Chemical Science - Particle-like Nature of Light - Lecture 4

Principles of Chemical Science/n * Email this page/nVideo Lectures - Lecture 4/nTopics covered: /nParticle-like Nature of Light/nInstructor: /nProf. Sylvia Ceyer/nTranscript - Lecture 4/nAll right. Let's just remind ourselves for a moment where we were, and that is that we were talking about the structure of the atom. And we talked about how classical physics, the classical mechanics, classical electromagnetism failed to explain how the electron and the nucleus hung together./nAnd I said that the clues about how to solve that conundrum actually came from a different area of discussion, and that is this wave particle duality. And so remember that I said we are going to put aside that discussion of the structure of the atom all the way until next Monday. Pardon? No class Monday? We're going to put it aside all the way until next Wednesday./nThank you. [LAUGHTER] You sound surprised? You're not. It's a student holiday on Monday. And we're going to talk about the wave particle duality of light and matter, get clues from that area of discussion, and we're going to apply those clues then to solving the structure of the atom next Wednesday.,/nNow, we started on this discussion by talking about the wave-like nature of radiation. And to illustrate that, we did this two-slit experiment where we saw an interference pattern. And we talked about how that interference pattern arose because of the property of superposition of waves. That is if you have waves in the two same points in space their amplitude superpose or they add./nAnd because some of those amplitudes are positive and some are negative, depending on the relative phase of the waves, you have constructive or you have destructive interference. And specifically what we were trying to understand is, in the two-slit experiment, why we had this line of bright spots, dark spots, bright spots, dark spots./nAnd we analyzed it by drawing these semicircles which represent the wave front kind of sneaking through this little pinhole here, this slit, emanating equally in all directions. These semicircles here represent the crest of the wave. The distance here between these crests is 1?, is one wavelength. And can we dim the lights in the front here?/nIf you closely, what we discovered is that all points along this line constructively interfere. That is we have waves that are at the same place in space and the maxima and the minima are in the same place in space, and so that's going to lead to a very bright spot here on the screen. The amplitude is going to be twice as high as it was in each of the individual waves./nAnd we get this very bright spot. And we realize that along this line the difference in the distance traveled by the two waves that are constructively interfering was 1?. And then we did so for this line here. And we discovered that all along this line we had these two waves in phase./nAll along this line we had constructive interference which, of course, imaged as another bright spot. And when we analyzed it, we found that the difference in the distance traveled by the two waves was 2?. And likewise here. All along this line we had constructive interference imaged as a bright spot. And we recognize that in this case the difference in the distance traveled by the two waves was 0?./nAnd given that pattern, we then concluded that there was a general condition for constructive interference. That is if you had maximum constructive interference, the general condition that had to apply was this, that the difference in the distance traveled by the two waves was an integral multiple of the wavelength where this integer N could be 0 or 1 or 2 or 3 or 4./nAnd it's actually symmetric here along the center of the two slits. So, you would see the same kind of interference pattern up here. And so we call this interference feature right here, we often call it the zero-order interference feature. We call this one, when N is equal to 1 the first-order. We call this one the second-order interference feature./nAnd in between the very bright spots on that screen, what we have are dark spots here, here and here. And what I am going to ask you to do in recitation section, actually next Tuesday, is to look at what condition has to apply in order to have these dark spots which result from destructive interference./nAnd, of course, when you analyze that you're going to see, say right here, the crest, the maximum of one wave at the same point in space as the minimum of the other wave. And you get maximum destructive interference everywhere along that line. And what you will do in recitation next week is actually to figure out what the condition is for destructive interference. And that is going to be N + Ω?./nSo, it is this observation of these interference phenomena that are characteristics of the wave-like nature of anything. In this case, the wave-like nature of light, of radiation. On Friday, what we are going to do is we're going to see this same sort of interference phenomena, a property of waves, be observed from particles, from electrons. This diagram you are going to see once again on Friday. And we will use that to demonstrate the wave-like nature of particles, electrons. That is our evidence for the wave-like nature of light./nNow what we are going to do is we're going to talk about the other behavior of light, and that is the particle light behavior of radiation. And I forgot to get out my notes. What is that evidence? Well, soon after Thompson's discovery the electron there were some observations being made of something called the photoelectric