• Energy Of Hydrogen Atom In Ground State In Joules
• Energy Of Hydrogen Atom At N 3

The total energy of any level in Hdrogen atom is E = -13.6/n2E (ground state) = -13.6evE (first exited state) = -13.6/4 ev potential of ground state = -27.2evif potential of ground state is 0 then we have to add +27.2, by this method wecan make g. HYDROGEN IS THE universe's simplest atom: a single electron orbiting a single proton. In a fuel cell, incoming hydrogen gas is separated by a catalyst at the anode into protons and electrons. Estimate the Hydrogen Ground State Energy The reason the Hydrogen atom (and other atoms) is so large is the essentially uncertainty principle. If the electron were confined to a smaller volume, would increase, causing to increase on average. The energy would increase not decrease.

Recommended ionization energy: Δ f H (+) ion,0K: Enthalpy of formation of positive ion at 0K: Δ r G° Free energy of reaction at standard conditions: Δ r H° Enthalpy of reaction at standard conditions.

Learning Outcomes

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

Following the work of Ernest Rutherford and his colleagues in the early twentieth century, the picture of atoms consisting of tiny dense nuclei surrounded by lighter and even tinier electrons continually moving about the nucleus was well established. This picture was called the planetary model, since it pictured the atom as a miniature “solar system” with the electrons orbiting the nucleus like planets orbiting the sun.

The simplest atom is hydrogen, consisting of a single proton as the nucleus about which a single electron moves. The electrostatic force attracting the electron to the proton depends only on the distance between the two particles. This classical mechanics description of the atom is incomplete, however, since an electron moving in an elliptical orbit would be accelerating (by changing direction) and, according to classical electromagnetism, it should continuously emit electromagnetic radiation. This loss in orbital energy should result in the electron’s orbit getting continually smaller until it spirals into the nucleus, implying that atoms are inherently unstable.

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

[latex]mid Delta Emid =mid {E}_{text{f}}-{E}_{text{i}}mid =hnu =dfrac{hc}{lambda }[/latex]

In this equation, h is Planck’s constant and E_i and E_f are the initial and final orbital energies, respectively. The absolute value of the energy difference is used, since frequencies and wavelengths are always positive. Instead of allowing for continuous values for the angular momentum, energy, and orbit radius, Bohr assumed that only discrete values for these could occur (actually, quantizing any one of these would imply that the other two are also quantized). Bohr’s expression for the quantized energies is:

[latex]{E}_{n}=-dfrac{k}{{n}^{2}},n=1,2,3,dots [/latex]

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

[latex]Delta E=kleft(dfrac{1}{{n}_{1}^{2}}-dfrac{1}{{n}_{2}^{2}}right)=dfrac{hc}{lambda }[/latex]

or

[latex]dfrac{1}{lambda }=dfrac{k}{hc}left(dfrac{1}{{n}_{1}^{2}}-dfrac{1}{{n}_{2}^{2}}right)[/latex]

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

Figure 1. Quantum numbers and energy levels in a hydrogen atom. The more negative the calculated value, the lower the energy.

The lowest few energy levels are shown in Figure 1. One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the n = 1 orbit, the orbit in which it has the lowest energy. When the electron is in this lowest energy orbit, the atom is said to be in its ground electronic state (or simply ground state). If the atom receives energy from an outside source, it is possible for the electron to move to an orbit with a higher n value and the atom is now in an excited electronic state (or simply an excited state) with a higher energy. When an electron transitions from an excited state (higher energy orbit) to a less excited state, or ground state, the difference in energy is emitted as a photon. Similarly, if a photon is absorbed by an atom, the energy of the photon moves an electron from a lower energy orbit up to a more excited one. We can relate the energy of electrons in atoms to what we learned previously about energy. The law of conservation of energy says that we can neither create nor destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, that same amount of energy will be liberated when the electron returns to its initial state (Figure 2). In effect, an atom can “store” energy by using it to promote an electron to a state with a higher energy and release it when the electron returns to a lower state. The energy can be released as one quantum of energy, as the electron returns to its ground state (say, from n = 5 to n = 1), or it can be released as two or more smaller quanta as the electron falls to an intermediate state, then to the ground state (say, from n = 5 to n = 4, emitting one quantum, then to n = 1, emitting a second quantum).

