bohr experiment hydrogen

30.3 Bohr’s Theory of the Hydrogen Atom

Learning objectives.

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

Describe the mysteries of atomic spectra.
Explain Bohr’s theory of the hydrogen atom.
Explain Bohr’s planetary model of the atom.
Illustrate energy state using the energy-level diagram.
Describe the triumphs and limits of Bohr’s theory.

The great Danish physicist Niels Bohr (1885–1962) made immediate use of Rutherford’s planetary model of the atom. ( Figure 30.13 ). Bohr became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on the planetary model of the atom. For decades, many questions had been asked about atomic characteristics. From their sizes to their spectra, much was known about atoms, but little had been explained in terms of the laws of physics. Bohr’s theory explained the atomic spectrum of hydrogen and established new and broadly applicable principles in quantum mechanics.

Mysteries of Atomic Spectra

As noted in Quantization of Energy , the energies of some small systems are quantized. Atomic and molecular emission and absorption spectra have been known for over a century to be discrete (or quantized). (See Figure 30.14 .) Maxwell and others had realized that there must be a connection between the spectrum of an atom and its structure, something like the resonant frequencies of musical instruments. But, in spite of years of efforts by many great minds, no one had a workable theory. (It was a running joke that any theory of atomic and molecular spectra could be destroyed by throwing a book of data at it, so complex were the spectra.) Following Einstein’s proposal of photons with quantized energies directly proportional to their wavelengths, it became even more evident that electrons in atoms can exist only in discrete orbits.

In some cases, it had been possible to devise formulas that described the emission spectra. As you might expect, the simplest atom—hydrogen, with its single electron—has a relatively simple spectrum. The hydrogen spectrum had been observed in the infrared (IR), visible, and ultraviolet (UV), and several series of spectral lines had been observed. (See Figure 30.15 .) These series are named after early researchers who studied them in particular depth.

The observed hydrogen-spectrum wavelengths can be calculated using the following formula:

where λ λ is the wavelength of the emitted EM radiation and R R is the Rydberg constant , determined by the experiment to be

The constant n f n f is a positive integer associated with a specific series. For the Lyman series, n f = 1 n f = 1 ; for the Balmer series, n f = 2 n f = 2 ; for the Paschen series, n f = 3 n f = 3 ; and so on. The Lyman series is entirely in the UV, while part of the Balmer series is visible with the remainder UV. The Paschen series and all the rest are entirely IR. There are apparently an unlimited number of series, although they lie progressively farther into the infrared and become difficult to observe as n f n f increases. The constant n i n i is a positive integer, but it must be greater than n f n f . Thus, for the Balmer series, n f = 2 n f = 2 and n i = 3, 4, 5, 6, ... n i = 3, 4, 5, 6, ... . Note that n i n i can approach infinity. While the formula in the wavelengths equation was just a recipe designed to fit data and was not based on physical principles, it did imply a deeper meaning. Balmer first devised the formula for his series alone, and it was later found to describe all the other series by using different values of n f n f . Bohr was the first to comprehend the deeper meaning. Again, we see the interplay between experiment and theory in physics. Experimentally, the spectra were well established, an equation was found to fit the experimental data, but the theoretical foundation was missing.

Example 30.1

Calculating wave interference of a hydrogen line.

What is the distance between the slits of a grating that produces a first-order maximum for the second Balmer line at an angle of 15º 15º ?

Strategy and Concept

For an Integrated Concept problem, we must first identify the physical principles involved. In this example, we need to know (a) the wavelength of light as well as (b) conditions for an interference maximum for the pattern from a double slit. Part (a) deals with a topic of the present chapter, while part (b) considers the wave interference material of Wave Optics .

Solution for (a)

Hydrogen spectrum wavelength . The Balmer series requires that n f = 2 n f = 2 . The first line in the series is taken to be for n i = 3 n i = 3 , and so the second would have n i = 4 n i = 4 .

The calculation is a straightforward application of the wavelength equation. Entering the determined values for n f n f and n i n i yields

Inverting to find λ λ gives

Discussion for (a)

This is indeed the experimentally observed wavelength, corresponding to the second (blue-green) line in the Balmer series. More impressive is the fact that the same simple recipe predicts all of the hydrogen spectrum lines, including new ones observed in subsequent experiments. What is nature telling us?

Solution for (b)

where d d is the distance between slits and θ θ is the angle from the original direction of the beam. The number m m is the order of the interference; m = 1 m = 1 in this example. Solving for d d and entering known values yields

Discussion for (b)

This number is similar to those used in the interference examples of Introduction to Quantum Physics (and is close to the spacing between slits in commonly used diffraction glasses).

Bohr’s Solution for Hydrogen

Bohr was able to derive the formula for the hydrogen spectrum using basic physics, the planetary model of the atom, and some very important new proposals. His first proposal is that only certain orbits are allowed: we say that the orbits of electrons in atoms are quantized . Each orbit has a different energy, and electrons can move to a higher orbit by absorbing energy and drop to a lower orbit by emitting energy. If the orbits are quantized, the amount of energy absorbed or emitted is also quantized, producing discrete spectra. Photon absorption and emission are among the primary methods of transferring energy into and out of atoms. The energies of the photons are quantized, and their energy is explained as being equal to the change in energy of the electron when it moves from one orbit to another. In equation form, this is

Here, Δ E Δ E is the change in energy between the initial and final orbits, and hf hf is the energy of the absorbed or emitted photon. It is quite logical (that is, expected from our everyday experience) that energy is involved in changing orbits. A blast of energy is required for the space shuttle, for example, to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. This is not observed for satellites or planets, which can have any orbit given the proper energy. (See Figure 30.16 .)

Figure 30.17 shows an energy-level diagram , a convenient way to display energy states. In the present discussion, we take these to be the allowed energy levels of the electron. Energy is plotted vertically with the lowest or ground state at the bottom and with excited states above. Given the energies of the lines in an atomic spectrum, it is possible (although sometimes very difficult) to determine the energy levels of an atom. Energy-level diagrams are used for many systems, including molecules and nuclei. A theory of the atom or any other system must predict its energies based on the physics of the system.

where L L is the angular momentum, m e m e is the electron’s mass, r n r n is the radius of the n n th orbit, and h h is Planck’s constant. Note that angular momentum is L = Iω L = Iω . For a small object at a radius r , I = mr 2 r , I = mr 2 and ω = v / r ω = v / r , so that L = mr 2 v / r = mvr L = mr 2 v / r = mvr . Quantization says that this value of mvr mvr can only be equal to h / 2, 2 h / 2, 3 h / 2 h / 2, 2 h / 2, 3 h / 2 , etc. At the time, Bohr himself did not know why angular momentum should be quantized, but using this assumption he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time.

From Bohr’s assumptions, we will now derive a number of important properties of the hydrogen atom from the classical physics we have covered in the text. We start by noting the centripetal force causing the electron to follow a circular path is supplied by the Coulomb force. To be more general, we note that this analysis is valid for any single-electron atom. So, if a nucleus has Z Z protons ( Z = 1 Z = 1 for hydrogen, 2 for helium, etc.) and only one electron, that atom is called a hydrogen-like atom . The spectra of hydrogen-like ions are similar to hydrogen, but shifted to higher energy by the greater attractive force between the electron and nucleus. The magnitude of the centripetal force is m e v 2 / r n m e v 2 / r n , while the Coulomb force is k Zq e q e / r n 2 k Zq e q e / r n 2 . The tacit assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. This is consistent with the planetary model of the atom. Equating these,

Angular momentum quantization is stated in an earlier equation. We solve that equation for v v , substitute it into the above, and rearrange the expression to obtain the radius of the orbit. This yields:

where a B a B is defined to be the Bohr radius , since for the lowest orbit n = 1 n = 1 and for hydrogen Z = 1 Z = 1 , r 1 = a B r 1 = a B . It is left for this chapter’s Problems and Exercises to show that the Bohr radius is

These last two equations can be used to calculate the radii of the allowed (quantized) electron orbits in any hydrogen-like atom . It is impressive that the formula gives the correct size of hydrogen, which is measured experimentally to be very close to the Bohr radius. The earlier equation also tells us that the orbital radius is proportional to n 2 n 2 , as illustrated in Figure 30.18 .

To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy:

Kinetic energy is the familiar KE = 1 / 2 m e v 2 KE = 1 / 2 m e v 2 , assuming the electron is not moving at relativistic speeds. Potential energy for the electron is electrical, or PE = q e V PE = q e V , where V V is the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge Zq e Zq e ; thus, V = kZq e / r n V = kZq e / r n , recalling an earlier equation for the potential due to a point charge. Since the electron’s charge is negative, we see that PE = − kZq e / r n PE = − kZq e / r n . Entering the expressions for KE KE and PE PE , we find

Now we substitute r n r n and v v from earlier equations into the above expression for energy. Algebraic manipulation yields

for the orbital energies of hydrogen-like atoms . Here, E 0 E 0 is the ground-state energy n = 1 n = 1 for hydrogen Z = 1 Z = 1 and is given by

Thus, for hydrogen,

Figure 30.19 shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.

Electron total energies are negative, since the electron is bound to the nucleus, analogous to being in a hole without enough kinetic energy to escape. As n n approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since r n r n gets very large for large n n , and the electric potential energy thus becomes zero. Thus, 13.6 eV is needed to ionize hydrogen (to go from –13.6 eV to 0, or unbound), an experimentally verified number. Given more energy, the electron becomes unbound with some kinetic energy. For example, giving 15.0 eV to an electron in the ground state of hydrogen strips it from the atom and leaves it with 1.4 eV of kinetic energy.

Finally, let us consider the energy of a photon emitted in a downward transition, given by the equation to be

Substituting E n = ( – 13.6 eV / n 2 ) E n = ( – 13.6 eV / n 2 ) , we see that

Dividing both sides of this equation by hc hc gives an expression for 1 / λ 1 / λ :

It can be shown that

is the Rydberg constant . Thus, we have used Bohr’s assumptions to derive the formula first proposed by Balmer years earlier as a recipe to fit experimental data.

