Understanding Waves & the Discovery of Quantum Theory: Amplitude, Frequency, & Wavelength, Study notes of Chemistry

Download Understanding Waves & the Discovery of Quantum Theory: Amplitude, Frequency, & Wavelength and more Study notes Chemistry in PDF only on Docsity! -1.5 -1 -0.5 0 0.5 1 1.5 TimeA m pl itu de , o r st re ng th o f s ig na l λ ∆t Lecture Set #2: The Paradoxes of Classical Mechanics and the Origins of Quantum Theory If we are to ever understanding chemical bonding, then it is important that we first understand the structure of the atom. Unlike most of the principles involving conservation of mass, volumes of gases, etc., the structure of the atom is a truly 20th century discovery. As the matter of fact, about 100 years ago many scientists felt that they pretty much understood the physical universe – there were only one or two nagging problems, and those problems would soon work themselves out – no doubt about it. The solutions to those nagging problems turned out to completely revolutionize both science and society. Let’s go over these problems as they existed 100 years ago, and try to understand how they were solved. In the latter half of the 19th century, James Clarke Maxwell unified the study of electricity with the study of magnetism. Although Oersted had shown, in 1820, that electricity and magnetism were related, it was Maxwell who really unified the two phenomena. What Maxwell showed is that an oscillating electric (magnetic) field must produce a complimentary and orthogonal magnetic (electric) field. The term orthogonal refers to a magnetic field that is at right angles to an electric field. For example, a z-axis is orthogonal to both the x- and y-axis. To understand his model, let’s first describe what a wave is: This Figure is a periodic function. If we were watching some phenomena, such as waves at an ocean beach, or if we were listening to some particular musical note, then we might receive a signal such as this one. The x-axis here is time, and the y-axis represents the strength of the signal. There are three properties of this wave that we need to worry about. First is the amplitude – or how strong is the signal? Second is the frequency of the wave (in s-1, denoted by the symbol ν) – i.e. how many crests pass by a fixed point in a given period of time? Finally, we have the wavelength of the wave, which we denote by the symbol λ. λ is defined as the velocity of the wave (in m/s, for example) divided by the frequency of the wave, and so its units are those of length. For example, assume that we have an acoustic wave, travelling through some material that is characterized by a speed of sound of 1 x 103 m/s. Let’s further assume, for the above graph, the entire x-axis spans 0.1 second. We have about 8.5 complete waves that pass us in 0.1 second. Now we can calculate frequency and wavelength. ν = 8.5 waves/0.1 second = 85 s-1, or 85 Hz. current loop Magnetic field Figure 2. A current loop generating a magnetic field. The arrows indicate the direction of current flow and the direction of the generated field. λ = (1 x 103 m/s)/(85 s-1) = 11.5 m – a pretty long wavelength Let’s try to develop a feeling for waves. Waves at the beach: λ = 20 m; ν = 0.1 s-1; velocity = 2 m/s Audible Sound: λ (in air) = 16 m to .016 m; ν = 20 s-1 to 20,000 s-1; velocity = 331 m/s in air at 25oC; 1500 m/s in H2O at 25oC Light or electromagnetic radiation λ = 10-16 to 108 m (spanning γ-rays to radio waves); ν = 1024 s-1 to 10o s-1 (10o = 1, by the way) velocity = 3 x 108 m/s in vacuum. Velocity in air is very similar. Visible Light λ = 750 nm to 400 nm (1 nm = 10-9 m) Note that electromagnetic radiation spans wavelengths covering 24 orders of magnitude! This is a lot. We can break this down into: γ-rays (Gamma Rays) λ = 10-16 to 10-11 m x-rays: λ = 10-11 to 10-8 m UV (ultraviolet): λ = 10-8 to 4 x 10-8 m Vis (visible): λ = 4 x 10-8 to 7.5 x 10-8 m IR (infrared): λ = 7.5 x 10-8 to 10-3 m Microwave: λ = 10-3 to 100 m Longer wavelengths are FM, the AM radio waves, and the Long radio waves which you may have heard before referenced as k-band, or x-band, or some such type of band. A major difference between all these waves is that acoustic waves, ocean waves, and similar phenomena propagate through some definite medium, but electromagnetic (EM) waves can propagate through vacuum. This was a rather bizarre concept around the turn of the century – many people searched for the ‘ether’, which was the supposed omnipresent medium that supported electromagnetic wave propagation. The are a couple of other differences between EM and other waves, as well. Above we