9

Atomic Spectra, Bohr Model

Bohr ModelAtomic Spectra
This line spectrum was simple enough that an empirical formula called the Rydberg formula could be derived from the spectrum. Three historically important paradoxes from the late 19th and early 20th centuries that could not be explained within the existing framework of classical mechanics and classical electromagnetism were the blackbody problem, the photoelectric effect, and the discrete spectra of atoms. The resolution of these paradoxes ultimately led to quantum theories that superseded the classical theories. Bohr incorporated Planck’s and Einstein’s quantization ideas into a model of the hydrogen atom that resolved the paradox of atom stability and discrete spectra. The Bohr model of the hydrogen atom explains the connection between the quantization of photons and the quantized emission from atoms. Bohr described the hydrogen atom in terms of an electron moving in a circular orbit about a nucleus. He postulated that the electron was restricted to certain orbits characterized by discrete energies. Transitions between these allowed orbits result in the absorption or emission of photons. When an electron moves from a higher-energy orbit to a more stable one, energy is emitted in the form of a photon. To move an electron from a stable orbit to a more excited one, a photon of energy must be absorbed. Using the Bohr model, we can calculate the energy of an electron and the radius of its orbit in any one-electron system.

9.1 Line Spectra

Another paradox within the classical electromagnetic theory that scientists in the late nineteenth century struggled with concerned the light emitted from atoms and molecules. When solids, liquids, or condensed gases are heated sufficiently, they radiate some of the excess energy as light. Photons produced in this manner have a range of energies, and thereby produce a continuous spectrum in which an unbroken series of wavelengths is present. Most of the light generated from stars (including our sun) is produced in this fashion. You can see all the visible wavelengths of light present in sunlight by using a prism to separate them. As can be seen in Figure 9.1, sunlight also contains UV light (shorter wavelengths) and IR light (longer wavelengths) that can be detected using instruments but that are invisible to the human eye.

Figure 9.1

The spectral distribution (light intensity vs. wavelength) of sunlight reaches the Earth's atmosphere as UV light, visible light, and IR light. The unabsorbed sunlight at the top of the atmosphere has a distribution that approximately matches the theoretical distribution of a blackbody at 5250 °C, represented by the blue curve. (credit: modification of work by American Society for Testing and Materials (ASTM) Terrestrial Reference Spectra for Photovoltaic Performance Evaluation)

A graph is shown with a horizontal axis labeled, “Wavelength ( n m ),” and a vertical axis labeled, “Spectral irradiance ( W divided by m superscript 2 divided by n m ).” The horizontal axis begins at 250 and extends to 4000 with markings provided every 250 n m. Similarly, the vertical axis begins at 0.00 and extends to 2.00 with markings every 0.25 units. Two vertical dashed lines are drawn. The first appears at about 400 nanometers and the second at nearly 700 nanometers. To the left of the first of these lines, the label, “U V,” appears at the top of the graph. Between these lines, the label, “Visible,” appears at the top of the graph. To the right of the second of these lines, the label, “Infrared,” appears at the top of the graph. A grey curve begins on the vertical axis at about 0.10. This curve increases steeply to a maximum value between the two vertical line segments of approximately 1.75 at about 625 nanometers. This curve decreases rapidly at first, then tapers off to reach a value of about 0 at the far right end of the graph. A golden colored curve traces along the same path as the grey curve, but shows a significant degree of variation in the region of the peak of the graph. In this general region, the gold curve is jagged and somewhat erratic. This curve reaches a maximum over 2.00 at around 475 nanometers. A key provided in the open space of the graph shows that the gold graph represents sunlight at the top of the atmosphere, and the grey curve represents the 5250 degrees C Blackbody spectrum.

Incandescent (glowing) solids such as tungsten filaments in incandescent lights also give off light that contains all wavelengths of visible light. These continuous spectra can often be approximated by blackbody radiation curves at some appropriate temperature, such as those shown in Figure 9.2.

Figure 9.2

Blackbody spectral distribution curves are shown for some representative temperatures.

