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Table of Contents

Contents
Section 36.1 - Classifying states
Section 36.2 - Angular momentum in three dimensions
Section 36.3 - The hydrogen atom
Section 36.4 - Energies of states in hydrogen (optional)
Section 36.5 - Electron spin
Section 36.6 - Atoms with more than one electron
Section 36.7 - Summary

big-hydrogen-wavefunction

A wavefunction of an electron in a hydrogen atom.

Chapter 36. The atom

You can learn a lot by taking a car engine apart, but you will have learned a lot more if you can put it all back together again and make it run. Half the job of reductionism is to break nature down into its smallest parts and understand the rules those parts obey. The second half is to show how those parts go together, and that is our goal in this chapter. We have seen how certain features of all atoms can be explained on a generic basis in terms of the properties of bound states, but this kind of argument clearly cannot tell us any details of the behavior of an atom or explain why one atom acts differently from another.

The biggest embarrassment for reductionists is that the job of putting things back together is usually much harder than the taking them apart. Seventy years after the fundamentals of atomic physics were solved, it is only beginning to be possible to calculate accurately the properties of atoms that have many electrons. Systems consisting of many atoms are even harder. Supercomputer manufacturers point to the folding of large protein molecules as a process whose calculation is just barely feasible with their fastest machines. The goal of this chapter is to give a gentle and visually oriented guide to some of the simpler results about atoms.

36.1 Classifying states

wrapping-wave

a / Eight wavelengths fit around this circle: \(\ell=8\).

We'll focus our attention first on the simplest atom, hydrogen, with one proton and one electron. We know in advance a little of what we should expect for the structure of this atom. Since the electron is bound to the proton by electrical forces, it should display a set of discrete energy states, each corresponding to a certain standing wave pattern. We need to understand what states there are and what their properties are.

What properties should we use to classify the states? The most sensible approach is to used conserved quantities. Energy is one conserved quantity, and we already know to expect each state to have a specific energy. It turns out, however, that energy alone is not sufficient. Different standing wave patterns of the atom can have the same energy.

Momentum is also a conserved quantity, but it is not particularly appropriate for classifying the states of the electron in a hydrogen atom. The reason is that the force between the electron and the proton results in the continual exchange of momentum between them. (Why wasn't this a problem for energy as well? Kinetic energy and momentum are related by \(KE=p^2/2m\), so the much more massive proton never has very much kinetic energy. We are making an approximation by assuming all the kinetic energy is in the electron, but it is quite a good approximation.)

Angular momentum does help with classification. There is no transfer of angular momentum between the proton and the electron, since the force between them is a center-to-center force, producing no torque.

Like energy, angular momentum is quantized in quantum physics. As an example, consider a quantum wave-particle confined to a circle, like a wave in a circular moat surrounding a castle. A sine wave in such a “quantum moat” cannot have any old wavelength, because an integer number of wavelengths must fit around the circumference, \(C\), of the moat. The larger this integer is, the shorter the wavelength, and a shorter wavelength relates to greater momentum and angular momentum. Since this integer is related to angular momentum, we use the symbol \(\ell\) for it:

\[\begin{equation*} \lambda = \frac{C}{\ell} \end{equation*}\]

The angular momentum is

\[\begin{equation*} L = rp . \end{equation*}\]

Here, \(r=C/2\pi \), and \(p=h/\lambda=h\ell/C\), so

\[\begin{align*} L &= \frac{C}{2\pi}\cdot\frac{h\ell}{C} \\ &= \frac{h}{2\pi}\ell \end{align*}\]

In the example of the quantum moat, angular momentum is quantized in units of \(h/2\pi\), and this turns out to be a completely general fact about quantum physics. That makes \(h/2\pi\) a pretty important number, so we define the abbreviation \(\hbar=h/2\pi\). This symbol is read “h-bar.”

quantization of angular momentum

The angular momentum of a particle due to its motion through space is quantized in units of \(\hbar\).

self-check:

What is the angular momentum of the wavefunction shown on page 981?

(answer in the back of the PDF version of the book)

36.2 Angular momentum in three dimensions

top

b / The angular momentum vector of a spinning top.

particle-a-m-examples

c / 1. This particle is moving directly away from the axis, and has no angular momentum. 2. This particle has angular momentum.

Up until now we've only worked with angular momentum in the context of rotation in a plane, for which we could simply use positive and negative signs to indicate clockwise and counterclockwise directions of rotation. A hydrogen atom, however, is unavoidably three-dimensional. Let's first consider the generalization of angular momentum to three dimensions in the classical case, and then consider how it carries over into quantum physics.

