Lecture 5: Wave-Particle Duality and Quantum Mechanics

Lecture 5: Wave-Particle Duality and Quantum Mechanics Summary This chapter, we continued our discussion of ionization. We restricted our examples to ionization of hydrogen atoms in order to use Bohr’s model of the atom. An electron in the ground state, or lowest energy state, of a hydrogen atom has an energy of -13.6 eV. Therefore, this electron’s ionization energy, or the energy required to free it from the nuclear pull and effectively transition it to energy level n=∞, is 13.6 eV. The ionization energy of an electron in an atom with atomic number Z can be found with the following formula: ! 1 1 − 2 [eV ] ΔEionization = −13.6Z 2 n2f ni Ionization energy can be thought of in terms of Einstein’s photoelectric effect. In his groundbreaking experiment, Einstein shone light (streams of photons) of different energies onto slabs of metal. The photons had to reach a minimum threshold energy in order to ionize the metal atoms. It makes sense that metals with higher first ionization energies will be better able to hold on to their electrons. In other words, light with higher frequency will be necessary to ionize metals with higher first ionization energies. Photo-Electron Spectroscopy (PES) is a characterization technique that involves ionizing materials with photons and measuring the removed electrons’ kinetic energies. The energy of the bound state of each removed electron can be calculated by finding the difference between the incoming photon’s energy and the resulting kinetic energy of the freed electron. The Bohr model couldn’t explain quantization of electron levels; they were imposed in the model, but not understood. Through a series of experiments in the early 20th century, wave-particle duality was demonstrated. This meant that light waves were shown to behave as particles, and particles were shown to behave as waves. The most famous of these experiments was the doubleslit experiment, where electrons were shown to act like both waves and particles. Louis de Broglie showed that all matter acts like a wave and has a corresponding wavelength (λ). This wavelength can be found using the de Broglie relation, where p is the object’s momentum, which is equal to its mass times its velocity, and h is Planck’s constant: λ= h p Erwin Schrödinger, with The Schrödinger Equation, described the relation between a wavelike particle’s spatial distribution and its allowed energies. In the Schrödinger equation, quantization is a natural consequence of solving for a standing wave. This wavefunction ψ does not have direct physical meaning, at least that we know of, but its square |ψ|2 corresponds to the probability of finding the particle at any given location. Solving the Schrödinger equation for an electron in an atom gives a set of wavefunctions that correspond to the electron’s energy levels and spatial probability distributions. Note the dramatic difference from the Bohr picture: now the electron can be anywhere there is a nonzero |ψ|2 which is almost everywhere. The concept of exact position does not apply in the quantum world, since particles cannot have their position and momentum specified exactly, in accordance with the Heisenberg Uncertainty Principle. Why this matters Let’s apply the de Broglie wavelength equation to an electron. We know the mass of an electron is 9 × 10−31 kg. Now put the electron in an electric field caused by a voltage of 100 V. We know this will cause the electron to accelerate and gain some momentum. Specifically, the energy of the electron is: charge*voltage = e*V = 100 volts *1.6 × 10−19 Coulombs = 1.6 × 10−17 Joules. This is because the unit volts is equal to units of J/C. So now we have an energy of the electron accelerated across 100V, and from that we can get its velocity, v 6 × 106 m/s. Next we can get its wavelength: λ = h/mv = 6.6 × 10−34 J ∗ s/(9 × 10−31 kg ∗ 6 × 106 m/s), which when we realize the units of J = [kg][m2 ]/[s2 ] we see gives us λ = 1.2 × 10−10 m = 0.12nm for this electron. Remember the electromagnetic spectrum from last chapter? That’s light covering orders of magnitude of wavelength, only a sliver of which lies in the visible range (400-750 nm). But in order to use this light to see things, we must use a wavelength of light that is either equal or smaller in size to what we’re trying to see. Otherwise the light simply won’t pick it up (you This figure is in the public domain (created by NASA). can check out the physics of diffraction for more info on this, but for now, you can take my word on it). The correlation between the limit of what we can see and the wavelength of light we use to see it is why I like this other electromagnetic spectrum picture. On it is superimposed examples of objects the size of the given wavelength. If we want to use a type of light to see, say, a human, then we’d need to stay at or below wavelengths of 1 m since humans are about 1 m in size. That doesn’t mean we can see in the microwave region, but that’s just a limitation of our eyes not of what is possible with a type of light. With better eyes than ours, we’d be able to use microwaves to see people, but we wouldn’t, for example, be able to use radio waves since they have a longer wavelength than the thing we’re trying to see. Now let’s go to the atomic scale: at the x-ray part of the spectrum, we’d need to use x-rays or gamma rays to see objects at that scale. But it turns out that not only can our eyes not see in that part of the spectrum, but even the devices and detectors we’ve built cannot see all that well. We can take x-rays and do so all the time, but only to see images on the mm resolution as opposed to nm. For the latter, we’d need to see an x-ray reflecting off of a nm-sized object and be able to measure it with precision, something we’re not able to do. That said, we do use x-rays to get atomic-scale information, but that’s a different thing all together and it has to do with the spacing between atoms in solids. We’ll learn all about that later in the semester. But here we’re still trying to actually “see” the object and the object is an atom. We can’t use x-rays—even though they’re at the right wavelength—and forget about gamma rays, because they’re even harder to detect. But what else could we use? That’s why we did the electron wavelength calculation to start this Why This Matters section – using the fact that an electron is a wave, and by giving it the right velocity, we can make an electron have exactly the wavelength needed to see at the atomic