Niels Bohr (1885-1962), Danish physicist and winner of the 1922 Nobel Prize in physics, proposed in 1913 the first quantized model of the atom: in it, the electron occupies only discrete energy levels. This model explains the spectral lines of hydrogen but has since been superseded by wave quantum mechanics.
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How to measure it in class
Atomic energy levels cannot be measured directly with a smartphone, but their most visible consequence can be observed: some light sources emit a spectrum of discrete lines, the signature of quantized energy levels, whereas others emit a continuous spectrum.
Steps:
- Build a simple spectroscope: an opaque cardboard box pierced with a fine slit on one side, with a piece of a recorded CD or DVD acting as a diffraction grating on the other.
- Aim at an incandescent lamp or daylight through the slit and photograph the resulting spectrum with the smartphone camera: the spectrum is continuous, all the colors follow one another without interruption.
- Then aim at a fluorescent tube, an energy-saving lamp or sodium street lighting, and photograph in the same way: the spectrum is made of lines separated by dark zones. These are the transitions between discrete energy levels of the atoms in the vapor.
- Open FizziQ and use the colorimeter to compare the red, green and blue components of the different sources, and highlight the difference in spectral composition between a continuous source and a line source.
- Measure on the photographs the position of the lines relative to a continuous reference spectrum, and estimate their wavelength.
- Compare the estimated wavelengths to the tabulated values of the hydrogen lines (656 nm, 486 nm, 434 nm, 410 nm) and check that they are indeed predicted by Bohr’s energy level formula.
Scientific activities on this topic
Using the story of Niels Bohr as a pretext, physicist Julien Bobroff and his team studied multiple ways to calculate the height of a building with a smartphone (see the project).
You can redo them with FizziQ, but why not also think about other suggestions, for example by using FizziQ’s kinematics module?
Learn more
The problem Bohr wanted to solve
In 1911, Rutherford showed that the atom consists of a tiny, very dense nucleus surrounded by electrons. This model has a fatal flaw: according to classical electromagnetism, an accelerated charge radiates energy. An orbiting electron should therefore lose its energy in a spiral and crash into the nucleus in a fraction of a nanosecond. Yet matter is stable. Moreover, it had been known since Balmer (1885) that the hydrogen lines obey a simple numerical formula, without anyone knowing why.
The postulates of 1913
Bohr solved both problems at once, at the cost of postulates that he did not justify and that contradict classical physics. First, the electron can only occupy certain orbits, called stationary orbits, on which it does not radiate - which ensures the stability of the atom. Second, these orbits are those for which the angular momentum is an integer multiple of ħ = h/2π: this is quantization. Third, the atom emits or absorbs light only when passing from one level to another, and the energy of the exchanged photon is exactly the energy difference between the two levels.
From these postulates follows the expression for the energy levels of hydrogen, E_n = −13.6/n² eV, and therefore Balmer’s formula, this time recovered from fundamental constants. It is this quantitative success that established the model.
Why this model is wrong, and what to retain from it
Bohr’s model fails as soon as helium, which has two electrons: it does not correctly predict any spectrum other than that of hydrogen and of one-electron ions such as He⁺. It explains neither the relative intensity of the lines, nor chemical bonding, nor fine structure. Above all, it rests on an inaccurate picture: the electron does not have a circular trajectory. Heisenberg’s indeterminacy principle (1927) forbids attributing to it simultaneously a well-defined position and velocity. Quantum mechanics describes it by a wave function whose square gives the probability density of presence: these are orbitals, fuzzy regions of spherical, dumbbell or more complex shape, not orbits.
What survives of the model is essential: the energy levels of the atom are quantized, the principal quantum number n indexes them, and the relation ΔE = h·ν between transition and photon remains exact. It is on this basis that it appears in the high school physics curriculum, as a historical model and calculation tool, not as a description of reality.