Three-dimensional angular momentum in classical physics

If we are to completely specify the angular momentum of a classical object like a top, b, in three dimensions, it's not enough to say whether the rotation is clockwise or counterclockwise. We must also give the orientation of the plane of rotation or, equivalently, the direction of the top's axis. The convention is to specify the direction of the axis. There are two possible directions along the axis, and as a matter of convention we use the direction such that if we sight along it, the rotation appears clockwise.

Angular momentum can, in fact, be defined as a vector pointing along this direction. This might seem like a strange definition, since nothing actually moves in that direction, but it wouldn't make sense to define the angular momentum vector as being in the direction of motion, because every part of the top has a different direction of motion. Ultimately it's not just a matter of picking a definition that is convenient and unambiguous: the definition we're using is the only one that makes the total angular momentum of a system a conserved quantity if we let “total” mean the vector sum.

As with rotation in one dimension, we cannot define what we mean by angular momentum in a particular situation unless we pick a point as an axis. This is really a different use of the word “axis” than the one in the previous paragraphs. Here we simply mean a point from which we measure the distance \(r\). In the hydrogen atom, the nearly immobile proton provides a natural choice of axis.

Three-dimensional angular momentum in quantum physics

Once we start to think more carefully about the role of angular momentum in quantum physics, it may seem that there is a basic problem: the angular momentum of the electron in a hydrogen atom depends on both its distance from the proton and its momentum, so in order to know its angular momentum precisely it would seem we would need to know both its position and its momentum simultaneously with good accuracy. This, however, might seem to be forbidden by the Heisenberg uncertainty principle.

Actually the uncertainty principle does place limits on what can be known about a particle's angular momentum vector, but it does not prevent us from knowing its magnitude as an exact integer multiple of \(\hbar\). The reason is that in three dimensions, there are really three separate uncertainty principles:

\[\begin{align*} \Delta p_x\Delta x & \gtrsim h \\ \Delta p_y\Delta y & \gtrsim h \\ \Delta p_z\Delta z & \gtrsim h \end{align*}\]

Now consider a particle, c/1, that is moving along the \(x\) axis at position \(x\) and with momentum \(p_x\). We may not be able to know both \(x\) and \(p_x\) with unlimited accuracy, but we can still know the particle's angular momentum about the origin exactly. Classically, it is zero, because the particle is moving directly away from the origin: if it was to be nonzero, we would need both a nonzero \(x\) and a nonzero \(p_y\). In quantum terms, the uncertainty principle does not place any constraint on \(\Delta x \Delta p_y\).

Suppose, on the other hand, a particle finds itself, as in figure c/2, at a position \(x\) along the \(x\) axis, and it is moving parallel to the \(y\) axis with momentum \(p_y\). It has angular momentum \(xp_y\) about the \(z\) axis, and again we can know its angular momentum with unlimited accuracy, because the uncertainty principle only relates \(x\) to \(p_x\) and \(y\) to \(p_y\). It does not relate \(x\) to \(p_y\).

As shown by these examples, the uncertainty principle does not restrict the accuracy of our knowledge of angular momenta as severely as might be imagined. However, it does prevent us from knowing all three components of an angular momentum vector simultaneously. The most general statement about this is the following theorem, which we present without proof:

the angular momentum vector in quantum physics

The most that can be known about an angular momentum vector is its magnitude and one of its three vector components. Both are quantized in units of \(\hbar\).

36.3 The hydrogen atom

hydrogen-energy-levels

e / The energy of a state in the hydrogen atom depends only on its \(n\) quantum number.

Deriving the wavefunctions of the states of the hydrogen atom from first principles would be mathematically too complex for this book, but it's not hard to understand the logic behind such a wavefunction in visual terms. Consider the wavefunction from the beginning of the chapter, which is reproduced below. Although the graph looks three-dimensional, it is really only a representation of the part of the wavefunction lying within a two-dimensional plane. The third (up-down) dimension of the plot represents the value of the wavefunction at a given point, not the third dimension of space. The plane chosen for the graph is the one perpendicular to the angular momentum vector.

big-hydrogen-wavefunction

d / A wavefunction of a hydrogen atom.

Each ring of peaks and valleys has eight wavelengths going around in a circle, so this state has \(L=8\hbar\), i.e., we label it \(\ell=8\). The wavelength is shorter near the center, and this makes sense because when the electron is close to the nucleus it has a lower PE, a higher KE, and a higher momentum.

Between each ring of peaks in this wavefunction is a nodal circle, i.e., a circle on which the wavefunction is zero. The full three-dimensional wavefunction has nodal spheres: a series of nested spherical surfaces on which it is zero. The number of radii at which nodes occur, including \(r=\infty\), is called \(n\), and \(n\) turns out to be closely related to energy. The ground state has \(n=1\) (a single node only at \(r=\infty\)), and higher-energy states have higher \(n\) values. There is a simple equation relating \(n\) to energy, which we will discuss in section 36.4.

