30.3

The Bohr Model of the Hydrogen Atom

In 1913 Bohr presented a model that led to equations such as Balmer’s for the wavelengths that the hydrogen atom radiates. Bohr’s theory begins with Rutherford’s picture of an atom as a nucleus surrounded by electrons moving in circular orbits. In his theory, Bohr made a number of assumptions and combined the new quantum ideas of Planck and Einstein with the traditional description of a particle in uniform circular motion.
Adopting Planck’s idea of quantized energy levels (see Section 29.2), Bohr hypothesized that in a hydrogen atom there can be only certain values of the total energy (electron kinetic energy plus potential energy). These allowed energy levels correspond to different orbits for the electron as it moves around the nucleus, the larger orbits being associated with larger total energies. Figure 30-5 illustrates two of the orbits. In addition, Bohr assumed that an electron in one of these orbits does not radiate electromagnetic waves. For this reason, the orbits are called stationary orbits or stationary states. Bohr recognized that radiationless orbits violated the laws of physics, as they were then known. But the assumption of such orbits was necessary because the traditional laws indicated that an electron radiates electromagnetic waves as it accelerates around a circular path, and the loss of the energy carried by the waves would lead to the collapse of the orbit.

Figure 30-5 In the Bohr model, a photon is emitted when the electron drops from a larger, higher-energy orbit (energy = Ei) to a smaller, lower-energy orbit (energy = Ef).

To incorporate Einstein’s photon concept (see Section 29.3), Bohr theorized that a photon is emitted only when the electron changes orbits from a larger one with a higher energy to a smaller one with a lower energy, as Figure 30-5 indicates. But how do electrons get into the higher-energy orbits in the first place? They get there by picking up energy when atoms collide, which happens more often when a gas is heated, or by acquiring energy when a high voltage is applied to a gas.
When an electron in an initial orbit with a larger energy Ei changes to a final orbit with a smaller energy Ef, the emitted photon has an energy of , consistent with the law of conservation of energy. But according to Einstein, the energy of a photon is hf, where f is its frequency and h is Planck’s constant. As a result, we find that

| (30.4)

Since the frequency of an electromagnetic wave is related to the wavelength by , Bohr could use Equation 30.4 to determine the wavelengths radiated by a hydrogen atom. First, however, he had to derive expressions for the energies Ei and Ef.

| The Energies and Radii of the Bohr Orbits

CONCEPTS AT A GLANCE To derive an expression for the quantized energy levels, Bohr brought together concepts from the older classical physics and the newer modern physics that emerged at the beginning of the twentieth century. The Concepts-at-a-Glance chart in Figure 30-6 shows the classical and modern physics that he merged to obtain the correct expression for the electron energy levels in atomic hydrogen.

Figure 30-6 CONCEPTS AT A GLANCE By combining ideas from classical and modern physics, Niels Bohr was able to obtain values for the quantized energy levels of the electron in a hydrogen atom. Here, Bohr (on the left) is shown with another famous physicist, Max Planck. (Margrethe Bohr Collection/American Institute of Physics/SPL/Photo Researchers, Inc.)

Let’s now bring together the concepts outlined in Figure 30-6 to derive Bohr’s result. For an electron of mass m and speed v in an orbit of radius r, the total energy is the kinetic energy of the electron plus the electric potential energy EPE. The potential energy is the product of the charge (−e) on the electron and the electric potential produced by the positive nuclear charge, in accord with Equation 19.3. We assume that the nucleus contains Z protons,* for a total nuclear charge of +Ze. The electric potential at a distance r from a point charge of +Ze is given as by Equation 19.6, where the constant k is . The electric potential energy is, then, . Consequently, the total energy E of the atom is

| (30.5)

But a centripetal force of magnitude mv2/r (Equation 5.3) acts on a particle in uniform circular motion. As Figure 30-7 indicates, the centripetal force is provided by the electrostatic force of attraction that the protons in the nucleus exert on the electron. According to Coulomb’s law (Equation 18.1), the magnitude of the electrostatic force is . Therefore, , or

| (30.6)

We can use this relation to eliminate the term mv2 from Equation 30.5, with the result that

| (30.7)

The total energy of the atom is negative because the negative electric potential energy is larger in magnitude than the positive kinetic energy.

Figure 30-7 In the Bohr model, the electron is in uniform circular motion around the nucleus. The centripetal force is the electrostatic force of attraction that the positive nuclear charge exerts on the electron.