A graph is shown with a horizontal axis labeled, “Wavelength lambda (micrometers)” and a vertical axis labeled, “Intensity I (a r b. units).” The horizontal axis begins at 0 and extends to 3.0 with markings provided every 0.1 micrometer. Similarly, the vertical axis begins at 0 and extends to 10 with markings every 1 unit. Two vertical dashed lines are drawn. The first appears at about 0.39 micrometers and the second at about 0.72 micrometers. To the left of the first of these lines, the label, “Ultraviolet,” appears at the top of the graph. Between these lines, the label, “Visible,” appears at the top of the graph. To the right of the second of these lines, the label, “Infrared,” appears at the top of the graph. To the far right of the graph in open space a purple dot is placed which is labeled, “lambda maximum.” A “Temperature” label is located in a central region of the graph. A blue curve begins on the horizontal axis at about 0.05 micrometers. This curve increases steeply to a maximum value between the two vertical line segments of approximately 9.5 at about 0.55 micrometers. This curve decreases rapidly at first, then tapers off to reach a value of about 1.5 at the far right end of the graph. This blue curve is labeled 6000 K beneath the “Temperature” label. Curves are similarly drawn in green for 5000 K, orange for 4000 K, and red for 3000 K. As the temperature decreases, the height of the peak is lower and shifted right on the graph. The maximum value for the green curve is around 4.5 at 7.2 micrometers. This curve tapers at the right end of the graph to a value around 0.6. The maximum for the orange curve is around 2 at about 0.9 micrometers. This curve tapers at the right end of the graph to a value around 0.2. The maximum for the red curve is around 0.7 at about 1.2 micrometers. This curve tapers at the right end of the graph to a value around 0.1. The entire region under the blue curve that is between the two dashed lines, indicating the visible region, is shaded with vertical bands of color. The colors extending left to right across this region are violet, indigo, blue, green, yellow, orange, and red. A purple dot is placed at the peak of each of the four colored curves. These peaks are connected with a dashed curve.

In contrast to continuous spectra, light can also occur as discrete or line spectra having very narrow line widths interspersed throughout the spectral regions such as those shown in Figure 9.4. Exciting a gas at low partial pressure using an electrical current, or heating it, will produce line spectra. Fluorescent light bulbs and neon signs operate in this way (Figure 9.3). Each element displays its own characteristic set of lines, as do molecules, although their spectra are generally much more complicated.

Figure 9.3

Neon signs operate by exciting a gas at low partial pressure using an electrical current. This sign shows the elaborate artistic effects that can be achieved. (credit: Dave Shaver)

This figure shows a colorful neon sign. The tubes are bent into various shapes.

Each emission line consists of a single wavelength of light, which implies that the light emitted by a gas consists of a set of discrete energies. For example, when an electric discharge passes through a tube containing hydrogen gas at low pressure, the H2 molecules are broken apart into separate H atoms and we see a blue-pink color. Passing the light through a prism produces a line spectrum, indicating that this light is composed of photons of four visible wavelengths, as shown in Figure 9.4.

Figure 9.4

Compare the two types of emission spectra: continuous spectrum of white light (top) and the line spectra of the light from excited sodium, hydrogen, calcium, and mercury atoms.

An image is shown with 5 rows. Across the top and bottom of the image is a scale that begins at 4000 angstroms at the left and extends to 740 angstroms at the far right. The top row is a continuous band of the visible spectrum, showing the colors from violet at the far left through indigo, blue, green, yellow, orange, and red at the far right. The second row, labeled, “N a,” shows the emission spectrum for the element sodium, which includes two narrow vertical bands in the blue range, two narrow bands in the yellow-green range, two narrow bands in the yellow range, and one narrow band in the orange range. The third row, labeled, “H,” shows the emission spectrum for hydrogen. This spectrum shows single bands in the violet, indigo, blue, and orange regions. The fourth row, labeled, “C a,” shows the emission spectrum for calcium. This spectrum shows bands in the following colors and frequencies; one violet, five indigo, one blue, two green, two yellow-green, one yellow, two yellow-orange, one orange, and one red. The fifth row, labeled, “H g,” shows the emission spectrum for mercury. This spectrum shows bands in the following colors and frequencies; two violet, one indigo, two blue, one green, two yellow, two orange, and one orange-red. It is important to note that each of the color bands for the emission spectra of the elements matches to a specific wavelength of light. Extending a vertical line from the bands to the scale above or below the diagram will match the band to a specific measurement on the scale.

The origin of discrete spectra in atoms and molecules was extremely puzzling to scientists in the late nineteenth century, since according to classical electromagnetic theory, only continuous spectra should be observed. Even more puzzling, in 1885, Johann Balmer was able to derive an empirical equation that related the four visible wavelengths of light emitted by hydrogen atoms to whole integers. That equation is the following one, in which k is a constant:

1λ=k(141n2),n=3,4,5,61λ=k(141n2),n=3,4,5,6

Other discrete lines for the hydrogen atom were found in the UV and IR regions. Johannes Rydberg generalized Balmer's work and developed an empirical formula that predicted all of hydrogen's emission lines, not just those restricted to the visible range, where, n1 and n2 are integers, n1 < n2, and RR is the Rydberg constant (1.097×107 m−1).