The numbers \(n\) and \(\ell\), which identify the state, are called its quantum numbers. A state of a given \(n\) and \(\ell\) can be oriented in a variety of directions in space. We might try to indicate the orientation using the three quantum numbers \(\ell_x=L_x/\hbar\), \(\ell_y=L_y/\hbar\), and \(\ell_z=L_z/\hbar\). But we have already seen that it is impossible to know all three of these simultaneously. To give the most complete possible description of a state, we choose an arbitrary axis, say the \(z\) axis, and label the state according to \(n\), \(\ell\), and \(\ell_z\).

hydrogen-three-states

f / The three lowest-energy states of hydrogen.

Angular momentum requires motion, and motion implies kinetic energy. Thus it is not possible to have a given amount of angular momentum without having a certain amount of kinetic energy as well. Since energy relates to the \(n\) quantum number, this means that for a given \(n\) value there will be a maximum possible \(\ell\). It turns out that this maximum value of \(\ell\) equals \(n-1\).

In general, we can list the possible combinations of quantum numbers as follows:


n can equal 1, 2, 3, …
ell can range from 0 ton − 1, in steps of 1
ellz can range fromell toell, in steps of 1

Applying these rules, we have the following list of states:





n = 1,ell0, ellz0 one state
n = 2, ell0, ellz0 one state
n = 2, ell1,ellz1, 0, or 1three states




self-check:

Continue the list for \(n=3\).

(answer in the back of the PDF version of the book)

Figure f shows the lowest-energy states of the hydrogen atom. The left-hand column of graphs displays the wavefunctions in the \(x-y\) plane, and the right-hand column shows the probability distribution in a three-dimensional representation.

Discussion Questions

The quantum number \(n\) is defined as the number of radii at which the wavefunction is zero, including \(r=\infty\). Relate this to the features of the figures on the facing page.

Based on the definition of \(n\), why can't there be any such thing as an \(n=0\) state?

Relate the features of the wavefunction plots in figure f to the corresponding features of the probability distribution pictures.

How can you tell from the wavefunction plots in figure f which ones have which angular momenta?

Criticize the following incorrect statement: “The \(\ell=8\) wavefunction in figure d has a shorter wavelength in the center because in the center the electron is in a higher energy level.”

Discuss the implications of the fact that the probability cloud in of the \(n=2\), \(\ell=1\) state is split into two parts.

36.4 Energies of states in hydrogen (optional)

hydrogen-versus-box

g / The energy levels of a particle in a box, contrasted with those of the hydrogen atom.

History

The experimental technique for measuring the energy levels of an atom accurately is spectroscopy: the study of the spectrum of light emitted (or absorbed) by the atom. Only photons with certain energies can be emitted or absorbed by a hydrogen atom, for example, since the amount of energy gained or lost by the atom must equal the difference in energy between the atom's initial and final states. Spectroscopy had become a highly developed art several decades before Einstein even proposed the photon, and the Swiss spectroscopist Johann Balmer determined in 1885 that there was a simple equation that gave all the wavelengths emitted by hydrogen. In modern terms, we think of the photon wavelengths merely as indirect evidence about the underlying energy levels of the atom, and we rework Balmer's result into an equation for these atomic energy levels:

\[\begin{equation*} E_n = -\frac{2.2\times10^{-18}\ \text{J}}{n^2} , \end{equation*}\]

This energy includes both the kinetic energy of the electron and the electrical energy. The zero-level of the electrical energy scale is chosen to be the energy of an electron and a proton that are infinitely far apart. With this choice, negative energies correspond to bound states and positive energies to unbound ones.

Where does the mysterious numerical factor of \(2.2\times10^{-18}\ \text{J}\) come from? In 1913 the Danish theorist Niels Bohr realized that it was exactly numerically equal to a certain combination of fundamental physical constants:

\[\begin{equation*} E_n = -\frac{mk^2e^4}{2\hbar^2}\cdot\frac{1}{n^2} , \end{equation*}\]

where \(m\) is the mass of the electron, and \(k\) is the Coulomb force constant for electric forces.

Bohr was able to cook up a derivation of this equation based on the incomplete version of quantum physics that had been developed by that time, but his derivation is today mainly of historical interest. It assumes that the electron follows a circular path, whereas the whole concept of a path for a particle is considered meaningless in our more complete modern version of quantum physics. Although Bohr was able to produce the right equation for the energy levels, his model also gave various wrong results, such as predicting that the atom would be flat, and that the ground state would have \(\ell=1\) rather than the correct \(\ell=0\).

Approximate treatment

A full and correct treatment is impossible at the mathematical level of this book, but we can provide a straightforward explanation for the form of the equation using approximate arguments.