Figure 2. The horizontal lines show the relative energy of orbits in the Bohr model of the hydrogen atom, and the vertical arrows depict the energy of photons absorbed (left) or emitted (right) as electrons move between these orbits.

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

[latex]{E}_{n}=-dfrac{k{Z}^{2}}{{n}^{2}}[/latex]

The sizes of the circular orbits for hydrogen-like atoms are given in terms of their radii by the following expression, in which [latex]{a}_{0}[/latex] is a constant called the Bohr radius, with a value of 5.292 [latex]times [/latex] 10⁻¹¹ m:

[latex]r=dfrac{{n}^{2}}{Z}{a}_{0}[/latex]

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

[latex]Delta E={E}_{nlongrightarrow infty }-{E}_{1}=0+k=k[/latex]

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

You can view the transcript for “Bohr Model of the Hydrogen Atom, Electron Transitions, Atomic Energy Levels, Lyman & Balmer Series” here (opens in new window).

Energy Of Hydrogen Atom In Ground State In Joules

Example 1: Calculating the Energy of an Electron in a Bohr Orbit

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

The energy of the electron is given by this equation:

[latex]E=dfrac{-k{Z}^{2}}{{n}^{2}}[/latex]

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

[latex]E=dfrac{-left(2.179times {10}^{-18}text{J}right)times {left(1right)}^{2}}{{left(3right)}^{2}}=-2.421times {10}^{-19}text{J}[/latex]

Check Your Learning

The electron in Figure 2 is promoted even further to an orbit with n = 6. What is its new energy?

Example 2: Calculating the Energy and Wavelength of Electron Transitions in a One–electron (Bohr) System

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

In this case, the electron starts out with n = 4, so n₁ = 4. It comes to rest in the n = 6 orbit, so n₂ = 6. The difference in energy between the two states is given by this expression:

[latex]begin{array}{rcl}Delta E&=&{E}_{1}-{E}_{2}=2.179times {10}^{-18}left(dfrac{1}{{n}_{1}^{2}}-dfrac{1}{{n}_{2}^{2}}right)Delta E&=&2.179times {10}^{-18}left(dfrac{1}{{4}^{2}}-dfrac{1}{{6}^{2}}right)text{J}Delta E&=&2.179times {10}^{-18}left(dfrac{1}{16}-dfrac{1}{36}right)text{J}Delta E&=&7.566times {10}^{-20}text{J}end{array}[/latex]

This energy difference is positive, indicating a photon enters the system (is absorbed) to excite the electron from the n = 4 orbit up to the n = 6 orbit. The wavelength of a photon with this energy is found by the expression [latex]Etext{=}dfrac{hc}{lambda }[/latex]. Rearrangement gives:

[latex]begin{array}{rll}lambda&=&dfrac{hc}{E} &=&left(6.626times {10}^{-34}cancel{text{J}}cancel{text{s}}right)times dfrac{2.998times {10}^{8}text{m}{cancel{text{s}}}^{-1}}{7.566times {10}^{-20}cancel{text{J}}} &=&2.626times {10}^{-6}text{m}end{array}[/latex]

From Figure 3, we can see that this wavelength is found in the infrared portion of the electromagnetic spectrum.

Figure 3. Portions of the electromagnetic spectrum are shown in order of decreasing frequency and increasing wavelength. Examples of some applications for various wavelengths include positron emission tomography (PET) scans, X-ray imaging, remote controls, wireless Internet, cellular telephones, and radios. (credit “Cosmic ray”: modification of work by NASA; credit “PET scan”: modification of work by the National Institute of Health; credit “X-ray”: modification of work by Dr. Jochen Lengerke; credit “Dental curing”: modification of work by the Department of the Navy; credit “Night vision”: modification of work by the Department of the Army; credit “Remote”: modification of work by Emilian Robert Vicol; credit “Cell phone”: modification of work by Brett Jordan; credit “Microwave oven”: modification of work by Billy Mabray; credit “Ultrasound”: modification of work by Jane Whitney; credit “AM radio”: modification of work by Dave Clausen)

Check Your Learning

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

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

• The energies of electrons (energy levels) in an atom are quantized, described by quantum numbers: integer numbers having only specific allowed value and used to characterize the arrangement of electrons in an atom.
• An electron’s energy increases with increasing distance from the nucleus.
• The discrete energies (lines) in the spectra of the elements result from quantized electronic energies.

