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Chemistry 2e · Electronic Structure and Periodic Properties of Elements

The Bohr Model

7 min read
7 key termsSourcesReport an error
Physical constants (R, h, 1 eV, hydrogen ionization energy 13.6 eV) are standard textbook values; verify against current sources before use in assessments.
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On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

Pass an electric discharge through a tube of hydrogen gas and it glows pink — yet through a prism the light separates into a few sharp colored lines, not a rainbow. This is the puzzle Niels Bohr set out to solve in 1913: why does hydrogen emit only specific wavelengths, never the ones in between? Classical physics predicted the opposite: an orbiting electron should radiate continuously and spiral into the nucleus almost instantly. Bohr's answer was to impose quantization on the atom: electrons occupy only certain allowed orbits and radiate or absorb light only when jumping between them. His model reproduced the hydrogen with remarkable precision and introduced the energy-level picture all of quantum chemistry still uses — though the model itself was later superseded.

Why this matters

The Bohr model is the bridge between the photon idea and the quantum-mechanical atom. It explains why atoms emit fingerprint-like line spectra — the tool astronomers use to identify elements in stars, and the principle behind neon signs, flame tests, and lasers. Its key habit of thought — electrons occupy discrete energy levels, and light is emitted when they move between levels — survives unchanged in modern quantum mechanics. The model still works for single-electron species like H, He+, Li2+; knowing its successes AND limits prevents the classic mistake of applying it to multi-electron atoms.

The college version

Core Concepts

The puzzle: line spectra instead of rainbows

A hot solid emits a continuous spectrum; a heated gas emits only discrete wavelengths, appearing as sharp lines. For hydrogen the visible lines are the — red 656 nm, blue-green 486 nm, violet 434 nm — following a pattern encoded in the empirical that a successful atomic model had to explain.

Bohr's postulates

Bohr kept Rutherford's picture of a tiny, dense, positive nucleus but added three quantum rules:

  1. Quantized orbits. The electron orbits only where its angular momentum is an integer multiple of h2π: mvr = nh2π, n = 1, 2, 3, ….
  2. Stationary states. In an allowed orbit, the electron does NOT radiate — a direct violation of classical electromagnetism, justified by its success.
  3. Quantized transitions. Light is emitted or absorbed only during jumps between orbits, with photon energy equal to the level difference:

Ephoton = |ΔE| = hν

Energy levels of the hydrogen atom

Combining the postulates with Coulomb's law and circular motion gives the energy of level n:

En = -2.18 × 10-18 J × Z2n2

For hydrogen (Z = 1), the lowest level is E1 = -2.18 × 10-18 J (-13.6 eV). The negative sign means the electron is bound — energy is required to remove it. As n grows, levels crowd closer together, approaching E = 0 at n = ∞, where the electron is free.

Emission, absorption, and spectral series

For a jump from level n2 (higher) to n1 (lower), the emitted wavelength obeys

1λ = R(1n12 - 1n22),   R = 1.097 × 107 m-1

Series are named for the final level: Lyman (n1 = 1, ultraviolet), Balmer (n1 = 2, visible), Paschen (n1 = 3, infrared). The Balmer series lands in the visible window — hence hydrogen's pink glow.

Successes and limits of the model

The Bohr model reproduces the hydrogen spectrum almost perfectly and gives the correct (13.6 eV), but fails for helium and all multi-electron atoms: it treats electrons as point particles, ignores electron–electron repulsion, and cannot explain line splitting in magnetic fields. It is a heroic first step — the next topic replaces orbits with probability clouds.

Common Confusions

Do Not ConfuseWithDifference
EmissionAbsorptionEmission: electron falls, photon released. Absorption: electron rises, photon consumed.
n1 in the Rydberg equationThe starting levelIn emission problems n1 is the FINAL (lower) level; n2 is the starting (higher) level
The Bohr modelThe modern quantum modelBohr = fixed circular orbits, one-electron species only; modern = probability orbitals, all atoms
Hydrogen's pink glowA continuous rainbowGases emit discrete lines; only hot solids emit continuous spectra
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine an electron is a kid on a staircase — it can stand on step 1, 2, or 3, but never hover between steps. Jumping down a step releases a flash of light; climbing up requires absorbing light of exactly the right color. Each jump gives one exact color — that is why hydrogen's glow shows sharp lines instead of a rainbow. The Bohr model first said "electrons live on staircases," and it worked beautifully for the simplest atom.

