Bohr Model of the Hydrogen Atom, Electron Transitions, Atomic Energy Levels, Lyman & Balmer Series INSTRUCTOR: In this video, I'm going to focus on Bohr's model of the atom. Now, he proposed that the electrons move around the nucleus in circular orbits. So let's say this is the nucleus. And so this is the first energy level. This is the second. This is the third, and so forth. Now, the electrons can only be in these energy levels. It could be in the fourth, the fifth, but it can't be like anywhere in between. And so the energy levels of electrons are quantized because they can only have discrete values. The electron can't be in the 1.4 energy level or 1.86 energy level. It has to be in the first, second, or third, or fourth energy level. So these energy levels, they're integer-based values. So the electrons, they can only occupy certain orbits. And so in that sense, they're quantized. Now, I need to know that when an electron falls from a high energy level to a low energy level, it's going to emit energy in the form of a photon. Now, when an electron jumps, let's say, from a low energy level to a high energy level, it can only do so if it absorbs a photon with the right energy. Now, in this problem, we're going to talk about how to calculate how much energy is released or absorbed when an electron moves from one energy level to another. So let's read the first question. How much energy is released when an electron falls from the n equal 4 to the n equal 2 energy level inside a hydrogen atom? Now, the equation that you need to calculate it is this equation. E is equal to negative 2.178 times 10 to the minus 18 joules multiplied by 1 over n final squared minus 1 over n initial squared. n initial is the original energy level, which is the fourth energy level. And final is where the electron is going to. That is the second energy level. So all we need to do is plug those numbers into the equation, and we can find out how much energy is released from an electron within a hydrogen atom. So let's go ahead and do that. So n final, in this problem, is 2. Don't forget to square it. And n initial is 4. So 1 over 2 squared, 2 squared is 4. And 4 squared is 16. So we got 1/4 minus 1/16, and that's equal to 3/16, which is, as a decimal, 0.1875. So let's go ahead and multiply these two numbers. So you should get negative 4.084 times 10 to the negative 19 joules. So that's how much energy is released when an electron falls from the fourth to the second energy level. Now, here's a question for you. Why is this answer negative? Recall that when it falls from, let's say, a high energy level to a low energy level, of photon is emitted. So the electron is losing energy. And that's why it's negative. If the electron jumped from, let's say, the second to the fourth, it would be absorbing energy, so E would be positive. So that's what we have a negative sign, to show that energy is being released as the electron falls to a lower energy level. Now, let's move on to part b. How can we calculate the frequency now that we have the energy of the photon? Now, if you recall from earlier videos, the energy of a photon is equal to the product of Planck's constant and the frequency of that photon. So to calculate the frequency, it's simply the energy divided by Planck's constant. So it's going to be 4.084 times 10 to the negative 19 joules. You don't need the negative sign because frequency is always positive. The purpose of the negative sign is to tell us if a photon is being absorbed or released by an electron. So let's go ahead and divide those two numbers. So you should get 6.164 times 10 to the 14 hertz, or 1 over seconds. So that's the frequency. Now that we have the frequency, let's move on to part c. Let's calculate the wavelength of the photon in nanometers. Now, the speed of light is the product of the wavelength and the frequency. So the wavelength is the speed of light divided by the frequency. The speed of light is 3 times 10 to the 8 meters per second. And the frequency is 6.164 times 10 to the 14 with the units 1 over seconds. Now, let's go ahead and put that in the calculator. So I got 4.87 times 10 to the negative 7 meters. Now, we need to convert this to nanometers. So how can we do that? Now, you need to know one nanometer is equivalent to 10 to the minus 9 meters. So you can take the answer that we have, 4.87 times 10 to the minus 7, and divide it by 1 times 10 to the negative 9. And so this is equal to 487 nanometers. So that's how you can calculate the wavelength of this photon. Number two, an electron in the third energy level, or n equal 3 state, absorbs a photon with a wavelength of 1283.45 nanometers. Into what energy level will the electron jump to? So we know what equation we need to use, but we're given the wavelength. The first thing we need to do is calculate the energy of the photon using the wavelength. Now, the energy of a photon is Planck's constant times of frequency. And we know the frequency is the speed of light divided by the wavelength. So that equation comes from this equation if you divide both sides by lamda. So what I'm going to do is replace the frequency with this expression. So in terms of the wavelength, the energy of the photon is Planck's constant times the speed of light divided by the wavelength. So let's use that equation. Now, we need to convert nanometers to meters. So just replace nm with 10 to minus 9 meters. And this will give you the energy of the photon. So the energy of the photon is 1.549 times 10 and negative 19 joules. So now that we have the energy of the photon, we can now use this equation. Our goal is to find the value of n final. And it should be an integer value, like 2, 3, 4, 5, 6, 7. It shouldn't be like 6.4 or 8.7. We should get a whole number. So we have n initial. It's 3. And we have the energy. Now, what we need to determine is, is the energy positive or negative? Because if you get this wrong, it can mess up the entire problem. So our key expression is absorbs. The electron absorbs a photon. So it's absorbing energy, which means the energy of the electron is increasing, so E is positive. So we're going to replace E with this number with a positive sign. So 1.549 times 10 to the negative 19, and that's a joule symbol, is equal to this number multiplied by 1 over n final squared minus 1 over n initial squared, which n initial is 3. So all we got to do at this point is we need to solve for the missing variable. So I'm going to walk you through this step-by-step. So the first thing you want to do is take this number and divide it by that number. So 1.549 times 10 to the negative 19 divided by negative 2.178 times 10 to the negative 18. You should get negative 0.07112, and that's equal to 1 over n final squared minus-- 3 squared is 9-- so that's 1/9. So next, let's add 1/9 to both sides. So if you have your calculator, take a negative 0.07112 plus 1/9. And you should get 0.03999, which should probably be 0.04. Now, let's put this over 1. And let's cross-multiply. 