Samuel J. Ling; Jeff Sanny; William Moebs

Learning Objectives

By the end of this section, you will be able to:

Describe the hydrogen atom in terms of wave function, probability density, total energy, and orbital angular momentum
Identify the physical significance of each of the quantum numbers ( $n, l, m$ ) of the hydrogen atom
Distinguish between the Bohr and Schrödinger models of the atom
Use quantum numbers to calculate important information about the hydrogen atom

The hydrogen atom is the simplest atom in nature and, therefore, a good starting point to study atoms and atomic structure. The hydrogen atom consists of a single negatively charged electron that moves about a positively charged proton (Figure 8.2). In Bohr’s model, the electron is pulled around the proton in a perfectly circular orbit by an attractive Coulomb force. The proton is approximately 1800 times more massive than the electron, so the proton moves very little in response to the force on the proton by the electron. (This is analogous to the Earth-Sun system, where the Sun moves very little in response to the force exerted on it by Earth.) An explanation of this effect using Newton’s laws is given in Photons and Matter Waves.

The Bohr model of the hydrogen atom has the proton, charge q = plus e, at the center and the electron, charge q = minus e, in a circular orbit centered on the proton.

Figure 8.2 A representation of the Bohr model of the hydrogen atom.

With the assumption of a fixed proton, we focus on the motion of the electron.

In the electric field of the proton, the potential energy of the electron is

U (r) = − k \frac{e^{2}}{r},

8.1

where $k = 1 / 4 π ε_{0}$ and r is the distance between the electron and the proton. As we saw earlier, the force on an object is equal to the negative of the gradient (or slope) of the potential energy function. For the special case of a hydrogen atom, the force between the electron and proton is an attractive Coulomb force.

Notice that the potential energy function U(r) does not vary in time. As a result, Schrödinger’s equation of the hydrogen atom reduces to two simpler equations: one that depends only on space (x, y, z) and another that depends only on time (t). (The separation of a wave function into space- and time-dependent parts for time-independent potential energy functions is discussed in Quantum Mechanics.) We are most interested in the space-dependent equation:

\frac{− ℏ^{2}}{2 m_{e}} (\frac{\partial^{2} ψ}{\partial x^{2}} + \frac{\partial^{2} ψ}{\partial y^{2}} + \frac{\partial^{2} ψ}{\partial z^{2}}) - k \frac{e^{2}}{r} ψ = E ψ,

8.2

where $ψ = ψ (x, y, z)$ is the three-dimensional wave function of the electron, $m_{e}$ is the mass of the electron, and E is the total energy of the electron. Recall that the total wave function $Ψ (x, y, z, t),$ is the product of the space-dependent wave function $ψ = ψ (x, y, z)$ and the time-dependent wave function $φ = φ (t)$ .

In addition to being time-independent, U(r) is also spherically symmetrical. This suggests that we may solve Schrödinger’s equation more easily if we express it in terms of the spherical coordinates $(r, θ, ϕ)$ instead of rectangular coordinates $(x, y, z)$ . A spherical coordinate system is shown in Figure 8.3. In spherical coordinates, the variable r is the radial coordinate, $θ$ is the polar angle (relative to the vertical z-axis), and $ϕ$ is the azimuthal angle (relative to the x-axis). The relationship between spherical and rectangular coordinates is $x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ .$

An x y z coordinate system is shown, along with a point P and the vector r from the origin to P. In this figure, the point P has positive x, y, and z coordinates. The vector r is inclined by an angle theta from the positive z axis. Its projection on the x y plane makes an angle theta from the positive x axis toward the positive y axis.

Figure 8.3 The relationship between the spherical and rectangular coordinate systems.

The factor $r sin θ$ is the magnitude of a vector formed by the projection of the polar vector onto the xy-plane. Also, the coordinates of x and y are obtained by projecting this vector onto the x- and y-axes, respectively. The inverse transformation gives

r = \sqrt{x^{2} + y^{2} + z^{2}}, θ = {cos}^{−1} (\frac{z}{r}), ϕ = {cos}^{−1} (\frac{x}{\sqrt{x^{2} + y^{2}}}) .

