This section examines the critically important case of the hydrogen atom. The hydrogen atom consists of a nucleus which is just a single proton, and an electron encircling that nucleus. The nucleus, being much heavier than the electron, can be assumed to be at rest, and only the motion of the electron is of concern.
The energy levels of the electron determine the photons that the atom will absorb or emit, allowing the powerful scientific tool of spectral analysis. The electronic structure is also essential for understanding the properties of the other elements and of chemical bonds.
The first step is to find the Hamiltonian of the electron. The
electron experiences an electrostatic Coulomb attraction to the oppositely charged nucleus. The
corresponding potential energy is
(4.28) |
(4.29) |
(4.30) |
Unlike for the harmonic oscillator discussed earlier, this potential energy cannot be split into separate parts for Cartesian coordinates , , and . To do the analysis for the hydrogen atom, you must put the nucleus at the origin of the coordinate system and use spherical coordinates (the distance from the nucleus), (the angle from an arbitrarily chosen -axis), and (the angle around the -axis); see figure 4.7. In terms of spherical coordinates, the potential energy above depends on just the single coordinate .
To get the Hamiltonian, you need to add to this potential energy the kinetic energy operator . Chapter 3.3 gave this
operator as
(4.32) |
It may be noted that the small proton motion can be corrected for by slightly adjusting the mass of the electron to be an effective 9.104,4 10 kg, {A.5}. This makes the solution exact, except for extremely small errors due to relativistic effects. (These are discussed in addendum {A.38}.)
Key Points
- To analyze the hydrogen atom, you must use spherical coordinates.
- The Hamiltonian in spherical coordinates has been written down. It is (4.31).
This subsection describes in general lines how the eigenvalue problem for the electron of the hydrogen atom is solved. The basic ideas are like those used to solve the particle in a pipe and the harmonic oscillator, but in this case, they are used in spherical coordinates rather than Cartesian ones. Without getting too much caught up in the mathematical details, do not miss the opportunity of learning where the hydrogen energy eigenfunctions and eigenvalues come from. This is the crown jewel of quantum mechanics; brilliant, almost flawless, critically important; one of the greatest works of physical analysis ever.
The eigenvalue problem for the Hamiltonian, as formulated in the
previous subsection, can be solved by searching for solutions
that take the form of a product of functions of each of the three
coordinates: . More
concisely, . The problem now is to
find separate equations for the individual functions ,
, and from which they can then be identified.
The arguments are similar as for the harmonic oscillator, but messier,
since the coordinates are more entangled. First, substituting
into the Hamiltonian eigenvalue problem
, with the Hamiltonian as given in the previous
subsection and the energy eigenvalue, produces:
In fact, the solution to this final problem has already been given,
since the operator involved is just the square of the angular momentum
operator of section 4.2.2:
The eigenvalue problem (4.34) for is even
easier; it is exactly the one for the square angular momentum of
section 4.2.3. (So, no, there was not really a need to
solve for separately.) Its eigenfunctions are therefore the
spherical harmonics,
Returning now to the solution of the original eigenvalue problem (4.33), replacement of the angular terms by turns it into an ordinary differential equation problem for the radial factor in the energy eigenfunction. As usual, this problem is a pain to solve, so that is again shoved away in a note, {D.15}.
It turns out that the solutions of the radial problem can be numbered
using a third quantum number, , called the “principal quantum number”. It is larger than the azimuthal
quantum number , which in turn must be at least as large as the
absolute value of the magnetic quantum number:
(4.35) |
In terms of these three quantum numbers, the final energy
eigenfunctions of the hydrogen atom are of the general form:
Bohr radiusand has the value
The energy eigenvalues are much simpler and more interesting than the
eigenfunctions; they are
You may wonder why the energy only depends on the principal quantum number , and not also on the azimuthal quantum number and the magnetic quantum number . Well, the choice of -axis was arbitrary, so it should not seem that strange that the physics would not depend on the angular momentum in that direction. But that the energy does not depend on is nontrivial: if you solve the simpler problem of a particle stuck inside an impenetrable spherical container, using procedures from {A.6}, the energy values depend on both and . So, that is just the way it is. (It stops being true anyway if you include relativistic effects in the Hamiltonian.)
Since the lowest possible value of the principal quantum number is one, the ground state of lowest energy is eigenfunction .
Key Points
- Skipping a lot of math, energy eigenfunctions and their energy eigenvalues have been found.
- There is one eigenfunction for each set of three integer quantum numbers , , and satisfying . The number is called the principal quantum number.
- The typical length scale in the solution is called the Bohr radius , which is about half an Ångstrom.
- The derived eigenfunctions are eigenfunctions of
- angular momentum, with eigenvalue ;
- square angular momentum, with eigenvalue ;
- energy, with eigenvalue .
- The energy values only depend on the principal quantum number .
- The ground state is .
Use the tables for the radial wave functions and the spherical harmonics to write down the wave function
Check that the state is normalized. Note: .
