common one you guys are presented with alongside this. Cool, so now we've done some calculations here we've got to talk about one other aspect of this and so The absorption spectra and the emission spectra. And so it turns out if you've got white light, white light is typically what we have is a continuous spectrum of all the colors, or at least almost all the colors. And if you use a prism, you can spread that white light out into a rainbow of all the different colors. And it's just kind of this continuous spectrum. Well, what you find is if you take that white light coming off the sun, is that there's gonna be certain colors that are missing. And so in this continuous spectrum from, you know, Red, you know, ROYGBIV, red, orange, yellow, blue, green, indigo, violet. ROYGBIV, as you go for that continuous spectrum, you're going to see certain black lines missing. And those correspond to wavelengths that have been absorbed. And so the white light coming off the Sun is created at the core of the Sun, and these nuclear reactions going on at the core of the Sun. But it has to pass through the rest of the Sun, if you will, that's not the core, which contains a lot of hydrogen gas. And that hydrogen gas is going to absorb. Very specific wavelengths like this one at 103 nanometers along with a whole host of others. And it's not just like starting at shell number one. You might in some cases have the electron starting at shell number two and going from two to three or two to four or two to five. And so you're going to see a series of different wavelengths that are going to be absorbed. And it turns out those are exactly the black lines that show up from the white light coming off the sun. And this is how we know that the sun is made up of predominantly hydrogen gas. So this absorption spectrum, we call it, serves as a signature to show us, you know, what elements are there. And it turns out that the sun is, you know, roughly three-fourths hydrogen and one-fourth helium and trace amounts of other elements and stuff like this. But we know that not because we went to the sun and took a sample and brought it back to Earth, but because we've examined the light coming off the sun and we have looked at the absorption spectra to see which colors are missing. And the missing colors just happen to correspond perfectly to the colors that are missing when we shine. white light right here on Earth through either hydrogen or helium gas. And so that's how we figure out what stars are composed of. It's just by examining the light coming off of them. So that's the absorption spectrum, but we want to talk a little more about the emission spectrum, and I need a fresh board here. So now we want to take a look at the various emission spectra. We're going to look at the Bohr model just a little bit different. Instead of looking at the circular orbits and stuff like that, we're just going to organize the energies of the different orbits in terms of increasing energy. So a couple things you notice, n equals 1 is still the lowest energy, and as you get to higher and higher values of n, which again would correspond to further and further away from the nucleus, you get to higher energy. But one thing you'll notice is that, we said earlier here that the orbits get closer together the further you go out. Well, that's true in terms of energy, therefore, as well. And so like notice the difference in energy between n equals 5 and n equals 6 is smaller than the difference between 4 and 5, which is smaller than the difference between 3 and 4, and so on and so forth. So the biggest gap we got here is between n equals 1 and and n equals two. Now it turns out if we take a sample of a gas and we apply a voltage to it, we zap it with electricity so to speak, what happens is you end up causing all the lower energy electrons to jump up to higher shells, higher n values, higher orbits if you will. And then we're going to watch them fall back down and as they fall back down they will emit photons of light. Now with the absorption spectra we were looking for the missing colors and those were just the ones that got absorbed. Well now we don't have white light anymore in this case. We don't have light at all until we see the emitted photons. And so now the colors that were missing the absorption spectra, those are the only ones we see. And they're the emitted photons. And so only certain colors are going to be present. And it turns out they form various series. Depending on how far they fall is going to be the type of series they are. And so they could just fall all the way down to n equals 1, the lowest energy orbit possible. So in which case... We have a series here, so we could go from some of them are, you know, got zapped up to two, so they fall from two to one, some got zapped up to three, so they fall from three to one. 4 to 1, 5 to 1, and 6 to 1. Now some of them could only fall, end up falling to 2. That can happen as well. And so in this case we have some falling from 3 to 2, some from 4 to 2, some from 5 to 2, 6 to 2. And notice I only showed up to n equals 6, but there's 7, 8, 9, there's a whole bunch of these still and things of a sort. And so we'd get another series. Now what you should notice though is that the difference in energy for this series is smaller than the differences in energy for this series. So, and it turns out that this particular series where they drop down to n equals 2 orbit is called the Balmer series and it gets a special name. So they all get special names, but this one gets special notoriety because it turns out that a lot of the lines in the Balmer series correspond to visible light, light that we can see with our eyes. Whereas here, This is called the Lyman series when they drop down to n equals one. So with that Lyman series, these aren't typically things, not for hydrogen especially, that we can detect with our eyes. It turns out with the bigger differences in energy, these are bigger, higher energy photons and they typically correspond to like the ultraviolet region of the spectrum, not the visible region of the spectrum. So then we'll move on to the next one here. So dropping down to n equals three. And if we drop down to n equals three, we can go four to three. five to three, six to three, and this is part of the PASCAN series. That's an E, I can't really draw right sideways very well at that height. So that's the PASCAN series. And notice these are smaller energy differences than our bomber, and if our bomber was visible, it turns out these often are in the infrared range. So something longer wavelength and lower energy than visible light. So, and then finally you could drop down to four. And we go like 5 to 4, 6 to 4, 7 to 4, 8 to 4 if I had enough diagram to show it. And this corresponds to the bracket. Cool. Now some of you might be on the hook for knowing all the names of these series, depending on how far you fall. But most of you probably just need to know the one I indicated in red here and highlighted. The Balmer series corresponds to n equals 2 as the final orbit they're dropped down to. And again, it's special because a lot of the lines in the Balmer series correspond to visible light. that we can see with our eyes. Now if you found this lesson helpful consider giving me a thumbs up. Best thing you can do to make sure YouTube shares this lesson with other students as well. And if you're looking for the study guide that goes with this lesson, if you are looking for general chemistry practice problems, quizzes, chapter tests, practice final exams, if you are looking for a final exam rapid reviews, check out my general chemistry master course. I'll leave a link in the description. Free trials available. Happy studying.