Worked example

Worked Example 1: Energy of the electron in level n = 2

Problem. Calculate the energy of the electron in the n = 2 level of hydrogen.

Strategy. Formula first, then substitute (Z = 1 for hydrogen):

En = -2.18 × 10-18 J × Z2n2

E2 = -2.18 × 10-18 J × 1222 = -2.18 × 10-18 J × 14 = -5.45 × 10-19 J

Dimensional analysis. Z2n2 is unitless, so the answer is in joules. Comparison. E1 is more negative than E2, so the ground state is more stable.

Worked Example 2: Wavelength of the first Balmer line

Problem. Hydrogen's most prominent visible line (H-alpha) comes from the n = 3 → n = 2 transition. Calculate its wavelength.

Strategy. Emission ends at n1 = 2, starts at n2 = 3. Write the Rydberg equation, then substitute:

1λ = R(1n12 - 1n22) = (1.097 × 107 m-1)(122 - 132) = (1.097 × 107 m-1)(536) = 1.524 × 106 m-1

λ= 11.524 × 106 m-1 = 6.56 × 10-7 m = 656 nm

Dimensional analysis. 1m-1 = m; 6.56 × 10-7 m × 109 nm1 m = 656 nm. Sanity check. 656 nm is visible and red — matching hydrogen's familiar red line.

Worked Example 3: Ionization energy of hydrogen

Problem. Calculate the energy needed to ionize a ground-state hydrogen atom (from n = 1 to n = ∞), in joules and in eV.

Strategy. Ionization is the n = 1 → n = ∞ transition, with E∞= 0:

ΔE = E∞- E1 = 0 - (-2.18 × 10-18 J) = +2.18 × 10-18 J

Convert with 1 eV = 1.602 × 10-19 J:

2.18 × 10-18 J × 1 eV1.602 × 10-19 J = 13.6 eV

Interpretation. The positive sign means energy must be supplied — the definition of ionization energy. The textbook value 13.6 eV is one of the model's celebrated successes.

Key takeaways

  • Energy of level n: En = -2.18 × 10-18 J × Z2n2; memorize E1 = -13.6 eV for H.
  • Negative energy = bound: n = 1 is ground state; n = ∞ is E = 0, the ionization limit.
  • Photon energy on any transition: hν= |Efinal - Einitial|.
  • Rydberg equation: 1λ = R(1n12 - 1n22), R = 1.097 × 107 m-1; for emission n1 is the LOWER (final) level.
  • Series: Lyman → UV, Balmer → visible, Paschen → infrared.
  • Ionization energy of H = 13.6 eV (from n = 1 to n = ∞).
  • Works ONLY for one-electron species (H, He+, Li2+).

Check yourself

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

  1. What are Bohr's three postulates, in your own words?

    Show answer

    (1) Only orbits with angular momentum mvr = nh2π are allowed; (2) electrons in allowed orbits do not radiate; (3) light is emitted/absorbed only during jumps, with hν= |ΔE|.

  2. Which is more stable: an electron in n = 1 or n = 4? Explain using the energy formula.

    Show answer

    n = 1. Its energy -2.18 × 10-18 J is more negative than E4 = -1.36 × 10-19 J; more negative = more stable.

  3. Why are the Balmer lines the visible ones while Lyman and Paschen lines are not?

    Show answer

    Balmer transitions end at n = 2 and produce wavelengths of 400–700 nm — the visible window. Lyman lines end at n = 1 (UV) and Paschen at n = 3 (infrared), so both are invisible.

  4. A hydrogen atom absorbs a photon and its electron moves from n = 1 to n = 3. What determines the photon's energy?

    Show answer

    The photon energy must exactly equal ΔE = E3 - E1, the level difference — no more, no less.

  5. Why does the Bohr model fail for helium?

    Show answer

    Helium has two electrons. The model treats one electron in a Coulomb field and ignores electron–electron repulsion, which is strong in multi-electron atoms.

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Line spectrum
Discrete wavelengths emitted by a gas, appearing as sharp lines
Stationary state
An allowed orbit in which the electron does not radiate
Principal quantum number (n)
Integer labeling the energy level: 1, 2, 3, ...
Ground state
The lowest energy level (n = 1 for hydrogen)
Balmer series
Transitions ending at n = 2, in the visible region
Rydberg equation
1λ = R(1n12 - 1n22)
Ionization energy
Energy to remove the electron completely

Sources & references

  1. openstax.org — Chemistry 2e

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

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