Bohr after 1913
In 1921 Bohr founded in Copenhagen the institute that would become the world center of quantum physics and through which Heisenberg, Pauli, Dirac and Landau would pass. There he developed the principle of complementarity, according to which the wave and particle descriptions are two mutually exclusive but equally necessary facets of the same object: it is the heart of what is called the Copenhagen interpretation. His debates with Einstein on the meaning of quantum mechanics, from 1927 to the end of their lives, remain famous. In 1939, with John Wheeler, he proposed the liquid drop model to explain nuclear fission. Jewish through his mother, he fled occupied Denmark in 1943 and briefly took part in the Manhattan Project, before campaigning after the war for international control of the atomic weapon.
The barometer anecdote
The story of the student who, asked to measure the height of a building with a barometer, proposes everything except the pressure measurement is regularly attributed to Bohr. It is apocryphal: its oldest known version is a text by the educator Alexander Calandra published in 1958, in which the student is anonymous. The attribution to Bohr came later. It remains an excellent illustration of the inquiry-based approach: the same problem admits several measurement protocols, and the student’s role is to compare them in terms of precision, cost and feasibility.
It is romanticized by Murray Gell-Mann in his book The Quark and the Jaguar as follows:
“I received a phone call from a colleague about a student. He felt he should give him a zero on a physics question, while the student was demanding a perfect score. The professor and student agreed to choose an impartial arbiter, and I was chosen. I read the exam question:
Show how it is possible to determine the height of a building using a barometer.
The student had answered: Take the barometer to the top of the building, tie a rope to it, lower it to the ground, then pull it back up and measure the length of the rope. The length of the rope gives the height of the building.
The student was right in that he had answered correctly and completely. On the other hand, I could not give him full marks: in that case, he would have received his physics degree without having shown me any knowledge of physics. I proposed to give the student another chance by giving him six minutes to answer the question with the warning that for the answer he had to use his knowledge of physics. After five minutes, he had not written anything yet. I asked him if he wanted to give up, but he replied that he had many answers to this problem and was looking for the best one. I apologized for interrupting him and asked him to continue. In the minute that followed, he hurried to answer me:
- Place the barometer at the height of the roof. Drop it while measuring its fall time with a stopwatch. Then, using the formula x = 1/2gt*t, you find the height of the building.
At this point, I asked my colleague if he wanted to give up. He replied yes and gave the student nearly full marks. Leaving his office, I called back the student because he had said he had several solutions to this problem.
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Well, he said, there are several ways to calculate the height of a building with a barometer. For example, place it outside when there is sun. Measure the height of the barometer, the length of its shadow, and the length of the building’s shadow. Then, with a simple proportion calculation, you find the height of the building.
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Good, I replied, and the others.
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There is a fairly basic method that you will appreciate. You climb the stairs with a barometer and at the same time mark the length of the barometer on the wall. By counting the number of marks, you have the height of the building in barometer lengths. It is a very direct method. Of course, if you want a more sophisticated method, you can hang the barometer from a rope, swing it like a pendulum, and determine the value of g at street level and at roof level. From the difference in g, the height of the building can be calculated. Similarly, attach it to a long rope and, while on the roof, lower it almost to street level. Swing it like a pendulum and calculate the height of the building from the period of oscillations.
Finally, he concluded:
- There are still other ways to solve this problem. Probably the best is to go to the basement, knock on the janitor’s door, and say: ‘I have a superb barometer for you if you tell me the height of the building.’
I then asked the student if he knew the answer I expected. He admitted that yes, but that he was fed up with the university and professors trying to teach him how he should think.”
Orders of magnitude
Ionization energy of hydrogen from the ground state: 13.6 eV, that is 2.18×10⁻¹⁸ J. Bohr radius, characteristic size of the hydrogen atom: a₀ = 52.9 pm, that is 5.29×10⁻¹¹ m. Radius of the nucleus: about 10⁻¹⁵ m, one hundred thousand times smaller. Wavelengths of the first four visible lines of hydrogen (Balmer series): 656 nm (red), 486 nm (blue-green), 434 nm and 410 nm (violet). Planck constant: h = 6.626×10⁻³⁴ J·s. Energy of a visible photon: from 1.6 eV (red) to 3.1 eV (violet).