You can calculate the energy of these absorbed photons, you can also calculate their frequencies or wavelengths. based on what we learned in the last lesson. Now we'll finish this lesson off by talking about emission line spectra, which can be used as a fingerprint for an atom as it turns out, and we'll look at the Lyman series, Balmer series, Paskin series, Brackett series very briefly, and talk about how the Balmer series is by far the most important.

My name is Chad, and welcome to Chad's Prep, where my goal is to take the stress out of learning science. In addition to high school and college science prep, we also do MCAT, DAT, and OAT prep as well. You can find those courses at chadsprep.com.

Now this lesson is part of my new general chemistry playlist. I'm releasing several lessons a week throughout the school year. So if you want to be notified every time I post a new one, subscribe to the channel and click the bell notification.

All right, so let's dive into this. And we first want to talk about the Bohr model of the atom. And it turns out his model worked very well for the hydrogen atom in certain respects and pretty much didn't work for any of the other atoms as we'll see.

So but for the hydrogen atom, it explained our emission spectra and our absorption spectra perfectly. just again didn't work for any multi-electron system. Well, hydrogen is the only neutral atom that has one electron, and that's why it didn't work for any of the other neutral atoms, it turns out. Now, what Mr. Bohr said is he agreed with Mr. Rutherford. He said that at the center of an atom is a nucleus, and then what he came up with is that electrons go around this nucleus at set orbits around this nucleus.

By having set orbits around the nucleus, they will have set energies. Now, we're going to talk about potential energies for just a brief moment. If you've had any sort of physics, we can talk about gravitational potential energy. And it turns out when something's above the ground, it has the potential to fall. And so we ascribe to it some measure of gravitational potential energy.

But it's attracted to the Earth by gravity. So any two objects having mass are attracted to each other by gravity. It's attracted to the Earth. And if I let it go, it will fall down to the Earth. And as it falls, it loses potential energy and gains kinetic energy.