stated that J.C. Maxwell had found the connection between electricity and magnetism. At left is shown this connection – a loop of current generates a magnetic field at right angles (orthogonal) to the current flow. The direction of the magnetic field is given by a cross product, which means that the magnetic field direction is that shown in the figure. Conversely, if we just had a wire loop, and we didn’t hook it up so that there was no supplied current, but rather we put the loop into a magnetic field such as that shown in Fig. 2, the field would exert an electromotive force (voltage, more or less) on the wire, and thus generate a current. One field does not exist without the other. You will learn about this in physics. One useful thing about Boltzmann’s constant is that it converts Temperature into an Energy by the equation E = kT. This is a very important relationship. Thus, if temperature can be considered to be an energy, then it is not surprise that a black body, held at a fixed temperature, will emit photons with a characteristic energy distribution that is related to the temperature of the box. The value of k is: k=1.38066x10-23 J K-1 We can make some connections here between energy, Temperature, and photon wavelength. The temperature of this radiation is 3000 Kelvin. We can convert this into an energy by using the equation E=kT, and show that 3000 K is equal to 4.14x10-20 J. Now, what is the energy of a photon with a wavelength of 1.8x10-6 m? We can get this by E = hν. Converting wavelength into frequency, we get a photon of energy E = (6.266x10-34 J s)(3x108m/s)/1.8x10-6m = 1.04E-19 J. We note that the energy that corresponds to 3000K, and the energy that corresponds to a 1.8x10-6 m photon are not the same, although they are reasonably close. These energies are not the same because the relationship between the energy curve and the wavelength is not a straight line relationship, or a ‘linear relationship’, in other words. However, they are only about a factor of 2.5 different. If we were to heat up the mass to temperatures above 3000K, the curve would shift to shorter wavelengths (higher energy photons), and if we were to cool it down, then the curve would shift to longer wavelengths (lower energy photons). So, what Planck did is to state that light came in discrete packets, and that the energy of such a discrete packet was proportional to its frequency, with a proportionality constant equal to h, now known as Planck’s constant. One question that bothered me for a long time had to do with the equation that Planck used to describe black body radiation. How did he come up with this? It turns out that your text has a really wonderful discussion of this on pages 527-529, and you should read it. It is a wonderful example of scientific reasoning. Paradox 2: Einstein and the Photoelectric Effect The was another glitch in the classical model of the universe, and it was the following: Shine light on a metal surface. As we discussed above, the more intense the light you shine onto the surface, the more energy you put into the metal. If you put energy into the metal, then you can presumably activate some chemical or physical process – and one such process is that you can ionize the metal – i.e. you supply enought energy so that electrons are ejected from the surface of the metal. Let’s assume that the energy required to eject 1 electron from the metal surface is 3.5 eV, and we’ll label this the Ionization Potential, or the IP. For bulk materials, the IP is also called the work function The eV is an energy unit we haven’t encountered yet. We use it here because we want a manageable unit of energy. For example, if we listed this energy in Joules, it would be 5.6x10-19J! Joules are useful when we discuss quantities such as moles. For a single molecule, or single electron energies, the eV is much better. 1J = 6.2x1018eV. We will use eV’s a lot. So now, if we shine 10 Watts of infrared light on the sample, in one second we are putting 10J, or 6.2x1019eV’s of energy into the sample. However, no electrons come off! What’s going on here? Recall that we have two energies that we are concerned with here – the intensity of the light beam (measured in Watts, or J s-1), and the energy of an individual photon = hν. We can increase the energy of a photon by increasing the frequency (decreasing the wavelength) of the light source. What we find is that when hν > 3.5 eV, we suddenly get a shower of electrons from the surface of the metal. As we increase the photon energy beyond 3.5 eV, then the electrons coming from the surface become more and more energetic, and it turns out