We see that Bohr’s theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to 1 / n 2 1 / n 2 , where n n is a non-negative integer. A downward transition releases energy, and so n i n i must be greater than n f n f . The various series are those where the transitions end on a certain level. For the Lyman series, n f = 1 n f = 1 — that is, all the transitions end in the ground state (see also Figure 30.19 ). For the Balmer series, n f = 2 n f = 2 , or all the transitions end in the first excited state; and so on. What was once a recipe is now based in physics, and something new is emerging—angular momentum is quantized.

Triumphs and Limits of the Bohr Theory

Bohr did what no one had been able to do before. Not only did he explain the spectrum of hydrogen, he correctly calculated the size of the atom from basic physics. Some of his ideas are broadly applicable. Electron orbital energies are quantized in all atoms and molecules. Angular momentum is quantized. The electrons do not spiral into the nucleus, as expected classically (accelerated charges radiate, so that the electron orbits classically would decay quickly, and the electrons would sit on the nucleus—matter would collapse). These are major triumphs.

But there are limits to Bohr’s theory. It cannot be applied to multielectron atoms, even one as simple as a two-electron helium atom. Bohr’s model is what we call semiclassical . The orbits are quantized (nonclassical) but are assumed to be simple circular paths (classical). As quantum mechanics was developed, it became clear that there are no well-defined orbits; rather, there are clouds of probability. Bohr’s theory also did not explain that some spectral lines are doublets (split into two) when examined closely. We shall examine many of these aspects of quantum mechanics in more detail, but it should be kept in mind that Bohr did not fail. Rather, he made very important steps along the path to greater knowledge and laid the foundation for all of atomic physics that has since evolved.

PhET Explorations

Models of the hydrogen atom.

How did scientists figure out the structure of atoms without looking at them? Try out different models by shooting light at the atom. Check how the prediction of the model matches the experimental results.

Click to view content .

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units

Authors: Paul Peter Urone, Roger Hinrichs
Publisher/website: OpenStax
Book title: College Physics 2e
Publication date: Jul 13, 2022
Location: Houston, Texas
Book URL: https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
Section URL: https://openstax.org/books/college-physics-2e/pages/30-3-bohrs-theory-of-the-hydrogen-atom

© Jul 9, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Chapter 30 Atomic Physics

30.3 Bohr’s Theory of the Hydrogen Atom

Learning objectives.

Describe the mysteries of atomic spectra.
Explain Bohr’s theory of the hydrogen atom.
Explain Bohr’s planetary model of the atom.
Illustrate energy state using the energy-level diagram.
Describe the triumphs and limits of Bohr’s theory.

The great Danish physicist Niels Bohr (1885–1962) made immediate use of Rutherford’s planetary model of the atom. ( Figure 1 ). Bohr became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on the planetary model of the atom. For decades, many questions had been asked about atomic characteristics. From their sizes to their spectra, much was known about atoms, but little had been explained in terms of the laws of physics. Bohr’s theory explained the atomic spectrum of hydrogen and established new and broadly applicable principles in quantum mechanics.

Mysteries of Atomic Spectra

As noted in Chapter 29.1 Quantization of Energy , the energies of some small systems are quantized. Atomic and molecular emission and absorption spectra have been known for over a century to be discrete (or quantized). (See Figure 2 .) Maxwell and others had realized that there must be a connection between the spectrum of an atom and its structure, something like the resonant frequencies of musical instruments. But, in spite of years of efforts by many great minds, no one had a workable theory. (It was a running joke that any theory of atomic and molecular spectra could be destroyed by throwing a book of data at it, so complex were the spectra.) Following Einstein’s proposal of photons with quantized energies directly proportional to their wavelengths, it became even more evident that electrons in atoms can exist only in discrete orbits.

This figure has two parts. Part a shows a discharge tube at the extreme left. Light from the discharge tube passes through a rectangular slit and a grating, going from left to right. From the grating, light of different colors falls on a photographic film. Part b of the figure shows the emission line spectrum for iron.

In some cases, it had been possible to devise formulas that described the emission spectra. As you might expect, the simplest atom—hydrogen, with its single electron—has a relatively simple spectrum. The hydrogen spectrum had been observed in the infrared (IR), visible, and ultraviolet (UV), and several series of spectral lines had been observed. (See Figure 3 .) These series are named after early researchers who studied them in particular depth.

The observed hydrogen-spectrum wavelengths can be calculated using the following formula:

where [latex]{\lambda}[/latex] is the wavelength of the emitted EM radiation and [latex]{R}[/latex] is the Rydberg constant , determined by the experiment to be

The constant nfnf is a positive integer associated with a specific series. For the Lyman series, [latex]{n_{\text{f}} = 1}[/latex]; for the Balmer series, [latex]{n_{\text{f}} = 2}[/latex]; for the Paschen series, [latex]{n_{\text{f}}=3}[/latex]; and so on. The Lyman series is entirely in the UV, while part of the Balmer series is visible with the remainder UV. The Paschen series and all the rest are entirely IR. There are apparently an unlimited number of series, although they lie progressively farther into the infrared and become difficult to observe as nfnf increases. The constant nini is a positive integer, but it must be greater than nfnf. Thus, for the Balmer series, [latex]{n_{\text{f}} = 2}[/latex] and [latex]{n_{\text{i}} = 3, 4, 5, 6, \dots}[/latex]. Note that [latex]text{n_{\text{i}}}[/latex] can approach infinity. While the formula in the wavelengths equation was just a recipe designed to fit data and was not based on physical principles, it did imply a deeper meaning. Balmer first devised the formula for his series alone, and it was later found to describe all the other series by using different values of [latex]{n_{\text{f}}}[/latex]. Bohr was the first to comprehend the deeper meaning. Again, we see the interplay between experiment and theory in physics. Experimentally, the spectra were well established, an equation was found to fit the experimental data, but the theoretical foundation was missing.

The figure shows three horizontal lines at small distances from each other. Between the two lower lines, the Lyman series, with four vertical red bands in compact form, is shown. The value of the constant n sub f is 1 and the wavelengths are ninety-one nanometers to one hundred nanometers. The Balmer series is shown to the right side of this series. The value of the constant n sub f is two, and the range of wavelengths is from three hundred sixty five to six hundred fifty six nanometers. At the right side of this, the Paschen series bands are shown. The value of the constant n sub f is three, and the range of the wavelengths is from eight hundred twenty nanometers to one thousand eight hundred and seventy five nanometers.

Calculating Wave Interference of a Hydrogen Line

What is the distance between the slits of a grating that produces a first-order maximum for the second Balmer line at an angle of [latex]{15 ^{\circ}}[/latex]?

Strategy and Concept

Solution for (a)

Hydrogen spectrum wavelength . The Balmer series requires that [latex]{n_{\text{f}} = 2}[/latex]. The first line in the series is taken to be for [latex]{n_{\text{i}}=3}[/latex], and so the second would have [latex]{n_{\text{i}}=4}[/latex].

The calculation is a straightforward application of the wavelength equation. Entering the determined values for [latex]{n_{\text{f}}}[/latex] and [latex]{n_{\text{i}}}[/latex] yields

Inverting to find [latex]{\lambda}[/latex] gives

Discussion for (a)

Solution for (b)

Double-slit interference ( Chapter 27 Wave Optics ). To obtain constructive interference for a double slit, the path length difference from two slits must be an integral multiple of the wavelength. This condition was expressed by the equation

where [latex]{d}[/latex] is the distance between slits and [latex]{\theta}[/latex] is the angle from the original direction of the beam. The number [latex]{m}[/latex] is the order of the interference; [latex]{m = 1}[/latex] in this example. Solving for [latex]{d}[/latex] and entering known values yields

Discussion for (b)

This number is similar to those used in the interference examples of Chapter 29 Introduction to Quantum Physics (and is close to the spacing between slits in commonly used diffraction glasses).

Bohr’s Solution for Hydrogen

Here, [latex]{\Delta E}[/latex] is the change in energy between the initial and final orbits, and [latex]{hf}[/latex] is the energy of the absorbed or emitted photon. It is quite logical (that is, expected from our everyday experience) that energy is involved in changing orbits. A blast of energy is required for the space shuttle, for example, to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. This is not observed for satellites or planets, which can have any orbit given the proper energy. (See Figure 4 .)

The orbits of Bohr’s planetary model of an atom; five concentric circles are shown. The radii of the circles increase from innermost to outermost circles. On the circles, labels E sub one, E sub two, up to E sub i are marked.

Figure 5 shows an energy-level diagram , a convenient way to display energy states. In the present discussion, we take these to be the allowed energy levels of the electron. Energy is plotted vertically with the lowest or ground state at the bottom and with excited states above. Given the energies of the lines in an atomic spectrum, it is possible (although sometimes very difficult) to determine the energy levels of an atom. Energy-level diagrams are used for many systems, including molecules and nuclei. A theory of the atom or any other system must predict its energies based on the physics of the system.

The energy level diagram is shown. A number of horizontal lines are shown. The lines are labeled from bottom to top as n is equal to one, n is equal to two and so on up to n equals infinity; the energy levels increase from bottom to top. The distance between the lines decreases from the bottom line to the top line. A vertical arrow shows an electron transitioning from n equals four to n equals two.

where [latex]{L}[/latex] is the angular momentum, [latex]{m_e}[/latex] is the electron’s mass, [latex]{r_n}[/latex] is the radius of the [latex]{n}[/latex] th orbit, and [latex]{h}[/latex] is Planck’s constant. Note that angular momentum is [latex]{L = I \omega}[/latex]. For a small object at a radius [latex]{r}[/latex], [latex]{I = mr^2}[/latex] and [latex]{\omega = v/r}[/latex], so that [latex]{L = (mr^2)(v/r) = mvr}[/latex]. Quantization says that this value of [latex]{mvr}[/latex] can only be equal to [latex]{h/2}[/latex], [latex]{2h/2}[/latex], [latex]{3h/2}[/latex], etc. At the time, Bohr himself did not know why angular momentum should be quantized, but using this assumption he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time.

