Bohr Model Calculator

Calculate orbital properties for hydrogen-like atoms using the Bohr model

Bohr Model Formulas

Radius: r_n = n² × a₀ / Z, where a₀ = 0.529 A

Energy: E_n = -13.6 × Z² / n² eV

Velocity: v_n = Z × α × c / n

Angular Momentum: L = n × ℏ

What Is the Bohr Model?

The Bohr model of the atom, proposed by Niels Bohr in 1913, was the first quantum model to successfully explain the discrete spectral lines of hydrogen. Bohr postulated that electrons orbit the nucleus in specific allowed circular orbits with quantized angular momentum L = nħ, where n is the principal quantum number. Electrons can only exist in these stationary states and emit or absorb photons when jumping between them, with the photon energy equal to the difference in energy levels.

Despite being superseded by full quantum mechanics (Schrödinger equation), the Bohr model remains invaluable for its physical intuitiveness and remarkably accurate predictions for hydrogen-like atoms. The energy levels E_n = −13.6 Z²/n² eV, orbital radii r_n = n²a₀/Z, and orbital velocities v_n = Zαc/n are all exactly correct for hydrogen. The model also beautifully connects to the de Broglie hypothesis: the allowed orbits are precisely those where an integer number of electron wavelengths fit around the circumference.

Bohr Model Formulas

The Bohr model predicts all orbital properties from just two inputs: the principal quantum number n and the atomic number Z:

Bohr Model Equations

r_n = n²a₀/Z (radius, a₀ = 0.529 Å) v_n = Zαc/n (velocity, α ≈ 1/137) E_n = −13.6 Z²/n² eV (energy) L = nħ (angular momentum)

Where:

  • n= Principal quantum number (1, 2, 3, ...)
  • Z= Atomic number (1 for hydrogen, 2 for He⁺, etc.)
  • a₀= Bohr radius = 5.29 × 10⁻¹¹ m = 0.529 Å
  • α= Fine-structure constant ≈ 1/137.036
  • c= Speed of light = 2.998 × 10⁸ m/s

Hydrogen Energy Levels

nEnergy (eV)Radius (Å)Velocity (% c)Ionization (eV)
1−13.600.5290.73%13.60
2−3.402.120.37%3.40
3−1.514.760.24%1.51

The ground state electron in hydrogen orbits at ~0.73% the speed of light — fast enough for relativistic corrections to be measurable. The de Broglie wavelength verification is a beautiful check: the circumference 2πr should exactly equal nλ, and this calculator confirms the result.

How to Use This Calculator

  1. Enter Principal Quantum Number n: Any positive integer (1, 2, 3...). Higher n means larger orbits, slower electrons, and weaker binding.
  2. Enter Atomic Number Z: 1 for hydrogen, 2 for He⁺ (singly ionized helium), 3 for Li²⁺, etc. The Bohr model applies to any one-electron (hydrogen-like) ion.
  3. Review Results: The calculator outputs orbital radius (in nm and Bohr radii), velocity (m/s and % of light speed), total/kinetic/potential energies, ionization energy, angular momentum, orbital period and frequency, and the de Broglie wavelength. The wavelengths-in-orbit check (should equal n exactly) confirms the self-consistency of Bohr's quantization condition.

Significance and Limitations

The Bohr model's greatest triumph was explaining the Balmer series of hydrogen spectral lines. Bohr predicted the Rydberg constant R = mₑe⁴/(8ε₀²h³c) from fundamental constants, matching experimental measurements to within 0.05%. The model correctly predicted that the Lyman series (transitions to n=1) would be ultraviolet and the Paschen series (to n=3) would be infrared — all before they were observed.

The model's limitations, however, are instructive. It cannot explain the spectra of multi-electron atoms, the relative intensities of spectral lines, the fine structure (splitting of lines due to spin-orbit coupling), or the Zeeman effect (magnetic splitting). These failures motivated the development of quantum mechanics — Schrödinger's wave equation provides the complete description that the Bohr model approximates. Modern quantum mechanics still uses Bohr's key insight of quantized angular momentum as its starting point.

Worked Examples

Hydrogen Ground State

Problem:

Calculate all orbital properties for hydrogen (Z=1) in its ground state (n=1).

Solution Steps:

  1. 1Radius: r₁ = 1² × 0.529/1 = 0.529 Å = 0.0529 nm
  2. 2Velocity: v₁ = 1 × (1/137) × c / 1 = 2.188 × 10⁶ m/s = 0.73% c
  3. 3Energy: E₁ = −13.6 × 1² / 1² = −13.60 eV
  4. 4Angular momentum: L = 1 × ħ = 1.055 × 10⁻³⁴ J·s
  5. 5De Broglie wavelength: λ = h/(mₑv) = 6.626×10⁻³⁴/(9.109×10⁻³¹×2.188×10⁶) = 3.325×10⁻¹⁰ m
  6. 6Circumference: 2πr = 2π×5.29×10⁻¹¹ = 3.324×10⁻¹⁰ m ≈ 1.00 λ → exactly one wavelength fits!