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

Think about It

How are the Bohr model and the Rutherford model of the atom similar? How are they different?

Key Concepts and Summary

Bohr incorporated Planck’s and Einstein’s quantization ideas into a model of the hydrogen atom that resolved the paradox of atom stability and discrete spectra. The Bohr model of the hydrogen atom explains the connection between the quantization of photons and the quantized emission from atoms. Bohr described the hydrogen atom in terms of an electron moving in a circular orbit about a nucleus. He postulated that the electron was restricted to certain orbits characterized by discrete energies. Transitions between these allowed orbits result in the absorption or emission of photons. When an electron moves from a higher-energy orbit to a more stable one, energy is emitted in the form of a photon. To move an electron from a stable orbit to a more excited one, a photon of energy must be absorbed. Using the Bohr model, we can calculate the energy of an electron and the radius of its orbit in any one-electron system.

Key Equations

• [latex]{E}_{n}=-dfrac{k{Z}^{2}}{{n}^{2}},n=1,2,3,dots [/latex]
• [latex]Delta E=k{Z}^{2}left(dfrac{1}{{n}_{1}^{2}}-dfrac{1}{{n}_{2}^{2}}right)[/latex]
• [latex]r=dfrac{{n}^{2}}{Z}{a}_{0}[/latex]

Try It

1. Why is the electron in a Bohr hydrogen atom bound less tightly when it has a quantum number of 3 than when it has a quantum number of 1?
2. What does it mean to say that the energy of the electrons in an atom is quantized?
3. Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.
4. The electron volt (eV) is a convenient unit of energy for expressing atomic-scale energies. It is the amount of energy that an electron gains when subjected to a potential of 1 volt; 1 eV = 1.602 [latex]times [/latex] 10^–19 J. Using the Bohr model, determine the energy, in electron volts, of the photon produced when an electron in a hydrogen atom moves from the orbit with n = 5 to the orbit with n = 2. Show your calculations.
5. Using the Bohr model, determine the lowest possible energy, in joules, for the electron in the Li²⁺ ion.
6. Using the Bohr model, determine the lowest possible energy for the electron in the He⁺ ion.
7. Using the Bohr model, determine the energy of an electron with n = 6 in a hydrogen atom.
8. Using the Bohr model, determine the energy of an electron with n = 8 in a hydrogen atom.
9. How far from the nucleus in angstroms (1 angstrom = 1 × 10^–10 m) is the electron in a hydrogen atom if it has an energy of –8.72 × 10^–20 J?
10. What is the radius, in angstroms, of the orbital of an electron with n = 8 in a hydrogen atom?
11. Using the Bohr model, determine the energy in joules of the photon produced when an electron in a He⁺ ion moves from the orbit with n = 5 to the orbit with n = 2.
12. Using the Bohr model, determine the energy in joules of the photon produced when an electron in a Li²⁺ ion moves from the orbit with n = 2 to the orbit with n = 1.
13. Consider a large number of hydrogen atoms with electrons randomly distributed in the n = 1, 2, 3, and 4 orbits.
    1. How many different wavelengths of light are emitted by these atoms as the electrons fall into lower-energy orbitals?
    2. Calculate the lowest and highest energies of light produced by the transitions described in part (a).
    3. Calculate the frequencies and wavelengths of the light produced by the transitions described in part (b).

2. Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted.