A value for the radius r is needed, if Equation 30.7 is to be useful. To determine r, Bohr made an assumption about the orbital angular momentum of the electron. The magnitude L of the angular momentum is given by Equation 9.10 as , where is the moment of inertia of the electron moving on its circular path and is the angular speed of the electron in radians per second. Thus, the angular momentum is . Bohr conjectured that the angular momentum can assume only certain discrete values; in other words, L is quantized. He postulated that the allowed values are integer multiples of Planck’s constant divided by :

| (30.8)

Solving this equation for vn and substituting the result into Equation 30.6 lead to the following expression for the radius rn of the nth Bohr orbit:

| (30.9)

With , , , and , this expression reveals that

| (30.10)

Therefore, in the hydrogen atom (Z = 1) the smallest Bohr orbit (n = 1) has a radius of . This particular value is called the Bohr radius. Figure 30-8 shows the first three Bohr orbits for the hydrogen atom.

Figure 30-8 The first Bohr orbit in the hydrogen atom has a radius . The second and third Bohr orbits have radii r2 = 4r1 and r3 = 9r1, respectively.

The expression for the radius of a Bohr orbit can be substituted into Equation 30.7 to show that the corresponding total energy for the nth orbit is

| (30.11)

Substituting values for h, m, k, and e into this expression yields

| (30.12)

Often, atomic energies are expressed in electron volts rather than joules. Since , Equation 30.12 can be rewritten as

| (30.13)

| Energy Level Diagrams

It is useful to represent the energy values given by Equation 30.13 on an energy level diagram, as in Figure 30-9. In this diagram, which applies to the hydrogen atom (Z = 1), the highest energy level corresponds to in Equation 30.13 and has an energy of 0 eV. This is the energy of the atom when the electron is completely removed () from the nucleus and is at rest. In contrast, the lowest energy level corresponds to n = 1 and has a value of − 13.6 eV. The lowest energy level is called the ground state, to distinguish it from the higher levels, which are called excited states. Observe how the energies of the excited states come closer and closer together as n increases.
The electron in a hydrogen atom at room temperature spends most of its time in the ground state. To raise the electron from the ground state (n = 1) to the highest possible excited state (), 13.6 eV of energy must be supplied. Supplying this amount of energy removes the electron from the atom, producing the positive hydrogen ion H+. This is the minimum energy needed to remove the electron and is called the ionization energy. Thus, the Bohr model predicts that the ionization energy of atomic hydrogen is 13.6 eV, in excellent agreement with the experimental value. In Example 3 the Bohr model is applied to doubly ionized lithium.

Figure 30-9 Energy level diagram for the hydrogen atom.

| Example 3 | | |

The Ionization Energy of Li2+ The Bohr model does not apply when more than one electron orbits the nucleus because it does not account for the electrostatic force that one electron exerts on another. For instance, an electrically neutral lithium atom (Li) contains three electrons in orbit around a nucleus that includes three protons (Z = 3), and Bohr’s analysis is not applicable. However, the Bohr model can be used for the doubly charged positive ion of lithium (Li2+) that results when two electrons are removed from the neutral atom, leaving only one electron to orbit the nucleus. Obtain the ionization energy that is needed to remove the remaining electron from Li2+.
Reasoning The lithium ion Li2+ contains three times the positive nuclear charge that the hydrogen atom contains. Therefore, the orbiting electron is attracted more strongly to the nucleus in Li2+ than to the nucleus in the hydrogen atom. As a result, we expect that more energy is required to ionize Li2+ than the 13.6 eV required for atomic hydrogen.
Solution The Bohr energy levels for Li2+ are obtained from Equation 30.13 with Z = 3; . Therefore, the ground state (n = 1) energy is

Removing the electron from Li2+ requires 122 eV of energy: . This value for the ionization energy agrees well with the experimental value of 122.4 eV and, as expected, is greater than the 13.6 eV required for atomic hydrogen.

| The Line Spectra of the Hydrogen Atom

CONCEPTS AT A GLANCE To predict the wavelengths in the line spectrum of the hydrogen atom, Bohr combined his ideas about atoms (electron orbits are stationary orbits and the angular momentum of an electron is quantized) with Einstein’s idea of the photon. This synthesis is illustrated in the Concepts-at-a-Glance chart in Figure 30-10, which is an extension of the one in Figure 30-6.