effect./nAnd this was very curious. The photoelectric effect is simply that you have some metal here and you shine some light on that metal. And what happens is that an electron is ejected. Metal. Light coming in. Electron comes out./nBut the frequency of this light that is coming in, this frequency nu, we talked about frequency the other day, has to be greater than or equal to some threshold frequency, some threshold frequency that I am going to call ?0. In other words, if I come in with radiation here that has got a frequency that is less than ?0, well, then no electrons come off of the metal./nMaybe we could raise the screen here for the moment. Well, another way to kind of plot that observation is to just plot number of electrons versus the frequency of the radiation./nAnd so if you're at low frequency there are going to be no electrons that come off, but then when you get to some frequency ?0 all of a sudden electrons are going to come off. And the number of electrons that come off is going to remain constant as you increase the frequency of the light. And typically this frequency of the light that causes these electrons to be emitted is in the ultraviolet range./nWell, this was a curious observation in the late 1800s, very early 1900s. And scientists just started to measure everything about these electrons they possibly could measure because they just didn't understand the effect. And one of the things that they measured is the kinetic energy of the electrons./nHere I am going to plot kinetic energy, KE. And they measure the kinetic energy of the electrons as the function of the frequency of the light. Again, at low frequencies there is no kinetic energy because there are no electrons. But then here at some frequency v0 all of a sudden electrons come out. And these are points just representing the individual measurements./nLow and behold, as you increase the frequency above v0 the kinetic energy increased. It increased and looked directly proportional. The kinetic energy looked directly proportional to the frequency of the light. Well, let me tell you, that was a big mystery as far as classical physical ideas were concerned. That was a big mystery./nBecause, in the classical physics that was understood at that time, the kinetic energy of the electrons, KE as a function of the frequency of light, well, that classical physics would predict that that kinetic energy were independent of the frequency of the light. There was nothing in the known classical physics, classical electromagnetism that would allow the frequency of the light to be related in any way to the kinetic energy of the electrons kicked off./nThere was no reason for why bluer light, which has a higher frequency, would eject electrons and redder light, which has a lower frequency would not eject the electron. And classical physics also predicted that the kinetic energy of the electrons should be dependent on the intensity of the light./nI am going to call it i. In kind of general this sort of dependence. That is the brighter the light, the more kinetic energy these electrons ought to be kicked out with. In other words, the brighter the light the more energy you were putting into that metal. And, therefore, the electrons should be kicked out with more energy./nThat's what classical physics was predicting. But, go in the lab and you do the measurement, the measurement showed that the kinetic energy was independent of these intensities of light. This is the experiment. It had nothing to do with the intensity of the light./nThe kinetic energy remained the same no matter how bright that light was, not matter how much energy you were sticking into the system. Now, as you put more and more energy or as you put more and more intensity, what changed was the number of electrons that came off but not their kinetic energy. So, this was another conundrum here in terms of being able to explain it with the known classical physics at that time./nThese data were around for a few years before Einstein took a look at it. And he looked at these plots of the kinetic energy versus the frequency of the light. He looked at it, say, this is the data for metal B./nHere is the data for metal A. In all cases, it kind of looked like it was linear dependent on it. And the only thing that was seemingly different, for the different metals, was the threshold frequency here. What Einstein did, a real leap here of intuition, was he went and he fit a straight line to this data./nWell, he did a better job fitting a straight line than I did. He fit a straight line here. y = mx + b, no problem, where kinetic energy is y and x is v, right? He looked at that and calculated the slope. And he looked at it and he looked at it./nAnd he goes finally, very interesting, because what was so interesting here was the value of the slow, the value of m. The value of this slope here was equal to 6.626 x 10-34 joule seconds./nNo matter what metal he looked at, the value of the slope was 6.626 x 10-34 joule seconds. So what you say? Well, it was very interesting to Einstein, because just a few years before there was a gentleman who was analyzing what's called black-body radiation data./nLet me just explain this a little bit to you, but you're not responsible for it. What is a black-body? Well, let's just, for a simple example, think about the burner on your electric stove. We will consider that to be a black-body. What you know is that when you turn up the voltage the temperature starts going up and at some point the temperature gets high enough such that the burner glows a little bit dull red./nAnd then if you turn it up