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6.4.3 Bohr Model

By the end of this lesson you can:

  • Describe the Bohr model of the hydrogen atom
  • Explain emission lines as energy-level transitions
  • Calculate the photon energy and wavelength for a transition
Find your course: NSW Year 11NSW Year 12: Module 8, IQ3QLD Unit 4: Quantum theoryVIC Unit 4: Light, matter and relativityAP Physics 2: Unit 15IB: E.1 Structure of the atom
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Learn the idea: a short explanation, then a check.

Following Ernest Rutherford’s 1911 alpha-scattering experiment, which established the planetary model of the atom, classical physics faced a critical flaw regarding atomic stability.

The Classical Dilemma: Radiation Collapse

According to Maxwell’s Electromagnetic Theory, any accelerating charged particle must continuously radiate energy in the form of electromagnetic waves.

Because an electron orbiting a positively charged nucleus undergoes continuous centripetal acceleration (ac = v2 / r), classical theory predicts that the electron must continuously lose kinetic energy. As its energy decreases, the orbital radius shrinks, causing the electron to spiral inward toward the nucleus and collapse the atom in a fraction of a second (∼ 10−11 s).

ac = v2⁄r = e2⁄4π ε0 me r2

Clearly, stable atoms exist—meaning classical electrodynamics fails at the atomic scale.

Bohr's Quantum Breakthrough

In 1913, Danish physicist Niels Bohr resolved this paradox by incorporating Max Planck’s concept of energy quantization into the atomic model.

Before examining Bohr’s key postulates and mathematical formulation, we must look at another critical piece of experimental evidence that shaped his work: discrete atomic spectral lines.


In the late 1800s, spectroscopy and spectral analysis advanced significantly following Joseph von Fraunhofer’s discovery of dark absorption lines in the solar spectrum.

However, a major question remained:

Was there a mathematical pattern to these discrete spectral lines?

The key breakthroughs came from Swedish physicist Johannes Rydberg and Swiss mathematician Johann Balmer, who derived empirical formulas to predict the exact wavelengths of light emitted by hydrogen atoms.

Watch the video to learn how their mathematical equations paved the way for Niels Bohr's quantum model of the atom.

Theory

We have now covered the groundwork behind Bohr's model: the behavior of atomic spectra, the empirical mathematical models of Balmer and Rydberg, the instability of the classical Rutherford model, and Planck’s quantum hypothesis that energy is quantized.

With these foundational concepts in place, Niels Bohr set out to construct a new atomic model founded on quantized energy states (though, as you will see, it still retained several classical elements).

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An emission spectrum of hydrogen reveals four lines in the visible range, called the Balmer Series. What frequency of light is emitted if the electron transition associated with this frequency moves from 5 to 2?

Practice Problems: The Bohr Model and the Hydrogen Spectrum

Rydberg formula: 1/λ = R (1/nf2 - 1/ni2), where R = 1.097 × 107 m-1. In the Bohr model the energy of level n in hydrogen is En = -13.6 / n2 eV.

1. Construct a diagram, using the Bohr model of the atom, of an excited electron dropping from the n = 5 level to the n = 2 level.

What to show: the nucleus, the circular orbits (levels) n = 1, 2, 3, 4 and 5, the electron starting in the n = 5 orbit, an arrow to the n = 2 orbit, and the photon emitted.

n = 2 n = 3 n = 4 n = 5 electron (n = 5) photon emitted

The orbit radii are drawn in proportion to n2.

Energy of the photon: In hydrogen the energy of level n is En = -13.6 / n2 eV. So E5 = -0.54 eV and E2 = -3.40 eV.

The energy lost by the electron is carried away by one photon: E = E5 - E2 = -0.54 - (-3.40) = 2.86 eV = 4.6 × 10-19 J, which is a photon of wavelength about 434 nm (violet-blue light in the Balmer series).

Answer: A diagram like the one above, with the electron jumping from n = 5 to n = 2 and a photon of energy 2.86 eV emitted
2. The Balmer series is the set of visible spectral lines of hydrogen. Use the Rydberg formula, where R = 1.097 × 107 m-1 and nf = 2, to calculate the wavelength of the lines (in metres) for ni = 3, 5 and 7.

Given: R = 1.097 × 107 m-1, nf = 2

Formula: 1/λ = R (1/nf2 - 1/ni2)

ni = 3: 1/λ = 1.097 × 107 × (1/4 - 1/9) = 1.097 × 107 × 0.1389 = 1.524 × 106 m-1, so λ = 6.56 × 10-7 m

ni = 5: 1/λ = 1.097 × 107 × (1/4 - 1/25) = 1.097 × 107 × 0.2100 = 2.304 × 106 m-1, so λ = 4.34 × 10-7 m

ni = 7: 1/λ = 1.097 × 107 × (1/4 - 1/49) = 1.097 × 107 × 0.2296 = 2.519 × 106 m-1, so λ = 3.97 × 10-7 m

Answer: 6.56 × 10-7 m (ni = 3, red), 4.34 × 10-7 m (ni = 5, violet-blue) and 3.97 × 10-7 m (ni = 7, violet)
3. Use the Rydberg equation to calculate the wavelength (in metres) of the radiation emitted when an excited electron drops from the n = 7 level to the n = 2 level.

Given: ni = 7, nf = 2, R = 1.097 × 107 m-1

Formula: 1/λ = R (1/nf2 - 1/ni2)

Calculation: 1/λ = 1.097 × 107 × (1/22 - 1/72) = 1.097 × 107 × (0.25 - 0.0204) = 2.519 × 106 m-1

λ = 1 / (2.519 × 106) = 3.97 × 10-7 m

This is the same line as the ni = 7 line in question 2. It is in the violet, near the edge of the visible spectrum.

Answer: 3.97 × 10-7 m (397 nm)
4. Identify TWO limitations of the Bohr model of the atom.

Limitation 1: It only works for hydrogen and other one-electron ions (such as He+). It cannot predict the spectra of atoms with more than one electron, because it does not include the repulsion between electrons.

Limitation 2: It cannot explain the relative intensities of the spectral lines, or why some lines are split into two or more close lines (fine structure, and splitting in a magnetic field).

Other accepted limitations: It treats the electron as a particle in a definite orbit, which breaks the uncertainty principle and does not fit the wave nature of the electron. It does not explain why the electron does not radiate energy in a stable orbit (it is simply assumed). It does not explain chemical bonding.

Answer: Any two, for example: it works only for hydrogen-like (one-electron) atoms, and it cannot explain the intensity or splitting of spectral lines
5. How does the Bohr model of the atom account for the emission spectrum of hydrogen?

Step 1 (Stationary orbits): The electron can only be in certain orbits, each with a fixed energy, En = -13.6 / n2 eV. While it stays in an orbit it does not radiate.

Step 2 (Excitation): When the gas is heated or an electric current passes through it, electrons gain energy and jump up to higher orbits (excited states).

Step 3 (Emission): An excited electron falls to a lower orbit and emits one photon whose energy equals the difference between the two levels: hf = Ei - Ef.

Step 4 (Lines): Because the energy levels are fixed, only certain energy differences are possible, so only certain frequencies are emitted. This gives a spectrum of separate bright lines instead of a continuous spectrum. Jumps to n = 2 give the visible Balmer series, and the wavelengths agree with the Rydberg formula.

Answer: Electrons move between fixed energy levels, and each drop emits a photon of energy hf = Ei - Ef, so only particular frequencies (lines) appear
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​From Walter Fendt - self explanatory. It does mention de Broglie , which is in the next lesson