But the key is, whenever two objects are attracted to each other, the closer they get together, the lower their potential energy, as is the case with this marker falling to the Earth. Well, the same thing is going to be true for an electron, which is negatively charged, being attracted to the nucleus, which is going to be positively charged, having protons. And so it turns out when an electron's in this first orbit, the one that's closest to the nucleus, that's when it'll have the lowest amount of potential energy, and we'll just simply at this point say lowest energy. And the further it gets out from the nucleus, the higher the energy it's going to have. So it turns out it follows a fairly simple equation to figure that out.

And so here's this lovely energy for an electron in any one of these orbits, and it's got a constant, negative 2.18 times 10 to the negative 18 joules, divided by n squared, where n happens to correspond to these orbits. The closest orbit in, we get n equals 1 value. So the next orbit, n equals 2, and the outermost orbit, n equals 3. And it just followed these lovely integers, and they just, these aren't just three orbits, there are an infinite number of orbits. But what you find though is that the further you go out, the closer these orbits get together. So n equals four would be even closer to three than three is to two.

And notice three is closer to two than two is to one, and so on and so forth. So they do reach an upper limit on how far that radius goes out. So but eventually you'll reach n equals infinity. And if you look, what would be the energy of an electron in that n equals infinity state? Well, you'd plug in infinity for n here and If infinity is infinity, then infinity squared is definitely still a bigger infinity, or just still infinity.

And anything divided by infinity is zero. And so, as you get to the highest energy shell possible, n equals infinity, you're essentially not even part of this atom anymore, and at that point the electron has zero potential energy. But as you get closer to the nucleus, that potential energy drops.

And so if the highest energy it can have is zero, then as it drops it's going to get more and more negative. and the most negative it can be is at n equals one here. And we can do this one in our head, right? So if n equals one, one squared is one, and this constant divided by one is still this constant. And so the energy corresponding to an electron in that first orbit is just gonna be negative 2.18 times 10 to the negative 18 joules.

Okay, and we can use this equation just to calculate the energy of an electron in any of these orbits. So if we wanna go to the second orbit, n equals two. Then we're going to divide by 2 squared, divide by 4 here. So in right off your study guide there, we can see that that's going to come out to negative 5.45 times 10 to the negative 19. So, and if you look here, so 10 to the negative 19 is smaller than 10 to the negative 18, so this right here is a less negative number, so in being less negative it's closer to zero. So this is lower, more negative, this is higher, less negative.

We go to n equals 3, same thing, 3 squared is 9, so this constant divided by 9, right off the study guide again is negative 2.42. times 10 to the negative 19 joules. And so for any orbit, you can use this lovely simple equation to calculate the energy of an electron that exists in that orbit according to Mr. Bohr. And you just got to plug the appropriate number of n. So n equals 4, n equals 5, n equals 6, any possible value of n, you can plug it right in and get the energy of an electron in that orbit.

Now it turns out what Mr. Bohr was really trying to explain is why atoms either absorb or emit photons. And the idea is that if you've got an electron here, let's say... So in the first orbit, it turns out it can jump out to, say, the second orbit, or it can jump out to the third orbit. But if it's going to do that, we're going to need energy. Because again, an electron in these orbits are higher energy than an electron in that first orbit, where it's lowest in energy.

And so where can an electron get energy? Well, it turns out that electrons can absorb photons of light, and they absorb the energy of that photon. And so it turns out for an electron to go from n equals 1 to n equals 2 orbit, it has to absorb exactly that difference in energy. And so what's interesting here is notice we have these orbits, but there's no orbits in between the first and the second. And this is what was revolutionary for Mr. Bohr.

He said there's only orbits for these electrons at certain distances away. And because it's only at certain distances, that means that there's only going to be certain energies for these electrons possible, these energies we just calculated. But there isn't an energy in between these two. And so what this means is that the electron's energy in an atom is quantized. It only exists at specific quantities.

And because of that, though, it turns out that only certain photons could get absorbed. So like when we go from an electron goes from n equals 1 to n equals 2, or n equals 1 to n equals 3, or n equals 1 to n equals 4, so on and so forth, that's going to correspond to a specific energy photon. as we'll see.

