Chemistry: Atoms First 2e · Electronic Structure and Periodic Properties of Elements

The Bohr Model

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

In 1913, Niels Bohr built the first working model of the hydrogen atom that explained why atoms emit light only at specific colors, not a continuous rainbow. His key idea was quantization: the electron in a hydrogen atom can occupy only certain allowed orbits (energy levels), never the space between them. Each orbit corresponds to an integer quantum number n = 1, 2, 3, …. The energy of the electron in orbit n is:

$E n = -2.18 \times 10 -18 J \times 1 n 2$

The negative sign means the electron is bound to the nucleus; the most negative energy (n = 1, the ) is the most stable. When an electron jumps between allowed orbits, the atom absorbs or emits exactly one photon whose energy equals the difference between the two levels:

$E photon = |E final - E initial | = h ν$

This single idea turned the mystery of atomic line spectra into a calculable problem. The Bohr model works beautifully for hydrogen, but it fails for atoms with more than one electron — a limitation that pushed science toward full quantum theory.

Why this matters

The Bohr model is the bridge between the particle picture of light (Topic 1) and modern quantum mechanics (Topic 3). It explains:

Why each element has a unique spectrum — a set of colored lines — used in flame tests, fireworks, neon signs, astronomy, and forensic analysis.
How spectral lines reveal atomic structure: each line corresponds to a specific electron transition.
The concept of quantized energy levels, which survives in modern theory even though the orbit picture does not.

The hydrogen energy equation E_n = -2.18 × 10^-18 J/n² is still used today to describe hydrogen-like species (ions with one electron, such as He⁺ and Li²⁺). In physics, astronomy, and chemistry, "hydrogen-like" calculations power models of stars, fusion research, and quantum chemistry approximations. Understanding where the model succeeds and where it breaks is a classic example of how science works: a good but incomplete model gets replaced, not because it was useless, but because better evidence demanded better theory.

The college version

Core Concepts

Quantized energy levels

The electron in hydrogen can only occupy orbits with n = 1, 2, 3, …, where n is the principal quantum number. The energy of each level:

$E n = -2.18 \times 10 -18 J \times 1 n 2$

For example:

$E 1 = -2.18 \times 10 -18 J, E 2 = -5.45 \times 10 -19 J, E 3 = -2.42 \times 10 -19 J$

As n increases, the levels get closer together (the energy spacing shrinks) and approach zero energy — the point at which the electron is free of the nucleus. The electron is in the ground state at n = 1 (lowest energy) and in an at any n > 1.

Absorption and emission of photons

: an electron absorbs a photon whose energy exactly matches the gap between its current level and a higher level, and jumps up.
Emission: an electron falls from a higher level to a lower one, releasing the energy difference as a photon.

The photon energy is:

$E photon = h ν = hc λ = |E higher - E lower |$

Because only certain energy gaps exist, only certain photon frequencies exist — hence discrete spectral lines instead of a continuous spectrum.

The spectral series of hydrogen

Transitions that end on the same lower level form a series with a characteristic region of the spectrum:

Series	Lower level	Region
Lyman	n = 1	Ultraviolet
Balmer	n = 2	Visible
Paschen	n = 3	Infrared

The Balmer series is why hydrogen gas glows pink-violet in discharge tubes: those transitions land in the visible range. The Lyman series is mostly UV and invisible to the eye but detectable with instruments.

Why the model fails for multi-electron atoms

The Bohr model ignores electron–electron repulsion entirely. It treats each electron as if it orbited alone, which is a reasonable approximation only when there is a single electron. For helium (two electrons), the electrons repel each other and screen the nucleus from each other, so the simple 1/n² formula no longer matches experiment. Bohr's circular orbits were also later shown to be wrong in detail — electrons do not travel on fixed paths at all (Topic 3). The lasting legacy is quantization of energy, not the orbits.

How It Works / Step-by-Step Process

To find the wavelength of light emitted when a hydrogen electron transitions from level n_i to lower level n_f:

Calculate the initial energy with E_{n_i} = -2.18 × 10^-18 J/n_i².
Calculate the final energy with E_{n_f} = -2.18 × 10^-18 J/n_f².
Take the difference: ΔE = E_{n_i} - E_{n_f} (positive, since n_i > n_f).
Set the photon energy equal to this difference: E_photon = ΔE = hc/λ.
Solve for wavelength: λ= hc/ΔE, then convert meters to nanometers ( × 10⁹).
Check the region: visible light is 400–700 nm; UV is shorter; IR is longer.

Common Confusions

Common Confusion	Correct Understanding
Higher n means higher energy.	Yes — but "higher energy" means less negative; E approaches zero as n grows, and n = 1 is the lowest (most negative) energy.
The electron can be anywhere between orbits.	No — Bohr's central claim is that only specific orbits exist; energies in between are forbidden.
Emission and absorption are the same process.	Emission releases a photon when the electron falls; absorption consumes a photon when the electron rises.
The Bohr model describes all atoms.	It is exact only for one-electron species; electron–electron repulsion breaks it for everything else.
The Balmer series is the whole hydrogen spectrum.	It is only the visible set (final level n = 2); Lyman (UV) and Paschen (IR) are other series.
Photon energy equals the final energy of the level.	Photon energy equals the difference between levels, (	E_f - E_i	).

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a staircase with very special steps: the electron can stand only on the steps, never between them. When the electron gets energy, it jumps up a step; when it falls down a step, it lets out a little flash of light. Each step is a different color flash, and hydrogen only has a few step sizes, so it only makes a few colors — like a secret code that tells us which element is glowing.

Worked example

Example 1: The first Balmer line (H-alpha, 656 nm)

Find the wavelength of light emitted when a hydrogen electron falls from n = 3 to n = 2.