that the energy carried by the photon was: hν – IP = ½ mev2 = the kinetic energy of the ejected electron. This is what Einstein explained. He borrowed Planck’s ideas for the energy of a photon, and stipulated that if hν was above the ionization threshold for the metal, then electrons would come off, now matter how intense the light beam was. More intense light beams produce more electrons, but even a single photon can produce an electron. For hν-IP < 0, no electrons are emitted from the metal, regardless of how intense the light source is. Einstein won the Nobel prize in physics for explaining the photoelectric effect. Let’s do an example here. Say that we have a H atom, and we want to ionize it. Its IP is 13.6 eV. What is the longest wavelength photon that will emit an electron? If we have a photon with λ=50 nm, then what is the velocity of the emitted electron? 13.6 eV x 1.6x10-19 J/eV = hν = (6.626x10-34 J s)( 3x108 m/s)(λ)-1 = 9.1x10-8 m, or 91 nm light is necessary So if we have 50 nm light, what is the velocity of the emitted electron? ½ mev2 = Kinetic Energy of electron = hν (λ=50 nm) – hν (λ=91 nm) = (6.626x10-34)(3x108m/s)[(1/50x10-9)-(1/91x10-9)]=1.8x10-18 J So, ½ mev2 = (0.5)(9.11x10-31kg)(v2) = 1.8x10-18 J, and v = 1.98x106 m/s Paradox 3: The structure of the Atom Recall from our previous lectures that we discussed that the structure of the atom appeared, from Rutherford’s experiments, to consist of a very dense and tiny nucleus, surrounded by a dilute and relatively large cloud of electrons. Recall also that we discussed Coulombic potentials – i.e. the potential energy curves that describe the interactions between charges. There was a nagging question here that we didn’t answer. If like charges repel each other (as we know that they do), and opposite charges attract, then: • Why are all the positive charges segregated from all the negative charges; and, In te ns ity o f E m itt ed R ad ia tio n Wavelength Figure 5. What the emission spectra of some arbitrary atom might look like. • Why doesn’t the atom collapse – i.e. why don’t electrons get sucked into the core of an atom? Good thing that this doesn’t happen, by the way. It turns out that the actual answer to this dilemma is related to what is known as the Heisenberg Uncertainty Principle – one of the most non-intuitive concepts in all of chemistry and physics. However, there are other, more intuitive models that steered people toward the right pathways, and the adoption of these models was aided not only by Rutherford’s experiments, but also by a set of recently collected emission spectra of various atoms. To understand what an emission spectrum is, just look at a neon sign. As electric current is run through the neon gas, the neon gas will convert that electric current into light – fluorescent lamps work in the same way. People had collected emission spectra from a whole number of atoms and ions, including H, He+ and Li2+ - an isoelectronic set of atoms and ions. Isoelectronic means that they each have the same number of electrons – exactly 1 each in this case. What they had found was very surprising: If emission intensity were plotted vs. wavelength, they would see something like what is shown in Fig. 5. So there were two clues to atomic structure: The electrons existed as diffuse clouds surrounding tiny, positively charged neuclei, and the emission spectrum of the atoms consisted of discrete lines. Niels Bohr rationalized these observations in the following way: the electrons and the nuclei had a relationship with each other that was similar to that of the sun and the planets. Notice that the gravitational attraction between the sun and the earth is tremendous. However, the earth doesn’t collapse into the sun, but rather it orbits around it – a planet orbiting the sun is sort of like a child spinning an airplane around on a string. The child is pulling on the string, but the airplane doesn’t turn inward and collide with the child. It has its on angular velocity that keeps it spinning around the child in balance with the child’s pull. In a similar way, Niels Bohr imagined the electrons rotating around the nuclei. The discrete lines in the emission spectrum resulted when an electron in one orbit switched to a different orbit – one that required a less energetic electron. The change in energy between the two orbits was then given off as a photon. This model turns out to be quite wrong. Nevertheless, it was the first time that discrete orbits of the electrons were proposed (a correct notion!), and it was the first model that could correctly account for the electron/nuclei model of the atom.