A typical standing-wave pattern for the electron consists of a central oscillating area surrounded by a region in which the wavefunction tails off. As discussed in section 35.6, the oscillating type of pattern is typically encountered in the classically allowed region, while the tailing off occurs in the classically forbidden region where the electron has insufficient kinetic energy to penetrate according to classical physics. We use the symbol \(r\) for the radius of the spherical boundary between the classically allowed and classically forbidden regions. Classically, \(r\) would be the distance from the proton at which the electron would have to stop, turn around, and head back in.

If \(r\) had the same value for every standing-wave pattern, then we'd essentially be solving the particle-in-a-box problem in three dimensions, with the box being a spherical cavity. Consider the energy levels of the particle in a box compared to those of the hydrogen atom, g. They're qualitatively different. The energy levels of the particle in a box get farther and farther apart as we go higher in energy, and this feature doesn't even depend on the details of whether the box is two-dimensional or three-dimensional, or its exact shape. The reason for the spreading is that the box is taken to be completely impenetrable, so its size, \(r\), is fixed. A wave pattern with \(n\) humps has a wavelength proportional to \(r/n\), and therefore a momentum proportional to \(n\), and an energy proportional to \(n^2\). In the hydrogen atom, however, the force keeping the electron bound isn't an infinite force encountered when it bounces off of a wall, it's the attractive electrical force from the nucleus. If we put more energy into the electron, it's like throwing a ball upward with a higher energy --- it will get farther out before coming back down. This means that in the hydrogen atom, we expect \(r\) to increase as we go to states of higher energy. This tends to keep the wavelengths of the high energy states from getting too short, reducing their kinetic energy. The closer and closer crowding of the energy levels in hydrogen also makes sense because we know that there is a certain energy that would be enough to make the electron escape completely, and therefore the sequence of bound states cannot extend above that energy.

When the electron is at the maximum classically allowed distance \(r\) from the proton, it has zero kinetic energy. Thus when the electron is at distance \(r\), its energy is purely electrical:

\[\begin{equation*} E = -\frac{ke^2}{r} \end{equation*}\]

Now comes the approximation. In reality, the electron's wavelength cannot be constant in the classically allowed region, but we pretend that it is. Since \(n\) is the number of nodes in the wavefunction, we can interpret it approximately as the number of wavelengths that fit across the diameter \(2r\). We are not even attempting a derivation that would produce all the correct numerical factors like 2 and \(\pi \) and so on, so we simply make the approximation

\[\begin{equation*} \lambda \sim \frac{r}{n} . \end{equation*}\]

Finally we assume that the typical kinetic energy of the electron is on the same order of magnitude as the absolute value of its total energy. (This is true to within a factor of two for a typical classical system like a planet in a circular orbit around the sun.) We then have \begin{subequations} \renewcommand{\theequation}{\theparentequation}

\[\begin{align*} \text{absolute}&\text{ value of total energy} \\ &= \frac{ke^2}{r} \notag \\ &\sim K \notag \ &= p^2/2m \notag \\ &= (h/\lambda)^2/2m \notag \\ &\sim h^2n^2/2mr^2 \notag \end{align*}\]

\end{subequations} We now solve the equation \(ke^2/r \sim h^2n^2 / 2mr^2\) for \(r\) and throw away numerical factors we can't hope to have gotten right, yielding

\[\begin{equation*} r \sim \frac{h^2n^2}{mke^2} . \end{equation*}\]

Plugging \(n=1\) into this equation gives \(r=2\) nm, which is indeed on the right order of magnitude. Finally we combine equations [4] and [1] to find

\[\begin{equation*} E \sim -\frac{mk^2e^4}{h^2n^2} , \end{equation*}\]

which is correct except for the numerical factors we never aimed to find.

Discussion Questions

States of hydrogen with \(n\) greater than about 10 are never observed in the sun. Why might this be?

Sketch graphs of \(r\) and \(E\) versus \(n\) for the hydrogen, and compare with analogous graphs for the one-dimensional particle in a box.

36.5 Electron spin

spin-vs-orbital

h / The top has angular momentum both because of the motion of its center of mass through space and due to its internal rotation. Electron spin is roughly analogous to the intrinsic spin of the top.

It's disconcerting to the novice ping-pong player to encounter for the first time a more skilled player who can put spin on the ball. Even though you can't see that the ball is spinning, you can tell something is going on by the way it interacts with other objects in its environment. In the same way, we can tell from the way electrons interact with other things that they have an intrinsic spin of their own. Experiments show that even when an electron is not moving through space, it still has angular momentum amounting to \(\hbar/2\).

This may seem paradoxical because the quantum moat, for instance, gave only angular momenta that were integer multiples of \(\hbar\), not half-units, and I claimed that angular momentum was always quantized in units of \(\hbar\), not just in the case of the quantum moat. That whole discussion, however, assumed that the angular momentum would come from the motion of a particle through space. The \(\hbar/2\) angular momentum of the electron is simply a property of the particle, like its charge or its mass. It has nothing to do with whether the electron is moving or not, and it does not come from any internal motion within the electron. Nobody has ever succeeded in finding any internal structure inside the electron, and even if there was internal structure, it would be mathematically impossible for it to result in a half-unit of angular momentum.