Energy of n = 3:

$E 3 = -2.18 \times 10 -18 J \times 132 = -2.42 \times 10 -19 J$

Energy of n = 2:

$E 2 = -2.18 \times 10 -18 J \times 122 = -5.45 \times 10 -19 J$

Energy difference (emitted photon):

$Δ E = E 3 - E 2 = (-2.42 \times 10 -19) - (-5.45 \times 10 -19) = 3.03 \times 10 -19 J$

Wavelength from E = hc/λ:

$λ = hc Δ E = (6.626 \times 10 -34 J\cdots)(3.00 \times 108 m/s) 3.03 \times 10 -19 J = 6.56 \times 10 -7 m = 656 nm$

Check the units: J·s × m/s / J = m. The answer, 656 nm, is deep red — this is the famous H-alpha line of hydrogen, visible in emission tubes and in the pink glow of many nebulae.

Example 2: Ionization energy of hydrogen

How much energy is required to ionize a hydrogen atom from its ground state, and what is this in kJ/mol?

Ionization means moving the electron from n = 1 to n = ∞, where E_∞= 0.

$Δ E = E \infty - E 1 = 0 - (-2.18 \times 10 -18 J) = 2.18 \times 10 -18 J per atom$

Convert to per mole:

$Δ E mol = (2.18 \times 10 -18 J)(6.022 \times 1023 mol -1) = 1.31 \times 106 J/mol = 1312 kJ/mol$

This matches the measured ionization energy of hydrogen (~1312 kJ/mol), a striking success of the model. It also explains why hydrogen is stable: pulling the electron away costs real energy, which must come from heat, electricity, or light.

Example 3: Recognizing the series

A photon of wavelength 97.2 nm is absorbed by hydrogen in its ground state. Which transition occurred, and is it part of the Lyman, Balmer, or Paschen series?

The wavelength 97.2 nm is ultraviolet, which suggests the Lyman series (transitions ending at n = 1). Check the energy of the photon:

$E = hc λ = (6.626 \times 10 -34 J\cdots)(3.00 \times 108 m/s) 9.72 \times 10 -8 m = 2.05 \times 10 -18 J$

Compare with the energy needed to reach n = 2 from n = 1:

$Δ E 2 \to 1 = E 2 - E 1 = (-5.45 \times 10 -19) - (-2.18 \times 10 -18) = 1.64 \times 10 -18 J$

The photon's energy (2.05 × 10⁻¹⁸ J) is larger than that, so the electron jumped higher — to n = ∞? No: to a finite higher level. The electron was ionized? Check n = 3: ΔE_3 → 1 = E₃ - E₁ = 1.94 × 10^-18 J, still less than 2.05 × 10⁻¹⁸ J. The photon's energy actually exceeds even the ionization energy (2.18 × 10⁻¹⁸ J)? It does not: 2.05 × 10⁻¹⁸ J is below 2.18 × 10⁻¹⁸ J, so the electron is excited to n = 4:

$Δ E 4 \to 1 = E 4 - E 1 = (-2.18 \times 10 -18 J \times 116) - (-2.18 \times 10 -18 J) = 2.04 \times 10 -18 J$

The match (2.05 vs 2.04 × 10⁻¹⁸ J, within rounding) confirms the transition n = 1 → n = 4. Because the absorption started from the ground state and the photon is UV, this is a Lyman-series absorption. This kind of calculation is how astronomers identify elements in distant gas clouds: match observed wavelengths to predicted transitions.

Key takeaways

Energy levels: E_n = -2.18 × 10^-18 J × (1/n²) for hydrogen; n = 1 is the ground state.
Photon energy equals the absolute difference between levels: E_photon = hν= |ΔE|.
Emission = electron falls down (photon out); absorption = electron jumps up (photon in).
Lyman series ends at n = 1 (UV); Balmer at n = 2 (visible); Paschen at n = 3 (IR).
The most negative E_n is the lowest energy; energy increases toward zero as n grows.
Ionization of hydrogen from the ground state requires 2.18 × 10^-18 J per atom (1312 kJ/mol).
The model applies exactly only to one-electron species (H, He⁺, Li²⁺).

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

Write the energy equation for hydrogen and state what n = 1 represents.
Show answer
E_n = -2.18 × 10^-18 J × (1/n²); n = 1 is the ground state, the lowest and most stable energy.
Which transition emits more energetic light: n = 3 → n = 2 or n = 3 → n = 1?
Show answer
n = 3 → n = 1: the energy gap (and hence photon energy) is larger for the bigger jump.
What is the energy of hydrogen in kJ/mol?
Show answer
2.18 × 10^-18 J per atom, or about 1312 kJ/mol.
To which does a transition ending at n = 2 belong?
Show answer
The Balmer series (visible region).
Why does the Bohr model fail for helium?
Show answer
Helium has two electrons that repel each other; the Bohr model ignores electron–electron interactions.
An electron falls from n = 4 to n = 2. Calculate the photon wavelength and state its color region.
Show answer
E₄ = -1.36 × 10^-19 J, E₂ = -5.45 × 10^-19 J, ΔE = 4.09 × 10^-19 J, λ= hc/ΔE = 4.86 × 10^-7 m = 486 nm — blue-green, a Balmer line (H-beta).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

quantum number n: Integer labeling the allowed energy levels of the electron
ground state: The lowest-energy level, n = 1
excited state: Any level above the ground state, n > 1
absorption: Electron takes in a photon and jumps to a higher level
emission: Electron falls to a lower level, releasing a photon
spectral series: A family of transitions sharing the same final level
ionization: Removing an electron completely from the atom

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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