The Hydrogen Atom

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 twodimensional 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.

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 = 8h[bar], i.e., we label it $ = 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.

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

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 = ∞, 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 = ∞), and higher-energy states have higher n values. There is a simple equation relating n to energy, which we will discuss in section 5.4.

The numbers n and $, which identify the state, are called its quantum numbers. A state of a given n and $ can be oriented in a variety of directions in space. We might try to indicate the orientation using the three quantum numbers $_x = L_x/h[bar], $_y = L_y/h[bar], and $_z = L_z/h[bar]. 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, $, and $_z.

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 $. It turns out that this maximum value of $ equals n - 1.

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

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

Self-Check	Continue the list for n = 3.
Answer	n = 3, $ = 0, $z = 0: one state n = 3, $=1, $z = -1, 0, or 1: three states n = 3, $ = 2, $z = -2, -1, 0, 1, or 2: five states

f / The three lowest-energy states of hydrogen.

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

Discussion Questions

A	The quantum number n is defined as the number of radii at which the wavefunction is zero, including r = ∞. Relate this to the features of the figures on the facing page.
B	Based on the definition of n, why can't there be any such thing as an n = 0 state?
C	Relate the features of the wavefunction plots in figure f to the corresponding features of the probability density pictures.
D	How can you tell from the wavefunction plots in figure f which ones have which angular momenta?
E	Criticize the following incorrect statement: "The $ = 8 wavefunction in figure d has a shorter wavelength in the center because in the center the electron is in a higher energy level."
F	Discuss the implications of the fact that the probability cloud in of the n = 2, $ = 1 state is split into two parts.