From Bohr’s assumptions, we will now derive a number of important properties of the hydrogen atom from the classical physics we have covered in the text. We start by noting the centripetal force causing the electron to follow a circular path is supplied by the Coulomb force. To be more general, we note that this analysis is valid for any single-electron atom. So, if a nucleus has [latex]{Z}[/latex] protons ([latex]{Z=1}[/latex] for hydrogen, 2 for helium, etc.) and only one electron, that atom is called a hydrogen-like atom . The spectra of hydrogen-like ions are similar to hydrogen, but shifted to higher energy by the greater attractive force between the electron and nucleus. The magnitude of the centripetal force is [latex]{m_ev^2/r_n}[/latex], while the Coulomb force is [latex]{k(Zq_e)(q_e)/r_n^2}[/latex]. The tacit assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. This is consistent with the planetary model of the atom. Equating these,

Angular momentum quantization is stated in an earlier equation. We solve that equation for [latex]{v}[/latex], substitute it into the above, and rearrange the expression to obtain the radius of the orbit. This yields:

where [latex]{a_B}[/latex] is defined to be the Bohr radius , since for the lowest orbit ([latex]{n = 1}[/latex]) and for hydrogen ([latex]{Z=1}[/latex]), [latex]{r_1 = a_\text{B}}[/latex]. It is left for this chapter’s Problems and Exercises to show that the Bohr radius is

The electron orbits are shown in the form of four concentric circles. The radius of each circle is marked as r sub one, r sub two, up to r sub four.

To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy:

Kinetic energy is the familiar [latex]{KE = (1/2)m_ev^2}[/latex], assuming the electron is not moving at relativistic speeds. Potential energy for the electron is electrical, or [latex]{PE = q_eV}[/latex], where [latex]{V}[/latex] is the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge [latex]{Zq_e}[/latex]; thus, [latex]{V = kZq_e/r_n}[/latex], recalling an earlier equation for the potential due to a point charge. Since the electron’s charge is negative, we see that [latex]{PE = -kZq_e/r_n}[/latex]. Entering the expressions for [latex]{KE}[/latex] and [latex]{PE}[/latex], we find

Now we substitute [latex]{r_n}[/latex] and [latex]{v}[/latex] from earlier equations into the above expression for energy. Algebraic manipulation yields

for the orbital energies of hydrogen-like atoms . Here, [latex]{E_0}[/latex] is the ground-state energy ([latex]{n=1}[/latex]) for hydrogen ([latex]{Z=1}[/latex]) and is given by

Thus, for hydrogen,

Figure 7 shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.

An energy level diagram is shown. At the left, there is a vertical arrow showing the energy levels increasing from bottom to top. At the bottom, there is a horizontal line showing the energy levels of Lyman series, n is one. The energy is marked as negative thirteen point six electron volt. Then, in the upper half of the figure, another horizontal line showing Balmer series is shown when the value of n is two. The energy level is labeled as negative three point four zero electron volt. Above it there is another horizontal line showing Paschen series. The energy level is marked as negative one point five one electron volt. Above this line, some more lines are shown in a small area to show energy levels of other values of n.

Electron total energies are negative, since the electron is bound to the nucleus, analogous to being in a hole without enough kinetic energy to escape. As [latex]{n}[/latex] approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since [latex]{r_n}[/latex] gets very large for large [latex]{n}[/latex], and the electric potential energy thus becomes zero. Thus, 13.6 eV is needed to ionize hydrogen (to go from –13.6 eV to 0, or unbound), an experimentally verified number. Given more energy, the electron becomes unbound with some kinetic energy. For example, giving 15.0 eV to an electron in the ground state of hydrogen strips it from the atom and leaves it with 1.4 eV of kinetic energy.

Finally, let us consider the energy of a photon emitted in a downward transition, given by the equation to be

Substituting [latex]{E_n = (-13.6 \;\text{eV/n}^2)}[/latex], we see that

Dividing both sides of this equation by [latex]{hc}[/latex] gives an expression for [latex]{1/ \lambda}[/latex]:

It can be shown that

is the Rydberg constant . Thus, we have used Bohr’s assumptions to derive the formula first proposed by Balmer years earlier as a recipe to fit experimental data.

We see that Bohr’s theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to [latex]{1/n^2}[/latex], where [latex]{n}[/latex] is a non-negative integer. A downward transition releases energy, and so [latex]{n_i}[/latex] must be greater than [latex]{n_f}[/latex]. The various series are those where the transitions end on a certain level. For the Lyman series, [latex]{n_f = 1}[/latex] — that is, all the transitions end in the ground state (see also Figure 7 ). For the Balmer series, [latex]{n_f = 2}[/latex], or all the transitions end in the first excited state; and so on. What was once a recipe is now based in physics, and something new is emerging—angular momentum is quantized.

Triumphs and Limits of the Bohr Theory

PhET Explorations: Models of the Hydrogen Atom

Section Summary

where [latex]{\lambda}[/latex] is the wavelength of the emitted EM radiation and [latex]{R}[/latex] is the Rydberg constant, which has the value

The constants [latex]{n_i}[/latex] and [latex]{n_f}[/latex] are positive integers, and [latex]{n_i}[/latex] must be greater than [latex]{n_f}[/latex].

where [latex]{\Delta E}[/latex] is the change in energy between the initial and final orbits and [latex]{hf}[/latex] is the energy of an absorbed or emitted photon. It is useful to plot orbital energies on a vertical graph called an energy-level diagram.

where [latex]{L}[/latex] is the angular momentum, [latex]{r_n}[/latex] is the radius of the [latex]{n \text{th}}[/latex]orbit, and [latex]{h}[/latex] is Planck’s constant. For all one-electron (hydrogen-like) atoms, the radius of an orbit is given by

[latex]{Z}[/latex] is the atomic number of an element (the number of electrons is has when neutral) and [latex]{a_{\text{B}}}[/latex] is defined to be the Bohr radius, which is

where [latex]{E_0}[/latex] is the ground-state energy and is given by

The Bohr Theory gives accurate values for the energy levels in hydrogen-like atoms, but it has been improved upon in several respects.

Conceptual Questions

1: How do the allowed orbits for electrons in atoms differ from the allowed orbits for planets around the sun? Explain how the correspondence principle applies here.

2: Explain how Bohr’s rule for the quantization of electron orbital angular momentum differs from the actual rule.

3: What is a hydrogen-like atom, and how are the energies and radii of its electron orbits related to those in hydrogen?

Problems & Exercises

1: By calculating its wavelength, show that the first line in the Lyman series is UV radiation.

2: Find the wavelength of the third line in the Lyman series, and identify the type of EM radiation.

3: Look up the values of the quantities in [latex]{a_{\text{B}} = \frac{h^2}{4 \pi ^2m_ekq_e^2}}[/latex], and verify that the Bohr radius [latex]{a_{\text{B}}}[/latex] is [latex]{0.529 \times 10^{-10} \;\text{m}}[/latex].

4: Verify that the ground state energy [latex]{E_0}[/latex] is 13.6 eV by using [latex]{E_0 = \frac{2 \pi ^2q_e^4m_ek^2}{h^2}}[/latex].

5: If a hydrogen atom has its electron in the [latex]{n = 4}[/latex] state, how much energy in eV is needed to ionize it?

6: A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is [latex]{n}[/latex]for a hydrogen atom if 0.850 eV of energy can ionize it?

7: Find the radius of a hydrogen atom in the [latex]{n=2}[/latex] state according to Bohr’s theory.

8: Show that [latex]{(13.6 \;\text{eV})/hc = 1.097 \times 10^7 \;\text{m} = R}[/latex] (Rydberg’s constant), as discussed in the text.

9: What is the smallest-wavelength line in the Balmer series? Is it in the visible part of the spectrum?

10: Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.

11: Do the Balmer and Lyman series overlap? To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line.

12: (a) Which line in the Balmer series is the first one in the UV part of the spectrum?

(b) How many Balmer series lines are in the visible part of the spectrum?

13: A wavelength of[/latex]late {4.653 \;\mu \text{m}}[/latex] is observed in a hydrogen spectrum for a transition that ends in the [latex]{n_f = 5}[/latex] level. What was [latex]{n_i}[/latex] for the initial level of the electron?

14: A singly ionized helium ion has only one electron and is denoted [latex]{\text{He}+}[/latex]. What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?

15: A beryllium ion with a single electron (denoted [latex]{\text{Be}^{3+}}[/latex]) is in an excited state with radius the same as that of the ground state of hydrogen.

(a) What is [latex]{n}[/latex] for the [latex]{\text{Be}^{3+}}[/latex] ion?

(b) How much energy in eV is needed to ionize the ion from this excited state?

16: Atoms can be ionized by thermal collisions, such as at the high temperatures found in the solar corona. One such ion is [latex]{\text{C}^{+5}}[/latex], a carbon atom with only a single electron.

(a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen?

(b) What is the wavelength of the first line in this ion’s Paschen series?

17: Verify Equations [latex]{r_n = \frac{n^2}{Z}a_{\text{B}}}[/latex] and [latex]{a_{\text{B}} = \frac{h^2}{4 \pi ^2m_ekq_e^2} = 0.529 \times 10^{-10} \;\text{m}}[/latex] using the approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization.

18: The wavelength of the four Balmer series lines for hydrogen are found to be 410.3, 434.2, 486.3, and 656.5 nm. What average percentage difference is found between these wavelength numbers and those predicted by [latex]{\frac{1}{\lambda} = R (\frac{1}{n_f^2} - \frac{1}{n_i^2})}[/latex] ? It is amazing how well a simple formula (disconnected originally from theory) could duplicate this phenomenon.

1: [latex]{\frac{1}{\lambda} = R (\frac{1}{n_f^2} - \frac{1}{n_i^2}) \Rightarrow \lambda = \frac{1}{R} [\frac{(n_i \cdot n_f)^2}{n_i^2 - n_f^2}] \; n_i = 2, \; n_f = 1}[/latex], so that

[latex]{\lambda = (\frac{m}{1.097 \times 10^7}) [\frac{(2 \times 1)^2}{2^2 - 1^2}] = 1.22 \times 10^{-7} \;\text{m} = 122 \;\text{nm}}[/latex]

3: [latex]{a_{\text{B}} = \frac{h^2}{4 \pi ^2m_ekZq_e^2} = \frac{(6.626 \times 10^{-34} \;\text{J} \cdot \text{s})^2}{4 \pi ^2 (9.109 \times 10^{-31} \;\text{kg})(8.988 \times 10^9 \;\text{N} \cdot \text{m}^2/ \text{C}^2)(1)(1.602 \times 10^{-19} \;\text{C})^2} = 0.529 \times 10^{-10} \;\text{m}}[/latex]

5: 0.850 eV

7: [latex]{2.12 \times 10^{-10} \;\text{m}}[/latex]

It is in the ultraviolet.