1λ=R(1n121n22)1λ=R(1n121n22)

Even in the late nineteenth century, spectroscopy was a very precise science, and so the wavelengths of hydrogen were measured to very high accuracy, which implied that the Rydberg constant could be determined very precisely as well. That such a simple formula as the Rydberg formula could account for such precise measurements seemed astounding at the time, but it was the eventual explanation for emission spectra by Neils Bohr in 1913 that ultimately convinced scientists to abandon classical physics and spurred the development of modern quantum mechanics.

Link to Supplemental Exercises

Supplemental exercises are available if you would like more practice with these concepts.

9.2 The Bohr Model

Learning Objectives

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

  • Describe the Bohr model of the hydrogen atom
  • Use the Rydberg equation to calculate energies of light emitted or absorbed by hydrogen atoms

Following the work of Ernest Rutherford and his colleagues in the early twentieth century, the picture of atoms consisting of tiny dense nuclei surrounded by lighter and even tinier electrons continually moving about the nucleus was well established. This picture was called the planetary model, since it pictured the atom as a miniature “solar system” with the electrons orbiting the nucleus like planets orbiting the sun. The simplest atom is hydrogen, consisting of a single proton as the nucleus about which a single electron moves. The electrostatic force attracting the electron to the proton depends only on the distance between the two particles. This classical mechanics description of the atom is incomplete, however, since an electron moving in an elliptical orbit would be accelerating (by changing direction) and, according to classical electromagnetism, it should continuously emit electromagnetic radiation. This loss in orbital energy should result in the electron’s orbit getting continually smaller until it spirals into the nucleus, implying that atoms are inherently unstable.

In 1913, Niels Bohr attempted to resolve the atomic paradox by ignoring classical electromagnetism’s prediction that the orbiting electron in hydrogen would continuously emit light. Instead, he incorporated into the classical mechanics description of the atom Planck’s ideas of quantization and Einstein’s finding that light consists of photons whose energy is proportional to their frequency. Bohr assumed that the electron orbiting the nucleus would not normally emit any radiation (the stationary state hypothesis), but it would emit or absorb a photon if it moved to a different orbit. The energy absorbed or emitted would reflect differences in the orbital energies according to this equation:

ΔE=EfEi=hν=hcλΔE=EfEi=hν=hcλ

In this equation, h is Planck’s constant and Ei and Ef are the initial and final orbital energies, respectively. The absolute value of the energy difference is used, since frequencies and wavelengths are always positive. Instead of allowing for continuous values of energy, Bohr assumed the energies of these electron orbitals were quantized:

En=kn2,n=1,2,3,En=kn2,n=1,2,3,

In this expression, k is a constant comprising fundamental constants such as the electron mass and charge and Planck’s constant. Inserting the expression for the orbit energies into the equation for ΔE gives

ΔE=k(1n121n22)=hcλΔE=k(1n121n22)=hcλ

or

1λ=khc(1n121n22)1λ=khc(1n121n22)

which is identical to the Rydberg equation in which R=khc.R=khc. When Bohr calculated his theoretical value for the Rydberg constant, R,R, and compared it with the experimentally accepted value, he got excellent agreement. Since the Rydberg constant was one of the most precisely measured constants at that time, this level of agreement was astonishing and meant that Bohr’s model was taken seriously, despite the many assumptions that Bohr needed to derive it.

The lowest few energy levels are shown in Figure 9.5. One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n = 1 orbit, the orbit in which it has the lowest energy. When the electron is in this lowest energy orbit, the atom is said to be in its ground electronic state (or simply ground state). If the atom receives energy from an outside source, it is possible for the electron to move to an orbit with a higher n value and the atom is now in an excited electronic state (or simply an excited state) with a higher energy. When an electron transitions from an excited state (higher energy orbit) to a less excited state, or ground state, the difference in energy is emitted as a photon. Similarly, if a photon is absorbed by an atom, the energy of the photon moves an electron from a lower energy orbit up to a more excited one. We can relate the energy of electrons in atoms to what we learned previously about energy. The law of conservation of energy says that we can neither create nor destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, that same amount of energy will be liberated when the electron returns to its initial state (Figure 9.6).