Result:

Ground state energy = −13.60 eV. The electron orbits at 0.73% light speed with exactly one de Broglie wavelength fitting the orbit — the Bohr quantization condition in action.

He⁺ Ion (n=2)

Problem:

Singly ionized helium He⁺ has Z=2. Find properties for the n=2 state.

Solution Steps:

  1. 1Radius: r₂ = 2² × 0.529/2 = 2 × 0.529 = 1.058 Å
  2. 2Velocity: v₂ = 2 × (1/137) × c / 2 = (1/137) × c = 2.188 × 10⁶ m/s (same as H n=1!)
  3. 3Energy: E₂ = −13.6 × 2² / 2² = −13.60 eV (same as H n=1!)
  4. 4This is why He⁺ n=2 → n=1 transition produces photons close to the hydrogen Lyman-alpha line

Result:

E₂ = −13.60 eV. The n=2 state of He⁺ has the same energy and velocity as the ground state of hydrogen — a consequence of the Z²/n² scaling.

Transition Energy: H-α Line

Problem:

Calculate the photon energy when a hydrogen electron drops from n=3 to n=2 (Balmer H-α line).

Solution Steps:

  1. 1E₃ = −13.6/9 = −1.511 eV, E₂ = −13.6/4 = −3.400 eV
  2. 2ΔE = E₃ − E₂ = −1.511 − (−3.400) = 1.889 eV
  3. 3Wavelength: λ = 1240/1.889 = 656.3 nm
  4. 4This is the famous red H-α line at 656.3 nm

Result:

Photon energy = 1.889 eV, wavelength = 656.3 nm (deep red). This is the brightest line in the Balmer series and gives emission nebulae like the Orion Nebula their characteristic red color.

Tips & Best Practices

  • The Bohr radius a₀ = 0.529 Å = 5.29×10⁻¹¹ m — memorize this fundamental length scale of atomic physics
  • Energy scales as Z²/n² — a hydrogen-like uranium ion (Z=92) has ground state energy ~115 keV, compared to 13.6 eV for hydrogen
  • Angular momentum quantization L = nħ was Bohr's key postulate — the de Broglie standing wave explanation came later
  • Circular orbit circumference should exactly equal n × de Broglie wavelength — use this calculator's 'Wavelengths in Orbit' result to verify
  • The velocity at n=1 is αc ≈ c/137 — the fine-structure constant α directly measures the electron's speed in hydrogen

Frequently Asked Questions

The Bohr model gets the energy levels, orbital radii, and angular momentum exactly right for hydrogen — it's not 'wrong' in its predictions, just limited in scope. More importantly, it introduces the critical quantum concepts of discrete energy states and quantization in an intuitive, visualizable way. Students can draw orbits and calculate transitions before tackling the mathematical complexity of wavefunctions. It's the conceptual stepping stone to quantum mechanics, much like Newton's laws precede relativity.
Louis de Broglie (1924) provided a physical justification for Bohr's quantization: the electron's matter wave must form a standing wave around the orbit, meaning the circumference must equal an integer number of wavelengths: 2πr = nλ. Substituting λ = h/p = h/mv gives 2πr = nh/mv, which rearranges to mvr = nħ — exactly Bohr's angular momentum quantization. This elegant connection between wave-particle duality and atomic structure was a key insight in the development of quantum mechanics.
Hydrogen-like ions are atoms with exactly one electron (He⁺, Li²⁺, Be³⁺, etc.). The Bohr model applies perfectly because there's only one electron — no electron-electron interactions to complicate the picture. The Coulomb potential is simply V = −kZe²/r, and the model's predictions scale exactly with Z. For helium (two electrons), the model fails because it cannot account for electron-electron repulsion and the Pauli exclusion principle.
In hydrogen's ground state, the electron orbits at v = αc ≈ c/137 ≈ 2.18 × 10⁶ m/s, about 0.73% the speed of light. For higher n, velocity decreases as 1/n. For heavier hydrogen-like ions, velocity increases with Z: a uranium U⁹¹⁺ ion (Z=92) in n=1 has v ≈ 92αc ≈ 0.67c — well into the relativistic regime where the Bohr model breaks down and the Dirac equation is needed.

Sources & References

Last updated: 2026-06-06

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Editorial Note

MyCalcBuddy Editorial Team

This page is maintained as an educational calculator reference.

Source

Formula Source: University Physics

by Young & Freedman

UpdatedLast reviewed: May 2026
CheckedFormula checks are based on standard references and internal QA review.