 We've seen that the solution of the Schrodinger equation for a hydrogen atom consists of families of electron orbitals of varying energies and spatial distributions. The number of orbitals per energy level, when multiplied by two, gives the sequence of numbers two, eight, eighteen, and thirty-two, and these correspond to the number of elements of the various periods of the periodic table. Let's examine this in more detail by imagining the process of building the periodic table, determined by element. We start with the simplest nucleus, a proton having charge plus one. We add an electron of charge minus one to form a hydrogen atom. In its normal state, the electron occupies the lowest possible energy level, the tightly packed one-ass orbital. The combination of positive and negative charge means that the hydrogen atom itself has no net charge. Now we add another proton to the nucleus, and technically we also need a couple of uncharged neutrons. The result is a system with net charge of plus one. If we bring a second electron into the system, it will, roughly speaking, interact with this net charge as it would with a hydrogen nucleus. In its lowest energy state, it will therefore also be in a one-ass orbital, and we obtain the helium atom, which has no net charge. We've now completed the first period of elements, and according to our bookkeeping it seems, filled up the one-ass orbital. Bringing another proton to the nucleus, along with one or two neutrons, again leaves us with a net charge of plus one. So again, an additional electron introduced to the system will, roughly speaking, act as if it's in a hydrogen atom. And since the one-ass orbital is filled by the other two electrons, this third electron has to enter the next available orbital, the larger two-ass orbital, and we obtain the neutral lithium atom. This process goes on, one electron at a time, as we build the entire periodic table, by filling each orbital with two electrons, and then proceeding to the next orbital. Why do two electrons, quote, fill an orbital, and what is the mechanism by which the other electrons are excluded? We don't know, but this is the conclusion we're forced to accept given the correspondence between the number of orbitals and the structure of the periodic table. We'll come back to this point in a future video. Let's now take a more detailed look at the lithium atom. The lithium nucleus has a charge of plus three. Packed tightly around this in the 1s orbital are two electrons with total charge minus two. The result is a net charge of plus one. The third electron of lithium will interact with this net charge more or less as it would with a hydrogen nucleus, except that for some reason, it's excluded from the 1s orbital. The next energy level has four orbitals, the 2s and three 2p orbitals. In a hydrogen atom, these would all have the same energy. Let's take a look at this. Here we plot the second, third, and fourth hydrogen energy levels. They don't depend on the type of orbital, spd or f. When the energy of essentially the third electron of lithium is measured, we find that the results do depend on the type of orbital. The p, d, and f orbitals have nearly the same energy they would have in hydrogen, but the s orbitals have lower energies. In particular, the 2s orbital has a significantly lower energy. Let's see if we can figure this out. The lithium nucleus has charge plus three. It's surrounded by a cloud of 2 1s electrons with total charge minus two. These electrons cancel, or screen, two of the three nuclear charges. The result is that if our third electron is outside this 1s cloud, it sees a net charge of only plus one, and it behaves essentially as if it were in a hydrogen atom. But if it penetrates the 1s cloud and approaches the nucleus, the screening effect is much less. Therefore, the electron sees almost the entire nuclear charge of plus three, and it will feel almost three times the force it would in a hydrogen atom. This leads us to predict that an orbital with a higher probability of the electron being near the nucleus is more tightly bound than it would be in a hydrogen atom, that is it has a lower energy. The 2s and 2p orbitals have very different behavior near the nucleus. The 2s orbital has its largest probability density at the nucleus, while the 2p orbitals have essentially zero probability there. So we would expect the 2s orbital in lithium to have a lower energy than the 2p orbitals, and that is indeed what we observe. As we continue building the periodic table, this effect explains why, for example, 4s orbitals are occupied before 3d orbitals. These results strongly suggest that the shapes of these orbitals, these electron clouds we find as solutions to the Schrodinger equation, are not just abstract mathematical expressions, they correspond in some real sense to the actual internal structure of atoms. The more people learned about the Schrodinger equation and its solutions, the more apparent this became. As another example, let's consider four atoms that play a critical role in biology, hydrogen, carbon, nitrogen, and oxygen. The last three are in the second period of elements, after lithium. So their outer electrons should occupy 2s and 2p orbitals. But as was pointed out most notably by Linus Pauling, these hydrogen orbitals are stationary states with the same energy. Therefore any combination of them will also be a stationary state with the same energy. These are so-called hybrid orbitals. If we sum all four of the S and P orbitals, we get an asymmetric shape with a small lobe near the nucleus and a large lobe sticking out in one direction. By negating one or more of the S and P orbitals before summing, we obtain the same asymmetric orbital pointing in different directions. These four hybrids are valid solutions of Schrodinger's equation, and conceivably in some situations they might be more stable than the S and P orbitals themselves. Let's represent the hybrid orbitals by four bulges pointing off in four different directions. These define the vertices of a tetrahedron, and