Note: In the indented paragraphs of this chapter I describe some important fundamental concepts that will help in understanding the physics of the atom. And I also use a few numbers here and there in these indented sections to provide a sense of magnitudes. These concepts are central to all of this Part Four, so it's worth spending a little extra time here to take in what is presented.
IT ALL DEPENDS ON ENERGY
Remember the various spatial-state solutions to Schrödinger's equation for the hydrogen atom that were described in Chapter 3. Well, one might infer from the pronounced differences in the sizes of spatial-state cloud cross sections shown in Figure 3.8 that there are also pronounced differences in the energy levels of these states. And that is true. Schrödinger calculated these energy differences in the process of solving for the spatial states. But we need to understand why his calculations got these results. And before that we need to define what energy means in the context of the planets and the atom, as follows.
Objects that move around freely have a positive kinetic energy of motion depending on how heavy they are and how fast they are moving. In contrast, an object under some force of attraction is defined to have a negative potential energy.
The potential energy is made more negative if the object is placed closer to the source of attraction where the attraction is stronger. The object could be a planet attracted by the gravity of the sun or an electron attracted by the pull of its negative charge toward the positive charges of the protons in an atomic nucleus.
If an object is in some sense moving (i.e., has kinetic energy) in the presence of an attraction, then it can break free of the attraction if its positive kinetic energy is larger than its negative potential energy so that the sum of the two energies is greater than zero. If this sum is less than zero, negative (that is, if the negative potential energy of attraction overcomes the positive kinetic energy of motion), then the object will be in a bound state, perhaps still moving around in some sense but unable to break completely free of the attraction. The more negative this total energy, the more tightly the object is bound.
So, it's quite in keeping with the forces of nature that Schrödinger found that the spatial states of lower total energy would also more tightly bind the electron to the nucleus and have smaller probability-cloud cross sections. (Remember, though, what is special about his solutions is that they allow only certain discrete bound states, that is, states having only certain discrete quantized energies, not the continuum of energies and orbit sizes that would be available from classical physics.)
The potential energies of the planets in the solar system are very large negative numbers, large enough to overcome the very large positive kinetic energies of these rapidly moving huge masses and to bind them in their orbits. In contrast, the energies of the very small and light electrons in the atom are quite small. For the hydrogen atom in particular, the energies of the bound states of the electrons are small enough so that we could (in principle) remove hydrogen's one electron from any of its bound states using a couple of nine-volt batteries.
We are now going to demonstrate this using a thought experiment. Imagine that we hook the two batteries together in series (negative terminal of one to the positive terminal of the other) so that with the two of them we have a total of eighteen volts. Scientists call that an “electric potential” of eighteen volts. Now if we could attract a single electron to move from the unconnected negative terminal of one of the batteries to the positive terminal of the other battery (at its electric potential of eighteen volts), we could impart an energy to the electron of eighteen electron volts; in scientific shorthand, that is 18 eV. If we moved two electrons, we would provide an energy of 36 eV. It's that simple. (The e in “eV” refers to the charge on the electron. When a charge is moved by an electric potential, whether it's the charge on an electron or something else, it acquires energy that can be measured in electron volts.)
The solution to Schrödinger's equation gives –13.60 eV as the n = 1 energy level of the electron in hydrogen's fundamental most tightly bound, 1s state, its ground state (probability cloud shown at the bottom left of Fig. 3.8). With our batteries, we could provide an 18eV energy boost that would not only lift the electron free of the hydrogen atom but also leave it with an extra 4.4 eV of kinetic energy so that it could speed off to somewhere else.
Now let's examine the energy levels in the rest of the spatial states of the electron in the hydrogen atom. These include the bound spatial states, some of which are represented by the probability-cloud cross sections displayed in Figure 3.8.
As part of the solutions to his equation, Schrödinger found that each bound state for the electron in hydrogen has a total negative energy at only one of an infinite number of possible discrete allowed energy levels characterized by the so-called primary quantum numbers labeled generally by the symbol n. His solutions provide that n can only equal the integers 1, 2, 3, or 4, and so on. These are the numbers shown before the letters above the probability-cloud cross section representations for some of the individual states in Figure 3.8. (They are also the energy quantum numbers for the atomic orbits of the Bohr model constructed in the early days of the development of the theory.) Note that there can be more than one state at each energy level, as will be discussed a bit further on in this chapter.
The solved-for energy of each energy level is –13.60 eV divided by n2, so at each successive energy level the energy gets to be less negative. (That is because negative 13.60 eV is being divided by an increasingly larger squared integer.) So, calculating the energies for the first seven levels, we find the following, starting with the highest level (having the smallest negative energy) with states that least tightly bind the electron. For energy level:
n = 7 the energy of every state is (–13.60)/(7×7) = –0.28 eV
n = 6 the energy of every state is (–13.60)/(6×6) = –0.38 eV
n = 5 the energy of every state is (–13.60)/(5×5) = –0.54 eV
n = 4 the energy of every state is (–13.60)/(4×4) = –0.85 eV
n = 3 the energy of every state is (–13.60)/(3×3) = –1.51 eV
n = 2 the energy of every state is (–13.60)/(2×2) = –3.40 eV
n = 1 the energy of every state is (–13.60)/(1×1) = –13.60 eV
You can see the trend: the higher and higher the n value, the closer and closer to zero is the negative energy level, at which point the electron becomes free and is not bound at all. States with very high n have very small negative energies, and electrons in these states are therefore not very tightly bound and are correspondingly relatively large in size. The trend in size as n increases is illustrated in Figure 3.8.