4. The answer can be found as follows:

[latex]begin{array}{ll}Ehfill & ={E}_{2}-{E}_{5}=2.179times {10}^{-18}left(frac{1}{{n}_{2}^{2}}-frac{1}{{n}_{5}^{2}}right)text{J}hfill hfill & =2.179times {10}^{-18}left(frac{1}{{2}^{2}}-frac{1}{{5}^{2}}right)=4.576times {10}^{-19}text{J}hfill hfill & =frac{4.576times {10}^{-19}cancel{text{J}}}{1.602times {10}^{-19}cancel{text{J}}{text{eV}}^{-1}}=2.856text{eV}hfill end{array}[/latex]

6. −8.716 × 10⁻¹⁸ J

8. −3.405 × 10⁻²⁰ J

10. 33.9 Å

12. 1.471 × 10⁻¹⁷ J

Glossary

Bohr’s model of the hydrogen atom: structural model in which an electron moves around the nucleus only in circular orbits, each with a specific allowed radius; the orbiting electron does not normally emit electromagnetic radiation, but does so when changing from one orbit to another.

excited state: state having an energy greater than the ground-state energy

ground state: state in which the electrons in an atom, ion, or molecule have the lowest energy possible

quantum number: integer number having only specific allowed values and used to characterize the arrangement of electrons in an atom

As you may remember from chemistry, an atom consists of electrons orbiting around a nucleus. However, the electrons cannot choose any orbit they wish. They are restricted to orbits with only certain energies. Electrons can jump from one energy level to another, but they can never have orbits with energies other than the allowed energy levels.

Let's look at the simplest atom, a neutral hydrogen atom. Its energy levels are given in the diagram below. The x-axis shows the allowed energy levels of electrons in a hydrogen atom, numbered from 1 to 5. The y-axis shows each level's energy in electron volts (eV). One electron volt is the energy that an electron gains when it travels through a potential difference of one volt (1 eV = 1.6 x 10^-19 Joules).

Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy. If it is in the second energy level, it must have -3.4 eV of energy. An electron in a hydrogen atom cannot have -9 eV, -8 eV or any other value in between.

Let's say the electron wants to jump from the first energy level, n = 1, to the second energy level n = 2. The second energy level has higher energy than the first, so to move from n = 1 to n = 2, the electron needs to gain energy. It needs to gain (-3.4) - (-13.6) = 10.2 eV of energy to make it up to the second energy level.

The electron can gain the energy it needs by absorbing light. If the electron jumps from the second energy level down to the first energy level, it must give off some energy by emitting light. The atom absorbs or emits light in discrete packets called photons, and each photon has a definite energy. Only a photon with an energy of exactly 10.2 eV can be absorbed or emitted when the electron jumps between the n = 1 and n = 2 energy levels.

The energy that a photon carries depends on its wavelength. Since the photons absorbed or emitted by electrons jumping between the n = 1 and n = 2 energy levels must have exactly 10.2 eV of energy, the light absorbed or emitted must have a definite wavelength. This wavelength can be found from the equation

E = hc/l,

where E is the energy of the photon (in eV), h is Planck's constant (4.14 x 10^-15 eV s) and c is the speed of light (3 x 10⁸ m/s). Rearranging this equation to find the wavelength gives

l = hc/E.

A photon with an energy of 10.2 eV has a wavelength of 1.21 x 10^-7 m, in the ultraviolet part of the spectrum. So when an electron wants to jump from n = 1 to n = 2, it must absorb a photon of ultraviolet light. When an electron drops from n = 2 to n = 1, it emits a photon of ultraviolet light.

The step from the second energy level to the third is much smaller. It takes only 1.89 eV of energy for this jump. It takes even less energy to jump from the third energy level to the fourth, and even less from the fourth to the fifth.

What would happen if the electron gained enough energy to make it all the way to 0eV? The electron would then be free of the hydrogen atom. The atom would be missing an electron, and would become a hydrogen ion.

The table below shows the first five energy levels of a hydrogen atom.

Energy Of Hydrogen Atom At N 3

You can use this method to find the wavelengths emitted by electrons jumping between energy levels in various elements. However, finding the correct energy levels gets much more difficult for larger atoms with many electrons. In fact, the energy levels of neutral helium are different from the energy levels of singly ionized helium! Therefore, we will skip how to calculate all the energy levels for different atoms for now. The energy levels are published in the CRC Handbook of Chemistry and Physics if you want to look them up.