the angle between any two of these directions is about 109.5 degrees. If each orbital contained two electrons, this would account for all eight electrons added in the second period, which ends with the practically inert element neon. Carbon has four electrons fewer, nitrogen three electrons fewer, and oxygen two electrons fewer than the highly stable configuration of neon. Representing two electron orbitals as opaque lobes and one electron orbitals as translucent lobes, we can imagine oxygen, nitrogen, and carbon with the hybrid orbital configurations shown here. Assuming nature likes filled orbitals, we can imagine each one electron orbital and a hydrogen atom mutually sharing their two combined electrons. This results in three critically important compounds, water, ammonia, and methane. Methane is precisely a tetrahedral molecule, while ammonia and water are approximately the same. As people continue to study the Schrodinger equation and its solutions, they found it precisely explained an ever-expanding range of physical phenomena. So by 1929, physicist Paul Dirac was justified in writing, The fundamental laws necessary for the mathematical treatment of a large part of physics and the whole of chemistry are thus completely known. And the difficulty lies only in the fact that application of these laws leads to equations that are too complex to be solved. This difficulty has lessened in the decades since, especially as ever more powerful computers have enabled increasingly sophisticated numerical calculations. Today using the techniques of quantum chemistry, it is possible to solve the Schrodinger equation for quite large systems and understand in detail the chemistry of a wide range of molecules. A fundamental observation that quantum mechanics was developed to explain is the discrete emission spectra of atoms and molecules. Recall how in the Bohr model the hydrogen spectrum was explained by imagining an electron jumping from one allowed orbit to another of lower energy with the emission of a discrete photon. The frequency of the emitted radiation depends on the energy difference of the two orbits. As we've discussed in previous videos, this idea of a quantum jump was incompatible with classical physics. Moreover, we've seen that the emitted frequencies don't equal the frequencies of the orbits themselves as would be required by a classical theory. The Schrodinger description of hydrogen appeared to overcome both of these issues. Imagine an electron in a 2P orbital. We want to know how it can transition to the lower energy 1S orbital with the emission of light. In our discussion of hybrid orbitals, we've seen that combining stationary states of the same energy results in a new stationary state of the same energy. But two orbitals of different energies are characterized by different frequencies. If they're combined, the result is a non-stationary state, a state in which the electron probability distribution changes with time. And the frequency of this change is precisely the difference of the frequencies of the orbitals that is precisely the frequency of the light we expect to be emitted. We end up with a picture that looks very much like a little antenna with oscillating electric charge, which, according to classical electromagnetic theory, would emit a light wave at the oscillation frequency. This is why Schrodinger was led to conclude, The wave function physically means and determines a continuous distribution of electricity in space, the fluctuations of which determine the radiation by the laws of ordinary electrodynamics. Schrodinger believed that he had found a way to explain the discrete nature of emission lines without the need of quantum jumps or wave particle duality and the associated lack of determinism. The electron was not a particle but a cloud of electric charge governed by the Schrodinger equation. Two states of different energy could combine to produce an oscillating charge cloud that would radiate away energy as a classical electromagnetic wave in a continuous and fully deterministic manner. The energy of the higher state would gradually be lost and the electron would completely transition to the lower energy state. But Schrodinger's view was not compatible with the overwhelming evidence for the particle nature of both electric charge and radiation. Measurements always detect discrete electrons or photons. So the continuous charge clouds and electromagnetic waves Schrodinger envisioned are instead taken to represent probabilities of finding a discrete electron or photon at a given position in space at a given time. The discrete nature of radiation still required the concept of a quantum jump just as it did in the Bohr model. At one moment the probability distribution of the electron is described by some orbital. Then at random, but with a probability determined by Schrodinger's calculations, the atom jumps to a lower energy orbital and a discrete photon is emitted in some direction into space. In fact, the, quote, bullet from a gun nature of photon emission actually produces an observable recoil effect, as one might expect. This cannot be explained in Schrodinger's classical picture. The ironic result was that although Schrodinger's equation was a monumental triumph for physics, Schrodinger himself intensely disliked the conceptual framework in which it found its great success. He's quoted as having told Niels Bohr, if we are still going to put up with these damn quantum jumps, I am sorry that I ever had anything to do with quantum theory. Bohr said to have responded, but the rest of us are extremely grateful that you did. Decades later Schrodinger was still arguing against the quantum jump idea demanded by wave particle duality. He even went so far as comparing it to the preco-pernic and geocentric model of the solar system, which assumed the planets traveled in complex spiral orbits described by so-called epicycles. He wrote, I am thinking of the theory of epicycles, I confess to the heretical view that their modern counterpart in physical theory are the quantum jumps.