We simply have to accept this \(\hbar/2\) angular momentum, called the “spin” of the electron --- Mother Nature rubs our noses in it as an observed fact.

Protons and neutrons have the same \(\hbar/2\) spin, while photons have an intrinsic spin of \(\hbar\). In general, half-integer spins are typical of material particles. Integral values are found for the particles that carry forces: photons, which embody the electric and magnetic fields of force, as well as the more exotic messengers of the nuclear and gravitational forces.

As was the case with ordinary angular momentum, we can describe spin angular momentum in terms of its magnitude, and its component along a given axis. We notate these quantities, in units of \(\hbar\), as \(s\) and \(s_z\), so an electron has \(s=1/2\) and \(s_z=+1/2\) or -1/2.

Taking electron spin into account, we need a total of four quantum numbers to label a state of an electron in the hydrogen atom: \(n\), \(\ell\), \(\ell_z\), and \(s_z\). (We omit \(s\) because it always has the same value.) The symbols \(\ell\) and \(\ell_z\) include only the angular momentum the electron has because it is moving through space, not its spin angular momentum. The availability of two possible spin states of the electron leads to a doubling of the numbers of states:






n = 1,ell0, ellz0, sz = + 1 / 2 or − 1 / 2two states
n = 2, ell0, ellz0, sz = + 1 / 2 or − 1 / 2two states
n = 2, ell1,ellz1, 0, or 1,sz = + 1 / 2 or − 1 / 2six states





36.6 Atoms with more than one electron

small-periodic-table

i / The beginning of the periodic table.

hindenburg

j / Hydrogen is highly reactive.

What about other atoms besides hydrogen? It would seem that things would get much more complex with the addition of a second electron. A hydrogen atom only has one particle that moves around much, since the nucleus is so heavy and nearly immobile. Helium, with two, would be a mess. Instead of a wavefunction whose square tells us the probability of finding a single electron at any given location in space, a helium atom would need to have a wavefunction whose square would tell us the probability of finding two electrons at any given combination of points. Ouch! In addition, we would have the extra complication of the electrical interaction between the two electrons, rather than being able to imagine everything in terms of an electron moving in a static field of force created by the nucleus alone.

Despite all this, it turns out that we can get a surprisingly good description of many-electron atoms simply by assuming the electrons can occupy the same standing-wave patterns that exist in a hydrogen atom. The ground state of helium, for example, would have both electrons in states that are very similar to the \(n=1\) states of hydrogen. The second-lowest-energy state of helium would have one electron in an \(n=1\) state, and the other in an \(n=2\) states. The relatively complex spectra of elements heavier than hydrogen can be understood as arising from the great number of possible combinations of states for the electrons.

A surprising thing happens, however, with lithium, the three-electron atom. We would expect the ground state of this atom to be one in which all three electrons settle down into \(n=1\) states. What really happens is that two electrons go into \(n=1\) states, but the third stays up in an \(n=2\) state. This is a consequence of a new principle of physics:

the Pauli exclusion principle

Only one electron can ever occupy a given state.

There are two \(n=1\) states, one with \(s_z=+1/2\) and one with \(s_z=-1/2\), but there is no third \(n=1\) state for lithium's third electron to occupy, so it is forced to go into an \(n=2\) state.

It can be proved mathematically that the Pauli exclusion principle applies to any type of particle that has half-integer spin. Thus two neutrons can never occupy the same state, and likewise for two protons. Photons, however, are immune to the exclusion principle because their spin is an integer. Material objects can't pass through each other, but beams of light can. With a little oversimplification, we can say that the basic reason is that the exclusion principle applies to one but not to the other.1

Photons are made out electric and magnetic fields, which are directly measurable, but the wavefunction of a spin-1/2 particle is not observable (p. 959). The exclusion principle offers a more fundamental explanation of this difference between light and matter. We saw in example 2 on p. 943 that in a typical light wave, a huge number of photons overlap one another within a volume of one cubic wavelength, and this is what allows us to measure the amplitude and phase of the wave with a device like an antenna. But for electrons, the exclusion principle prevents us from having more than one particle in such a volume, so we can't perform this type of classical measurement of the wave.