11: No overlap

(b) 54.4 eV

17: [latex]{\frac{kZq_e^2}{r_n^2} = \frac{m_eV^2}{r_n}}[/latex], so that [latex]{r_n = \frac{kZq_e^2}{m_eV^2} = \frac{kZq_e^2}{m_e} \frac{1}{V^2}}[/latex]. From the equation [latex]{m_evr_n = n \frac{h}{2 \pi}}[/latex], we can substitute for the velocity, giving: [latex]{r_n = \frac{kZq_e^2}{m_e} \cdot \frac{4 \pi ^2m_e^2r_n^2}{n^2h^2}}[/latex] so that [latex]{r_n = \frac{n^2}{Z} \frac{h^2}{4 \pi ^2m_ekq_e^2} = \frac{n^2}{Z} a_{\text{B}}}[/latex], where [latex]{a_{\text{B}} = \frac{h^2}{4 \pi ^2m_ekq_e^2}}[/latex].

Share This Book

Atomic Physics

Bohr’s theory of the hydrogen atom, learning objectives.

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

Describe the mysteries of atomic spectra.
Explain Bohr’s theory of the hydrogen atom.
Explain Bohr’s planetary model of the atom.
Illustrate energy state using the energy-level diagram.
Describe the triumphs and limits of Bohr’s theory.

The great Danish physicist Niels Bohr (1885–1962) made immediate use of Rutherford’s planetary model of the atom. (Figure 1). Bohr became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on the planetary model of the atom. For decades, many questions had been asked about atomic characteristics. From their sizes to their spectra, much was known about atoms, but little had been explained in terms of the laws of physics. Bohr’s theory explained the atomic spectrum of hydrogen and established new and broadly applicable principles in quantum mechanics.

Figure 1. Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen atom. His many contributions to the development of atomic physics and quantum mechanics, his personal influence on many students and colleagues, and his personal integrity, especially in the face of Nazi oppression, earned him a prominent place in history. (credit: Unknown Author, via Wikimedia Commons)

Mysteries of Atomic Spectra

As noted in Quantization of Energy , the energies of some small systems are quantized. Atomic and molecular emission and absorption spectra have been known for over a century to be discrete (or quantized). (See Figure 2.) Maxwell and others had realized that there must be a connection between the spectrum of an atom and its structure, something like the resonant frequencies of musical instruments. But, in spite of years of efforts by many great minds, no one had a workable theory. (It was a running joke that any theory of atomic and molecular spectra could be destroyed by throwing a book of data at it, so complex were the spectra.) Following Einstein’s proposal of photons with quantized energies directly proportional to their wavelengths, it became even more evident that electrons in atoms can exist only in discrete orbits.

Figure 2. Part (a) shows, from left to right, a discharge tube, slit, and diffraction grating producing a line spectrum. Part (b) shows the emission line spectrum for iron. The discrete lines imply quantized energy states for the atoms that produce them. The line spectrum for each element is unique, providing a powerful and much used analytical tool, and many line spectra were well known for many years before they could be explained with physics. (credit for (b): Yttrium91, Wikimedia Commons)

In some cases, it had been possible to devise formulas that described the emission spectra. As you might expect, the simplest atom—hydrogen, with its single electron—has a relatively simple spectrum. The hydrogen spectrum had been observed in the infrared (IR), visible, and ultraviolet (UV), and several series of spectral lines had been observed. (See Figure 3.) These series are named after early researchers who studied them in particular depth.

The observed hydrogen-spectrum wavelengths can be calculated using the following formula:

[latex]\displaystyle\frac{1}{\lambda}=R\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\\[/latex],

where λ is the wavelength of the emitted EM radiation and R is the Rydberg constant , determined by the experiment to be R = 1.097 × 10 7 / m (or m −1 ).

The constant n f is a positive integer associated with a specific series. For the Lyman series, n f = 1; for the Balmer series, n f = 2; for the Paschen series, n f = 3; and so on. The Lyman series is entirely in the UV, while part of the Balmer series is visible with the remainder UV. The Paschen series and all the rest are entirely IR. There are apparently an unlimited number of series, although they lie progressively farther into the infrared and become difficult to observe as n f increases. The constant n i is a positive integer, but it must be greater than n f . Thus, for the Balmer series, n f = 2 and n i = 3, 4, 5, 6, …. Note that n i can approach infinity. While the formula in the wavelengths equation was just a recipe designed to fit data and was not based on physical principles, it did imply a deeper meaning. Balmer first devised the formula for his series alone, and it was later found to describe all the other series by using different values of n f . Bohr was the first to comprehend the deeper meaning. Again, we see the interplay between experiment and theory in physics. Experimentally, the spectra were well established, an equation was found to fit the experimental data, but the theoretical foundation was missing.

Figure 3. A schematic of the hydrogen spectrum shows several series named for those who contributed most to their determination. Part of the Balmer series is in the visible spectrum, while the Lyman series is entirely in the UV, and the Paschen series and others are in the IR. Values of n f and n i are shown for some of the lines.

Example 1. Calculating Wave Interference of a Hydrogen Line

What is the distance between the slits of a grating that produces a first-order maximum for the second Balmer line at an angle of 15º?

Strategy and Concept

For an Integrated Concept problem, we must first identify the physical principles involved. In this example, we need to know two things:

the wavelength of light
the conditions for an interference maximum for the pattern from a double slit

Part 1 deals with a topic of the present chapter, while Part 2 considers the wave interference material of Wave Optics.

Solution for Part 1

Hydrogen spectrum wavelength . The Balmer series requires that n f = 2. The first line in the series is taken to be for n i = 3, and so the second would have n i = 4.

The calculation is a straightforward application of the wavelength equation. Entering the determined values for n f and n i yields

[latex]\begin{array}{lll}\frac{1}{\lambda}&=&R\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\\\text{ }&=&\left(1.097\times10^7\text{ m}^-1\right)\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\\\text{ }&=&2.057\times10^6\text{ m}^{-1}\end{array}\\[/latex]

Inverting to find λ gives

[latex]\begin{array}{lll}\lambda&=&\frac{1}{2.057\times10^6\text{ m}^-1}=486\times10^{-9}\text{ m}\\\text{ }&=&486\text{ nm}\end{array}\\[/latex]

Discussion for Part 1

Solution for Part 2

Double-slit interference (Wave Optics). To obtain constructive interference for a double slit, the path length difference from two slits must be an integral multiple of the wavelength. This condition was expressed by the equation d sin θ = mλ , where d is the distance between slits and θ is the angle from the original direction of the beam. The number m is the order of the interference; m =1 in this example. Solving for d and entering known values yields

[latex]\displaystyle{d}=\frac{\left(1\right)\left(486\text{ nm}\right)}{\sin15^{\circ}}=1.88\times10^{-6}\text{ m}\\[/latex]

Discussion for Part 2

This number is similar to those used in the interference examples of Introduction to Quantum Physics (and is close to the spacing between slits in commonly used diffraction glasses).

Bohr’s Solution for Hydrogen

Figure 4. The planetary model of the atom, as modified by Bohr, has the orbits of the electrons quantized. Only certain orbits are allowed, explaining why atomic spectra are discrete (quantized). The energy carried away from an atom by a photon comes from the electron dropping from one allowed orbit to another and is thus quantized. This is likewise true for atomic absorption of photons.

Here, Δ E is the change in energy between the initial and final orbits, and hf is the energy of the absorbed or emitted photon. It is quite logical (that is, expected from our everyday experience) that energy is involved in changing orbits. A blast of energy is required for the space shuttle, for example, to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. This is not observed for satellites or planets, which can have any orbit given the proper energy. (See Figure 4.)

Figure 5. An energy-level diagram plots energy vertically and is useful in visualizing the energy states of a system and the transitions between them. This diagram is for the hydrogen-atom electrons, showing a transition between two orbits having energies E 4 and E 2 .

Bohr was clever enough to find a way to calculate the electron orbital energies in hydrogen. This was an important first step that has been improved upon, but it is well worth repeating here, because it does correctly describe many characteristics of hydrogen. Assuming circular orbits, Bohr proposed that the angular momentum L of an electron in its orbit is quantized , that is, it has only specific, discrete values. The value for L is given by the formula [latex]L=m_{e}vr_{n}=n\frac{h}{2\pi}\left(n=1,2,3,\dots\right)\\[/latex], where L is the angular momentum, m e is the electron’s mass, r n is the radius of the n th orbit, and h is Planck’s constant. Note that angular momentum is L = Iω . For a small object at a radius r , I = mr 2 and [latex]\omega=\frac{v}{r}\\[/latex], so that [latex]L=\left(mr^2\right)\frac{v}{r}=mvr\\[/latex]. Quantization says that this value of mvr can only be equal to [latex]\frac{h}{2},\frac{2h}{2},\frac{3h}{2}\\[/latex], etc. At the time, Bohr himself did not know why angular momentum should be quantized, but using this assumption he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time.

From Bohr’s assumptions, we will now derive a number of important properties of the hydrogen atom from the classical physics we have covered in the text. We start by noting the centripetal force causing the electron to follow a circular path is supplied by the Coulomb force. To be more general, we note that this analysis is valid for any single-electron atom. So, if a nucleus has Z protons ( Z = 1 for hydrogen, 2 for helium, etc.) and only one electron, that atom is called a hydrogen-like atom . The spectra of hydrogen-like ions are similar to hydrogen, but shifted to higher energy by the greater attractive force between the electron and nucleus. The magnitude of the centripetal force is [latex]\frac{m_{e}v^2}{r_n}\\[/latex], while the Coulomb force is [latex]k\frac{\left(Zq_{e}\right)\left(q_e\right)}{r_n^2}\\[/latex]. The tacit assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. This is consistent with the planetary model of the atom. Equating these,

[latex]k\frac{Zq_{e}^2}{r_n^2}=\frac{m_{e}v^2}{r_n}\text{ (Coulomb = centripetal)}\\[/latex].