Now if we want to go from one to three here, and that's the question on your handouts, what's delta E for that? So we could, in this case, just do E3 minus E1 is going to equal our delta E. And just take the numbers we'd already calculated and just subtract the difference, final minus initial. And so we could just go negative 2.42 times 10 to the negative 19 joules minus negative 2.18 times 10 to the negative 18 joules.

So we can calculate delta E here. Let me grab my handy-dandy calculator. Alright, so negative 2.42 times 10 to the negative 19 minus a negative, i.e. plus, 2.18 times 10 to the negative 18. And we're going to get 1.938.

times 10 to the negative 18 joules. Now, it turns out we could have calculated this another way, but that is the energy difference between, again, n equals 1 and n equals 3. And so if an electron is going to jump from this first orbit to this third orbit, it needs to... absorb exactly that amount of energy. And so what's true then is that the photon it's going to absorb has to have exactly that much energy. And so it turns out this difference in energy between three and one here has to equal the energy of the photon.

Now in this case it was convenient that we'd already calculated, you know, n equals 1, the energy level, and the n equals 3, energy value as well, and we could just take the difference. But had we not, it turns out we can just get a simple relation between these as well. And again, e3 minus e1, using this level of formula, they're both going to have this constant.

What's going to be different is the n value, that squared that we're dividing by. And so it turns out you can actually calculate delta E using this lovely formula right here as well, where you've just got the two different n values, but the constant here is just factored right out front. And it is this delta E that is also, again, equal to the energy of the photon.

So, but if we look here, if we calculate the same thing out here, we could say, using that equation now, delta E equals negative 2.18 times 10 to the negative 18. Joules times 1 over n1 squared and in this case We're going to make that 3 squared minus 1 over n2 squared, which is now 1 squared. Now one thing to note about this, when you're setting this equal to the energy of a photon, so n1 and n2, so some people will tell you how you define these, and some people will say n initial, n final, which is convenient as well. Now one thing to note, the energy of a photon can never be a negative number. And so depending on what order you do these, you might come out with a negative number, and you just realize you need to take the absolute value in that case.

But whether or not you come out with a positive or negative number actually is going to determine whether or not a photon of light is either absorbed or emitted. So what we're going to find out is that if you go from a lower energy orbit, one closer in, out to a higher energy orbit, one further away, you need to absorb a photon. But if the electron actually moves the opposite direction and goes from a higher energy orbit to a lower energy orbit, it actually emits a photon instead.

And you'll find out that when you calculate delta E, you'd probably get a negative number. with the negative meaning nothing more than that energy is being emitted instead of absorbed, kind of like what we did with delta H is negative for exothermic, delta H is positive for endothermic, same kind of thing. But when you go to plug it in for the energy of the photon, always just take the absolute value because there's no such thing as a negative energy for a photon.

Again, whether delta E is positive or negative just refers to whether it's absorbed or emitted. Cool, but if we calculate this out here once again, and we're going to take negative 2.18 times 10 to the negative 18, times, and I'll use parentheses, 1 9th minus 1, and we get 1.938 times 10 to the negative 18 joules, the exact same number. So two different ways to calculate this, but this equation is often given as well.

And once again, it's this energy, 1.938 times 10 to the negative 18 joules, that's equal to the energy of a photon. And if you recall from the last chapter, the last lesson, I should say. The energy of a photon was also equal to h nu or hc over lambda. And so in getting this delta E, which is the energy of the photon, you could then use it to find either the frequency or the wavelength.

And the question on your handout here was to find that energy and wavelength. So we want the wavelength, which means we're going to use this relation right here to pull it off. And so from here, we'll take this. Energy of the photon equals hc over lambda. We can rearrange that to be lambda equals hc over the energy over the photon.

And h, Planck's constant, 6.626 times 10 to the negative 34 joule seconds. We learned that one in the last lesson. Probably not something you need to memorize.

Probably something you'll be provided with. And the speed of light, c, 3.0 times 10 to the eighth. meters per second, also a constant you're often provided with, but some of you'll have to memorize that one. And then divided by the energy of the photon, we found to be 1.938 times 10 to the negative 18 joules.