1 times 1 is 1. And then we have 0.03999 times n final squared. Next, divide both sides by 0.03999. So n final squared is 1 divided by 0.03999. So that's basically about 25. It's like 25.006. And if we take the square root of both sides, n final is 5. If you get a whole number as an answer, then chances are you have the right answer. If you don't get a whole number, somewhere along the lines you might have made a mistake. But this answer is reasonable because if it absorbs a photon, we should get an energy value that's greater than 3. And 5 is greater than 3, so this makes sense. And that's the answer. It's going to jump into the fifth energy level. So here's a multiple choice problem for you. Which electron transition involves the greatest release of energy? Is it a, b, c, d, or e? So let's focus on the term release of energy. That means that the electron has to fall from a high energy level to a low energy level. So n has to be decreasing, not increasing, if energy is being released. So we can eliminate answer choice a because it's going from low energy level to a high energy level. And we don't want that. And we could eliminate answer choice c as well. So now, is it going to be b, d, or e? A quick way to tell is to look at the final energy level. Here, the final energy level is 4. Here, it's 1. And here, it's 2. Typically, the answer is going to be the one with the lowest energy level. So it's going to be d. Because n equals 1 is less than 2 or 4. And it really doesn't depend on the difference between the energy levels. If you're looking at e, if you'd pick e as an answer, it's not correct. But if you're thinking because 7 to 2, that's a difference of 5. That must be greater than 3 to 1, which is a difference of 2, it's not going to work out that way. Now, instead of drawing a circular orbit, I'm going to draw a horizontal line to represent the energy levels. So let's say if this is the first energy level, the second would probably be somewhere over here. And then this would be like the third. And then this would be like the fourth. And then this would be like the fifth. Do you see what's happening? As you move further away from the nucleus, the distance between the energy levels, they get smaller and smaller and smaller. So a transition from 2 to 1 involves a lot more energy than 3 to 2. It even involves a lot more energy than 4 to 2. Now, granted this is not drawn to scale. But here's how you can find out for sure which one is going to be the highest. You can use the energy equation that we've talked about. But you could ignore the 2.178 times 10 to the negative 18 value. Focus on this portion of the equation, 1 over n final squared minus 1 over n initial squared. So the choice that has the greatest value of this expression will be the one that's going to release the greatest amount of energy. So let's calculate it for b. So that's 1 over 6 squared, actually, n final's 4. So to 1 over 4 squared minus 1 over 6 squared. Go ahead and type that in your calculator and get the decimal value for it. So you should get 0.03472. Now, let's do the same for answer choice d. So it's going to be 1 over 1 squared minus 1 over 3 squared. So this is going to give you 0.889. It's basically 0.8 repeating. And then for answer choice e, it's going to be 1/2 squared minus 1/7 squared. And so this is going to be 0.2296. So as you can see, the one with the largest energy value is the one with the lowest final state. That's answer choice d. The second highest was the one with the next lowest final energy level, which was n equals 2. The lowest was the one that had the highest final energy level, which was n equals 4. So a quick way to get the answer is to look at the final state. The one with the lowest final state is the one that's going to release the most energy. And it usually works out that way. But if you feel dubious about that, just do what I did here. Use this equation and see which one gives you the largest value. And so in this problem, our answer is answer choice d. It's going to release the greatest amount of energy. Which of the following electron transitions will emit a photon in the visible light spectrum? So how can we find the answer to this question? The only way to do it is just to know your stuff. So let's go over some stuff that you need to know. So just like before, I'm going to draw the different energy levels. So this is going to be the first energy level. Let's call this one the second. And this is going to be the third and then the fourth, and let's put the fifth here. Now, any time an electron falls to the first energy level, it can go from 2 to 1. It can go from 3 to 1, 4 to 1, or 5 to 1. The final energy level has to be 1. So any time it falls to the first energy level, this is known as the Lyman series. Now, any time it falls to the second energy level-- it could go from 3 to 2. It can go from 4 to 2 or 5 to 2 or 6 to 2. This is known as the Balmer series. And if the final state is the third energy level, let's say from 5 to 3 or 4 to 3, this is known as the Paschen series, or pashken, or I'm not even sure exactly how you say its name. I think is passion. You can look it up though. And then if it falls, let's say, from 6 to 4 or 5 to 4, that's known as the Bracket series. And then if it goes down to the fifth-- I'm going to put this in yellow-- that's known as the PFund series. Now, the most common ones that you're going to deal with are the first three. Your teacher may expect you to know all five, but, typically, you're probably going to be tested on the first three, if you have to choose which ones you're going to memorize. Now, here's what you need to know. The Lyman series is associated with the ultraviolet spectrum. The Balmer series is associated with the visible light spectrum. Most of it is visible light, but I think one, though, might be UV, but the majority of it is visible light. And everything else, like the Paschen, the bracket, the PFund, It's all in the infrared or the IR region. So to answer this question, all we got to do was look for a photon that's in the Balmer series. So the final state has to be n equal 2. So it's not a. It's not b. It's c. Because the final state is n equals 2, so c is the answer. So answer choice a is associated with the Paschen series. So this would emit a photon in the infrared vision. b is associated with the Lyman series. It ends at n equals 1. So it's going to emit a photon in the UV spectrum. c is going to be visible light. d is associated with the Bracket series, so that's going to be the infrared region. And e is associated with the PFund series, which is also in the infrared region.