Schrödinger’s wave equation for the hydrogen atom in spherical coordinates is discussed in more advanced courses in modern physics, so we do not consider it in detail here. However, due to the spherical symmetry of U(r), this equation reduces to three simpler equations: one for each of the three coordinates $(r, θ, and ϕ) .$ Solutions to the time-independent wave function are written as a product of three functions:

ψ (r, θ, ϕ) = R (r) Θ (θ) Φ (ϕ),

where R is the radial function dependent on the radial coordinate r only; $Θ$ is the polar function dependent on the polar coordinate $θ$ only; and $Φ$ is the phi function of $ϕ$ only. Valid solutions to Schrödinger’s equation $ψ (r, θ, ϕ)$ are labeled by the quantum numbers n, l, and m.

\begin{matrix} n : & principal quantum number \\ l : & angular momentum quantum number \\ m : & angular momentum projection quantum number \end{matrix}

(The reasons for these names will be explained in the next section.) The radial function R depends only on n and l; the polar function $Θ$ depends only on l and m; and the phi function $Φ$ depends only on m. The dependence of each function on quantum numbers is indicated with subscripts:

ψ_{n l m} (r, θ, ϕ) = R_{n l} (r) Θ_{l m} (θ) Φ_{m} (ϕ) .

Not all sets of quantum numbers (n, l, m) are possible. For example, the orbital angular quantum number l can never be greater or equal to the principal quantum number $n (l < n)$ . Specifically, we have

\begin{matrix} n & = & 1, 2, 3, … \\ l & = & 0, 1, 2, …, (n - 1) \\ m & = & − l, (− l + 1), …, 0, …, (+ l - 1), + l \end{matrix}

Notice that for the ground state, $n = 1$ , $l = 0$ , and $m = 0$ . In other words, there is only one quantum state with the wave function for $n = 1$ , and it is $ψ_{100}$ . However, for $n = 2$ , we have

\begin{matrix} l = 0, m = 0 \\ l = 1, m = −1, 0, 1 . \end{matrix}

Therefore, the allowed states for the $n = 2$ state are $ψ_{200}$ , $ψ_{21 - 1}, ψ_{210}$ , and $ψ_{211}$ . Example wave functions for the hydrogen atom are given in Table 8.1. Note that some of these expressions contain the letter i, which represents $\sqrt{−1}$ . When probabilities are calculated, these complex numbers do not appear in the final answer.

$n = 1, l = 0, m_{l} = 0$	$ψ_{100} = \frac{1}{\sqrt{π}} \frac{1}{a_{0}^{3 / 2}} e^{− r / a_{0}}$
$n = 2, l = 0, m_{l} = 0$	$ψ_{200} = \frac{1}{4 \sqrt{2 π}} \frac{1}{a_{0}^{3 / 2}} (2 - \frac{r}{a_{0}}) e^{− r / 2 a_{0}}$
$n = 2, l = 1, m_{l} = −1$	$ψ_{21 - 1} = \frac{1}{8 \sqrt{π}} \frac{1}{a_{0}^{3 / 2}} \frac{r}{a_{0}} e^{− r / 2 a_{0}} sin θ e^{− i ϕ}$
$n = 2, l = 1, m_{l} = 0$	$ψ_{210} = \frac{1}{4 \sqrt{2 π}} \frac{1}{a_{0}^{3 / 2}} \frac{r}{a_{0}} e^{− r / 2 a_{0}} cos θ$
$n = 2, l = 1, m_{l} = 1$	$ψ_{211} = \frac{1}{8 \sqrt{π}} \frac{1}{a_{0}^{3 / 2}} \frac{r}{a_{0}} e^{− r / 2 a_{0}} sin θ e^{i ϕ}$