Angular momentum quantization is stated in an earlier equation. We solve that equation for v , substitute it into the above, and rearrange the expression to obtain the radius of the orbit. This yields:

[latex]\displaystyle{r}_{n}=\frac{n^2}{Z}a_{\text{B}},\text{ for allowed orbits }\left(n=1,2,3\dots\right)\\[/latex],

where a B is defined to be the Bohr radius , since for the lowest orbit ( n = 1) and for hydrogen ( Z = 1), r 1 = a B . It is left for this chapter’s Problems and Exercises to show that the Bohr radius is

[latex]\displaystyle{a}_{\text{B}}=\frac{h^2}{4\pi^2m_{e}kq_{e}^{2}}=0.529\times10^{-10}\text{ m}\\[/latex].

Figure 6. The allowed electron orbits in hydrogen have the radii shown. These radii were first calculated by Bohr and are given by the equation [latex]r_n=\frac{n^2}{Z}a_{\text{B}}\\[/latex]. The lowest orbit has the experimentally verified diameter of a hydrogen atom.

To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy: E n = KE + PE.

Kinetic energy is the familiar [latex]KE=\frac{1}{2}m_{e}v^2\\[/latex], assuming the electron is not moving at relativistic speeds. Potential energy for the electron is electrical, or PE = q e V , where V is the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge Zq e ; thus, [latex]V=\frac{kZq_e}{r_n}\\[/latex], recalling an earlier equation for the potential due to a point charge. Since the electron’s charge is negative, we see that [latex]PE=-\frac{kZq_e}{r_n}\\[/latex]. Entering the expressions for KE and PE , we find

[latex]\displaystyle{E}_{n}=\frac{1}{2}m_{e}v^2-k\frac{Zq_{e}^{2}}{r_{n}}\\[/latex].

Now we substitute r n and v from earlier equations into the above expression for energy. Algebraic manipulation yields

[latex]\displaystyle{E}_{n}=-\frac{Z^2}{n^2}E_0\left(n=1,2,3,\dots\right)\\[/latex]

for the orbital energies of hydrogen-like atoms . Here, E 0 is the ground-state energy ( n = 1) for hydrogen ( Z = 1) and is given by

[latex]\displaystyle{E}_{0}=\frac{2\pi{q}_{e}^{4}m_{e}k^{2}}{h^2}=13.6\text{ eV}\\[/latex]

Thus, for hydrogen,

[latex]\displaystyle{E}_n=-\frac{13.6\text{ eV}}{n^2}\left(n=1,2,3\dots\right)\\[/latex]

Figure 7. Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. The orbital energies are calculated using the above equation, first derived by Bohr.

Figure 7 shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.

Electron total energies are negative, since the electron is bound to the nucleus, analogous to being in a hole without enough kinetic energy to escape. As n approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since r n gets very large for large n , and the electric potential energy thus becomes zero. Thus, 13.6 eV is needed to ionize hydrogen (to go from –13.6 eV to 0, or unbound), an experimentally verified number. Given more energy, the electron becomes unbound with some kinetic energy. For example, giving 15.0 eV to an electron in the ground state of hydrogen strips it from the atom and leaves it with 1.4 eV of kinetic energy.

Finally, let us consider the energy of a photon emitted in a downward transition, given by the equation to be ∆ E = hf = E i − E f .

Substituting E n = (–13.6 eV/ n 2 ), we see that

[latex]\displaystyle{hf}=\left(13.6\text{ eV}\right)\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\\[/latex]

Dividing both sides of this equation by hc gives an expression for [latex]\frac{1}{\lambda}\\[/latex]:

[latex]\displaystyle\frac{hf}{hc}=\frac{f}{c}=\frac{1}{\lambda}=\frac{\left(13.6\text{ eV}\right)}{hc}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\\[/latex]

It can be shown that

[latex]\displaystyle\left(\frac{13.6\text{ eV}}{hc}\right)=\frac{\left(13.6\text{ eV}\right)\left(1.602\times10^{-19}\text{ J/eV}\right)}{\left(6.626\times10^{-34}\text{ J }\cdot\text{ s}\right)\left(2.998\times10^{8}\text{ m/s}\right)}=1.097\times10^7\text{ m}^{-1}=R\\[/latex]

is the Rydberg constant . Thus, we have used Bohr’s assumptions to derive the formula first proposed by Balmer years earlier as a recipe to fit experimental data.

[latex]\displaystyle\frac{1}{\lambda}=R\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)\\[/latex]

We see that Bohr’s theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to [latex]\frac{1}{n^2}\\[/latex], where n is a non-negative integer. A downward transition releases energy, and so n i must be greater than n f . The various series are those where the transitions end on a certain level. For the Lyman series, n f = 1—that is, all the transitions end in the ground state (see also Figure 7). For the Balmer series, n f = 2, or all the transitions end in the first excited state; and so on. What was once a recipe is now based in physics, and something new is emerging—angular momentum is quantized.

Triumphs and Limits of the Bohr Theory

PhET Explorations: Models of the Hydrogen Atom

Click to download the simulation. Run using Java.

Section Summary

The planetary model of the atom pictures electrons orbiting the nucleus in the way that planets orbit the sun. Bohr used the planetary model to develop the first reasonable theory of hydrogen, the simplest atom. Atomic and molecular spectra are quantized, with hydrogen spectrum wavelengths given by the formula [latex]\frac{1}{\lambda }=R\left(\frac{1}{{n}_{\text{f}}^{2}}-\frac{1}{{n}_{\text{i}}^{2}}\right)\\[/latex], where λ is the wavelength of the emitted EM radiation and R is the Rydberg constant, which has the value R = 1.097 × 10 7 m −1 .
The constants n i and n f are positive integers, and n i must be greater than n f .
Bohr correctly proposed that the energy and radii of the orbits of electrons in atoms are quantized, with energy for transitions between orbits given by ∆ E = hf = E i − E f , where ∆ E is the change in energy between the initial and final orbits and hf is the energy of an absorbed or emitted photon. It is useful to plot orbital energies on a vertical graph called an energy-level diagram.
Bohr proposed that the allowed orbits are circular and must have quantized orbital angular momentum given by [latex]L={m}_{e}{\text{vr}}_{n}=n\frac{h}{2\pi }\left(n=1, 2, 3 \dots \right)\\[/latex], where L is the angular momentum, r n is the radius of the n th orbit, and h is Planck’s constant. For all one-electron (hydrogen-like) atoms, the radius of an orbit is given by [latex]{r}_{n}=\frac{{n}^{2}}{Z}{a}_{\text{B}}\left(\text{allowed orbits }n=1, 2, 3, ...\right)\\[/latex], Z is the atomic number of an element (the number of electrons is has when neutral) and a B is defined to be the Bohr radius, which is [latex]{a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}=\text{0.529}\times {\text{10}}^{-\text{10}}\text{ m}\\[/latex].
Furthermore, the energies of hydrogen-like atoms are given by [latex]{E}_{n}=-\frac{{Z}^{2}}{{n}^{2}}{E}_{0}\left(n=1, 2, 3 ...\right)\\[/latex], where E 0 is the ground-state energy and is given by [latex]{E}_{0}=\frac{{2\pi }^{2}{q}_{e}^{4}{m}_{e}{k}^{2}}{{h}^{2}}=\text{13.6 eV}\\[/latex]. Thus, for hydrogen, [latex]{E}_{n}=-\frac{\text{13.6 eV}}{{n}^{2}}\left(n,=,1, 2, 3 ...\right)\\[/latex].
The Bohr Theory gives accurate values for the energy levels in hydrogen-like atoms, but it has been improved upon in several respects.

Conceptual Questions

How do the allowed orbits for electrons in atoms differ from the allowed orbits for planets around the sun? Explain how the correspondence principle applies here.
Explain how Bohr’s rule for the quantization of electron orbital angular momentum differs from the actual rule.
What is a hydrogen-like atom, and how are the energies and radii of its electron orbits related to those in hydrogen?

Problems & Exercises

By calculating its wavelength, show that the first line in the Lyman series is UV radiation.
Find the wavelength of the third line in the Lyman series, and identify the type of EM radiation.
Look up the values of the quantities in [latex]{a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}\\[/latex] , and verify that the Bohr radius a B is 0.529 × 10 −10 m.
Verify that the ground state energy E 0 is 13.6 eV by using [latex]{E}_{0}=\frac{{2\pi }^{2}{q}_{e}^{4}{m}_{e}{k}^{2}}{{h}^{2}}\\[/latex].
If a hydrogen atom has its electron in the n = 4 state, how much energy in eV is needed to ionize it?
A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is n for a hydrogen atom if 0.850 eV of energy can ionize it?
Find the radius of a hydrogen atom in the n = 2 state according to Bohr’s theory.
Show that [latex]\frac{\left(13.6 \text{eV}\right)}{hc}=1.097\times10^{7}\text{ m}=R\\[/latex] (Rydberg’s constant), as discussed in the text.
What is the smallest-wavelength line in the Balmer series? Is it in the visible part of the spectrum?
Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.
Do the Balmer and Lyman series overlap? To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line.
(a) Which line in the Balmer series is the first one in the UV part of the spectrum? (b) How many Balmer series lines are in the visible part of the spectrum? (c) How many are in the UV?
A wavelength of 4.653 µm is observed in a hydrogen spectrum for a transition that ends in the n f = 5 level. What was n i for the initial level of the electron?
A singly ionized helium ion has only one electron and is denoted He + . What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?
A beryllium ion with a single electron (denoted Be 3+ ) is in an excited state with radius the same as that of the ground state of hydrogen. (a) What is n for the Be 3+ ion? (b) How much energy in eV is needed to ionize the ion from this excited state?
Atoms can be ionized by thermal collisions, such as at the high temperatures found in the solar corona. One such ion is C +5 , a carbon atom with only a single electron. (a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen? (b) What is the wavelength of the first line in this ion’s Paschen series? (c) What type of EM radiation is this?
Verify Equations [latex]{r}_{n}=\frac{{n}^{2}}{Z}{a}_{\text{B}}\\[/latex] and [latex]{a}_{B}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{kq}_{e}^{2}}=0.529\times{10}^{-10}\text{ m}\\[/latex] using the approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization.
The wavelength of the four Balmer series lines for hydrogen are found to be 410.3, 434.2, 486.3, and 656.5 nm. What average percentage difference is found between these wavelength numbers and those predicted by [latex]\frac{1}{\lambda}=R\left(\frac{1}{{n}_{\text{f}}^{2}}-\frac{1}{{n}_{\text{i}}^{2}}\right)\\[/latex]? It is amazing how well a simple formula (disconnected originally from theory) could duplicate this phenomenon.