Use the generic expression
Plug numbers into the generic expression for the energy eigenvalues,
The only energy values that the electron in the hydrogen atom can have
are the “Bohr energies” derived in the previous subsection:
To aid the discussion, the allowed energies are plotted in the form of an energy spectrum in figure 4.8. To the right of the lowest three energy levels the values of the quantum numbers that give rise to those energy levels are listed.
The first thing that the energy spectrum illustrates is that the energy levels are all negative, unlike the ones of the harmonic oscillator, which were all positive. However, that does not mean much; it results from defining the potential energy of the harmonic oscillator to be zero at the nominal position of the particle, while the hydrogen potential is instead defined to be zero at large distance from the nucleus. (It will be shown later, chapter 7.2, that the average potential energy is twice the value of the total energy, and the average kinetic energy is minus the total energy, making the average kinetic energy positive as it should be.)
A more profound difference is that the energy levels of the hydrogen atom have a maximum value, namely zero, while those of the harmonic oscillator went all the way to infinity. It means physically that while the particle can never escape in a harmonic oscillator, in a hydrogen atom, the electron escapes if its total energy is greater than zero. Such a loss of the electron is called “ionization” of the atom.
There is again a ground state of lowest energy; it has total energy
electron voltis 1.6 10 J). The ground state is the state in which the hydrogen atom will be at absolute zero temperature. In fact, it will still be in the ground state at room temperature, since even then the energy of heat motion is unlikely to raise the energy level of the electron to the next higher one, .
The ionization energy of the hydrogen atom is 13.6 eV; this is the minimum amount of energy that must be added to raise the electron from the ground state to the state of a free electron.
If the electron is excited from the ground state to a higher but still bound energy level, (maybe by passing a spark through hydrogen gas), it will in time again transition back to a lower energy level. Discussion of the reasons and the time evolution of this process will have to wait until chapter 7. For now, it can be pointed out that different transitions are possible, as indicated by the arrows in figure 4.8. They are named by their final energy level to be Lyman, Balmer, or Paschen series transitions.
The energy lost by the electron during a transition is emitted as a quantum of electromagnetic radiation called a photon. The most energetic photons, in the ultraviolet range, are emitted by Lyman transitions. Balmer transitions emit visible light and Paschen ones infrared.
The photons emitted by isolated atoms at rest must have an energy very
precisely equal to the difference in energy eigenvalues; anything else
would violate the requirement of the orthodox interpretation that only
the eigenvalues are observable. And according to the “Planck-Einstein relation,” the photon’s energy equals the
angular frequency of its electromagnetic vibration times
:
(To be sure, the spectral frequencies are not truly mathematically
exact numbers. A slight spectral broadening
is
unavoidable because no atom is truly isolated as assumed here; there
is always some radiation that perturbs it even in the most ideal empty
space. In addition, thermal motion of the atom causes Doppler shifts.
In short, only the energy eigenvalues are observable, but exactly what
those eigenvalues are for a real-life atom can vary slightly.)
Atoms and molecules may also absorb electromagnetic energy of the same frequencies that they can emit. That allows them to enter an excited state. The excited state will eventually emit the absorbed energy again in a different direction, and possibly at different frequencies by using different transitions. In this way, in astronomy atoms can remove specific frequencies from light that passes them on its way to earth, resulting in an absorption spectrum. Or instead atoms may scatter specific frequencies of light in our direction that was originally not headed to earth, producing an emission spectrum. Doppler shifts can provide information about the thermal and average motion of the atoms. Since hydrogen is so prevalent in the universe, its energy levels as derived here are particularly important in astronomy. Chapter 7 will address the mechanisms of emission and absorption in much greater detail.
Key Points
- The energy levels of the electron in a hydrogen atom have a highest value. This energy is by convention taken to be the zero level.
- The ground state has a energy 13.6 eV below this zero level.
- If the electron in the ground state is given an additional amount of energy that exceeds the 13.6 eV, it has enough energy to escape from the nucleus. This is called ionization of the atom.
- If the electron transitions from a bound energy state with a higher principal quantum number to a lower one , it emits radiation with an angular frequency given by
- Similarly, atoms with energy may absorb electromagnetic energy of such a frequency.
If there are infinitely many energy levels , , , , , , ..., where did they all go in the energy spectrum?
What is the value of energy level ? And ?
Based on the results of the previous question, what is the color of the light emitted in a Balmer transition from energy to ? The Planck-Einstein relation says that the angular frequency of the emitted photon is its energy divided by , and the wave length of light is where is the speed of light. Typical wave lengths of visible light are: violet 400 nm, indigo 445 nm, blue 475 nm, green 510 nm, yellow 570 nm, orange 590 nm, red 650 nm.
What is the color of the light emitted in a Balmer transition from an energy level with a high value of to ?
The appearance of the energy eigenstates will be of great interest in understanding the heavier elements and chemical bonds. This subsection describes the most important of them.
It may be recalled from subsection 4.3.2 that there is one
eigenfunction for each set of three integer quantum numbers.