Table 8.1 Wave Functions of the Hydrogen Atom

Physical Significance of the Quantum Numbers

Each of the three quantum numbers of the hydrogen atom (n, l, m) is associated with a different physical quantity. The principal quantum number n is associated with the total energy of the electron, $E_{n}$ . According to Schrödinger’s equation:

E_{n} = − (\frac{m_{e} k^{2} e^{4}}{{2 ħ}^{2}}) (\frac{1}{n^{2}}) = − E_{0} (\frac{1}{n^{2}}),

8.3

where $E_{0} = −13.6 eV .$ Notice that this expression is identical to that of Bohr’s model. As in the Bohr model, the electron in a particular state of energy does not radiate.

Example 8.1

How Many Possible States?

For the hydrogen atom, how many possible quantum states correspond to the principal number

n = 3

? What are the energies of these states?

Strategy

For a hydrogen atom of a given energy, the number of allowed states depends on its orbital angular momentum. We can count these states for each value of the principal quantum number,

n = 1, 2, 3 .

However, the total energy depends on the principal quantum number only, which means that we can use Equation 8.3 and the number of states counted.

Solution

If

n = 3

, the allowed values of l are 0, 1, and 2. If

l = 0

,

m = 0

(1 state). If

l = 1

,

m = - 1, 0, + 1

(3 states); and if

l = 2

,

m = - 2, - 1, 0, + 1, + 2

(5 states). In total, there are

1 + 3 + 5 = 9

allowed states. Because the total energy depends only on the principal quantum number,

n = 3

, the energy of each of these states is

E_{n 3} = − E_{0} (\frac{1}{n^{2}}) = \frac{−13.6 eV}{9} = −1.51 eV .

Significance

An electron in a hydrogen atom can occupy many different angular momentum states with the very same energy. As the orbital angular momentum increases, the number of the allowed states with the same energy increases.

The angular momentum orbital quantum number l is associated with the orbital angular momentum of the electron in a hydrogen atom. Quantum theory tells us that when the hydrogen atom is in the state $ψ_{n l m}$ , the magnitude of its orbital angular momentum is

L = \sqrt{l (l + 1)} ℏ,

8.4

where

l = 0, 1, 2, …, (n - 1) .

This result is slightly different from that found with Bohr’s theory, which quantizes angular momentum according to the rule $L ℏ = n, where n = 1, 2, 3, ... .$

Quantum states with different values of orbital angular momentum are distinguished using spectroscopic notation (Table 8.2). The designations s, p, d, and f result from early historical attempts to classify atomic spectral lines. (The letters stand for sharp, principal, diffuse, and fundamental, respectively.) After f, the letters continue alphabetically.

The ground state of hydrogen is designated as the 1s state, where “1” indicates the energy level $(n = 1)$ and “s” indicates the orbital angular momentum state ( $l = 0$ ). When $n = 2$ , l can be either 0 or 1. The $n = 2$ , $l = 0$ state is designated “2s.” The $n = 2$ , $l = 1$ state is designated “2p.” When $n = 3$ , l can be 0, 1, or 2, and the states are 3s, 3p, and 3d, respectively. Notation for other quantum states is given in Table 8.3.

The angular momentum projection quantum number m is associated with the azimuthal angle $ϕ$ (see Figure 8.3) and is related to the z-component of orbital angular momentum of an electron in a hydrogen atom. This component is given by

L_{z} = m ℏ,

8.5

where

m = − l, − l + 1, …, 0, …, + l - 1, l .

The z-component of angular momentum is related to the magnitude of angular momentum by

L_{z} = L c o s θ,

8.6

where $θ$ is the angle between the angular momentum vector and the z-axis. Note that the direction of the z-axis is determined by experiment—that is, along any direction, the experimenter decides to measure the angular momentum. For example, the z-direction might correspond to the direction of an external magnetic field. The relationship between $L_{z} and L$ is given in Figure 8.4.