hydrogen spectrum wavelengths: the wavelengths of visible light from hydrogen; can be calculated by

[latex]\displaystyle\frac{1}{\lambda }=R\left(\frac{1}{{n}_{\text{f}}^{2}}-\frac{1}{{n}_{\text{i}}^{2}}\right)\\[/latex]

Rydberg constant: a physical constant related to the atomic spectra with an established value of 1.097 × 10 7 m −1

double-slit interference: an experiment in which waves or particles from a single source impinge upon two slits so that the resulting interference pattern may be observed

energy-level diagram: a diagram used to analyze the energy level of electrons in the orbits of an atom

Bohr radius: the mean radius of the orbit of an electron around the nucleus of a hydrogen atom in its ground state

hydrogen-like atom: any atom with only a single electron

energies of hydrogen-like atoms: Bohr formula for energies of electron states in hydrogen-like atoms: [latex]{E}_{n}=-\frac{{Z}^{2}}{{n}^{2}}{E}_{0}\left(n=\text{1, 2, 3,}\dots \right)\\[/latex]

Selected Solutions to Problems & Exercises

1. [latex]\displaystyle\frac{1}{\lambda}=R\left(\frac{1}{{n}_{\text{f}}^{2}}-\frac{1}{{n}_{\text{i}}^{2}}\right)\Rightarrow \lambda =\frac{1}{R}\left[\frac{\left({n}_{\text{i}}\cdot{n}_{\text{f}}\right)^{2}}{{n}_{\text{i}}^{2}-{n}_{\text{f}}^{2}}\right];{n}_{\text{i}}=2,{n}_{\text{f}}=1\\[/latex], so that

[latex]\displaystyle\lambda =\left(\frac{m}{1.097\times {\text{10}}^{7}}\right)\left[\frac{\left(2\times1\right)^{2}}{{2}^{2}-{1}^{2}}\right]=1\text{.}\text{22}\times {\text{10}}^{-7}\text{m}=\text{122 nm}\\[/latex] , which is UV radiation.

3. [latex]\begin{array}{lll}{a}_{\text{B}}&=&\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kZq}}_{e}^{2}}\\\text{ }&=&\frac{\left(\text{6.626}\times {\text{10}}^{-\text{34}}\text{J }\cdot\text{ s}\right)^{2}}{{4\pi }^{2}\left(9.109\times {\text{10}}^{-\text{31}}\text{kg}\right)\left(8.988\times {\text{10}}^{9}\text{N}\cdot{\text{m}}^{2}/{C}^{2}\right)\left(1\right)\left(1.602\times {\text{10}}^{-\text{19}}\text{C}\right)^{2}}\\\text{ }&=&\text{0.529}\times {\text{10}}^{-\text{10}}\text{m}\end{array}\\[/latex]

5. 0.850 eV

7. 2.12 × 10 −10 m

9. 365 nm; it is in the ultraviolet.

11. No overlap; 365 nm; 122 nm

15. (a) 2; (b) 54.4 eV

17. [latex]\displaystyle\frac{{\text{kZq}}_{e}^{2}}{{r}_{n}^{2}}=\frac{{m}_{e}{V}^{2}}{{r}_{n}}\\[/latex], so that [latex]\displaystyle{r}_{n}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}{V}^{2}}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}}\frac{1}{{V}^{2}}\\[/latex]. From the equation [latex]\displaystyle{m}_{e}{vr}_{n}=n\frac{h}{2\pi}\\[/latex], we can substitute for the velocity, giving:

[latex]\displaystyle{r}_{n}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}}\cdot \frac{{4\pi }^{2}{m}_{e}^{2}{r}_{n}^{2}}{{n}^{2}{h}^{2}}\\[/latex]

[latex]\displaystyle{r}_{n}=\frac{{n}^{2}}{Z}\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}=\frac{{n}^{2}}{Z}{a}_{\text{B}}\\[/latex],

[latex]\displaystyle{a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}\\[/latex].

Candela Citations

College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License
PhET Interactive Simulations . Provided by : University of Colorado Boulder . Located at : http://phet.colorado.edu . License : CC BY: Attribution

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

169 Bohr’s Theory of the Hydrogen Atom

[latexpage]

Learning Objectives

Describe the mysteries of atomic spectra.
Explain Bohr’s theory of the hydrogen atom.
Explain Bohr’s planetary model of the atom.
Illustrate energy state using the energy-level diagram.
Describe the triumphs and limits of Bohr’s theory.

The great Danish physicist Niels Bohr (1885–1962) made immediate use of Rutherford’s planetary model of the atom. ( (Figure) ). Bohr became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on the planetary model of the atom. For decades, many questions had been asked about atomic characteristics. From their sizes to their spectra, much was known about atoms, but little had been explained in terms of the laws of physics. Bohr’s theory explained the atomic spectrum of hydrogen and established new and broadly applicable principles in quantum mechanics.

Mysteries of Atomic Spectra

As noted in Quantization of Energy , the energies of some small systems are quantized. Atomic and molecular emission and absorption spectra have been known for over a century to be discrete (or quantized). (See (Figure) .) Maxwell and others had realized that there must be a connection between the spectrum of an atom and its structure, something like the resonant frequencies of musical instruments. But, in spite of years of efforts by many great minds, no one had a workable theory. (It was a running joke that any theory of atomic and molecular spectra could be destroyed by throwing a book of data at it, so complex were the spectra.) Following Einstein’s proposal of photons with quantized energies directly proportional to their wavelengths, it became even more evident that electrons in atoms can exist only in discrete orbits.

In some cases, it had been possible to devise formulas that described the emission spectra. As you might expect, the simplest atom—hydrogen, with its single electron—has a relatively simple spectrum. The hydrogen spectrum had been observed in the infrared (IR), visible, and ultraviolet (UV), and several series of spectral lines had been observed. (See (Figure) .) These series are named after early researchers who studied them in particular depth.

The observed hydrogen-spectrum wavelengths can be calculated using the following formula:

where \(\lambda \) is the wavelength of the emitted EM radiation and \(R\) is the Rydberg constant , determined by the experiment to be

The constant \({n}_{\text{f}}\) is a positive integer associated with a specific series. For the Lyman series, \({n}_{\text{f}}=1\); for the Balmer series, \({n}_{\text{f}}=2\); for the Paschen series, \({n}_{\text{f}}=3\); and so on. The Lyman series is entirely in the UV, while part of the Balmer series is visible with the remainder UV. The Paschen series and all the rest are entirely IR. There are apparently an unlimited number of series, although they lie progressively farther into the infrared and become difficult to observe as \({n}_{\text{f}}\) increases. The constant \({n}_{\text{i}}\) is a positive integer, but it must be greater than \({n}_{\text{f}}\). Thus, for the Balmer series, \({n}_{\text{f}}=2\) and \({n}_{\text{i}}=3, 4, 5, 6, …\text{}\). Note that \({n}_{\text{i}}\) can approach infinity. While the formula in the wavelengths equation was just a recipe designed to fit data and was not based on physical principles, it did imply a deeper meaning. Balmer first devised the formula for his series alone, and it was later found to describe all the other series by using different values of \({n}_{\text{f}}\). Bohr was the first to comprehend the deeper meaning. Again, we see the interplay between experiment and theory in physics. Experimentally, the spectra were well established, an equation was found to fit the experimental data, but the theoretical foundation was missing.

What is the distance between the slits of a grating that produces a first-order maximum for the second Balmer line at an angle of \(\text{15º}\)?

Strategy and Concept

Solution for (a)

Hydrogen spectrum wavelength . The Balmer series requires that \({n}_{\text{f}}=2\). The first line in the series is taken to be for \({n}_{\text{i}}=3\), and so the second would have \({n}_{\text{i}}=4\).

The calculation is a straightforward application of the wavelength equation. Entering the determined values for \({n}_{\text{f}}\) and \({n}_{\text{i}}\) yields

Inverting to find \(\lambda \) gives

Discussion for (a)

Solution for (b)

where \(d\) is the distance between slits and \(\theta \) is the angle from the original direction of the beam. The number \(m\) is the order of the interference; \(m=1\) in this example. Solving for \(d\) and entering known values yields

Discussion for (b)

This number is similar to those used in the interference examples of Introduction to Quantum Physics (and is close to the spacing between slits in commonly used diffraction glasses).

Bohr’s Solution for Hydrogen

Here, \(\Delta E\) is the change in energy between the initial and final orbits, and \(\text{hf}\) is the energy of the absorbed or emitted photon. It is quite logical (that is, expected from our everyday experience) that energy is involved in changing orbits. A blast of energy is required for the space shuttle, for example, to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. This is not observed for satellites or planets, which can have any orbit given the proper energy. (See (Figure) .)

(Figure) shows an energy-level diagram , a convenient way to display energy states. In the present discussion, we take these to be the allowed energy levels of the electron. Energy is plotted vertically with the lowest or ground state at the bottom and with excited states above. Given the energies of the lines in an atomic spectrum, it is possible (although sometimes very difficult) to determine the energy levels of an atom. Energy-level diagrams are used for many systems, including molecules and nuclei. A theory of the atom or any other system must predict its energies based on the physics of the system.

where \(L\) is the angular momentum, \({m}_{e}\) is the electron’s mass, \({r}_{n}\) is the radius of the \(n\) th orbit, and \(h\) is Planck’s constant. Note that angular momentum is \(L=\mathrm{I\omega }\). For a small object at a radius \(r,\phantom{\rule{0.25em}{0ex}}I={\text{mr}}^{2}\) and \(\omega =v/r\), so that \(L=\left({\text{mr}}^{2}\right)\left(v/r\right)=\text{mvr}\). Quantization says that this value of \(\text{mvr}\) can only be equal to \(h/2,\phantom{\rule{0.25em}{0ex}}2h/2,\phantom{\rule{0.25em}{0ex}}3h/2\), etc. At the time, Bohr himself did not know why angular momentum should be quantized, but using this assumption he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time.

