Hydrogen Atom Energy Levels Calculator
Computes the energy of any electron shell in a hydrogen atom using Bohr's model, given the principal quantum number n. Useful for understanding atomic spectra, photon emission energies, and introductory quantum chemistry.
About this calculator
The Bohr model of the hydrogen atom gives the energy of an electron in the nth shell as: E_n = −13.6 eV / n², where n = 1, 2, 3, … is the principal quantum number and 13.6 eV is the hydrogen ionisation energy. Converting to joules: E_n = −13.6 × 1.602×10⁻¹⁹ / n² J. The negative sign indicates a bound state; energy is measured relative to a free electron at rest (n → ∞). When an electron transitions from a higher level n₂ to a lower level n₁, it emits a photon whose energy equals ΔE = 13.6 eV × (1/n₁² − 1/n₂²). This gives rise to the Lyman, Balmer, and Paschen spectral series. Although the Bohr model is superseded by quantum mechanics for multi-electron atoms, it remains exact for hydrogen-like ions such as He⁺ and Li²⁺ with a scaled nuclear charge Z.
How to use
Find the energy of the n = 3 shell of hydrogen. Step 1 — square the quantum number: n² = 9. Step 2 — apply the formula: E₃ = −13.6 × 1.602×10⁻¹⁹ / 9 = −2.179×10⁻¹⁸ / 9 ≈ −2.421×10⁻¹⁹ J (or −1.511 eV). Step 3 — to find the photon energy for a transition from n = 3 to n = 2 (the red Hα line): ΔE = |E₂ − E₃| = |−5.45×10⁻¹⁹ − (−2.421×10⁻¹⁹)| = 3.03×10⁻¹⁹ J ≈ 1.89 eV, corresponding to a wavelength of λ = hc/ΔE ≈ 656 nm (red light).
Frequently asked questions
Why are hydrogen atom energy levels negative in the Bohr model?
Energy levels are defined relative to the ionisation threshold, where the electron is at infinite distance from the nucleus with zero kinetic energy. Work must be done against the attractive Coulomb force to remove the electron, so bound states have lower (negative) energy than the free state. The ground state at n = 1 has the most negative energy, −13.6 eV, meaning 13.6 eV of energy is required to ionise hydrogen from its ground state. This convention is standard across atomic physics and quantum chemistry.
How do you calculate the wavelength of light emitted during a hydrogen electron transition?
When an electron drops from level n₂ to level n₁ (n₁ < n₂), the emitted photon energy is ΔE = 13.6 eV × (1/n₁² − 1/n₂²). The photon wavelength is then λ = hc / ΔE, where h = 6.626×10⁻³⁴ J·s and c = 3×10⁸ m/s. For the Balmer series (n₁ = 2), transitions fall in the visible range: n₂ = 3 gives 656 nm (red), n₂ = 4 gives 486 nm (blue-green). Transitions to n₁ = 1 (Lyman series) produce ultraviolet photons, while n₁ = 3 (Paschen series) produces infrared.
Does the Bohr model work for atoms other than hydrogen?
The Bohr model works exactly only for hydrogen and hydrogen-like ions — species with a single electron, such as He⁺ (Z = 2), Li²⁺ (Z = 3), or Be³⁺ (Z = 4). For these ions, the formula is modified to E_n = −13.6 × Z² / n² eV, where Z is the atomic number. For multi-electron atoms, electron–electron repulsion and shielding effects make the Bohr model inadequate. These systems require the full quantum mechanical treatment using the Schrödinger equation and methods such as Hartree–Fock or density functional theory.