Deriving the periodic table

We can now account for the structure of the periodic table, which seemed so mysterious even to its inventor Mendeleev. The first row consists of atoms with electrons only in the \(n=1\) states:

H

1 electron in ann = 1 state

He

2 electrons in the twon = 1 states

The next row is built by filling the \(n=2\) energy levels:

Li

2 electrons inn = 1 states, 1 electron in ann = 2 state

Be

2 electrons inn = 1 states, 2 electrons inn = 2 states

O

2 electrons inn = 1 states, 6 electrons inn = 2 states

F

2 electrons inn = 1 states, 7 electrons inn = 2 states

Ne

2 electrons inn = 1 states, 8 electrons inn = 2 states

In the third row we start in on the \(n=3\) levels:

Na

2 electrons inn = 1 states, 8 electrons inn = 2 states, 1 electron in ann = 3 state

...

We can now see a logical link between the filling of the energy levels and the structure of the periodic table. Column 0, for example, consists of atoms with the right number of electrons to fill all the available states up to a certain value of \(n\). Column I contains atoms like lithium that have just one electron more than that.

This shows that the columns relate to the filling of energy levels, but why does that have anything to do with chemistry? Why, for example, are the elements in columns I and VII dangerously reactive? Consider, for example, the element sodium (Na), which is so reactive that it may burst into flames when exposed to air. The electron in the \(n=3\) state has an unusually high energy. If we let a sodium atom come in contact with an oxygen atom, energy can be released by transferring the \(n=3\) electron from the sodium to one of the vacant lower-energy \(n=2\) states in the oxygen. This energy is transformed into heat. Any atom in column I is highly reactive for the same reason: it can release energy by giving away the electron that has an unusually high energy.

Column VII is spectacularly reactive for the opposite reason: these atoms have a single vacancy in a low-energy state, so energy is released when these atoms steal an electron from another atom.

It might seem as though these arguments would only explain reactions of atoms that are in different rows of the periodic table, because only in these reactions can a transferred electron move from a higher-\(n\) state to a lower-\(n\) state. This is incorrect. An \(n=2\) electron in fluorine (F), for example, would have a different energy than an \(n=2\) electron in lithium (Li), due to the different number of protons and electrons with which it is interacting. Roughly speaking, the \(n=2\) electron in fluorine is more tightly bound (lower in energy) because of the larger number of protons attracting it. The effect of the increased number of attracting protons is only partly counteracted by the increase in the number of repelling electrons, because the forces exerted on an electron by the other electrons are in many different directions and cancel out partially.

Example 1: Neutron stars

Here's an exotic example that doesn't even involve atoms. When a star runs out of fuel for its nuclear reactions, it begins to collapse under its own weight. Since Newton's law of gravity depends on the inverse square of the distance, the gravitational forces become stronger as the star collapses, which encourages it to collapse even further. The final result depends on the mass of the star, but let's consider a star that's only a little more massive than our own sun. Such a star will collapse to the point where the gravitational energy being released is sufficient to cause the reaction \(\text{p}+\text{e}^{-}\rightarrow \text{n}+\nu\) to occur. (As you found in homework problem 10 on page 785, this reaction can only occur when there is some source of energy, because the mass-energy of the products is greater than the mass-energy of the things being consumed.) The neutrinos fly off and are never heard from again, so we're left with a star consisting only of neutrons!

Now the exclusion principle comes into play. The collapse can't continue indefinitely. The situation is in fact closely analogous to that of an atom. A lead atom's cloud of 82 electrons can't shrink down to the size of a hydrogen atom, because only two electrons can have the lowest-energy wave pattern. The same happens with the neutron star. No physical repulsion keeps the neutrons apart. They're electrically neutral, so they don't repel or attract one another electrically. The gravitational force is attractive, and as the collapse proceeds to the point where the neutrons come within range of the strong nuclear force (\(\sim10^{-15}\ \text{m}\)), we even start getting nuclear attraction. The only thing that stops the whole process is the exclusion principle. The star ends up being only a few kilometers across, and has the same billion-ton-per-teaspoon density as an atomic nucleus. Indeed, we can think of it as one big nucleus (with atomic number zero, because there are no protons).

As with a spinning figure skater pulling her arms in, conservation of angular momentum makes the star spin faster and faster. The whole object may end up with a rotational period of a fraction of a second! Such a star sends out radio pulses with each revolution, like a sort of lighthouse. The first time such a signal was detected, radio astronomers thought that it was a signal from aliens.