Cool. And we'll let our calculator do the work here for us. So 6.626 times 10 to the negative 34 times 3 times 10 to the eighth. Divided by the last answer we just had, the 1.938 times 10 to the negative 18. Oh, I just screwed that up.

Let's try this again. 6.626 times 10 to the negative 34 times 3 times 10 to the eighth divided by 1.938 times 10 to the negative 18 equals 1.02 times 10 to the negative 7. Actually, 1.03. Round that up.

1.03 times 10 to the negative 7, and that's going to be in units of meters. Notice our joules are going to cancel right here, our seconds are going to cancel right here, and we're left with an answer in meters. But this is more commonly probably going to be expressed in nanometers, in which case we'd multiply by 10 to the 9th and get 103 nanometers.

So notice that's a shorter wavelength than the 400 nanometer violet light, so it's definitely shorter. wavelength and the visible spectrum, so it's probably ultraviolet, maybe even x-rays, but probably ultraviolet would be my guess. Cool, but this is kind of how you calculate the energy of photons that are either absorbed or emitted, and again, you can do it if you just want to calculate the energy of whatever two levels you're given. Given this expression, you can and then take the difference, but again, there is a nice lovely plug-and-chug calculation for it just using the two different n values like so. Get that energy of the photon.

and then either set it equal to h nu or hc over lambda. Now one thing to note, turns out early on, it was more common to see the wavelengths of these different emitted or absorbed photons. And so Mr. Rydberg came up with this lovely expression right here with this number right here. being what's called the Rydberg constant. And so he turns out, he figured out and said just going straight to wavelength, not worrying about the energy of photons or frequency of photons, just for the wavelength alone, that 1 over lambda here was equal to his Rydberg constant, and again times the same difference right here.

So 1 over n1 squared, 1 over n2 squared. And so he kind of simplified the process. If all you want to get to is just the wavelength, you can go straight through this one equation, the Rydberg equation here.

So that's kind of the deal there. Again for his equation to work so that you don't come out with a negative number, n1 has to be the smaller n value and n2 has to be the larger n value fyi. Cool, but we would have got exactly the same wavelength had we gone through that equation as well, but common one you guys are presented with alongside this.

Cool, so now we've done some calculations here we've got to talk about one other aspect of this and so The absorption spectra and the emission spectra. And so it turns out if you've got white light, white light is typically what we have is a continuous spectrum of all the colors, or at least almost all the colors. And if you use a prism, you can spread that white light out into a rainbow of all the different colors.

And it's just kind of this continuous spectrum. Well, what you find is if you take that white light coming off the sun, is that there's gonna be certain colors that are missing. And so in this continuous spectrum from, you know, Red, you know, ROYGBIV, red, orange, yellow, blue, green, indigo, violet. ROYGBIV, as you go for that continuous spectrum, you're going to see certain black lines missing.

And those correspond to wavelengths that have been absorbed. And so the white light coming off the Sun is created at the core of the Sun, and these nuclear reactions going on at the core of the Sun. But it has to pass through the rest of the Sun, if you will, that's not the core, which contains a lot of hydrogen gas.

And that hydrogen gas is going to absorb. Very specific wavelengths like this one at 103 nanometers along with a whole host of others. And it's not just like starting at shell number one.

You might in some cases have the electron starting at shell number two and going from two to three or two to four or two to five. And so you're going to see a series of different wavelengths that are going to be absorbed. And it turns out those are exactly the black lines that show up from the white light coming off the sun. And this is how we know that the sun is made up of predominantly hydrogen gas.

So this absorption spectrum, we call it, serves as a signature to show us, you know, what elements are there. And it turns out that the sun is, you know, roughly three-fourths hydrogen and one-fourth helium and trace amounts of other elements and stuff like this. But we know that not because we went to the sun and took a sample and brought it back to Earth, but because we've examined the light coming off the sun and we have looked at the absorption spectra to see which colors are missing. And the missing colors just happen to correspond perfectly to the colors that are missing when we shine.

white light right here on Earth through either hydrogen or helium gas. And so that's how we figure out what stars are composed of. It's just by examining the light coming off of them. So that's the absorption spectrum, but we want to talk a little more about the emission spectrum, and I need a fresh board here. So now we want to take a look at the various emission spectra.