AND ANGULAR MOMENTUM
Another property of each state that results from the solution of Schrödinger's equation is the state's angular momentum. Angular momentum tends to keep a rotating object rotating or moving in a circular or elliptical motion, just as linear momentum tends to keep an object moving at its same speed and in its same direction. If an object rotates one way, the angular momentum is said to be positive; if it rotates the other way, the angular momentum is said to be negative.
While Schrödinger's probability-cloud cross sections don't rotate or spin, the states that they represent nevertheless have spatial-state angular momentum that is indicated by the letters above the cross sections. (There are units for angular momentum, just as there are units like eV for energy. And nature's basic unit of angular momentum is [you guessed it] Planck's constant divided by 2π, in scientific shorthand h/2π.)
Schrödinger's results tell us that, like the state's energy, the magnitude1 of its spatial-state angular momentum is quantized, as noted by the angular-momentum quantum numbers labeled generally by the letter ℓ, where ℓ can equal only the integers 0, 1, 2, or 3, and so on, providing that ℓ is always less than n. Spectral lines resulting from states having these ℓ numbers are found to have sharp, principal, diffuse, and fundamental characteristics found in spectra, so the spatial-state angular-momentum quantum numbers ℓ = 0, 1, 2, or 3 have historically come to be represented, respectively, by the letters s, p, d, or f.
And so it is that the clouds’ cross section representations shown in the first “column” of Figure 3.8 for the spatial states with ℓ = 0 and energies indicated by n = 1 and n = 2 are labeled 1s and 2s, respectively. And the cloud cross sections shown in the second “column” of Figure 3.8, each with ℓ = 1 but, respectively with n = 2 and n = 3, are labeled 2p and 3p, while the single-cloud cross section of the third “column,” with n = 3 and ℓ = 2, is labeled 3d.
Now we can understand an interesting result observed in hydrogen spectra. According to quantum mechanics, a photon has either +1 or –1 units of angular momentum, depending on the polarization of the photon that carries away the energy given off in the transition. (Strange “beasts” these photons: as determined by quantum mechanics they all have only this plus or minus one unit of angular momentum, even though they have no mass [i.e., weight in a gravitational field]! And despite this they can have widely different energies, depending on the energy levels of the states involved in the transition that produces them!)
So, for the transition of an electron from an excited state in a hydrogen atom, the angular momentum of the excited state and the angular momentum of the lower energy state that the electron transitions into must differ by the one unit of angular momentum that is carried away by the photon emitted in the transition. Just as energy is conserved in a transition, and the energy of the initial state equals the energy of the final state plus the energy of the photon; so angular momentum is also conserved, so that the angular momentum of the initial state equals the angular momentum of the final state plus the angular momentum of the photon. This means that s states can transition only into p states, p states can transition only into s or d states, d states can transition only into p or f states, and so on. There are thus fewer lines predicted for the hydrogen spectrum than otherwise would have been expected, and this is precisely what has been observed. And the same holds true for the transitions of the electron between states in the rest of the elements.
Angular momentum can be symbolized by an arrow (a vector) with length in proportion to the angular momentum's magnitude and pointed in a direction along an axis of rotation. Such a vector can be thought of as representing angular momentum for each of the spatial states. (Yes, it's strange, there is angular momentum but no evidence of orbit or rotation.) This vector can be broken down into two parts, that is, two components: one representing the portion of the vector that might lie in the direction of a magnetic field (if a magnetic field were applied) and the second representing the portion that would then lie at a right (90-degree) angle to the magnetic field. (An atom might find itself in the earth's magnetic field or in an experiment between the “jaws” of a horseshoe magnet. Even if the electron is not in a magnetic field, its possible components of angular momentum are defined by what would happen if it were in a magnetic field.) The component of the angular momentum that would line up in the direction of a magnetic field is also quantized as a part of each spatial-state solution to Schrödinger's equation. This component is represented by a particular integer (which we refer to generally as m) times one basic unit of angular momentum: that is, m × h/2π. And m can have any integral value in the limited range from –ℓ to +ℓ.2 For reasons that will be explained in Chapter 12, m is known as the magnetic quantum number.
Note that m values are also shown just above the cross sections of the probability clouds in Figure 3.8. The +1 or –1 values that are shown for several of the cross sections indicate that these cloud cross sections may represent either of two spatial states, one with m = +1 or one with m = –1, where +1 is for a state with the cross section shown and –1 is for a state that has a similar cross section that lies in a plane that is perpendicular to the plane of the page.
So, we've defined the three properties inherent in Schrödinger's spatial-state solutions for the hydrogen atom. Every spatial state for the electron is characterized by its own set of these properties, labeled for each state by the quantum numbers n, ℓ and m, plus one more: its intrinsic angular momentum, otherwise simply called spin, described as follows in Chapter 12.