Summary

Vocabulary

quantum number — a numerical label used to classify a quantum state

spin — the built-in angular momentum possessed by a particle even when at rest

Notation

\(n\) — the number of radial nodes in the wavefunction, including the one at \(r = \infty\) \(\hbar\) — \(h/2\pi\) \notationitem{\(\mathbf{L}\)}{the angular momentum vector of a particle, not including its spin} \(\ell\) — the magnitude of the \(L\) vector, divided by \(\hbar\) \(\ell_z\) — the \(z\) component of the \(L\) vector, divided by \(\hbar\); this is the standard notation in nuclear physics, but not in atomic physics \(s\) — the magnitude of the spin angular momentum vector, divided by \(\hbar\) \(s_z\) — the \(z\) component of the spin angular momentum vector, divided by \(\hbar\); this is the standard notation in nuclear physics, but not in atomic physics

Other Notation

\(m_\ell\) — a less obvious notation for \(\ell_z\), standard in atomic physics \(m_s\) — a less obvious notation for \(s_z\), standard in atomic physics

Summary

{}

Hydrogen, with one proton and one electron, is the simplest atom, and more complex atoms can often be analyzed to a reasonably good approximation by assuming their electrons occupy states that have the same structure as the hydrogen atom's. The electron in a hydrogen atom exchanges very little energy or angular momentum with the proton, so its energy and angular momentum are nearly constant, and can be used to classify its states. The energy of a hydrogen state depends only on its \(n\) quantum number.

In quantum physics, the angular momentum of a particle moving in a plane is quantized in units of \(\hbar\). Atoms are three-dimensional, however, so the question naturally arises of how to deal with angular momentum in three dimensions. In three dimensions, angular momentum is a vector in the direction perpendicular to the plane of motion, such that the motion appears clockwise if viewed along the direction of the vector. Since angular momentum depends on both position and momentum, the Heisenberg uncertainty principle limits the accuracy with which one can know it. The most that can be known about an angular momentum vector is its magnitude and one of its three vector components, both of which are quantized in units of \(\hbar\).

In addition to the angular momentum that an electron carries by virtue of its motion through space, it possesses an intrinsic angular momentum with a magnitude of \(\hbar/2\). Protons and neutrons also have spins of \(\hbar/2\), while the photon has a spin equal to \(\hbar\).

Particles with half-integer spin obey the Pauli exclusion principle: only one such particle can exist is a given state, i.e., with a given combination of quantum numbers.

We can enumerate the lowest-energy states of hydrogen as follows:






n = 1,ell0, ellz0, sz = + 1 / 2 or − 1 / 2two states
n = 2, ell0, ellz0, sz = + 1 / 2 or − 1 / 2two states
n = 2, ell1,ellz1, 0, or 1,sz = + 1 / 2 or − 1 / 2six states





The periodic table can be understood in terms of the filling of these states. The nonreactive noble gases are those atoms in which the electrons are exactly sufficient to fill all the states up to a given \(n\) value. The most reactive elements are those with one more electron than a noble gas element, which can release a great deal of energy by giving away their high-energy electron, and those with one electron fewer than a noble gas, which release energy by accepting an electron.

Homework Problems

hw-h-transitions

k / Problem 2.

\begin{homeworkforcelabel}{hydrogenscale}{1}{}{1}(a) A distance scale is shown below the wavefunctions and probability densities illustrated in figure f on page 987. Compare this with the order-of-magnitude estimate derived in section 36.4 for the radius \(r\) at which the wavefunction begins tailing off. Was the estimate in section 36.4 on the right order of magnitude?
(b) Although we normally say the moon orbits the earth, actually they both orbit around their common center of mass, which is below the earth's surface but not at its center. The same is true of the hydrogen atom. Does the center of mass lie inside the proton or outside it? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{hydrogenlevels}{1}{}{2}The figure shows eight of the possible ways in which an electron in a hydrogen atom could drop from a higher energy state to a state of lower energy, releasing the difference in energy as a photon. Of these eight transitions, only D, E, and F produce photons with wavelengths in the visible spectrum.
(a) Which of the visible transitions would be closest to the violet end of the spectrum, and which would be closest to the red end? Explain.
(b) In what part of the electromagnetic spectrum would the photons from transitions A, B, and C lie? What about G and H? Explain.
(c) Is there an upper limit to the wavelengths that could be emitted by a hydrogen atom going from one bound state to another bound state? Is there a lower limit? Explain. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{energysum}{1}{}{3}Before the quantum theory, experimentalists noted that in many cases, they would find three lines in the spectrum of the same atom that satisfied the following mysterious rule: \(1/\lambda_1=1/\lambda_2+1/\lambda_3\). Explain why this would occur. Do not use reasoning that only works for hydrogen --- such combinations occur in the spectra of all elements. [Hint: Restate the equation in terms of the energies of photons.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{hydrogenphoton}{1}{}{4} Find an equation for the wavelength of the photon emitted when the electron in a hydrogen atom makes a transition from energy level \(n_1\) to level \(n_2\). [You will need to have read optional section 36.4.](answer check available at lightandmatter.com)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{basketball}{1}{}{5}Estimate the angular momentum of a spinning basketball, in units of \(\hbar\). Explain how this result relates to the correspondence principle. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{hydrogennonrelativistic}{1}{}{6} Assume that the kinetic energy of an electron in the \(n=1\) state of a hydrogen atom is on the same order of magnitude as the absolute value of its total energy, and estimate a typical speed at which it would be moving. (It cannot really have a single, definite speed, because its kinetic and potential energy trade off at different distances from the proton, but this is just a rough estimate of a typical speed.) Based on this speed, were we justified in assuming that the electron could be described nonrelativistically? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{electroninproton}{1}{}{7} The wavefunction of the electron in the ground state of a hydrogen atom, shown in the top left of figure f on p. 987, is