From Bohr’s assumptions, we will now derive a number of important properties of the hydrogen atom from the classical physics we have covered in the text. We start by noting the centripetal force causing the electron to follow a circular path is supplied by the Coulomb force. To be more general, we note that this analysis is valid for any single-electron atom. So, if a nucleus has \(Z\) protons (\(Z=1\) for hydrogen, 2 for helium, etc.) and only one electron, that atom is called a hydrogen-like atom . The spectra of hydrogen-like ions are similar to hydrogen, but shifted to higher energy by the greater attractive force between the electron and nucleus. The magnitude of the centripetal force is \({m}_{e}{v}^{2}/{r}_{n}\), while the Coulomb force is \(k\left({\text{Zq}}_{e}\right)\left({q}_{e}\right)/{r}_{n}^{2}\). The tacit assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. This is consistent with the planetary model of the atom. Equating these,

Angular momentum quantization is stated in an earlier equation. We solve that equation for \(v\), substitute it into the above, and rearrange the expression to obtain the radius of the orbit. This yields:

where \({a}_{\text{B}}\) is defined to be the Bohr radius , since for the lowest orbit \(\left(n=1\right)\) and for hydrogen \(\left(Z=1\right)\), \({r}_{1}={a}_{\text{B}}\). It is left for this chapter’s Problems and Exercises to show that the Bohr radius is

To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy:

Kinetic energy is the familiar \(\text{KE}=\left(1/2\right){m}_{e}{v}^{2}\), assuming the electron is not moving at relativistic speeds. Potential energy for the electron is electrical, or \(\text{PE}={q}_{e}V\), where \(V\) is the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge \({\text{Zq}}_{e}\) ; thus, \(V={\text{kZq}}_{e}/{r}_{n}\), recalling an earlier equation for the potential due to a point charge. Since the electron’s charge is negative, we see that \(\text{PE}=-{\text{kZq}}_{e}/{r}_{n}\) . Entering the expressions for \(\text{KE}\) and \(\text{PE}\), we find

Now we substitute \({r}_{n}\) and \(v\) from earlier equations into the above expression for energy. Algebraic manipulation yields

for the orbital energies of hydrogen-like atoms . Here, \({E}_{0}\) is the ground-state energy \(\left(n=1\right)\) for hydrogen \(\left(Z=1\right)\) and is given by

Thus, for hydrogen,

(Figure) shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.

Electron total energies are negative, since the electron is bound to the nucleus, analogous to being in a hole without enough kinetic energy to escape. As \(n\) approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since \({r}_{n}\) gets very large for large \(n\), and the electric potential energy thus becomes zero. Thus, 13.6 eV is needed to ionize hydrogen (to go from –13.6 eV to 0, or unbound), an experimentally verified number. Given more energy, the electron becomes unbound with some kinetic energy. For example, giving 15.0 eV to an electron in the ground state of hydrogen strips it from the atom and leaves it with 1.4 eV of kinetic energy.

Finally, let us consider the energy of a photon emitted in a downward transition, given by the equation to be

Substituting \({E}_{n}=\left(–\text{13.6 eV}/{n}^{2}\right)\), we see that

Dividing both sides of this equation by \(\text{hc}\) gives an expression for \(1/\lambda \):

It can be shown that

is the Rydberg constant . Thus, we have used Bohr’s assumptions to derive the formula first proposed by Balmer years earlier as a recipe to fit experimental data.

We see that Bohr’s theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to \(1/{n}^{2}\), where \(n\) is a non-negative integer. A downward transition releases energy, and so \({n}_{i}\) must be greater than \({n}_{\text{f}}\). The various series are those where the transitions end on a certain level. For the Lyman series, \({n}_{\text{f}}=1\) — that is, all the transitions end in the ground state (see also (Figure) ). For the Balmer series, \({n}_{\text{f}}=2\), or all the transitions end in the first excited state; and so on. What was once a recipe is now based in physics, and something new is emerging—angular momentum is quantized.

Triumphs and Limits of the Bohr Theory

Section Summary

where \(\lambda \) is the wavelength of the emitted EM radiation and \(R\) is the Rydberg constant, which has the value

The constants \({n}_{i}\) and \({n}_{f}\) are positive integers, and \({n}_{i}\) must be greater than \({n}_{f}\).

where \(L\) is the angular momentum, \({r}_{n}\) is the radius of the \(n\text{th}\) orbit, and \(h\) is Planck’s constant. For all one-electron (hydrogen-like) atoms, the radius of an orbit is given by

\(Z\) is the atomic number of an element (the number of electrons is has when neutral) and \({a}_{\text{B}}\) is defined to be the Bohr radius, which is

where \({E}_{0}\) is the ground-state energy and is given by

The Bohr Theory gives accurate values for the energy levels in hydrogen-like atoms, but it has been improved upon in several respects.

Conceptual Questions

How do the allowed orbits for electrons in atoms differ from the allowed orbits for planets around the sun? Explain how the correspondence principle applies here.

Explain how Bohr’s rule for the quantization of electron orbital angular momentum differs from the actual rule.

What is a hydrogen-like atom, and how are the energies and radii of its electron orbits related to those in hydrogen?

Problems & Exercises

By calculating its wavelength, show that the first line in the Lyman series is UV radiation.

\(\frac{1}{\lambda }=R\left(\frac{1}{{n}_{\text{f}}^{2}}-\frac{1}{{n}_{\text{i}}^{2}}\right)⇒\lambda =\frac{1}{R}\left[\frac{\left({n}_{\text{i}}\cdot {n}_{\text{f}}{\right)}^{2}}{{n}_{\text{i}}^{2}-{n}_{\text{f}}^{2}}\right];\phantom{\rule{0.25em}{0ex}}{n}_{\text{i}}=2,\phantom{\rule{0.25em}{0ex}}{n}_{\text{f}}=1,\phantom{\rule{0.25em}{0ex}}\) so that

\(\lambda =\left(\frac{m}{1.097×{\text{10}}^{7}}\right)\left[\frac{\left(2×1{\right)}^{2}}{{2}^{2}-{1}^{2}}\right]=1\text{.}\text{22}×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{m}=\text{122 nm}\) , which is UV radiation.

Find the wavelength of the third line in the Lyman series, and identify the type of EM radiation.

Look up the values of the quantities in \({a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}{}^{}\), and verify that the Bohr radius \({a}_{\text{B}}\) is \(\text{0.529}×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{m}\).

\({a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kZq}}_{e}^{2}}=\frac{\left(\text{6.626}×{\text{10}}^{-\text{34}}\phantom{\rule{0.25em}{0ex}}\text{J·s}{\right)}^{2}}{{4\pi }^{2}\left(9.109×{\text{10}}^{-\text{31}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(8.988×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\text{·}{\text{m}}^{2}/{C}^{2}\right)\left(1\right)\left(1.602×{\text{10}}^{-\text{19}}\phantom{\rule{0.25em}{0ex}}\text{C}{\right)}^{2}}=\text{0.529}×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{m}\)

Verify that the ground state energy \({E}_{0}\) is 13.6 eV by using \({E}_{0}=\frac{{2\pi }^{2}{q}_{e}^{4}{m}_{e}{k}^{2}}{{h}^{2}}\text{.}\)

If a hydrogen atom has its electron in the \(n=4\) state, how much energy in eV is needed to ionize it?

A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is \(n\) for a hydrogen atom if 0.850 eV of energy can ionize it?

Find the radius of a hydrogen atom in the \(n=2\) state according to Bohr’s theory.

\(\text{2.12}×{\text{10}}^{\text{–10}}\phantom{\rule{0.25em}{0ex}}\text{m}\)

Show that \(\left(13.6 eV\right)/\text{hc}=\text{1.097}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{m}=R\) (Rydberg’s constant), as discussed in the text.

What is the smallest-wavelength line in the Balmer series? Is it in the visible part of the spectrum?

It is in the ultraviolet.

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.

Do the Balmer and Lyman series overlap? To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line.

(a) Which line in the Balmer series is the first one in the UV part of the spectrum?

(b) How many Balmer series lines are in the visible part of the spectrum?

A wavelength of \(4\text{.}\text{653 μm}\) is observed in a hydrogen spectrum for a transition that ends in the \({n}_{\text{f}}=5\) level. What was \({n}_{\text{i}}\) for the initial level of the electron?

A singly ionized helium ion has only one electron and is denoted \({\text{He}}^{+}\). What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?

A beryllium ion with a single electron (denoted \({\text{Be}}^{3+}\)) is in an excited state with radius the same as that of the ground state of hydrogen.

(a) What is \(n\) for the \({\text{Be}}^{3+}\) ion?

(b) How much energy in eV is needed to ionize the ion from this excited state?

(b) 54.4 eV

Atoms can be ionized by thermal collisions, such as at the high temperatures found in the solar corona. One such ion is \({C}^{+5}\), a carbon atom with only a single electron.

(a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen?

(b) What is the wavelength of the first line in this ion’s Paschen series?

Verify Equations \({r}_{n}=\frac{{n}^{2}}{Z}{a}_{\text{B}}\) and \({a}_{B}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}=\text{0.529}×{\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{m}\) using the approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization.

\(\frac{{\text{kZq}}_{e}^{2}}{{r}_{n}^{2}}=\frac{{m}_{e}{V}^{2}}{{r}_{n}}\text{,}\) so that \({r}_{n}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}{V}^{2}}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}}\frac{1}{{V}^{2}}\text{.}\) From the equation \({m}_{e}{\text{vr}}_{n}=n\frac{h}{2\pi }\text{,}\) we can substitute for the velocity, giving: \({r}_{n}=\frac{{\text{kZq}}_{e}^{2}}{{m}_{e}}\cdot \frac{{4\pi }^{2}{m}_{e}^{2}{r}_{n}^{2}}{{n}^{2}{h}^{2}}\) so that \({r}_{n}=\frac{{n}^{2}}{Z}\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}=\frac{{n}^{2}}{Z}{a}_{\text{B}},\) where \({a}_{\text{B}}=\frac{{h}^{2}}{{4\pi }^{2}{m}_{e}{\text{kq}}_{e}^{2}}\).