Formula
Energy of the levels of the hydrogen atom, in Bohr’s model:
E_n = −13.6 / n² (in eV)
where:
- E_n: energy of level n (eV)
- n: principal quantum number, non-zero natural integer (n = 1 for the ground state)
The energy is negative because it is measured from the free electron at rest (E = 0). It is therefore necessary to supply 13.6 eV to remove the electron from the ground state.
Energy of the photon emitted or absorbed during a transition between two levels:
ΔE = E_p − E_n = h · ν = h · c / λ
where:
- ΔE: energy variation of the atom (J)
- h: Planck constant, 6.626×10⁻³⁴ J·s
- ν: frequency of the photon (Hz)
- c: speed of light in vacuum, 3.00×10⁸ m/s
- λ: wavelength of the photon (m)
Beware of units: 1 eV = 1.602×10⁻¹⁹ J.
Quantization condition on the angular momentum, the central postulate of 1913:
m · v · r = n · ħ with ħ = h / 2π
where:
- m: mass of the electron, 9.109×10⁻³¹ kg
- v: speed of the electron on the orbit (m/s)
- r: radius of the orbit (m)
- n: non-zero natural integer
This formulation assumes a well-defined orbit: it is convenient for calculation, but it does not describe the reality of the electron in the atom.
Application examples
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The transition from level n = 3 to n = 2 releases ΔE = 13.6 × (1/4 − 1/9) = 1.89 eV, that is a photon with a wavelength of 656 nm: this is the red Hα line, visible in nebulae and responsible for their color.
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Ionizing the hydrogen atom from its ground state requires 13.6 eV, which corresponds to an ultraviolet photon of 91 nm: visible light cannot ionize hydrogen.
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A sodium streetlamp emits a characteristic orange light of 589 nm, produced by a single, well-defined electronic transition of the sodium atom.
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Astronomers identify the chemical composition of a star thousands of light-years away by recognizing, in its spectrum, the absorption lines specific to each element.
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A flame test in the laboratory gives a yellow flame for sodium, carmine red for lithium, violet for potassium: each element has its own set of energy levels, hence its colored signature.
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A helium-neon laser emits at 632.8 nm because this wavelength corresponds exactly to a transition between two energy levels of neon.
FAQ
Q: Is Bohr’s model still valid? A: No, not as a description of the atom. It only works for hydrogen and one-electron ions, and it was replaced as early as 1925 by wave quantum mechanics. It is taught because it introduces quantization in a simple way and gives the correct energies for hydrogen, but it must be presented as a historical model.
Q: Does the electron really orbit the nucleus like a planet? A: No, and that is the most widespread error. The electron has no trajectory: one cannot say where it is and where it is going at the same instant. Quantum mechanics describes it by an orbital, a cloud of probability of presence. The planetary picture is a convenient diagram, not a photograph of the atom.
Q: Why are the energies negative? A: By convention, the energy of the free electron, motionless and infinitely far from the nucleus, is set to zero. A bound electron is in a more stable state, hence of lower energy, that is, negative. The more negative the energy, the more strongly bound the electron.
Q: Why is the hydrogen spectrum made of lines and not continuous? A: Because the atom can only occupy discrete energy levels. The possible transitions are therefore limited in number, and each produces a photon of a very precise energy, hence of a very precise wavelength. A line spectrum is direct proof that energy in the atom is quantized.
Q: Is the barometer story true? A: No. It circulates in a dozen versions and its oldest known form is a text by Alexander Calandra from 1958, in which the student has no name. It was attributed to Bohr afterwards, without any source. It retains all its pedagogical value, but it must not be presented as an episode from Bohr’s life.
Related concepts
Visible Spectrum - Spectral Analysis - Wavelength - Frequency