An x y z coordinate system is shown. The vector L is at an angle theta to the positive z axis and has positive z component L sub z equal to m times h bar. The x and y components are positive but not specified.

Figure 8.4 The z-component of angular momentum is quantized with its own quantum number m.

Orbital Quantum Number l	Angular Momentum	State	Spectroscopic Name
0	0	s	Sharp
1	$\sqrt{2} h$	p	Principal
2	$\sqrt{6} h$	d	Diffuse
3	$\sqrt{12} h$	f	Fundamental
4	$\sqrt{20} h$	g
5	$\sqrt{30} h$	h

Table 8.2 Spectroscopic Notation and Orbital Angular Momentum

	$l = 0$	$l = 1$	$l = 2$	$l = 3$	$l = 4$	$l = 5$
$n = 1$	1s
$n = 2$	2s	2p
$n = 3$	3s	3p	3d
$n = 4$	4s	4p	4d	4f
$n = 5$	5s	5p	5d	5f	5g
$n = 6$	6s	6p	6d	6f	6g	6h

Table 8.3 Spectroscopic Description of Quantum States

The quantization of $L_{z}$ is equivalent to the quantization of $θ$ . Substituting $\sqrt{l (l + 1)} ℏ$ for L and m for $L_{z}$ into this equation, we find

m ℏ = \sqrt{l (l + 1)} ℏ cos θ .

8.7

Thus, the angle $θ$ is quantized with the particular values

θ = {cos}^{−1} (\frac{m}{\sqrt{l (l + 1)}}) .

8.8

Notice that both the polar angle ( $θ$ ) and the projection of the angular momentum vector onto an arbitrary z-axis ( $L_{z}$ ) are quantized.

The quantization of the polar angle for the $l = 3$ state is shown in Figure 8.5. The orbital angular momentum vector lies somewhere on the surface of a cone with an opening angle $θ$ relative to the z-axis (unless $m = 0,$ in which case $θ = 90 °$ and the vector points are perpendicular to the z-axis).

Seven vector, all of the same length L, are drawn at 7 different angles to the z axis. The z components of the vectors are indicated both by horizontal lines from the tip of the vector to the z axis and by labels on the z axis. For four of the vectors, the angle between the z axis and the vector is also labeled. The z component values are 3 h bar at angle theta sub three, 2 h bar at angle theta sub two, h bar at angle theta sub one, zero at angle theta sub zero, minus h bar, minus 2 h bar, and minus 3 h bar.

Figure 8.5 The quantization of orbital angular momentum. Each vector lies on the surface of a cone with axis along the z-axis.

A detailed study of angular momentum reveals that we cannot know all three components simultaneously. In the previous section, the z-component of orbital angular momentum has definite values that depend on the quantum number m. This implies that we cannot know both x- and y-components of angular momentum, $L_{x}$ and $L_{y}$ , with certainty. As a result, the precise direction of the orbital angular momentum vector is unknown.

Example 8.2

What Are the Allowed Directions?

Calculate the angles that the angular momentum vector

\vec{L}

can make with the z-axis for

l = 1

, as shown in Figure 8.6.

The image shows three possible values of a component of a given angular momentum along z-axis. The upper circular orbit is shown for m sub t = 1 at a distance L sub z above the origin. The vector L makes an angle of theta one with the z axis. The radius of the orbit is the component of L perpendicular to the z axis. The middle circular orbit is shown for m sub t = 0. It is in the x y plane. The vector L makes an angle of theta two of 90 degrees with the z axis. The radius of the orbit is L. The lower circular orbit is shown for m sub t = -1 at a distance L sub z below the origin. The vector L makes an angle of theta three with the z axis. The radius of the orbit is the component of L perpendicular to the z axis.

Figure 8.6 The component of a given angular momentum along the z-axis (defined by the direction of a magnetic field) can have only certain values. These are shown here for

l = 1

, for which

m = −1, 0, and + 1 .

The direction of

\vec{L}

is quantized in the sense that it can have only certain angles relative to the z-axis.