The wavelength of the four Balmer series lines for hydrogen are found to be 410.3, 434.2, 486.3, and 656.5 nm. What average percentage difference is found between these wavelength numbers and those predicted by \(\frac{1}{\lambda }=R\left(\frac{1}{{n}_{\text{f}}^{2}}-\frac{1}{{n}_{\text{i}}^{2}}\right)\)? It is amazing how well a simple formula (disconnected originally from theory) could duplicate this phenomenon.

Share This Book

Nuclear Physics
Bohr Model Of The Hydrogen Atom

Bohr Model of the Hydrogen Atom

Bohr model of the hydrogen atom was the first atomic model to successfully explain the radiation spectra of atomic hydrogen. Niels Bohr introduced the atomic Hydrogen model in the year 1913. Bohr’s Model of the hydrogen atom attempts to plug in certain gaps as suggested by Rutherford’s model. It holds a special place in history as it gave rise to quantum mechanics by introducing the quantum theory.

Planetary Model of the Atom

Quantum mechanics emerged in the mid-1920s. Neil Bohr, one of the founders of quantum mechanics , was interested in the much-debated topic of the time – the structure of the atom. Numerous atomic models, including the theory postulated by J.J Thompson and the discovery of the nucleus by Ernest Rutherford, had emerged. But Bohr supported the planetary model, which asserted that electrons revolved around a positively charged nucleus just like the planets around the sun.

Planetary Model of The Atom

Nevertheless, scientists still had many unanswered questions such as

Why didn’t the electrons drop into the nucleus as foretold by classical physics?
Where are the electrons and what do they do there?
How is the discrete emission lines produced by excited elements correlated to the internal structure of the atom?

Bohr addressed all these questions using a seemingly simple assumption: What if electron orbits and energies, could exhibit only specific values? You can check Atomic Theory to learn about the various atomic theory put forward by scientists in the early 20 th century.

You may also want to check out these topics given below!

Energy level
Bohr’s Theory of Hydrogen Atoms

Bohr’s Equation

Bohr Model of the hydrogen atom first proposed the planetary model, but later an assumption concerning the electrons was made. The assumption was the quantization of the structure of atoms. Bohr’s proposed that electrons orbited the nucleus in specific orbits or shells with a fixed radius. Only those shells with a radius provided by the equation below were allowed, and it was impossible for electrons to exist between these shells.

Mathematically, the allowed value of the atomic radius is given by the equation:

n is a positive integer
r(1) is the smallest allowed radius for the hydrogen atom also known as the Bohr’s radius

The Bohr’s radius has a value of: \(\begin{array}{l}r(1)=0.529\times 10^{-10}\,m\end{array} \) .

Bohr calculated the energy of an electron in the nth level of hydrogen by considering the electrons in circular, quantized orbits as:

13.6 eV is the lowest possible energy of a hydrogen electron E(1).

The energy obtained is always a negative number and the ground state n = 1, has the most negative value. The reason being that the energy of an electron in orbit is relative to the energy of an electron that is entirely separated from its nucleus, \(\begin{array}{l}n=\infty\end{array} \) and it is recognised to have an energy of 0 eV. Since the electron in a fixed orbit around the nucleus is more stable than an electron that is extremely far from its nucleus, the energy of the electron in orbit is always negative.

Absorption and Emission

According to Bohr’s model, an electron would absorb energy in the form of photons to get excited to a higher energy level . After escaping to the higher energy level, also known as the excited state, the excited electron is less stable, and therefore, would rapidly emit a photon to come back to a lower, more stable energy level. The energy of the emitted photon is equal to the difference in energy between the two energy levels for a specific transition. The energy can be calculated using the equation

Atomic Excitation and Atomic De-excitation

Atomic Excitation and De-excitation

Limitations of the Bohr Model of the Hydrogen Atom:

Bohr’s model doesn’t work well for complex atoms.
It couldn’t explain why some spectral lines are more intense than others.
It could not explain why some spectral lines split into multiple lines in the presence of a magnetic field.
Heisenberg’s uncertainty principle contradicts Bohr’s idea of electrons existing in specific orbits with a known radius and velocity.

Although the modern quantum mechanical model and the Bohr Model of the Hydrogen Atom may seem vastly different, the fundamental idea is the same in both. Classical physics isn’t sufficient to describe all the phenomena that occur on an atomic level. But, Bohr was the first to realise the quantization of electronic shells by fusing the idea of quantization into the electronic structure of the hydrogen atom and was successfully able to explain the emission spectra of hydrogen as well as other one-electron systems.

Bohr proposed that electrons travel in specific orbits, shells around the nucleus.
According to Bohr’s calculation, the energy for an electron in the shell is given by the expression:
The hydrogen spectrum is explained in terms of electrons absorbing and emitting photons to change energy levels, where the photon energy is:
Bohr’s Model of the Hydrogen Atom isn’t applicable for systems with more than one electron.

If you wish to learn more Physics-concepts with the help of interactive video lessons, download BYJU’S – The Learning App.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

Visit BYJU’S for all Physics related queries and study materials

Your result is as below

Request OTP on Voice Call

Register with BYJU'S & Download Free PDFs

IMAGES

Niels Bohr Experiment With Hydrogen Spectrum
PPT
Niels Bohr Experiment With Hydrogen Spectrum
Niels Bohr Experiment With Hydrogen Spectrum
Niels Bohr Experiment With Hydrogen Spectrum
Bohr Model: Definition, Features, and Limitations

VIDEO

Hydrogen Gas Experiment with bottle #shorts #surajkeexperiment
Biography of Niels Bohr
GAS BALLOON MAKING AT HOME
Exploring the Bohr-Einstein Debate
Structure Of Atoms And Nuclei Full chapter notes
H2 Hydrogen Gas ⛽ Parikshan By Mr Jitesh Sir Chemistry ⚗️🧪

COMMENTS

6.4 Bohr's Model of the Hydrogen Atom
Historically, Bohr's model of the hydrogen atom is the very first model of atomic structure that correctly explained the radiation spectra of atomic hydrogen. The model has a special place in the history of physics because it introduced an early quantum theory, which brought about new developments in scientific thought and later culminated in ...
Bohr model
Bohr model, description of the structure of atoms, especially that of hydrogen, proposed (1913) by the Danish physicist Niels Bohr.The Bohr model of the atom, a radical departure from earlier, classical descriptions, was the first that incorporated quantum theory and was the predecessor of wholly quantum-mechanical models. The Bohr model and all of its successors describe the properties of ...
Bohr model
The Bohr model of the hydrogen atom (Z = 1) or a hydrogen-like ion (Z > 1), where the negatively charged electron confined to an atomic shell encircles a small, positively charged atomic nucleus and where an electron jumps between orbits, is accompanied by an emitted or absorbed amount of electromagnetic energy (hν). [1] The orbits in which the electron may travel are shown as grey circles ...
30.3 Bohr's Theory of the Hydrogen Atom
We see that Bohr's theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to 1 / n 2 1 / n 2, where n n is a non-negative integer. A downward transition releases energy, and so n i n i must be greater than n f n f. The various ...
Bohr's Model of the Hydrogen Atom
Describe the Rutherford gold foil experiment and the discovery of the atomic nucleus; Explain the atomic structure of hydrogen; ... The Bohr model of hydrogen is a semi-classical model because it combines the classical concept of electron orbits with the new concept of quantization. The remarkable success of this model prompted many physicists ...
30.3 Bohr's Theory of the Hydrogen Atom
Bohr's theory explained the atomic spectrum of hydrogen and established new and broadly applicable principles in quantum mechanics. Figure 1. Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen atom.
Bohr's Theory of the Hydrogen Atom
Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen atom. His many contributions to the development of atomic physics and quantum mechanics, his personal influence on many students and colleagues, and his personal integrity, especially in the face of Nazi oppression, earned him a prominent place in history.
Bohr's Theory of the Hydrogen Atom
We see that Bohr's theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to [latex]\frac{1}{n^2}\\[/latex], where n is a non-negative integer. A downward transition releases energy, and so n i must be greater than n f. The ...
169 Bohr's Theory of the Hydrogen Atom
Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen atom. His many contributions to the development of atomic physics and quantum mechanics, his personal influence on many students and colleagues, and his personal integrity, especially in the face of Nazi oppression, earned him a prominent place in history.
Bohr Model of the Hydrogen Atom
Bohr's Theory of Hydrogen Atoms; Bohr's Equation. Bohr Model of the hydrogen atom first proposed the planetary model, but later an assumption concerning the electrons was made. The assumption was the quantization of the structure of atoms. Bohr's proposed that electrons orbited the nucleus in specific orbits or shells with a fixed radius.

30.3 Bohr’s Theory of the Hydrogen Atom

Mysteries of Atomic Spectra

Example 30.1

Strategy and Concept

Solution for (a)

Discussion for (a)

Solution for (b)

Discussion for (b)

Bohr’s Solution for Hydrogen

Triumphs and Limits of the Bohr Theory

PhET Explorations

30.3 Bohr’s Theory of the Hydrogen Atom

Mysteries of Atomic Spectra

Calculating Wave Interference of a Hydrogen Line

Bohr’s Solution for Hydrogen

Triumphs and Limits of the Bohr Theory

PhET Explorations: Models of the Hydrogen Atom

Section Summary

Conceptual Questions

Share This Book

Atomic Physics

Mysteries of Atomic Spectra

Example 1. Calculating Wave Interference of a Hydrogen Line

Strategy and Concept

Solution for Part 1

Discussion for Part 1

Solution for Part 2

Discussion for Part 2

Bohr’s Solution for Hydrogen

Triumphs and Limits of the Bohr Theory

PhET Explorations: Models of the Hydrogen Atom

Section Summary

Conceptual Questions

Problems & Exercises

Selected Solutions to Problems & Exercises

Candela Citations

169 Bohr’s Theory of the Hydrogen Atom

Learning Objectives

Mysteries of Atomic Spectra

Bohr’s Solution for Hydrogen

Triumphs and Limits of the Bohr Theory

Section Summary

Conceptual Questions

Problems & Exercises

Share This Book

Bohr Model of the Hydrogen Atom

Planetary Model of the Atom

Bohr’s Equation

Absorption and Emission

Limitations of the Bohr Model of the Hydrogen Atom:

Leave a Comment Cancel reply

Register with BYJU'S & Download Free PDFs

IMAGES

VIDEO

COMMENTS