Since Bohr’s model involved only a single electron, it could also be applied to the single electron ions He+, Li2+, Be3+, and so forth, which differ from hydrogen only in their nuclear charges, and so one-electron atoms and ions are collectively referred to as hydrogen-like atoms. The energy expression for hydrogen-like atoms is a generalization of the hydrogen atom energy, in which Z is the nuclear charge (+1 for hydrogen, +2 for He, +3 for Li, and so on) and k has a value of 2.179×10–18 J.

En=kZ2n2En=kZ2n2

The sizes of the circular orbits for hydrogen-like atoms are given in terms of their radii by the following expression, in which a0a0 is a constant called the Bohr radius, with a value of 5.292×10−11 m:

r=n2Za0r=n2Za0

The equation also shows us that as the electron’s energy increases (as n increases), the electron is found at greater distances from the nucleus. This is implied by the inverse dependence of electrostatic attraction on distance, since, as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases and it is held less tightly in the atom. Note that as n gets larger and the orbits get larger, their energies get closer to zero, and so the limits nn and rr imply that E = 0 corresponds to the ionization limit where the electron is completely removed from the nucleus. Thus, for hydrogen in the ground state n = 1, the ionization energy would be:

ΔE=EnE1=0+k=kΔE=EnE1=0+k=k

With three extremely puzzling paradoxes now solved (blackbody radiation, the photoelectric effect, and the hydrogen atom), and all involving Planck’s constant in a fundamental manner, it became clear to most physicists at that time that the classical theories that worked so well in the macroscopic world were fundamentally flawed and could not be extended down into the microscopic domain of atoms and molecules. Unfortunately, despite Bohr’s remarkable achievement in deriving a theoretical expression for the Rydberg constant, he was unable to extend his theory to the next simplest atom, He, which only has two electrons. Bohr’s model was severely flawed, since it was still based on the classical mechanics notion of precise orbits, a concept that was later found to be untenable in the microscopic domain, when a proper model of quantum mechanics was developed to supersede classical mechanics.

Figure 9.5

Quantum numbers and energy levels in a hydrogen atom. The more negative the calculated value, the lower the energy.

The figure includes a diagram representing the relative energy levels of the quantum numbers of the hydrogen atom. An upward pointing arrow at the left of the diagram is labeled, “E.” A grey shaded vertically-oriented rectangle is placed just right of the arrow. The rectangle height matches the arrow length. Colored horizontal line segments are placed inside the rectangle and labels are placed to the right of the box and arranged in a column with the heading, “Energy, n.” At the very base of the rectangle, a purple horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 2.18 times 10 superscript negative 18 J, 1.” At a level approximately three-quarters of the distance to the top of the rectangle, a blue horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 5.45 times 10 superscript negative 19 J, 2.” At a level approximately seven-eighths the distance from the base of the rectangle, a green horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 2.42 times 10 superscript negative 19 J, 3.” Just a short distance above this segment, an orange horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 1.36 times 10 superscript negative 19 J, 4.” Just above this segment, a red horizontal line segment is drawn. A black line segment extends to the right to the label, “negative 8.72 times 10 superscript negative 20 J, 5.” Just a short distance above this segment, a brown horizontal line segment is drawn. A black line segment extends to the right to the label, “0.00 J, infinity.”

Example 9.1

Calculating the Energy of an Electron in a Bohr Orbit

Early researchers were very excited when they were able to predict the energy of an electron at a particular distance from the nucleus in a hydrogen atom. If a spark promotes the electron in a hydrogen atom into an orbit with n = 3, what is the calculated energy, in joules, of the electron?

Solution

The energy of the electron is given by this equation:
E=kZ2n2E=kZ2n2

The atomic number, Z, of hydrogen is 1; k = 2.179 ×× 10–18 J; and the electron is characterized by an n value of 3. Thus,

E=(2.179×10−18J)×(1)2(3)2=−2.421×10−19JE=(2.179×10−18J)×(1)2(3)2=−2.421×10−19J

Check Your Learning

The electron in Figure 9.6 is promoted even further to an orbit with n = 6. What is its new energy?

−6.053 ×× 10–20 J

Figure 9.6

The horizontal lines show the relative energy of orbits in the Bohr model of the hydrogen atom, and the vertical arrows depict the energy of photons absorbed (left) or emitted (right) as electrons move between these orbits.