scale. Electrons can be our “flashlight”! It also turns out they’re a lot easier to capture and use to construct images than x-rays. This discovery jump-started the whole field of electron microscopy and changed the way we see matter. It was Don Eigler and Erhard Schweizer who showed in 1989 that by seeing individual Xe atoms, they could arrange 35 of them to spell their company’s logo. This is definitely an expensive way to write a logo, but the point is that the accomplishment realized Richard Feynman’s dream of being able to put an atom wherever you want. Speaking of Feynman, he was an incredible scientist and a masterful teacher. If you want to see an amazing lecture entirely devoted to the double-slit experiment, find and watch Feynman’s 1959 speech, “There’s Plenty of Room at the Bottom,” and you won’t regret it. This is where he laid out the dream of putting atoms anywhere you want, and that was truly the first vision of the field we now call nanotechnology. The reason it took 30 years to realize Feynman’s dream is that we were working in the dark, literally. If you can’t see atoms and molecules, it’s hard to control them or at least know that something you’ve done led to an atomic-scale control that you wanted. This type of control led to the nanotechnology revolution and it was entirely enabled by the ability of scientists to see materials at the nm scale, which as you now know was made possible by the wave nature of the electron. Why this employs It is true that we put humans on the human without resorting to quantum mechanics, evidence to how powerful and useful the classical view of the world is. But in order to truly understand the world, from the atom to the oceans and mountains to the universe to life itself, the quantum nature of it all must be understood and in some cases quantum effects completely dominate. Electrons and their interaction with light fall into this category: without a quantum description there’s no hope in explaining the interactions that take place. So let’s talk about jobs related to lasers. Stimulated emission (first discussed in a paper by Einstein in 1917) involves the interaction between a photon and an atom that has electrons in an excited state. If the photon has just the right wavelength when it hits the excited atom, then the atom emits a second photon identical to the first. The LASER is “Light Amplification by Stimulated Emission of Radiation,” and it is entirely and only described by quantum mechanics since it relies on knowing the discrete energies levels of electrons in the atom. And by the way, any time you’re online, you’re using a laser since that’s how information travels through fiber optic cables. There are so many job postings for “laser engineer” that it wouldn’t be possible to even begin to list (one site I just checked has 7000+ listing!). The reason is that lasers are used today in every walk of life. There are jobs with lasers in medicine, like in eye surgery, to industry where lasers cut and weld, to tracking systems for all sorts of applications, to research in new types of spectroscopy, to so many many more. So if you want a job that uses wave-particle duality to make coherent light, then that’s an awesome first step, but the next step would be to get more specific. Take LIDAR as an example: this stands for Light Detection and Ranging, which is a remote sensing method that uses laser pulses to measure distances. LIDAR has completely revolutionized the ability of robots, like self-driving cars for example, to see. It’s inside those bulky boxes on top of a self-driving car that spins continuously, giving 360 degrees of visibility as well as extremely accurate depth information (to +/- 2 cm). It’s still expensive ($75K a pop for the car-top systems!), but because of how enabling LIDAR is, and how much the demand for this technology will grow, there are many jobs opportunities. These range from jobs at big player companies like Velodyne that’s already selling systems to many markets, to smaller companies trying to innovate the technology like Luminar that promises a $500 LIDAR system in the size of a soda can. There’s a whole range in between including self-driving car companies themselves like Waymo that have decided to build their own LIDAR systems. And this was just one tiny example of the vast number of LASER-based job opportunities. Example Problems 1. You observe hydrogen electrons transitioning from n=3 to n=1. a) What is the energy change in this transition? Is it absorption or emission? b) What color light do you observe? c) How many possible transitions could an electron falling from =3 to n=1 complete? Which transition is responsible for the orange color? 2. A red laser, a green laser, and a blue laser shine on different pieces of the same kind of metal, and detectors are set up to measure the presence of emitted electrons. For the red laser, no electrons are detected. For the green laser and the blue laser, electrons are observed flying off of the metal. Which of the following statements must be true? i) The number of electrons emitted from the metal using the blue laser is greatest ii) The kinetic energy of electrons emitted from the metal using the blue laser is highest iii) The red laser would lead to electron emission if the laser intensity was increased Solutions 1. a) The electron is falling to a lower energy state from a higher energy state, so it must be emission. ΔE = −13.6Z 2 ( 1 1 − 2 ) = −12.09eV 2 3 1 We can double check our intuition: since the energy value we found is negative, the transition must correspond to emission. b) λ= hc = 1.22 × 10−7 m = 102.6nm 10.2[eV ] This transition is in the UV. c) There are two possible paths: straight from n=3 to n=1 or n=3 to n=2 + n=2 to n=1. Next, calculate the wavelength of the three possible photon emissions. We already know that from n=3 to n=1, the wavelength of the emitted photon was 102.6nm. From n=2 to n=1: ΔE = −13.6Z 2 ( λ= 1 1 − 2 ) = −10.2eV 2 2 1 hc = 1.22 × 10−7 m = 122nm 10.2eV This transition is also in the UV. From n=3 to n=2: ΔE = −13.6Z 2 ( 1 1 − 2 ) = −1.89eV 32 2 hc = 6.57 × 10−7 m = 657nm 1.89eV This transition corresponds to orange light. λ= 2. ii) is true, because the blue light has higher energy than green and red light, and the energy of electrons emitted is proportional to the energy of the photons hitting the metal surface i) and iii) are incorrect because the number of electrons is related to the number of photons = light intensity, and light intensity does not affect energy of electrons emitted.