Figure 30-10 CONCEPTS AT A GLANCE This chart extends the chart in Figure 30-6 and shows that Bohr was able to predict the line spectrum of the hydrogen atom by combining his quantized energy level expression with Einstein’s photon concept. The visible wavelengths in the line spectra of atoms are used to achieve special effects, such as the colors in these signs. (Veronica Jones/Taxi/Getty Images)

As applied by Bohr, the photon concept is inherent in Equation 30.4, , which states that the frequency f of the photon is proportional to the difference between two energy levels of the hydrogen atom. If we substitute Equation 30.11 for the total energies Ei and Ef into Equation 30.4 and recall from Equation 16.1 that , we obtain the following result:

| (30.14)

With the known values for h, m, k, e, and c, it can be seen that , in agreement with the Rydberg constant R that appears in Equations 30.1, 30.2 and 30.3. The agreement between the theoretical and experimental values of the Rydberg constant was a major accomplishment of Bohr’s theory.
With Z = 1 and nf = 1, Equation 30.14 reproduces Equation 30.1 for the Lyman series. Thus, Bohr’s model shows that the Lyman series of lines occurs when electrons make transitions from higher energy levels with ni = 2, 3, 4, … to the first energy level where nf = 1. Figure 30-11 shows these transitions. Notice that when an electron makes a transition from ni = 2 to nf = 1, the longest wavelength photon in the Lyman series is emitted, since the energy change is the smallest possible. When an electron makes a transition from the highest level where to the lowest level where nf = 1, the shortest wavelength is emitted, since the energy change is the largest possible. Since the higher energy levels are increasingly close together, the lines in the series become more and more crowded toward the short wavelength limit, as can be seen in Figure 30-4. Figure 30-11 also shows the energy level transitions for the Balmer series, where ni = 3, 4, 5,…, and nf = 2. In the Paschen series (see Figure 30-4) ni = 4, 5, 6,…, and nf = 3. The next example deals further with the line spectrum of the hydrogen atom.

Figure 30-11 The Lyman and Balmer series of lines in the hydrogen atom spectrum correspond to transitions that the electron makes between higher and lower energy levels, as indicated here.

| Example 4 | | | The Brackett Series for Atomic Hydrogen

In the line spectrum of atomic hydrogen there is also a group of lines known as the Brackett series. These lines are produced when electrons, excited to high energy levels, make transitions to the n = 4 level. Determine (a) the longest wavelength in this series and (b) the wavelength that corresponds to the transition from ni = 6 to nf = 4. (c) Refer to Figure 24-10 and identify the spectral region in which these lines are found.
Reasoning The longest wavelength corresponds to the transition that has the smallest energy change, which is between the ni = 5 and nf = 4 levels in Figure 30-9. The wavelength for this transition, as well as that for the transition from ni = 6 to nf = 4, can be obtained from Equation 30.14.
Solution

a. | Using Equation 30.14 with Z = 1, ni = 5, and nf = 4, we find that

b. | The calculation here is similar to that in part (a):

c. | According to Figure 24-10, these lines lie in the of the spectrum.

Problem solving insight In the line spectrum of atomic hydrogen, all lines in a given series (e.g., the Brackett series) are identified by a single value of the quantum number nf for the lower energy level into which an electron falls. Each line in a given series, however, corresponds to a different value of the quantum number ni for the higher energy level where an electron originates.

The various lines in the hydrogen atom spectrum are produced when electrons change from higher to lower energy levels and photons are emitted. Consequently, the spectral lines are called emission lines. Electrons can also make transitions in the reverse direction, from lower to higher levels, in a process known as absorption. In this case, an atom absorbs a photon that has precisely the energy needed to produce the transition. Thus, if photons with a continuous range of wavelengths pass through a gas and then are analyzed with a grating spectroscope, a series of dark absorption lines appears in the continuous spectrum. The dark lines indicate the wavelengths that have been removed by the absorption process. Such absorption lines can be seen in Figure 30-3 in the spectrum of the sun, where they are called Fraunhofer lines, after their discoverer. They are due to atoms, located in the outer and cooler layers of the sun, that absorb radiation coming from the interior. The interior portion of the sun is too hot for individual atoms to retain their structures and, therefore, emits a continuous spectrum of wavelengths.

The physics of absorption lines in the sun’s spectrum.

The Bohr model provides a great deal of insight into atomic structure. However, this model is now known to be oversimplified and has been superseded by a more detailed picture provided by quantum mechanics and the Schrödinger equation (see Section 30.5).

Need more practice?

Interactive LearningWare30.1
A stationary hydrogen atom () emits a photon by making a transition from the ni = 3 to the nf = 1 state, (a) What is the momentum of the emitted photon? (b)Determine the recoil speed of the hydrogen atom . Related Homework: Problem 18

Check Your Understanding 1

An electron in the hydrogen atom is in the n = 4 energy level. When this electron makes a transition to a lower energy level, is the wavelength of the photon emitted in (a) the Lyman series only, (b) the Balmer series only, (c) the Paschen series only, or (d) could it be in the Lyman, the Balmer, or the Paschen series? (The answer is given at the end of the book.)
Background: The Bohr model of the hydrogen atom explains the discrete nature of the wavelengths of the photons that can be emitted by the atom.
For similar questions (including calculational counterparts), consult Self-Assessment Test 30.1, which is described at the end of Section 30.5.

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