The figure includes a diagram representing the relative energy levels of the quantum numbers of the hydrogen atom. An upward pointing arrow at the left of the diagram is labeled, “E.” A grey shaded vertically oriented rectangle is placed just right of the arrow. The rectangle height matches the arrow length. Colored, horizontal line segments are placed inside the rectangle and labels are placed to the right of the box, arranged in a column with the heading, “Energy, n.” At the very base of the rectangle, a purple horizontal line segment is drawn. A black line extends to the right to the label, “1.” At a level approximately three-quarters of the distance to the top of the rectangle, a blue horizontal line segment is drawn. A black line extends to the right to the label, “2.” At a level approximately seven-eighths the distance from the base of the rectangle, a green horizontal line segment is drawn. A black line extends to the right to the label, “3.” Just a short distance above this segment, an orange horizontal line segment is drawn. A black line segment extends to the right to the label, “4.” Just above this segment, a red horizontal line segment is drawn. A black line extends to the right to the label, “5.” Just a short distance above this segment, a brown horizontal line segment is drawn. A black line extends to the right to the label, “infinity.” Arrows are drawn to depict energies of photons absorbed, as shown by upward pointing arrows on the left, or released as shown by downward pointing arrows on the right side of the diagram between the colored line segments. The label, “Electron moves to higher energy as light is absorbed,” is placed beneath the upward pointing arrows. Similarly, the label, “Electron moves to lower energy as light is emitted,” appears beneath the downward pointing arrows. Moving left to right across the diagram, arrows extend from one colored line segment to the next in the following order: purple to blue, purple to green, purple to orange, purple to red, purple to brown, blue to green, blue to orange, and blue to red. The arrows originating from the same colored segment are grouped together by close placement of the arrows. Similarly, the downward arrows follow in this sequence; brown to purple, red to purple, orange to purple, green to purple, blue to purple, red to blue, orange to blue, and green to blue. Arrows are again grouped by close placement according to the color at which the arrows end.

Example 9.2

Calculating the Energy and Wavelength of Electron Transitions in a One–electron (Bohr) System

What is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 4 to the orbit with n = 6? In what part of the electromagnetic spectrum do we find this radiation?

Solution

In this case, the electron starts out with n = 4, so n1 = 4. It comes to rest in the n = 6 orbit, so n2 = 6. The difference in energy between the two states is given by this expression:
ΔE=E1E2=2.179×10−18(1n121n22)ΔE=E1E2=2.179×10−18(1n121n22)
ΔE=2.179×10−18(142162)JΔE=2.179×10−18(142162)J
ΔE=2.179×10−18(116136)JΔE=2.179×10−18(116136)J
ΔE=7.566×10−20JΔE=7.566×10−20J

This energy difference is positive, indicating a photon enters the system (is absorbed) to excite the electron from the n = 4 orbit up to the n = 6 orbit. The wavelength of a photon with this energy is found by the expression E=hcλ.E=hcλ. Rearrangement gives:

λ=hcEλ=hcE
=(6.626×10−34Js)×2.998×108ms−17.566×10−20J=2.626×10−6m=(6.626×10−34Js)×2.998×108ms−17.566×10−20J=2.626×10−6m

From the illustration of the electromagnetic spectrum in Electromagnetic Energy, we can see that this wavelength is found in the infrared portion of the electromagnetic spectrum.

Check Your Learning

What is the energy in joules and the wavelength in meters of the photon produced when an electron falls from the n = 5 to the n = 3 level in a He+ ion (Z = 2 for He+)?

6.198 ×× 10–19 J; 3.205 ×× 10−7 m

Bohr’s model of the hydrogen atom provides insight into the behavior of matter at the microscopic level, but it does not account for electron–electron interactions in atoms with more than one electron. It does introduce several important features of all models used to describe the distribution of electrons in an atom. These features include the following:

Of these features, the most important is the postulate of quantized energy levels for an electron in an atom. As a consequence, the model laid the foundation for the quantum mechanical model of the atom. Bohr won a Nobel Prize in Physics for his contributions to our understanding of the structure of atoms and how that is related to line spectra emissions.

Link to Supplemental Exercises

Supplemental exercises are available if you would like more practice with these concepts.

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Previous Citation(s)
Flowers, P., Neth, E. J., Robinson, W. R., Theopold, K., & Langley, R. (2019). Chemistry in Context. In Chemistry: Atoms First 2e. OpenStax. https://openstax.org/books/chemistry-atoms-first-2e/pages/3-introduction

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