We're going to look at the Bohr model just a little bit different. Instead of looking at the circular orbits and stuff like that, we're just going to organize the energies of the different orbits in terms of increasing energy. So a couple things you notice, n equals 1 is still the lowest energy, and as you get to higher and higher values of n, which again would correspond to further and further away from the nucleus, you get to higher energy.

But one thing you'll notice is that, we said earlier here that the orbits get closer together the further you go out. Well, that's true in terms of energy, therefore, as well. And so like notice the difference in energy between n equals 5 and n equals 6 is smaller than the difference between 4 and 5, which is smaller than the difference between 3 and 4, and so on and so forth. So the biggest gap we got here is between n equals 1 and and n equals two.

Now it turns out if we take a sample of a gas and we apply a voltage to it, we zap it with electricity so to speak, what happens is you end up causing all the lower energy electrons to jump up to higher shells, higher n values, higher orbits if you will. And then we're going to watch them fall back down and as they fall back down they will emit photons of light. Now with the absorption spectra we were looking for the missing colors and those were just the ones that got absorbed.

Well now we don't have white light anymore in this case. We don't have light at all until we see the emitted photons. And so now the colors that were missing the absorption spectra, those are the only ones we see. And they're the emitted photons.

And so only certain colors are going to be present. And it turns out they form various series. Depending on how far they fall is going to be the type of series they are.

And so they could just fall all the way down to n equals 1, the lowest energy orbit possible. So in which case... We have a series here, so we could go from some of them are, you know, got zapped up to two, so they fall from two to one, some got zapped up to three, so they fall from three to one. 4 to 1, 5 to 1, and 6 to 1. Now some of them could only fall, end up falling to 2. That can happen as well.

And so in this case we have some falling from 3 to 2, some from 4 to 2, some from 5 to 2, 6 to 2. And notice I only showed up to n equals 6, but there's 7, 8, 9, there's a whole bunch of these still and things of a sort. And so we'd get another series. Now what you should notice though is that the difference in energy for this series is smaller than the differences in energy for this series.

So, and it turns out that this particular series where they drop down to n equals 2 orbit is called the Balmer series and it gets a special name. So they all get special names, but this one gets special notoriety because it turns out that a lot of the lines in the Balmer series correspond to visible light, light that we can see with our eyes. Whereas here, This is called the Lyman series when they drop down to n equals one.

So with that Lyman series, these aren't typically things, not for hydrogen especially, that we can detect with our eyes. It turns out with the bigger differences in energy, these are bigger, higher energy photons and they typically correspond to like the ultraviolet region of the spectrum, not the visible region of the spectrum. So then we'll move on to the next one here.

So dropping down to n equals three. And if we drop down to n equals three, we can go four to three. five to three, six to three, and this is part of the PASCAN series.

That's an E, I can't really draw right sideways very well at that height. So that's the PASCAN series. And notice these are smaller energy differences than our bomber, and if our bomber was visible, it turns out these often are in the infrared range.

So something longer wavelength and lower energy than visible light. So, and then finally you could drop down to four. And we go like 5 to 4, 6 to 4, 7 to 4, 8 to 4 if I had enough diagram to show it. And this corresponds to the bracket.

Cool. Now some of you might be on the hook for knowing all the names of these series, depending on how far you fall. But most of you probably just need to know the one I indicated in red here and highlighted.

The Balmer series corresponds to n equals 2 as the final orbit they're dropped down to. And again, it's special because a lot of the lines in the Balmer series correspond to visible light. that we can see with our eyes. Now if you found this lesson helpful consider giving me a thumbs up.

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Transcript for:Understanding Electronic Transitions and Bohr Model

Transcript for:
Understanding Electronic Transitions and Bohr Model