some more it gets hotter, the burner glows a little bit redder. That's the black-body radiation. If you turn it up some more the burner glows orange. Frequency going up. If you turn it up some more, and you better not do this, the burner starts glowing yellow. And then, if you could, and you probably cannot turn it up some more, it ultimately is white. All frequencies./nNow, what scientists had been doing is they had been analyzing the intensity of that radiation from a black-body as a function of the frequency. And they did it for lots of different temperatures. This is our black-body radiation. I am plotting the intensity here./nAnd so they dispersed the radiation, they analyzed it as a function of frequency, and what they saw was a curve that looked kind of like this. And this temperature here, we will call it T1, that was for some temperature T1, a low temperature. This isn't in your notes because you're not responsible for this. I'm just trying to make the importance here of this slope, or I'm just trying to illustrate the importance of this value for the slope in the photoelectric experiment./nAnd then at a higher temperature here, T2, the intensity got bigger, but also the maximum shifted to higher frequency. And at temperature T3, here now T3 is greater than T2 and greater than T1, frequency shifts even to a higher frequency./nThat was the observation. This is dull red, red, orange, etc. Well, Planck, along with a lot of other people, was trying to understand those curves. And he came up with a model. And the model said that materials, these black-bodies, had oscillators in them. And these oscillators were emitting radiation./nThese oscillators were responsible for giving off this radiation. And he also added another twist to those oscillators in that those oscillators were quantize, that they could only emit discrete packets of energy. And using that basic idea and something called statistical mechanics he was able to calculate the shapes of these curves./nA lot of his calculation. And so this green here is going to be a calculated curve here for T = 1, a calculated curve here for T = 2, a calculated curve here for T + 3. He got the shape right, but he didn't get the magnitude right. And he said how do I get the magnitude right?/nWell, what he realized is that in that formulation what he had to do was multiply the frequency of the oscillators by this constant m or this constant that is called h, Planck's constant. Where does Planck's constant come from? From this. Hey, that's it. It is, in a sense, a fitting constant./nThere isn't any derivation of Planck's constant. You cannot sit down and derive Planck's constant. It is our observation of the way nature works. So, that is where Planck's constant comes from. And so you can imagine Einstein's surprise when he analyzed these photoelectron data. And, low and behold, out from the data comes the same constant that Planck had found./nAnd Einstein realized that there must be something really very fundamental about this h, about what is now known as Planck's constant. So, he pursued. And what he did then was he just wrote down the equations here for that straight line./nThat's kinetic energy is equal to nu, that is my x, the slope is h, and he found that the intercept here was -hv0. So, this will be then plus the intercept, or -hv0. That was the equation of the line he fitted through the data./nNow, he also realized he has energy units on this side so he better have energy units on this side. If this is a frequency then h must have the unit such that hv0, this has got to be energy. And since this is the frequency of the light coming in, this energy E must be the energy of the incident light./nAnd here is where this equation E = hv comes from. This is it. This is where this equation comes from. What it is telling you is that the energy of the light is proportional to the frequency. This was the first time that there was any indication that the energy of the light had anything to do with the frequency of the light./nThe very first time. And what E = hv ended up being thought of is as a particle of energy, as a packet of energy, as a quantum of energy, eventually going to be called a photon./nThe photon, this particle or packet of energy is given by its frequency times this constant, this natural constant h, Planck's constant. You could have here any frequency of light you wanted, but the energy of that light only comes in chunks given by this quantization constant h times that frequency./nI will explain that a little more in just a moment. Well, likewise, this hv0 here, that better be an energy also. And so what that hv0 is, is the threshold energy. Since v0 was a threshold frequency hv0 better be the threshold energy./nAnd let's try to understand that in terms of an energy level diagram. I am going to plot here energy. This electron that is bound to the metal, before we shine any light on it, I am going to let its energy level be represented here by this line. This is the electron bound to the metal way down here./nAnd what we know now is that it takes some energy, hv0, to rip that electron apart, to pull it off. And so I am going to draw another energy level up here which is going to represent the energy of the electron free from the metal. So, I will call it electron free. And the amount of energy that we have to put in to get it out is this threshold energy, hv0./nThis energy here is like an ionization potential, which you are familiar with. The energy required to pull an electron off an atom or a molecule, the ionization potential. Well, this is like an ionization potential. It really