\[\begin{equation*} \Psi = \pi^{-1/2} a^{-3/2} e^{-r/a} , \end{equation*}\]

where \(r\) is the distance from the proton, and \(a=\hbar^2/kme^2=5.3\times10^{-11}\ \text{m}\) is a constant that sets the size of the wave. The figure doesn't show the proton; let's take the proton to be a sphere with a radius of \(b=0.5\) fm.
(a) Reproduce figure f in a rough sketch, and indicate, relative to the size of your sketch, some idea of how big \(a\) and \(b\) are.
(b) Calculate symbolically, without plugging in numbers, the probability that at any moment, the electron is inside the proton. [Hint: Does it matter if you plug in \(r=0\) or \(r=b\) in the equation for the wavefunction?](answer check available at lightandmatter.com)
(c) Calculate the probability numerically.(answer check available at lightandmatter.com)
(d) Based on the equation for the wavefunction, is it valid to think of a hydrogen atom as having a finite size? Can \(a\) be interpreted as the size of the atom, beyond which there is nothing? Or is there any limit on how far the electron can be from the proton?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{hydrogenlike}{2}{}{8}Use physical reasoning to explain how the equation for the energy levels of hydrogen,

\[\begin{equation*} E_n = -\frac{mk^2e^4}{2\hbar^2}\cdot\frac{1}{n^2} , \end{equation*}\]

should be generalized to the case of an atom with atomic number \(Z\) that has had all its electrons removed except for one. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{muonic-spectrum}{1}{}{9} This question requires that you read optional section 36.4. A muon is a subatomic particle that acts exactly like an electron except that its mass is 207 times greater. Muons can be created by cosmic rays, and it can happen that one of an atom's electrons is displaced by a muon, forming a muonic atom. If this happens to a hydrogen atom, the resulting system consists simply of a proton plus a muon.
(a) How would the size of a muonic hydrogen atom in its ground state compare with the size of the normal atom?
(b) If you were searching for muonic atoms in the sun or in the earth's atmosphere by spectroscopy, in what part of the electromagnetic spectrum would you expect to find the absorption lines?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{hydrogen-dipole}{1}{}{10}Consider a classical model of the hydrogen atom in which the electron orbits the proton in a circle at constant speed. In this model, the electron and proton can have no intrinsic spin. Using the result of problem 14 in ch. 24, show that in this model, the atom's magnetic dipole moment \(D_m\) is related to its angular momentum by \(D_m=(-e/2m)L\), regardless of the details of the orbital motion. Assume that the magnetic field is the same as would be produced by a circular current loop, even though there is really only a single charged particle. [Although the model is quantum-mechanically incorrect, the result turns out to give the correct quantum mechanical value for the contribution to the atom's dipole moment coming from the electron's orbital motion. There are other contributions, however, arising from the intrinsic spins of the electron and proton.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{hydrogen-ratio}{1}{}{11}Hydrogen is the only element whose energy levels can be expressed exactly in an equation. Calculate the ratio \(\lambda_E/\lambda_F\) of the wavelengths of the transitions labeled E and F in problem 2 on p. 998. Express your answer as an exact fraction, not a decimal approximation. In an experiment in which atomic wavelengths are being measured, this ratio provides a natural, stringent check on the precision of the results.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{handson}{}{Quantum versus classical randomness}{\onecolumn}

1. Imagine the classical version of the particle in a one-dimensional box. Suppose you insert the particle in the box and give it a known, predetermined energy, but a random initial position and a random direction of motion. You then pick a random later moment in time to see where it is. Sketch the resulting probability distribution by shading on top of a line segment. Does the probability distribution depend on energy?

2. Do similar sketches for the first few energy levels of the quantum mechanical particle in a box, and compare with 1.

3. Do the same thing as in 1, but for a classical hydrogen atom in two dimensions, which acts just like a miniature solar system. Assume you're always starting out with the same fixed values of energy and angular momentum, but a position and direction of motion that are otherwise random. Do this for \(L=0\), and compare with a real \(L=0\) probability distribution for the hydrogen atom.

4. Repeat 3 for a nonzero value of \(L\), say L=\(\hbar\).

5. Summarize: Are the classical probability distributions accurate? What qualitative features are possessed by the classical diagrams but not by the quantum mechanical ones, or vice-versa?

\end{handson}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

Footnotes
[1] There are also electrical forces between atoms, but as argued on page 956, the attractions and repulsions tend to cancel out.