They are the principal quantum number (determining the energy of the
state), the azimuthal quantum number (determining the square
angular momentum), and the magnetic quantum number (determining the
angular momentum in the chosen -direction.) They must satisfy the
requirements that
For the ground state, with the lowest energy , 1 and hence according to the conditions above both and must be zero. So the ground state eigenfunction is ; it is unique.
The expression for the wave function of the ground state is (from the
results of subsection 4.3.2):
(4.41) |
The square magnitude of the energy states will again be displayed as grey tones, darker regions corresponding to regions where the electron is more likely to be found. The ground state is shown this way in figure 4.9; the electron may be found within a blob size that is about three times the Bohr radius, or roughly an Ångstrom, (10 m), in diameter.
It is the quantum mechanical refusal of electrons to restrict themselves to a single location that gives atoms their size. If Planck's constant would have been zero, so would have been the Bohr radius, and the electron would have been in the nucleus. It would have been a very different world.
The ground state probability distribution is spherically symmetric: the probability of finding the electron at a point depends on the distance from the nucleus, but not on the angular orientation relative to it.
The excited energy levels , , ...are all degenerate; as the spectrum figure 4.8 indicated, there is more than one eigenstate producing each level. Let’s have a look at the states at energy level now.
Figure 4.10 shows energy eigenfunction . Like , it is spherically symmetric. In fact, all eigenfunctions are spherically symmetric. However, the wave function has blown up a lot, and now separates into a small, more or less spherical region in the center, surrounded by a second region that forms a spherical shell. Separating the two is a radius at which there is zero probability of finding the electron.
The state is commonly referred to as the
2s
state. The 2 indicates that it is a state with
energy . The “s” indicates that the azimuthal quantum number is zero; just
think spherically symmetric.
Similarly,
the ground state is commonly indicated as
1s
, having the lowest energy .
States which have azimuthal quantum number 1 are called
“p” states, for some historical reason. Historically,
physicists have always loved confusing and inconsistent notations. In
particular, the states are called 2p
states. As first example of such a state, figure 4.11 shows
. This wave function squeezes itself close to the
-axis, which is plotted horizontally by convention. There is zero
probability of finding the electron at the vertical symmetry
plane, and maximum probability at two symmetric points on the
-axis.
Since the wave function squeezes close to the -axis, this
state is often more specifically referred to as the
2p
state. Think “points
along the -axis.”
Figure 4.12 shows the other two 2p
states,
and . These two states look exactly
the same as far as the probability density is concerned. It is
somewhat hard to see in the figure, but they really take the shape of
a torus around the left-to-right -axis.
Eigenfunctions , , , and are degenerate: they all four have the same energy 3.4 eV. The consequence is that they are not unique. Combinations of them can be formed that have the same energy. These combination states may be more important physically than the original eigenfunctions.
In particular, the torus-shaped eigenfunctions and
are often not very useful for descriptions of heavier
elements and chemical bonds. Two states that are more likely to be
relevant here are called 2p and 2p; they are
the combination states:
(4.42) |
These two states are shown in figure 4.13; they look exactly
like the pointer
state 2p of figure
4.11, except that they squeeze along the -axis,
respectively the -axis, instead of along the -axis.
(Since the -axis is pointing towards you, 2p looks
rotationally symmetric. Seen from the side, it would look like
p in figure 4.11.)
Note that unlike the two original states and , the states 2p and 2p do not have a definite value of the -component of angular momentum; the -component has a 50/50 uncertainty of being either or . But that is not important in most circumstances. What is important is that when multiple electrons occupy the p states, mutual repulsion effects tend to push them into the p, p, and p states.
So, the four independent eigenfunctions at energy level are best thought of as consisting of one spherically symmetrical 2s state, and three directional states, 2p, 2p, and 2p, pointing along the three coordinate axes.
But even that is not always ideal; as discussed in chapter
5.11.4, for many chemical bonds, especially those involving the
important element carbon, still different combination states called
hybrids
show up. They involve combinations of
the 2s and the 2p states and therefore have uncertain
square angular momentum as well.
Key Points
- The typical size of eigenstates is given by the Bohr radius, making the size of the atom of the order of an Å.
- The ground state , or 1s state, is nondegenerate: no other set of quantum numbers produces energy .
- All higher energy levels are degenerate, there is more than one eigenstate producing that energy.
- All states of the form , including the ground state, are spherically symmetric, and are called s states. The ground state is the 1s state, is the 2s state, etcetera.
- States of the form are called p states. The basic 2p states are , , and .
- The state is also called the 2p state, since it squeezes itself around the -axis.
- There are similar 2p and 2p states that squeeze around the and axes. Each is a combination of and .
- The four spatial states at the energy level can therefore be thought of as one spherically symmetric 2s state and three 2p pointer states along the axes.
- However, since the energy level is degenerate, eigenstates of still different shapes are likely to show up in applications.
At what distance from the nucleus does the square of the ground state wave function become less than one percent of its value at the nucleus? Express it both as a multiple of the Bohr radius and in Å.
Check from the conditions
basicsolutions , , , and .)
Check that the states