is the ionization potential, except since it's coming off of a metal historically we call it something else./nIt's called the work function. And we give it often the symptom phi. But F and hv0, they are the same thing, the energy required to rip the electron off. And so what we know is that the packet of energy that you send in here has to be at least this amount./nIn other words, if you come in with a photon and the photon has this much energy, Ei = h0, you can ripe the electron off, but you have to have at least this. Now, you can come in with a photon that has more energy than that, Ei, from here to here./nBut if you do then the rest of this energy here goes to the kinetic energy of the electron. So, if you come in with exactly the energy, the electron is ripped off but it doesn't move away from the metal. If you come in with a little bit more, the electron is ripped of the metal, but it actually moves away then from the metal./nAnd so, in this case and in all cases, we have got to conserve energy. The incident energy of the light, from here to here, has to be equal to hv0 plus the kinetic energy, or if turn this around the kinetic energy is equal to the incident energy - hv0./nThis is an important expression. You need to know it. I won't give it to you on an exam. You don't have to memorize it, though. You just have to reason here conservation of energy. You can draw this diagram and figure it out./nNow, another very important point here that I am going to emphasize is that you absolutely, positively, no questions have to have a photon of at least this energy to get the electron off. What you cannot do is you cannot have, for example, a photon with half of this energy, Ωhv0./nOr, what you cannot do is have two photons come in with a half of this energy and make that electron come off. You have to have one photon to make this electron come off, and it has to have the frequency hv0. That's the particle-like nature of the radiation. Even though you have the same total amount of energy here, the fact that you cannot have two to pull the electron off, you have to have one and it has to have at least this quantum, at least that packet of energy./nThat's very important. All the problems you are going to do, if one photon comes in, you're going to get one electron off as long as that photon has hv0 of energy. Now what we've got to do is we have to do our own. Yes?/nIf the energy of the incident photon is twice the work function energy, yes, the electron will come off and the excess energy that doesn't go to the metal to pull the electron off goes to the kinetic energy of the electron./nAnd so if this incident energy is twice the work function, well, then the kinetic energy of that electron will be equal to the work function. I'm sorry? Two electrons won't come out. That's a good question. What she asked was if I put twice the amount of energy here with one photon, will I get two electrons out?/nNo, absolutely not. That is the quantum nature or the particle nature of the radiation. Let's do our own photoelectron experiment. Question?/nThose graphs up here, kinetic energy versus the frequency of light, those were the predictions from classical physics. Kinetic energy independent of frequency. Kinetic energy increasing with the intensity of light. Those are predictions from classical physics. How I got them, how I knew about them is something that we're not going to talk about because it's wrong./nIt's best at this point not to go through the logic of how I got that, but maybe after you have 8.02 it will become obvious to you. Good question. Let's do our own photoelectron experiment. Now we need the camera here. Let me get this started here./nWhat I also need are two volunteers to help us do this experiment. Come on down. Somebody else want to come? OK, come on down./nAll right. I have two volunteers here. Oh, I forgot to tell you that I need to do something so you've got to stay up here for a moment. I forgot to explain something. My fault. Sorry. Maybe you want to turn around. I forgot to tell you something here./nWhat we're going to do --/n-- is use this device here. This is an aluminum plate./nAnd you are going to see it here on the side walls. That is a picture of the device we've got up here. This is an aluminum plate. And that aluminum plate is mounted to a metal rod. That's that blue rod. And then there is a needle here in that blue rod and it's mounted on a pivot so it moves very easily. What we're going to do is we're going to put some excess charge on this metal plate./nAnd, when we do that, that needle is going to deflect. It's going to deflect because those excess charges, the electrons are going to run down that metal rod and then onto the needle. And since these electrons repel each other, and this is a very low friction pivot, this needle is going to move. And then what we're going to try to do is we're going to try to do the photoelectron experiment./nWe are going to come in and shine some UV radiation on this plate and see if we can eject electrons. That's what we're going to do. Now I've got my volunteers who are going to do this. Who wants to be the charger? Who wants to be the discharger? You'll be the charger, OK. This will be your weapon. You hold that. You hold that. Keep your fingers on the yellow plastic. This is going to be your weapon, the UV light./nI am going to need you in a moment so just stand by. What is your name? Don, OK. What you need to do is you need to take that Lucite rod and rub it really hard on the fur. And what that is going to do is the fur has some natural oil, and he is going to transfer a little bit of the natural oil onto that rod./nAnd there are some stray electrons around and some negative ions around, and they like to be on that oil. But then we're going to contact that rod with the aluminum plate and the electrons will go onto the aluminum plate. So, you've got to rub this really hard and then you've got to go and touch that and put the electrons on. And you should watch the needle move here. You've got to touch it on the edge, actually. Good./nWell, you charged it a little bit. That's not bad. Now take the rod again and rub it really, really hard. [LAUGHTER] Really hard. Touch the edge of the plate there. Touch the edge of the plate. [LAUGHTER]/nThis way. You've got it. Pretty charge there. Touch the edge of the plate. Fantastic. Now we need the discharger. You've got to come right here and shine the UV light on the base of it. You've got to get a little closer. Maybe turn your hand this way so that people can see it discharge. Fantastic./nTry it again. Come on, charge it up. [LAUGHTER] Really hard. Touch the edge of the plate. [LAUGHTER]/nThis way. This works the best. All right. Touch the edge of the plate. Let's get it. Fantastic./nMaybe just a little bit more. We can get more electrons. Hard, hard, hard, really hard. OK. Charge it up. Oh, OK. Discharge here. [LAUGHTER] Fantastic. What you saw was when the UV radiation then was incident on that aluminum plate the electrons were emitted and then the needle deflected back./nNow we've got to do this one more time. Why don't you come over here? What is your name? Jenny. Come on over here, if you can. And this time, after you charge it up, I am going to hold this glass Pyrex plate between the light and the aluminum plate./nAnd that is going to prevent the UV radiation from going through and so it shouldn't discharge. OK? All right. Go ahead. Let's charge it up one more time. You really do have to rub that oil on there. [LAUGHTER] Really hard. Touch it up here. Fantastic./nNow I am going to put the light there and now we are going to shine it. Oops, it's already discharging. You've got to do it one more time. You've got to do the control experiment here. Really hard. It's a little bit humid out today so it's a little bit harder to make this work. A little bit harder. Done. Get it up there./nAll right. Touch the side there. Good. That's pretty good. Fantastic. We've blocked the UV radiation. Electrons didn't jump off. Hey, this works. Thank you very much. Let's give them a hand. [APPLAUSE] Thank you very much. This photoelectron business, it's really real./nWhat I want to do now is I want to just work a couple of problems./nAnd here is the first problem. What it says is how many photons, at the radiation wavelength of one picometer, do you need to supply a pulse of energy that is a one joule pulse, say from a laser? It's a pulse laser. Suppose you want to supply one joule of energy in this light pulse./nThat is what the problem says, one joule in a pulse. Now, how do you work a problem? Lots of ways to work a problem. Sometimes I like to draw pictures. 1J, that's my one joule. I want 1J of energy. But I am told that I have radiation that has a wavelength of one picometer./nThat's 1 x 10-12 meters. That's the wavelength of my radiation. And I am asked how many photons do I need? Well, what do you know about a photon? We said that the energy of a photon is equal h0. That's one photon, the energy of a photon. The units here are joules./nImplied is joules per photon. And so I know the wavelength, I don't know the frequency here, but I know the relationship between wavelength and frequency. Frequency is equal to C over the wavelength. And so I can then calculate the energy, and when I do that it's 1.99 x 10-13 joules per photon./nCarrying one more digit than is significant because this is an intermediate step in our calculation. And now, if I want a plus of energy that is one joule and each photon has 10-13 joules per photon, well, that's 1.0J / 1.99 x 10-13 joules per photon./nWell, you can calculate that. It comes out to be 5 x 1012 photons. That's what I will need then. That's the answer then. Let me just write that down./nLet's work one more where we can understand./nThis is important because you need it for your homework. Here is the problem. You've got power of a radiation from a continuous laser. You turn on your laser, you've got power. What is power? Power is energy per unit time. We're going to use in our unit system the unit as power as watt./nA watt is a joule per second. It is how much energy is delivered per unit time. And we're told that we've got radiation of three milliwatts, that's 3.0 x 10-3 joules per second. The question is how long will it take for a total energy of one joule to be supplied?/nWell, 1.0J / 3.0 x 10-3 J/S = 330 S. And then, rushing through this, one more. How many photons at a radiation wavelength of one picometer again --/nHow many photons per second have to be delivered if the power of the radiation is three milliwatts? Power is 3.0 x 10-3 J/S. The radiation is one picometer, but we already calculated how many joules there are in radiation per photon with wavelengths of one picometer./nThat's 1.99 x 10-13 joules per photon. Joules cancel. And that means we're going to have to have 1.5 x 10-10 photons per second, because each photon has this much energy./nThis was the power. The rate at which we're delivering the energy, we're going to have the have this many photons per second coming out if the photons have that wavelength of one picometer. Great. See you later this afternoon or on Friday.

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